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GEjRKxAiRms6EYbD | maths | differentiation | differentiation-of-implicit-function | Let $$f\left( x \right)$$ be a polynomial function of second degree. If $$f\left( 1 \right) = f\left( { - 1} \right)$$ and $$a,b,c$$ are in $$A.P, $$ then $$f'\left( a \right),f'\left( b \right),f'\left( c \right)$$ are in | [{"identifier": "A", "content": "Arithmetic -Geometric Progression "}, {"identifier": "B", "content": "$$A.P$$"}, {"identifier": "C", "content": "$$G.P$$"}, {"identifier": "D", "content": "$$H.P$$ "}] | ["B"] | null | $$f\left( x \right) = a{x^2} + bx + c$$
<br><br>$$f\left( 1 \right) = f\left( { - 1} \right)$$
<br><br>$$ \Rightarrow a + b + c = a - b + c$$
<br><br>or $$b = 0$$
<br><br>$$\therefore$$ $$f\left( x \right) = a{x^2} + c$$
<br><br>or $$f'\left( x \right) = 2ax$$
<br><br>Now $$f'\left( a \right);f'\left( b \right);$$
<br><br>and $$f'\left( c \right)$$ are $$2a\left( a \right);2a\left( b \right);2a\left( c \right)$$
<br><br>i.e.$$\,2{a^2},\,2ab,\,2ac.$$
<br><br>$$ \Rightarrow $$ If $$a,b,c$$ are in $$A.P.$$ then
<br><br>$$f'\left( a \right);f'\left( b \right)$$ and
<br><br>$$f'\left( c \right)$$ are also in $$A.P.$$ | mcq | aieee-2003 | 1,700 |
lYKxaNHOCAaR6Yxj | maths | differentiation | differentiation-of-implicit-function | Let $$y$$ be an implicit function of $$x$$ defined by $${x^{2x}} - 2{x^x}\cot \,y - 1 = 0$$. Then $$y'(1)$$ equals | [{"identifier": "A", "content": "$$1$$ "}, {"identifier": "B", "content": "$$\\log \\,2$$"}, {"identifier": "C", "content": "$$-\\log \\,2$$ "}, {"identifier": "D", "content": "$$-1$$"}] | ["D"] | null | $${x^{2x}} - 2{x^x}\,\cot \,y - 1 = 0$$
<br><br>$$ \Rightarrow 2\,\cot \,y = {x^x} - {x^{ - x}}$$
<br><br>$$ \Rightarrow 2\,\cot \,y\, = u - {1 \over u}$$
<br><br>where $$u = {x^x}$$
<br><br>Differentiating both sides with respect to $$x,$$
<br><br>we get $$ \Rightarrow - 2\cos e{c^2}y{{dy} \over {dx}}$$
<br><br>$$ = \left( {1 + {1 \over {{u^2}}}} \right){{du} \over {dx}}$$
<br><br>where $$u = {x^x} \Rightarrow \log \,u = x\,\log \,x$$
<br><br>$$ \Rightarrow {1 \over u}{{du} \over {dx}} = 1 + \log \,x$$
<br><br>$$ \Rightarrow {{du} \over {dx}} = {x^x}\left( {1 + \log \,x} \right)$$
<br><br>$$\therefore$$ We get $$ - 2\cos e{c^2}y{{dy} \over {dx}}$$
<br><br>$$ = \left( {1 + {x^{ - 2x}}} \right){x^x}\left( {1 + \log \,x} \right)$$
<br><br>$$ \Rightarrow {{dy} \over {dx}} = {{\left( {{x^x} + {x^{ - x}}} \right)\left( {1 + \log x} \right)} \over { - 2\left( {1 + {{\cot }^2}y} \right)}}\,\,\,\,\,\,...\left( i \right)$$
<br><br>Now when
<br><br>$$x=1,$$ $${x^{2x}} - 2{x^x}\,\cot \,y - 1 = 0,$$
<br><br>gives $$1 - 2\,\cot y - 1 = 0$$
<br><br>$$ \Rightarrow \,\,\cot y\, = 0$$
<br><br>$$\therefore$$ From equation $$(i),$$ at $$x=1$$
<br><br>and $$\cot \,y = 0,$$ we get
<br><br>$$y'\left( 1 \right) = {{\left( {1 + 1} \right)\left( {1 + 0} \right)} \over { - 2\left( {1 + 0} \right)}} = - 1$$ | mcq | aieee-2009 | 1,701 |
a7PCNBHZY1NukfT9vwNCx | maths | differentiation | differentiation-of-implicit-function | If y = $${\left[ {x + \sqrt {{x^2} - 1} } \right]^{15}} + {\left[ {x - \sqrt {{x^2} - 1} } \right]^{15}},$$
<br/><br/> then (x<sup>2</sup> $$-$$ 1) $${{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}$$ is equal to : | [{"identifier": "A", "content": "125 y"}, {"identifier": "B", "content": "124 y<sup>2</sup>"}, {"identifier": "C", "content": "225 y<sup>2</sup>"}, {"identifier": "D", "content": "225 y"}] | ["D"] | null | <p>The given equation is</p>
<p>$$y = {({x^2} + \sqrt {{x^2} - 1} )^{15}} + {(x - \sqrt {{x^2} - 1} )^{15}}$$</p>
<p>Differentiating w.r.t. x, we get</p>
<p>$${{dy} \over {dx}} = 15{(x + \sqrt {{x^2} - 1} )^{14}}\left( {1 + {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right) + 15{(x - \sqrt {{x^2} - 1} )^{14}}\left( {1 - {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right)$$</p>
<p>Here, we have used the standard differentiatials</p>
<p>$${d \over {dx}}{x^n} = n\,{x^{n - 1}}$$</p>
<p>That is, $${d \over {dx}}(\sqrt {f(x)} ) = {1 \over {2\sqrt {f(x)} }} \times {d \over {dx}}(f(x))$$</p>
<p>Therefore</p>
<p>$${{dy} \over {dx}} = {{15{{(x + \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} + x)} \over {\sqrt {{x^2} - 1} }} + {{15{{(x - \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} - x)} \over {\sqrt {{x^2} - 1} }}$$</p>
<p>$$ \Rightarrow \sqrt {{x^2} - 1} {{dy} \over {dx}} = 15{(x + \sqrt {{x^2} - 1} )^{15}} - 15{(x - \sqrt {{x^2} - 1} )^{15}}$$</p>
<p>Differentiating w.r.t. x, we get</p>
<p>$${{1(2x)} \over {2\sqrt {{x^2} + 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}} = 15 \times 15{(x + \sqrt {{x^2} - 1} )^{14}}\left( {1 + {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right) - 15 \times 15{(x - \sqrt {{x^2} - 1} )^{14}}\left( {{{1 - 1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right)$$</p>
<p>$$ \Rightarrow {x \over {\sqrt {{x^2} - 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}} = 225{(x + \sqrt {{x^2} - 1} )^{14}}{{(\sqrt {{x^2} - 1} + x)} \over {\sqrt {{x^2} - 1} }} - {{225{{(x - \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} - x)} \over {\sqrt {{x^2} - 1} }}$$</p>
<p>$$ \Rightarrow \sqrt {{x^2} - 1} \left[ {{x \over {\sqrt {{x^2} - 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}}} \right] = 225{(x + \sqrt {{x^2} - 1} )^{15}} + 225{(x - \sqrt {{x^2} - 1} )^{15}}$$</p>
<p>$$ \Rightarrow x{{dy} \over {dx}} + ({x^2} - 1){{{d^2}y} \over {d{x^2}}} = 225\left[ {{{(x + \sqrt {{x^2} - 1} )}^{15}} + {{(x - \sqrt {{x^2} - 1} )}^{15}}} \right]$$</p>
<p>Substituting $${(x + \sqrt {{x^2} - 1} )^{15}} + {(x - \sqrt {{x^2} - 1} )^{15}} = y$$, we get</p>
<p>$$({x^2} - 1){{{d^2}y} \over {d{x^2}}} + {{x\,dy} \over {dx}} = 225y$$</p> | mcq | jee-main-2017-online-8th-april-morning-slot | 1,702 |
DhiE82brjsITwGqGoJmYl | maths | differentiation | differentiation-of-implicit-function | If $$f\left( x \right) = \left| {\matrix{
{\cos x} & x & 1 \cr
{2\sin x} & {{x^2}} & {2x} \cr
{\tan x} & x & 1 \cr
} } \right|,$$ then $$\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$$ | [{"identifier": "A", "content": "does not exist. "}, {"identifier": "B", "content": "exists and is equal to 2. "}, {"identifier": "C", "content": "existsand is equal to 0."}, {"identifier": "D", "content": "exists and is equal to $$-$$ 2."}] | ["D"] | null | Given,<br><br>
$$f\left( x \right) = \left| {\matrix{
{\cos x} & x & 1 \cr
{2\sin x} & {{x^2}} & {2x} \cr
{\tan x} & x & 1 \cr
} } \right|$$<br><br>
= cosx(x<sup>2</sup> - 2x<sup>2</sup>) - x(2 sinx - 2x tanx) + (2x sinx - x<sup>2</sup> tanx)<br>
= x<sup>2</sup> (tanx - cosx)<br>
$$ \therefore $$ $${f^{'}}(x)$$ = 2x (tanx - cosx) + x<sup>2</sup>(sec<sup>2</sup>x + sinx)<br><br>
$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$$<br><br>
= $$\mathop {\lim }\limits_{x \to o} {{2x(\tan x - \cos x) + {x^2}({{\sec }^2}x + \sin x)} \over x}$$<br><br>
= $$\mathop {\lim }\limits_{x \to o} \,\,2(\tan x - \cos x) + x({\sec ^2}x + \sin x)$$<br><br>
= 2 (0-1) + 0<br><br>
= -2
| mcq | jee-main-2018-online-15th-april-morning-slot | 1,703 |
C8sdKeF0ryQ9Msh5eBUCc | maths | differentiation | differentiation-of-implicit-function | If x<sup>2</sup> + y<sup>2</sup> + sin y = 4, then the value of $${{{d^2}y} \over {d{x^2}}}$$ at the point ($$-$$2,0) is : | [{"identifier": "A", "content": "$$-$$ 34"}, {"identifier": "B", "content": "$$-$$ 32"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$-$$ 2"}] | ["A"] | null | Given, x<sup>2</sup> + y<sup>2</sup> + sin y = 4
<br><br>After differentiating the above equation w.r.t.x we get
<br><br>2x + 2y $${{dy} \over {dx}}$$ + cos y $${{dy} \over {dx}}$$ = 0 . . . . (1)
<br><br>$$ \Rightarrow $$ 2x + (2y + cos y) $${{dy} \over {dx}}$$ = 0
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${{ - 2x} \over {2y + \cos y}}$$
<br><br>At ($$-$$ 2, 0), $${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$$ = $${{ - 2x - 2} \over {2 \times 0 + \cos 0}}$$
<br><br>$$ \Rightarrow $$ $${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$$ = $${4 \over {0 + 1}}$$
<br><br>$$ \Rightarrow $$ $${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$$ = 4 . . . . .(2)
<br><br>Again differentiating equation (1) w.r.t to x, we get
<br><br>2 + 2 $${\left( {{{dy} \over {dx}}} \right)^2}$$ + 2y$${{{d^2}y} \over {d{x^2}}}$$ $$-$$ sin y $${\left( {{{dy} \over {dx}}} \right)^2}$$ + cos y $${{{d^2}y} \over {d{x^2}}}$$ = 0
<br><br>$$ \Rightarrow $$ 2 + (2 $$-$$ sin y) $${\left( {{{dy} \over {dx}}} \right)^2}$$ + (2y + cos y)$${{{d^2}y} \over {d{x^2}}}$$ = 0
<br><br>$$ \Rightarrow $$ (2y + cos y) $${{{d^2}y} \over {d{x^2}}}$$ = $$-$$ 2 $$-$$ (2 $$-$$ sin y)$${\left( {{{dy} \over {dx}}} \right)^2}$$
<br><br>$$ \Rightarrow $$ $${{{d^2}y} \over {d{x^2}}}$$ = $${{ - 2 - \left( {2 - \sin y} \right){{\left( {{{dy} \over {dx}}} \right)}^2}} \over {2y + \cos y}}$$
<br><br>So, at ($$-$$ 2, 0),
<br><br>$${{{d^2}y} \over {d{x^2}}}$$ = $${{ - 2 - \left( {2 - 0} \right) \times {4^2}} \over {2 \times 0 + 1}}$$
<br><br>$$ \Rightarrow $$ $${{{d^2}y} \over {d{x^2}}}$$ = $${{ - 2 - 2 \times 16} \over 1}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $${{{d^2}y} \over {d{x^2}}}$$ = $$-$$ 34 | mcq | jee-main-2018-online-15th-april-morning-slot | 1,704 |
Ne604P2uiWuSdG4XwJ3rsa0w2w9jx5deh3j | maths | differentiation | differentiation-of-implicit-function | If e<sup>y</sup>
+ xy = e, the ordered pair $$\left( {{{dy} \over {dx}},{{{d^2}y} \over {d{x^2}}}} \right)$$ at x = 0 is equal to : | [{"identifier": "A", "content": "$$\\left( {{1 \\over e}, - {1 \\over {{e^2}}}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - {1 \\over e},{1 \\over {{e^2}}}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - {1 \\over e}, - {1 \\over {{e^2}}}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over e},{1 \\over {{e^2}}}} \\right)$$"}] | ["B"] | null | y = 1 $$ \Rightarrow $$ x = 0<br><br>
$${e^y}{{dy} \over {dx}} + x{{dy} \over {dx}} + y = 0$$<br><br>
$$ \Rightarrow e{{dy} \over {dx}} + 1 = 0 \Rightarrow {{dy} \over {dx}} = - {1 \over e}$$<br><br>
$$ \Rightarrow {e^y}{{{d^2}y} \over {d{x^2}}} + {e^y}{\left( {{{dy} \over {dx}}} \right)^2} + x{{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} = 0$$<br><br>
x = 0, y = 1<br><br>
$$ \Rightarrow e{{{d^2}y} \over {d{x^2}}} + e{\left( { - {1 \over e}} \right)^2} + 0 + 2\left( { - {1 \over e}} \right) = 0$$<br><br>
$$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = {1 \over {{e^2}}}$$ | mcq | jee-main-2019-online-12th-april-morning-slot | 1,705 |
fd5wbl11yciwuHmLja7k9k2k5e29bcc | maths | differentiation | differentiation-of-implicit-function | Let x<sup>k</sup> + y<sup>k</sup> = a<sup>k</sup>, (a, k > 0 ) and $${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$$, then k is: | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${4 \\over 3}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["B"] | null | x<sup>k</sup> + y<sup>k</sup> = a<sup>k</sup>
<br><br>$$ \Rightarrow $$ kx<sup>k - 1</sup> + ky<sup>k - 1</sup>$${{{dy} \over {dx}}}$$ = 0
<br><br>$$ \Rightarrow $$ $${{{dy} \over {dx}} + {{\left( {{x \over y}} \right)}^{k - 1}}}$$ = 0 ...(1)
<br><br>Given $${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$$ ...(2)
<br><br>Comparing (1) and (2), we get
<br><br>k - 1 = $$ - {1 \over 3}$$
<br><br>$$ \Rightarrow $$ k = $${2 \over 3}$$ | mcq | jee-main-2020-online-7th-january-morning-slot | 1,706 |
kPxln5RHIGisQJ1WD17k9k2k5e4fsyg | maths | differentiation | differentiation-of-implicit-function | If $$y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}} ,\alpha \in \left( {{{3\pi } \over 4},\pi } \right)$$<br/><br/>
$${{dy} \over {d\alpha }}\,\,at\,\alpha = {{5\pi } \over 6}is$$ : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "-4"}, {"identifier": "C", "content": "$${4 \\over 3}$$"}, {"identifier": "D", "content": "-$${1 \\over 4}$$"}] | ["A"] | null | $$y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}}$$
<br><br>= $$\sqrt {2\left( {{{1 + {{\tan }^2}\alpha } \over {\tan \alpha \left( {1 + {{\tan }^2}\alpha } \right)}}} \right) + {1 \over {{{\sin }^2}\alpha }}} $$
<br><br>= $$\sqrt {2\left( {{{\cos \alpha } \over {\sin \alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}} $$
<br><br>= $$\sqrt {{{\sin 2\alpha + 1} \over {{{\sin }^2}\alpha }}} $$
<br><br>= $${{\left| {\sin \alpha + \cos \alpha } \right|} \over {\left| {\sin \alpha } \right|}}$$
<br><br>At $$\alpha $$ = $${{5\pi } \over 6}$$
<br><br>$${\left| {\sin \alpha + \cos \alpha } \right|}$$ = -(sin$$\alpha $$ + cos$$\alpha $$)
<br><br>and |sin$$\alpha $$| = sin$$\alpha $$
<br><br>$$ \therefore $$ y($$\alpha $$) = $${{ - \left( {\sin \alpha + \cos \alpha } \right)} \over {\sin \alpha }}$$
<br><br>= -1 - cot$$\alpha $$
<br><br>$$ \therefore $$ $${{dy} \over {d\alpha }}$$ = cosec<sup>2</sup>$$\alpha $$
<br><br>So $${{{dy} \over {d\alpha }}}$$ at $$\alpha $$ = $${{5\pi } \over 6}$$,
<br><br>= cosec<sup>2</sup>$${{5\pi } \over 6}$$ = 4 | mcq | jee-main-2020-online-7th-january-morning-slot | 1,707 |
jKg7va9mIxQ9FIreIH7k9k2k5fnompn | maths | differentiation | differentiation-of-implicit-function | Let y = y(x) be a function of x satisfying
<br/><br>$$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} $$ where k is a constant and
<br/><br>$$y\left( {{1 \over 2}} \right) = - {1 \over 4}$$. Then $${{dy} \over {dx}}$$ at x = $${1 \over 2}$$, is equal to :</br></br> | [{"identifier": "A", "content": "$${2 \\over {\\sqrt 5 }}$$"}, {"identifier": "B", "content": "$$ - {{\\sqrt 5 } \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt 5 } \\over 2}$$"}, {"identifier": "D", "content": "$$ - {{\\sqrt 5 } \\over 4}$$"}] | ["B"] | null | $$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} $$ ....(1)
<br><br>On differentiating both side of eq. (1) w.r.t. x we
get,
<br><br>$${{dy} \over {dx}}\sqrt {1 - {x^2}} - y{{2x} \over {2\sqrt {1 - {x^2}} }}$$
<br><br>= 0 - $$\sqrt {1 - {y^2}} + {{xy} \over {\sqrt {1 - {y^2}} }}{{dy} \over {dx}}$$
<br><br>Put x = $${1 \over 2}$$ and y = $$ - {1 \over 4}$$, we get
<br><br>$${{dy} \over {dx}}{{\sqrt 3 } \over 2} - \left( { - {1 \over 4}} \right){{{1 \over 2}} \over {{{\sqrt 3 } \over 2}}}$$
<br><br>= $$ - {{\sqrt {15} } \over 4} + {{ - {1 \over 8}} \over {{{\sqrt {15} } \over 4}}}.{{dy} \over {dx}}$$
<br><br>$$ \therefore $$ $${{dy} \over {dx}} = - {{\sqrt 5 } \over 2}$$ | mcq | jee-main-2020-online-7th-january-evening-slot | 1,708 |
Dv3KDCMR2csVxH0QHWjgy2xukf8zrqqn | maths | differentiation | differentiation-of-implicit-function | If $$\left( {a + \sqrt 2 b\cos x} \right)\left( {a - \sqrt 2 b\cos y} \right) = {a^2} - {b^2}$$<br/><br/>
where a > b > 0, then $${{dx} \over {dy}}\,\,at\left( {{\pi \over 4},{\pi \over 4}} \right)$$ is : | [{"identifier": "A", "content": "$${{a - 2b} \\over {a + 2b}}$$"}, {"identifier": "B", "content": "$${{a - b} \\over {a + b}}$$"}, {"identifier": "C", "content": "$${{a + b} \\over {a - b}}$$"}, {"identifier": "D", "content": "$${{2a + b} \\over {2a - b}}$$"}] | ["C"] | null | $$(a + \sqrt 2 b\cos x)(a - \sqrt 2 b\cos y) = {a^2} - {b^2}$$<br><br>$$ \Rightarrow {a^2} - \sqrt 2 ab\cos y + \sqrt 2 ab\cos x - 2{b^2}\cos x\cos y = {a^2} - {b^2}$$<br><br>Differentiating both sides :<br><br>$$0 - \sqrt 2 ab\left( { - \sin y{{dy} \over {dx}}} \right) + \sqrt 2 ab( - \sin x)$$<br><br>$$ - 2{b^2}\left[ {\cos x\left( { - \sin y{{dy} \over {dx}}} \right) + \cos y( - \sin x)} \right] = 0$$<br><br>At $$\left( {{\pi \over 4},{\pi \over 4}} \right)$$ :<br><br>$$ab{{dy} \over {dx}} - ab - 2{b^2}\left( { - {1 \over 2}{{dy} \over {dx}} - {1 \over 2}} \right) = 0$$<br><br>$$ \Rightarrow {{dx} \over {dy}} = {{ab + {b^2}} \over {ab - {b^2}}} = {{a + b} \over {a - b}}$$; a, b > 0 | mcq | jee-main-2020-online-4th-september-morning-slot | 1,709 |
1ldpt5swb | maths | differentiation | differentiation-of-implicit-function | <p>Let $$y=f(x)=\sin ^{3}\left(\frac{\pi}{3}\left(\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{\frac{3}{2}}\right)\right)\right)$$. Then, at x = 1,</p> | [{"identifier": "A", "content": "$$2 y^{\\prime}+\\sqrt{3} \\pi^{2} y=0$$"}, {"identifier": "B", "content": "$$y^{\\prime}+3 \\pi^{2} y=0$$"}, {"identifier": "C", "content": "$$\\sqrt{2} y^{\\prime}-3 \\pi^{2} y=0$$"}, {"identifier": "D", "content": "$$2 y^{\\prime}+3 \\pi^{2} y=0$$"}] | ["D"] | null | $f(x)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right)$
<br/><br/>$$
\begin{aligned}
& f^{\prime}(x)=3 \sin ^{2}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\
& \cos \left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\
& \frac{\pi}{3}\left(-\sin \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\
& \frac{\pi}{3 \sqrt{2}} \frac{3}{2}\left(-4 x^{3}+5 x^{3}+1\right)^{1 / 2}\left(-12 x^{2}+10 x\right)
\end{aligned}
$$
<br/><br/>$f^{\prime}(1)=\frac{3 \pi^{2}}{16}$
<br/><br/>$$
\begin{aligned}
& f(1)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}} 2 \sqrt{2}\right)\right) \\\\
&=\sin ^{3}\left(-\frac{\pi}{6}\right)=\frac{-1}{8} \\\\
& \therefore 2 f^{\prime}(1)+3 \pi^{2} f(1)=0
\end{aligned}
$$ | mcq | jee-main-2023-online-31st-january-morning-shift | 1,710 |
1lgoy3hxl | maths | differentiation | differentiation-of-implicit-function | <p>Let $$f(x)=\sum_\limits{k=1}^{10} k x^{k}, x \in \mathbb{R}$$. If $$2 f(2)+f^{\prime}(2)=119(2)^{\mathrm{n}}+1$$ then $$\mathrm{n}$$ is equal to ___________</p> | [] | null | 10 | Given, $f(x)=\sum_\limits{k=1}^{10} k x^{k}$
<br/><br/>$$
\begin{aligned}
& f(x)=x+2 x^2+\ldots \ldots \ldots+10 x^{10} \\\\
& f(x) . x=x^2+2 x^3+\ldots \ldots \ldots+9 x^{10}+10 x^{11} \\\\
& f(x)(1-x)=x+x^2+x^3+\ldots \ldots \ldots+x^{10}-10 x^{11} \\\\
& \therefore f(x)=\frac{x\left(1-x^{10}\right)}{(1-x)^2}-\frac{10 x^{11}}{(1-x)}
\end{aligned}
$$
<br/><br/>$$ \Rightarrow $$ $$
f(x)=\frac{x-x^{11}-10 x^{11}+10 x^{12}}{(1-x)^2} = \frac{10 x^{12}-11 x^{11}+x}{(1-x)^2}
$$
<br/><br/>So,
<br/><br/>$$
\begin{aligned}
& f(2)=2\left(1-2^{10}\right)+10 \cdot 2^{11} \\\\
& =2+18 \cdot 2^{10}
\end{aligned}
$$
<br/><br/>$$f'\left( x \right) = {{{{\left( {1 - x} \right)}^2}\left( {120{x^{11}} - 121{x^{10}} + 1} \right) + 2\left( {1 - x} \right)\left( {10{x^{12}} - {{11.x}^{11}} + 2} \right)} \over {{{\left( {1 - x} \right)}^4}}}$$
<br/><br/>So,
<br/><br/>$$f'\left( x \right) = {{1\left( {{{120.2}^{11}} - {{121.2}^{10}} + 1} \right) + 2\left( { - 1} \right)\left( {{{10.2}^{12}} - {{11.2}^{11}} + 2} \right)} \over {{{\left( { - 1} \right)}^4}}}$$
<br/><br/>= $${2^{10}}\left( {83} \right) - 3$$
<br/><br/>Hence $2 f(2)+f^{\prime}(2)=119.2^{10}+1$
<br/><br/>$$
\Rightarrow \text { So, } \mathrm{n}=10
$$ | integer | jee-main-2023-online-13th-april-evening-shift | 1,711 |
1lgzxhecr | maths | differentiation | differentiation-of-implicit-function | <p>Let $$f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}, x \in[0, \pi]-\left\{\frac{\pi}{4}\right\}$$. Then $$f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{2}{3 \\sqrt{3}}$$"}, {"identifier": "B", "content": "$$\\frac{2}{9}$$"}, {"identifier": "C", "content": "$$\\frac{-1}{3 \\sqrt{3}}$$"}, {"identifier": "D", "content": "$$\\frac{-2}{3}$$"}] | ["B"] | null | $$f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}$$
<br/><br/>$$
\begin{aligned}
& =\frac{\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x-1}{\frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x} \\\\
& =\frac{\cos \left(x-\frac{\pi}{4}\right)-1}{\sin \left(x-\frac{\pi}{4}\right)} \\\\
& =\frac{-2 \sin ^2\left(\frac{x}{2}-\frac{\pi}{8}\right)}{2 \sin \left(\frac{x}{2}-\frac{\pi}{8}\right) \cos \left(\frac{x}{2}-\frac{\pi}{8}\right)}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow f(x)=-\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \\\\
& \Rightarrow f^{\prime}(x)=-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right)
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow f^{\prime \prime}(x)=-\frac{1}{2} \cdot 2 \sec \left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \sec \left(\frac{x}{2}-\frac{\pi}{8}\right)\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \times \frac{1}{2} \\\\
& =-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \tan \left(\frac{x}{2}-\frac{\pi}{8}\right)
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Now, } f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right) \\\\
& =-\tan \left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \times \frac{-1}{2} \sec ^2\left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \times \tan \left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \\\\
& =\frac{1}{2} \tan ^2\left(\frac{\pi}{6}\right) \times \sec ^2 \frac{\pi}{6} \\\\
& =\frac{1}{2} \times \frac{1}{3} \times \frac{4}{3}=\frac{2}{9}
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-morning-shift | 1,712 |
cOntCBTD82jRclEC | maths | differentiation | differentiation-of-inverse-trigonometric-function | If $$y = \sec \left( {{{\tan }^{ - 1}}x} \right),$$ then $${{{dy} \over {dx}}}$$ at $$x=1$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "B", "content": "$${1 \\over 2}$$ "}, {"identifier": "C", "content": "$$1$$ "}, {"identifier": "D", "content": "$$\\sqrt 2 $$ "}] | ["A"] | null | Let $$y = \sec \left( {{{\tan }^{ - 1}}x} \right)$$
<br><br>and $${\tan ^{ - 1}}\,\,x = \theta .$$
<br><br>$$ \Rightarrow x = \tan \theta $$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267182/exam_images/gheykswaxc1dfck1kir0.webp" loading="lazy" alt="JEE Main 2013 (Offline) Mathematics - Differentiation Question 69 English Explanation">
<br><br>Thus, we have $$y = \sec \,\theta $$
<br><br>$$ \Rightarrow y = \sqrt {1 + {x^2}} $$
<br><br>$$\left( {\,\,} \right.$$ As $$\,\,\,\,\,\,{\sec ^2}\theta = 1 + {\tan ^2}\theta $$ $$\left. {\,\,} \right)$$
<br><br>$$ \Rightarrow {{dy} \over {dx}} = {1 \over {2\sqrt {1 + {x^2}} }}.2x$$
<br><br>At $$x = 1,\,\,{{dy} \over {dx}} = {1 \over {\sqrt 2 }}.$$ | mcq | jee-main-2013-offline | 1,713 |
yy401t5DTBxS3W73 | maths | differentiation | differentiation-of-inverse-trigonometric-function | If for $$x \in \left( {0,{1 \over 4}} \right)$$, the derivatives of
<br/><br/>$${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$ is $$\sqrt x .g\left( x \right)$$, then $$g\left( x \right)$$ equals | [{"identifier": "A", "content": "$${{{3x\\sqrt x } \\over {1 - 9{x^3}}}}$$"}, {"identifier": "B", "content": "$${{{3x} \\over {1 - 9{x^3}}}}$$"}, {"identifier": "C", "content": "$${{3 \\over {1 + 9{x^3}}}}$$"}, {"identifier": "D", "content": "$${{9 \\over {1 + 9{x^3}}}}$$"}] | ["D"] | null | Let y = $${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$
<br><br>= $${\tan ^{ - 1}}\left[ {{{2.\left( {3{x^{{3 \over 2}}}} \right)} \over {1 - {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}} \right]$$
<br><br>= 2$${\tan ^{ - 1}}\left( {3{x^{{3 \over 2}}}} \right)$$
<br><br>$$ \therefore $$ $${{dy} \over {dx}} = 2.{1 \over {1 + {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}.3 \times {3 \over 2}{\left( x \right)^{{1 \over 2}}}$$
<br><br>= $${9 \over {1 + 9{x^3}}}.\sqrt x $$
<br><br>$$ \therefore $$ g(x) = $${9 \over {1 + 9{x^3}}}$$ | mcq | jee-main-2017-offline | 1,714 |
WaMVJei76O03McRXFgNSd | maths | differentiation | differentiation-of-inverse-trigonometric-function | If f(x) = sin<sup>-1</sup> $$\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right),$$ then f'$$\left( { - {1 \over 2}} \right)$$ equals : | [{"identifier": "A", "content": "$$ - \\sqrt 3 {\\log _e}\\sqrt 3 $$"}, {"identifier": "B", "content": "$$ \\sqrt 3 {\\log _e}\\sqrt 3 $$"}, {"identifier": "C", "content": "$$ - \\sqrt 3 {\\log _e}\\, 3 $$"}, {"identifier": "D", "content": "$$ \\sqrt 3 {\\log _e}\\, 3 $$"}] | ["B"] | null | Since f(x) = sin$$\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right)$$
<br><br>Suppose 3<sup>x</sup> = tan t
<br><br> $$ \Rightarrow $$ f(x) = sin<sup>$$-$$1</sup> $$\left( {{{2\tan t} \over {1 + {{\tan }^2}t}}} \right)$$ = sin<sup>$$-$$1</sup> (sin2t) = 2t
<br><br> = 2tan<sup>$$-$$1</sup> (3x)
<br><br>So, f'(x) = $${2 \over {1 + {{\left( {{3^x}} \right)}^2}}} \times {3^x}.{\log _e}3$$
<br><br>$$ \therefore $$ f '$$\left( { - {1 \over 2}} \right)$$ = $${2 \over {1 + {{\left( {{3^{ - {1 \over 2}}}} \right)}^2}}} \times {3^{ - {1 \over 2}}}$$ . log<sub>e</sub> 3
<br><br>= $${1 \over 2} \times \sqrt 3 \times {\log _e}3$$ = $$\sqrt 3 \times {\log _e}\sqrt 3 $$ | mcq | jee-main-2018-online-15th-april-evening-slot | 1,715 |
9onUmZNU2npIbQ1VhNA44 | maths | differentiation | differentiation-of-inverse-trigonometric-function | If $$2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$$,
<br/><br/>x $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$ then $$dy \over dx$$ is equal to: | [{"identifier": "A", "content": "$$2x - {\\pi \\over 3}$$"}, {"identifier": "B", "content": "$${\\pi \\over 6} - x$$"}, {"identifier": "C", "content": "$${\\pi \\over 3} - x$$"}, {"identifier": "D", "content": "$$x - {\\pi \\over 6}$$"}] | ["D"] | null | $$2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$$
<br><br>$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 + \tan x} \over {1 - \sqrt 3 \tan x}}} \right)} \right)^2}$$
<br><br>$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\left( {{{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}} \right)} \right)^2}$$
<br><br>$$ \Rightarrow $$ 2y = $${\left( {{{\cot }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$$
<br><br>$$ \Rightarrow $$ 2y = $${\left( {{\pi \over 2} - {{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$$
<br><br>As x $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$ then
<br><br>$${{{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)}$$ = $${\left( {{\pi \over 3} + x} \right)}$$
<br><br>$$ \Rightarrow $$ 2y = $${\left( {{\pi \over 2} - \left( {{\pi \over 3} + x} \right)} \right)^2}$$
<br><br>$$ \Rightarrow $$ 2y = $${\left( {{\pi \over 6} - x} \right)^2}$$
<br><br>$$ \therefore $$ $$2{{dy} \over {dx}} = 2\left( {{\pi \over 6} - x} \right)\left( { - 1} \right)$$
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} = \left( {x - {\pi \over 6}} \right)$$ | mcq | jee-main-2019-online-8th-april-morning-slot | 1,716 |
9rwVQ5XrboppQiu5tE7k9k2k5gqvdkp | maths | differentiation | differentiation-of-inverse-trigonometric-function | Let ƒ(x) = (sin(tan<sup>–1</sup>x) + sin(cot<sup>–1</sup>x))<sup>2</sup> – 1, |x| > 1.
<br/>If $${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$$ and $$y\left( {\sqrt 3 } \right) = {\pi \over 6}$$,
then y($${ - \sqrt 3 }$$) is equal to : | [{"identifier": "A", "content": "$${{5\\pi } \\over 6}$$"}, {"identifier": "B", "content": "$$ - {\\pi \\over 6}$$"}, {"identifier": "C", "content": "$${\\pi \\over 3}$$"}, {"identifier": "D", "content": "$${{2\\pi } \\over 3}$$"}] | ["B"] | null | Given ƒ(x) = (sin(tan<sup>–1</sup>x) + sin(cot<sup>–1</sup>x))<sup>2</sup> – 1
<br><br> = (sin(tan<sup>–1</sup>x) + sin($${\pi \over 2}$$ - tan<sup>–1</sup>x))<sup>2</sup> – 1
<br><br> = (sin(tan<sup>–1</sup>x) + cos(tan<sup>–1</sup>x))<sup>2</sup> – 1
<br><br>= sin<sup>2</sup>(tan<sup>–1</sup>x) + cos<sup>2</sup>(tan<sup>–1</sup>x) + 2sin(tan<sup>–1</sup>x)cos(tan<sup>–1</sup>x) + 1
<br><br>= 1 + sin(2tan<sup>–1</sup>x) - 1
<br><br>= sin(2tan<sup>–1</sup>x)
<br><br>Also given $${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$$
<br><br> Integrating both sides we get
<br><br>y = $${1 \over 2}$$ sin<sup>-1</sup> (f(x)) + C
<br><br>= $${1 \over 2}$$ sin<sup>-1</sup> (sin(2tan<sup>–1</sup>x)) + C
<br><br>Given $$y\left( {\sqrt 3 } \right) = {\pi \over 6}$$ mean x = $$\sqrt 3 $$ and y = $${\pi \over 6}$$
<br><br>$$ \therefore $$ $${\pi \over 6}$$ = $${1 \over 2}$$ sin<sup>-1</sup> (sin(2tan<sup>–1</sup>$$\sqrt 3 $$)) + C
<br><br>$$ \Rightarrow $$ $${\pi \over 6}$$ = $${1 \over 2}$$ sin<sup>-1</sup> (sin(2$$ \times $$$${\pi \over 3}$$)) + C
<br><br>$$ \Rightarrow $$ $${\pi \over 6}$$ = $${1 \over 2}$$ sin<sup>-1</sup> ($${{\sqrt 3 } \over 2}$$) + C
<br><br>$$ \Rightarrow $$ $${\pi \over 6}$$ = $${1 \over 2}$$ $$ \times $$ $${\pi \over 3}$$ + C
<br><br>$$ \Rightarrow $$ C = 0
<br><br>Now y($${ - \sqrt 3 }$$) means when x = $${ - \sqrt 3 }$$ then find y.
<br><br>y = $${1 \over 2}$$ sin<sup>-1</sup> (sin(2tan<sup>–1</sup>x))
<br><br>= $${1 \over 2}$$ sin<sup>-1</sup> (sin(2tan<sup>–1</sup>($${ - \sqrt 3 }$$)))
<br><br>= $${1 \over 2}$$ sin<sup>-1</sup> (sin(-2tan<sup>–1</sup>($${ \sqrt 3 }$$)))
<br><br>= $${1 \over 2}$$ sin<sup>-1</sup> (sin(-2$$ \times $$$${\pi \over 3}$$))
<br><br>= $${1 \over 2}$$ sin<sup>-1</sup> (-sin(2$$ \times $$$${\pi \over 3}$$))
<br><br>= $${1 \over 2}$$ sin<sup>-1</sup> (-$${{\sqrt 3 } \over 2}$$)
<br><br>= $${1 \over 2}$$ $$ \times $$ -sin<sup>-1</sup> ($${{\sqrt 3 } \over 2}$$)
<br><br>= $${1 \over 2}$$ $$ \times $$ -$${\pi \over 3}$$
<br><br>= -$${\pi \over 6}$$ | mcq | jee-main-2020-online-8th-january-morning-slot | 1,717 |
o7hlhlRCMBt3fIKQdBjgy2xukezff9hs | maths | differentiation | differentiation-of-inverse-trigonometric-function | If y = $$\sum\limits_{k = 1}^6 {k{{\cos }^{ - 1}}\left\{ {{3 \over 5}\cos kx - {4 \over 5}\sin kx} \right\}} $$,
<br/><br/>then $${{dy} \over {dx}}$$ at x = 0 is _______. | [] | null | 91 | Put, $$\cos \alpha = {3 \over 5},\sin \alpha = {4 \over 5}$$<br><br>
$$ \therefore {3 \over 5}\cos kx - {4 \over 5}\sin \,kx$$<br><br>
$$ = \cos \alpha .\cos kx - \sin \alpha .\sin kx$$<br><br>
$$ = \cos \left( {\alpha + kx} \right)$$<br><br>
So, $$y = \sum\limits_{k = 1}^6 {k{{\cos }^{ - 1}}\left( {\cos \left( {\alpha + kx} \right)} \right)} $$<br><br>
$$ = \sum\limits_{k = 1}^6 {\left( {{k^2}x + kx} \right)} $$<br><br>
$$ \Rightarrow {{dy} \over {dx}} = \sum\limits_{k = 1}^6 {{k^2}} $$<br><br>
$$ = {{6 \times 7 \times 13} \over 6} = 91$$ | integer | jee-main-2020-online-2nd-september-evening-slot | 1,718 |
D1eT2RJT4etOfyTR0y1kmjcgbe6 | maths | differentiation | differentiation-of-inverse-trigonometric-function | If $$f(x) = \sin \left( {{{\cos }^{ - 1}}\left( {{{1 - {2^{2x}}} \over {1 + {2^{2x}}}}} \right)} \right)$$ and its first derivative with respect to x is $$ - {b \over a}{\log _e}2$$ when x = 1, where a and b are integers, then the minimum value of | a<sup>2</sup> $$-$$ b<sup>2</sup> | is ____________ . | [] | null | 481 | $$f(x) = \sin {\cos ^{ - 1}}\left( {{{1 - {{({2^x})}^2}} \over {1 + {{({2^x})}^2}}}} \right)$$<br><br>$$ = \sin (2{\tan ^{ - 1}}{2^x})$$<br><br>$$f'(x) = \cos (2{\tan ^{ - 1}}{2^x}).2.{1 \over {1 + {{({2^x})}^2}}} \times {2^x}.{\log _e}2$$<br><br>$$ \therefore $$ $$f'(1) = \cos (2{\tan ^{ - 1}}2).{2 \over {1 + 4}} \times 2 \times {\log _e}2$$<br><br>$$ \Rightarrow f'(1) = \cos {\cos ^{ - 1}}\left( {{{1 - {2^2}} \over {1 + {2^2}}}} \right).{4 \over 5}{\log _e}2$$<br><br>$$ = - {{12} \over {25}}{\log _e}2$$<br><br>$$ \Rightarrow a = 25,b = 12$$<br><br>$$|{a^2} - {b^2}| = |625 - 144| = 481$$ | integer | jee-main-2021-online-17th-march-morning-shift | 1,719 |
1ktbc8v27 | maths | differentiation | differentiation-of-inverse-trigonometric-function | Let $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0 < x < 1. Then : | [{"identifier": "A", "content": "$${(1 - x)^2}f'(x) - 2{(f(x))^2} = 0$$"}, {"identifier": "B", "content": "$${(1 + x)^2}f'(x) + 2{(f(x))^2} = 0$$"}, {"identifier": "C", "content": "$${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$"}, {"identifier": "D", "content": "$${(1 + x)^2}f'(x) - 2{(f(x))^2} = 0$$"}] | ["C"] | null | $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$<br><br>$${\cot ^{ - 1}}\sqrt {{{1 - x} \over x}} = {\sin ^{ - 1}}\sqrt x $$<br><br>or $$f(x) = \cos (2{\tan ^{ - 1}}\sqrt x )$$<br><br>$$ = \cos {\tan ^{ - 1}}\left( {{{2\sqrt x } \over {1 - x}}} \right)$$<br><br>$$f(x) = {{1 - x} \over {1 + x}}$$<br><br>Now, $$f'(x) = {{ - 2} \over {{{(1 + x)}^2}}}$$<br><br>or $$f'(x){(1 - x)^2} = - 2{\left( {{{1 - x} \over {1 + x}}} \right)^2}$$<br><br>or $${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$. | mcq | jee-main-2021-online-26th-august-morning-shift | 1,720 |
1ktg2zxf8 | maths | differentiation | differentiation-of-inverse-trigonometric-function | If $$y(x) = {\cot ^{ - 1}}\left( {{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \over {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right),x \in \left( {{\pi \over 2},\pi } \right)$$, then $${{dy} \over {dx}}$$ at $$x = {{5\pi } \over 6}$$ is : | [{"identifier": "A", "content": "$$ - {1 \\over 2}$$"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "0"}] | ["A"] | null | We have,
<br/><br/>$$
y(x)=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)
$$
<br/><br/>$$
=\cot ^{-1} \frac{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|+\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|-\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}
$$
<br/><br/>$$
\left[\text { as } \cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}=1\right.
$$
<br/><br/>and $\left.\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}\right]$
<br/><br/>$$
\begin{aligned}
& =\cot ^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}}\right) \forall x \in\left(\frac{\pi}{2}, \pi\right) \\
&
\end{aligned}
$$
<br/><br/>$$
=\cot ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)=\cot ^{-1}\left(\tan \frac{x}{2}\right)
$$
<br/><br/>$$
=\frac{\pi}{2}-\tan ^{-1}\left(\tan \frac{x}{2}\right)
$$
<br/><br/>$$
\begin{aligned}
& \therefore y^{\prime}(x)=\frac{\pi}{2}-\frac{x}{2} \\\\
& \Rightarrow \frac{d y}{d x}=y^{\prime}(x)=-\frac{1}{2}=y^{\prime}\left(\frac{5 \pi}{6}\right)
\end{aligned}
$$ | mcq | jee-main-2021-online-27th-august-evening-shift | 1,721 |
1l5banuqb | maths | differentiation | differentiation-of-inverse-trigonometric-function | <p>If $$y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),{\pi \over 2} < {x^3} < {{3\pi } \over 2}$$, then</p> | [{"identifier": "A", "content": "$$xy'' + 2y' = 0$$"}, {"identifier": "B", "content": "$${x^2}y'' - 6y + {{3\\pi } \\over 2} = 0$$"}, {"identifier": "C", "content": "$${x^2}y'' - 6y + 3\\pi = 0$$"}, {"identifier": "D", "content": "$$xy'' - 4y' = 0$$"}] | ["B"] | null | <p>Let $${x^3} = \theta \Rightarrow {\theta \over 2} \in \left( {{\pi \over 4},\,{{3\pi } \over 4}} \right)$$</p>
<p>$$\therefore$$ $$y = {\tan ^{ - 1}}(\sec \theta - \tan \theta )$$</p>
<p>$$ = {\tan ^{ - 1}}\left( {{{1 - \sin \theta } \over {\cos \theta }}} \right)$$</p>
<p>$$\therefore$$ $$y = {\pi \over 4} - {\theta \over 2}$$</p>
<p>$$y = {\pi \over 4} - {{{x^3}} \over 2}$$</p>
<p>$$\therefore$$ $$y' = {{ - 3{x^2}} \over 2}$$</p>
<p>$$y'' = - 3x$$</p>
<p>$$\therefore$$ $${x^2}y'' - 6y + {{3\pi } \over 2} = 0$$</p> | mcq | jee-main-2022-online-24th-june-evening-shift | 1,722 |
luxwdj6m | maths | differentiation | differentiation-of-inverse-trigonometric-function | <p>If $$\log _e y=3 \sin ^{-1} x$$, then $$(1-x^2) y^{\prime \prime}-x y^{\prime}$$ at $$x=\frac{1}{2}$$ is equal to</p> | [{"identifier": "A", "content": "$$9 e^{\\pi / 2}$$\n"}, {"identifier": "B", "content": "$$9 e^{\\pi / 6}$$\n"}, {"identifier": "C", "content": "$$3 e^{\\pi / 2}$$\n"}, {"identifier": "D", "content": "$$3 e^{\\pi / 6}$$"}] | ["A"] | null | <p>$$\begin{aligned}
&\log _e y=3 \sin ^{-1} x\\
&\begin{aligned}
& y=e^{3 \sin ^{-1} x} \\
& \frac{d y}{d x}=e^{3 \sin ^{-1} x} \cdot \frac{3}{\sqrt{1-x^2}}
\end{aligned}
\end{aligned}$$</p>
<p>$$\sqrt{1-x^2} \frac{d y}{d x}=3 y$$</p>
<p>Again differentiate</p>
<p>$$\begin{aligned}
& \sqrt{1-x^2} \cdot y^{\prime \prime}-\frac{2 x}{2 \sqrt{1-x^2}} y^{\prime}=3 y^{\prime} \\
& (1-x)^2 y^{\prime \prime}-x y^{\prime}=3 y^{\prime}\left(\sqrt{1-x^2}\right)
\end{aligned}$$</p>
<p>So value of $$3 y^{\prime}\left(\sqrt{1-x^2}\right)$$ at $$x=\frac{1}{2}$$</p>
<p>$$\begin{aligned}
& 3 \cdot \frac{3}{\sqrt{1-x^2}} e^{\sin ^{-1} x}\left(\sqrt{1-x^2}\right) \\
& =9 e^{3 \frac{\pi}{6}}=9 e^{\frac{\pi}{2}}
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift | 1,723 |
lvc57bcq | maths | differentiation | differentiation-of-inverse-trigonometric-function | <p>Let $$f:(-\infty, \infty)-\{0\} \rightarrow \mathbb{R}$$ be a differentiable function such that $$f^{\prime}(1)=\lim _\limits{a \rightarrow \infty} a^2 f\left(\frac{1}{a}\right)$$. Then $$\lim _\limits{a \rightarrow \infty} \frac{a(a+1)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+a^2-2 \log _e a$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{5}{2}+\\frac{\\pi}{8}$$\n"}, {"identifier": "B", "content": "$$\\frac{3}{8}+\\frac{\\pi}{4}$$\n"}, {"identifier": "C", "content": "$$\\frac{3}{4}+\\frac{\\pi}{8}$$\n"}, {"identifier": "D", "content": "$$\\frac{3}{2}+\\frac{\\pi}{4}$$"}] | ["A"] | null | <p>Let $$f^{\prime}(1)=k$$</p>
<p>$$\Rightarrow \quad \lim _\limits{x \rightarrow 0} \frac{f(x)}{x^2}=k \quad\left(\frac{0}{0}\right)$$</p>
<p>$$\begin{aligned}
& \lim _\limits{x \rightarrow 0} \frac{f^{\prime}(x)}{2 x}=\lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)}{2}=k \\
\Rightarrow & f^{\prime \prime}(0)=2 k
\end{aligned}$$</p>
<p>Given information is not complete.</p> | mcq | jee-main-2024-online-6th-april-morning-shift | 1,724 |
3BblMHpmENyy47EL | maths | differentiation | differentiation-of-logarithmic-function | If $$x = {e^{y + {e^y} + {e^{y + .....\infty }}}}$$ , $$x > 0,$$ then $${{{dy} \over {dx}}}$$ is | [{"identifier": "A", "content": "$${{1 + x} \\over x}$$ "}, {"identifier": "B", "content": "$${1 \\over x}$$ "}, {"identifier": "C", "content": "$${{1 - x} \\over x}$$"}, {"identifier": "D", "content": "$${x \\over {1 + x}}$$ "}] | ["C"] | null | $$x = {e^{y + {e^{y + .....\infty }}}}\,\, \Rightarrow x = {e^{y + x}}.$$
<br><br>Taking log.
<br><br>$$\log \,\,x = y + x$$
<br><br>$$ \Rightarrow {1 \over x} = {{dy} \over {dx}} + 1$$
<br><br>$$ \Rightarrow {{dy} \over {dx}} = {1 \over x} - 1 = {{1 - x} \over x}$$
| mcq | aieee-2004 | 1,725 |
jmcGBPjprXE1bdtL | maths | differentiation | differentiation-of-logarithmic-function | If $${x^m}.{y^n} = {\left( {x + y} \right)^{m + n}},$$ then $${{{dy} \over {dx}}}$$ is | [{"identifier": "A", "content": "$${y \\over x}$$ "}, {"identifier": "B", "content": "$${{x + y} \\over {xy}}$$ "}, {"identifier": "C", "content": "$$xy$$ "}, {"identifier": "D", "content": "$${x \\over y}$$"}] | ["A"] | null | $${x^m}.{y^n} = {\left( {x + y} \right)^{m + n}}$$
<br><br>$$ \Rightarrow m\ln x + n\ln y = \left( {m + n} \right)\ln \left( {x + y} \right)$$
<br><br>Differentiating both sides.
<br><br>$$\therefore$$ $${m \over x} + {n \over y}{{dy} \over {dx}} = {{m + n} \over {x + y}}\left( {1 + {{dy} \over {dx}}} \right)$$
<br><br>$$ \Rightarrow \left( {{m \over x} - {{m + n} \over {x + y}}} \right) = \left( {{{m + n} \over {x + y}} - {n \over y}} \right){{dy} \over {dx}}$$
<br><br>$$ \Rightarrow {{my - nx} \over {x\left( {x + y} \right)}} = \left( {{{my - nx} \over {y\left( {x + y} \right)}}} \right){{dy} \over {dx}}$$
<br><br>$$ \Rightarrow {{dy} \over {dx}} = {y \over x}$$ | mcq | aieee-2006 | 1,726 |
oMWsMZ2BcQMVesDBdN3ng | maths | differentiation | differentiation-of-logarithmic-function | If xlog<sub>e</sub>(log<sub>e</sub>x) $$-$$ x<sup>2</sup> + y<sup>2</sup> = 4(y > 0), then $${{dy} \over {dx}}$$ at x = e is equal to : | [{"identifier": "A", "content": "$${{\\left( {1 + 2e} \\right)} \\over {2\\sqrt {4 + {e^2}} }}$$"}, {"identifier": "B", "content": "$${{\\left( {1 + 2e} \\right)} \\over {\\sqrt {4 + {e^2}} }}$$"}, {"identifier": "C", "content": "$${{\\left( {2e - 1} \\right)} \\over {2\\sqrt {4 + {e^2}} }}$$"}, {"identifier": "D", "content": "$${e \\over {\\sqrt {4 + {e^2}} }}$$"}] | ["C"] | null | Differentiating with respect to x,
<br><br>$$x.{1 \over {\ell nx}}.{1 \over x} + \ell n(\ell nx) - 2x + 2y.{{dy} \over {dx}} = 0$$
<br><br>at $$x = e$$ we get
<br><br>$$1 - 2e + 2y{{dy} \over {dx}} = 0 \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2y}}$$
<br><br>$$ \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2\sqrt {4 + {e^2}} }}\,\,$$
<br><br>as $$y(e) = \sqrt {4 + {e^2}} $$ | mcq | jee-main-2019-online-11th-january-morning-slot | 1,727 |
ltTrASebDziGIvmUP3p9h | maths | differentiation | differentiation-of-logarithmic-function | For x > 1, if (2x)<sup>2y</sup> = 4e<sup>2x$$-$$2y</sup>,
<br/><br/>then (1 + log<sub>e</sub> 2x)<sup>2</sup> $${{dy} \over {dx}}$$ is equal to : | [{"identifier": "A", "content": "$${{x\\,{{\\log }_e}2x - {{\\log }_e}2} \\over x}$$"}, {"identifier": "B", "content": "log<sub>e</sub> 2x"}, {"identifier": "C", "content": "x log<sub>e</sub> 2x"}, {"identifier": "D", "content": "$${{x\\,{{\\log }_e}2x + {{\\log }_e}2} \\over x}$$"}] | ["A"] | null | (2x)<sup>2y</sup> = 4e<sup>2x-2y</sup>
<br><br>2y$$\ell $$n2x = $$\ell $$n4 + 2x $$-$$ 2y
<br><br>y = $${{x + \ell n2} \over {1 + \ell n2x}}$$
<br><br>y ' = $${{\left( {1 + \ell n2x} \right) - \left( {x + \ell n2} \right){1 \over x}} \over {{{\left( {1 + \ell n2x} \right)}^2}}}$$
<br><br>y '$${\left( {1 + \ell n2x} \right)^2} = \left[ {{{x\ell n2x - \ell n2} \over x}} \right]$$ | mcq | jee-main-2019-online-12th-january-morning-slot | 1,728 |
1ktbitz1b | maths | differentiation | differentiation-of-logarithmic-function | If y = y(x) is an implicit function of x such that log<sub>e</sub>(x + y) = 4xy, then $${{{d^2}y} \over {d{x^2}}}$$ at x = 0 is equal to ___________. | [] | null | 40 | ln(x + y) = 4xy (At x = 0, y = 1)<br><br>x + y = e<sup>4xy</sup><br><br>$$ \Rightarrow 1 + {{dy} \over {dx}} = {e^{4xy}}\left( {4x{{dy} \over {dx}} + 4y} \right)$$<br><br>At x = 0 <br><br>$${{dy} \over {dx}} = 3$$<br><br>$${{{d^2}y} \over {d{x^2}}} = {e^{4xy}}{\left( {4x{{dy} \over {dx}} + 4y} \right)^2} + {e^{4xy}}\left( {4x{{{d^2}y} \over {d{x^2}}} + 4y} \right)$$<br><br>At x = 0, $${{{d^2}y} \over {d{x^2}}} = {e^0}{(4)^2} + {e^0}(24)$$<br><br>$$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = 40$$ | integer | jee-main-2021-online-26th-august-morning-shift | 1,729 |
1l57o8xuz | maths | differentiation | differentiation-of-logarithmic-function | <p>If $${\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5},\,|y| < 2$$, then :</p> | [{"identifier": "A", "content": "$${x^2}y'' + xy' - 25y = 0$$"}, {"identifier": "B", "content": "$${x^2}y'' - xy' - 25y = 0$$"}, {"identifier": "C", "content": "$${x^2}y'' - xy' + 25y = 0$$"}, {"identifier": "D", "content": "$${x^2}y'' + xy' + 25y = 0$$"}] | ["D"] | null | <p>$${\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5}\,\,\,\,\,\,\,\,\,|y| < 2$$</p>
<p>Differentiating on both side</p>
<p>$$ - {1 \over {\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} }} \times {{y'} \over 2} = {5 \over {{x \over 5}}} \times {1 \over 5}$$</p>
<p>$${{ - xy'} \over 2} = 5\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} $$</p>
<p>Square on both side</p>
<p>$${{{x^2}y{'^2}} \over 4} = 25\left( {{{4 - {y^2}} \over 4}} \right)$$</p>
<p>Diff on both side</p>
<p>$$2xy{'^2} + 2y'y''{x^2} = - 25 \times 2yy'$$</p>
<p>$$xy' + y''{x^2} + 25y = 0$$</p> | mcq | jee-main-2022-online-27th-june-morning-shift | 1,730 |
1l58gt0p9 | maths | differentiation | differentiation-of-logarithmic-function | <p>Let f : R $$\to$$ R satisfy $$f(x + y) = {2^x}f(y) + {4^y}f(x)$$, $$\forall$$x, y $$\in$$ R. If f(2) = 3, then $$14.\,{{f'(4)} \over {f'(2)}}$$ is equal to ____________.</p> | [] | null | 248 | <p>$$\because$$ $$f(x + y) = {2^x}f(y) + {4^y}f(x)$$ ....... (1)</p>
<p>Now, $$f(y + x){2^y}f(x) + {4^x}f(y)$$ ...... (2)</p>
<p>$$\therefore$$ $${2^x}f(y) + {4^y}f(x) = {2^y}f(x) + {4^x}f(y)$$</p>
<p>$$({4^y} - {2^y})f(x) = ({4^x} - {2^x})f(y)$$</p>
<p>$${{f(x)} \over {{4^x} - {2^x}}} = {{f(y)} \over {{4^y} - {2^y}}} = k$$</p>
<p>$$\therefore$$ $$f(x) = k({4^x} - {2^x})$$</p>
<p>$$\because$$ $$f(2) = 3$$ then $$k = {1 \over 4}$$</p>
<p>$$\therefore$$ $$f(x) = {{{4^x} - {2^x}} \over 4}$$</p>
<p>$$\therefore$$ $$f'(x) = {{{4^x}\ln 4 - {2^x}\ln 2} \over 4}$$</p>
<p>$$f'(x) = {{({{2.4}^x} - {2^x})ln2} \over 4}$$</p>
<p>$$\therefore$$ $${{f'(4)} \over {f'(2)}} = {{2.256 - 16} \over {2.16 - 4}}$$</p>
<p>$$\therefore$$ $$14{{f'(4)} \over {f'(2)}} = 248$$</p> | integer | jee-main-2022-online-26th-june-evening-shift | 1,731 |
1l6hy9kbq | maths | differentiation | differentiation-of-logarithmic-function | <p>The value of $$\log _{e} 2 \frac{d}{d x}\left(\log _{\cos x} \operatorname{cosec} x\right)$$ at $$x=\frac{\pi}{4}$$ is</p> | [{"identifier": "A", "content": "$$-2 \\sqrt{2}$$"}, {"identifier": "B", "content": "$$2 \\sqrt{2}$$"}, {"identifier": "C", "content": "$$-4$$"}, {"identifier": "D", "content": "4"}] | ["D"] | null | <p>Let $$f(x) = {\log _{\cos x}}\cos ec\,x$$</p>
<p>$$ = {{\log \cos ec\,x} \over {\log \cos x}}$$</p>
<p>$$ \Rightarrow f'(x) = {{\log \cos x\,.\,\sin x\,.\,\left( { - \cos ec\,x\cot x - \log \cos ec\,x\,.\,{1 \over {\cos x}}\,.\, - \sin x} \right)} \over {{{(\log \cos x)}^2}}}$$</p>
<p>at $$x = {\pi \over 4}$$</p>
<p>$$f'\left( {{\pi \over 4}} \right) = {{ - \log \left( {{1 \over {\sqrt 2 }}} \right) + \log \sqrt 2 } \over {{{\left( {\log {1 \over {\sqrt 2 }}} \right)}^2}}} = {2 \over {\log \sqrt 2 }}$$</p>
<p>$$\therefore$$ $${\log _e}2f'(x)$$ at $$x = {\pi \over 4} = 4$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift | 1,732 |
1ldo5zk2p | maths | differentiation | differentiation-of-logarithmic-function | <p>If $$y(x)=x^{x},x > 0$$, then $$y''(2)-2y'(2)$$ is equal to</p> | [{"identifier": "A", "content": "$$4(\\log_{e}2)^{2}+2$$"}, {"identifier": "B", "content": "$$8\\log_{e}2-2$$"}, {"identifier": "C", "content": "$$4\\log_{e}2+2$$"}, {"identifier": "D", "content": "$$4(\\log_{e}2)^{2}-2$$"}] | ["D"] | null | $\begin{aligned} & y=x^x \\\\ & y^{\prime}=x^x(1+\ln x) \\\\ & y^{\prime \prime}=x^x(1+\ln x)^2+\frac{x^x}{x} \\\\ & f^{\prime \prime}(2)-2 f^{\prime}(2)=\left(4(1+\ln 2)^2+2\right)-(2)(4(1+\ln 2)) \\\\ & =4\left(1+(\ln 2)^2\right)+2-8 \\\\ & =4(\ln 2)^2-2 \\\\ & \end{aligned}$ | mcq | jee-main-2023-online-1st-february-evening-shift | 1,733 |
1lh23oisg | maths | differentiation | differentiation-of-logarithmic-function | <p>If $$2 x^{y}+3 y^{x}=20$$, then $$\frac{d y}{d x}$$ at $$(2,2)$$ is equal to :</p> | [{"identifier": "A", "content": "$$-\\left(\\frac{3+\\log _{e} 16}{4+\\log _{e} 8}\\right)$$"}, {"identifier": "B", "content": "$$-\\left(\\frac{2+\\log _{e} 8}{3+\\log _{e} 4}\\right)$$"}, {"identifier": "C", "content": "$$-\\left(\\frac{3+\\log _{e} 8}{2+\\log _{e} 4}\\right)$$"}, {"identifier": "D", "content": "$$-\\left(\\frac{3+\\log _{e} 4}{2+\\log _{e} 8}\\right)$$"}] | ["B"] | null | Given, $2 x^y+3 y^x=20$ ..........(i)
<br/><br/>Let $u=x^y$
<br/><br/>On taking log both sides, we get
<br/><br/>$\log u=y \log x$
<br/><br/>On differentiating both sides with respect to $x$, we get
<br/><br/>$$
\begin{array}{rlrl}
& \frac{1}{u} \frac{d u}{d x} =y \frac{1}{x}+\log x \frac{d y}{d x} \\\\
& \Rightarrow \frac{d u}{d x} =u\left(\frac{y}{x}+\log x \frac{d y}{d x}\right) \\\\
& \Rightarrow \frac{d u}{d x} =x^y\left(\frac{y}{x}+\log x \frac{d y}{d x}\right) .........(ii)
\end{array}
$$
<br/><br/>Also, let $v=y^x$
<br/><br/>On taking log both sides, we get
<br/><br/>$\log v=x \log y$
<br/><br/>On differentiating both sides, we get
<br/><br/>$$
\begin{array}{rlrl}
& \frac{1}{v} \frac{d v}{d x} =x \frac{1}{y} \frac{d y}{d x}+\log y \cdot 1 \\\\
&\Rightarrow \frac{d v}{d x} =v\left(\frac{x}{y} \frac{d y}{d x}+\log y\right) \\\\
&\Rightarrow \frac{d v}{d x} =y^x\left(\frac{x}{y} \frac{d y}{d x}+\log y\right) ..........(iii)
\end{array}
$$
<br/><br/>Now, from Equation (i), $2 u+3 v=20$
<br/><br/>$$
\begin{aligned}
& \Rightarrow 2 \frac{d u}{d x}+3 \frac{d v}{d x}=0 \\\\
& \Rightarrow 2 x^y\left(\frac{y}{x}+\log x \frac{d y}{d x}\right)+3 y^x\left(\frac{x}{y} \frac{d y}{d x}+\log y\right)=0 \text { [Using Eqs. (ii) and (iii)] }
\end{aligned}
$$
<br/><br/>On putting $x=2$ and $y=2$, we get
<br/><br/>$$
\begin{aligned}
& 2(4)\left(1+\log 2 \frac{d y}{d x}\right)+3(4)\left(\frac{d y}{d x}+\log 2\right)=0 \\\\
& \Rightarrow \frac{d y}{d x}(8 \log 2+12)+(8+12 \log 2)=0 \\\\
& \Rightarrow \frac{d y}{d x}=-\left(\frac{2+3 \log 2}{3+2 \log 2}\right) \\\\
& \Rightarrow \frac{d y}{d x}=-\left(\frac{2+\log 8}{3+\log 4}\right)
\end{aligned}
$$ | mcq | jee-main-2023-online-6th-april-morning-shift | 1,734 |
jaoe38c1lsfkl66w | maths | differentiation | differentiation-of-logarithmic-function | <p>$$\text { Let } y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1 < x<1 \text {. Then at } x=\frac{1}{2} \text {, the value of } 225\left(y^{\prime}-y^{\prime \prime}\right) \text { is equal to }$$</p> | [{"identifier": "A", "content": "732"}, {"identifier": "B", "content": "736"}, {"identifier": "C", "content": "742"}, {"identifier": "D", "content": "746"}] | ["B"] | null | <p>$$\begin{aligned}
& y=\log _e\left(\frac{1-x^2}{1+x^2}\right) \\
& \frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^4}
\end{aligned}$$</p>
<p>Again,</p>
<p>$$\frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}$$</p>
<p>Again</p>
<p>$$y^{\prime}-y^{\prime \prime}=\frac{-4 x}{1-x^4}+\frac{4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}$$</p>
<p>at $$\mathrm{x}=\frac{1}{2}$$,</p>
<p>$$y^{\prime}-y^{\prime \prime}=\frac{736}{225}$$</p>
<p>Thus $$225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift | 1,735 |
ghiwTEg0WvaIpfIxQkmJb | maths | differentiation | differentiation-of-parametric-function | If x $$=$$ 3 tan t and y $$=$$ 3 sec t, then the value of $${{{d^2}y} \over {d{x^2}}}$$ at t $$ = {\pi \over 4},$$ is : | [{"identifier": "A", "content": "$${1 \\over {3\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over {6\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${3 \\over {2\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${1 \\over 6}$$"}] | ["B"] | null | x = 3 tan t and y = 3 sec t
<br><br>So that $${{dx} \over {dt}}$$ = 3sec<sup>2</sup>t and $${{dy} \over {dt}}$$ = 3 sec t tan t
<br><br>$${{dy} \over {dx}}$$ = $${{dy/dt} \over {dx/dt}}$$ = sin t
<br><br>$${{{d^2}y} \over {d{x^2}}}$$ = (cos t)$$.{{dt} \over {dx}}$$
<br><br>$${{{d^2}y} \over {d{x^2}}} = \left( {\cos t} \right).{1 \over {3{{\sec }^2}t}}$$
<br><br>$${{{d^2}y} \over {d{x^2}}}$$ = $${1 \over 3}$$(cos<sup>3</sup> t)
<br><br>$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{t = \pi /4}}$$ = $${1 \over 3} \times {\left( {{1 \over {\sqrt 2 }}} \right)^3} = {1 \over {6\sqrt 2 }}$$ | mcq | jee-main-2019-online-9th-january-evening-slot | 1,736 |
JDlJzbuL7enwokKQJU7k9k2k5k6v2n8 | maths | differentiation | differentiation-of-parametric-function | If $$x = 2\sin \theta - \sin 2\theta $$ and $$y = 2\cos \theta - \cos 2\theta $$,<br/>
$$\theta \in \left[ {0,2\pi } \right]$$, then $${{{d^2}y} \over {d{x^2}}}$$ at $$\theta $$ = $$\pi $$ is : | [{"identifier": "A", "content": "$${3 \\over 8}$$"}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "$${3 \\over 4}$$"}, {"identifier": "D", "content": "-$${3 \\over 4}$$"}] | ["A"] | null | $$x = 2\sin \theta - \sin 2\theta $$
<br><br>$$ \Rightarrow $$ $${{dx} \over {d\theta }}$$ = $$2\cos \theta - 2\cos 2\theta $$
<br><br>$$y = 2\cos \theta - \cos 2\theta $$
<br><br>$$ \Rightarrow $$ $${{dy} \over {d\theta }}$$ = –2sin$$\theta $$ + 2sin2$$\theta $$
<br><br>$${{dy} \over {dx}} = {{{{dy} \over {d\theta }}} \over {{{dx} \over {d\theta }}}}$$ = $${{\sin 2\theta - \sin \theta } \over {\cos \theta - \cos 2\theta }}$$
<br/><br/>This expression is simplified by recognizing that $\sin 2 \theta = 2\sin \theta \cos \theta$ and $\cos 2 \theta = 2\cos^2 \theta -1$, leading to the expression:
<br/><br/>$$ \frac{dy}{dx} = \frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{3 \theta}{2}}{2 \sin \frac{\theta}{2} \cdot \sin \frac{3 \theta}{2}}=\cot \frac{3 \theta}{2}$$
<br/><br/>This is the rate of change of y with respect to x, as a function of θ.
<br/><br/>Next, we differentiate this function with respect to θ to find $\frac{d^2 y}{dx^2}$, yielding :
<br/><br/>$$ \frac{d^2 y}{dx^2}=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \frac{d \theta}{d x}=-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2} \cdot \frac{d \theta}{d x}$$
<br/><br/>Here, $\frac{d \theta}{d x}$ is the reciprocal of $\frac{dx}{d\theta}$, so the equation becomes :
<br/><br/>$$ \frac{d^2 y}{dx^2}=\frac{-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2}}{2\left(\cos \theta-\cos2 \theta\right)}$$
<br/><br/>Finally, we evaluate this expression at $\theta = \pi$, yielding :
<br/><br/>$$ \frac{d^2 y}{dx^2}(\pi)=\frac{-3}{4(-1-1)}=\frac{3}{8}$$
<br/><br/>Therefore, the correct answer is (A) $\frac{3}{8}$.
<br/><br/><b>Alternate Method :</b>
<br/><br/>First, let's find the derivatives of x and y with respect to θ :
<br/><br/>1) $$ \frac{dx}{d\theta} = 2\cos \theta - 2\cos 2\theta $$
<br/><br/>2) $$ \frac{dy}{d\theta} = -2\sin \theta + 2\sin 2\theta $$
<br/><br/>We know that $$ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} $$
<br/><br/>So, we substitute 1) and 2) into this equation :
<br/><br/>We have
<br/><br/>$$\frac{dy}{dx} = \frac{-2\sin \theta + 2\sin 2\theta}{2\cos \theta - 2\cos 2\theta} = \frac{\sin 2\theta - \sin \theta}{\cos \theta - \cos 2\theta}$$
<br/><br/>For simplification, let's denote the numerator as $$N = \sin 2\theta - \sin \theta$$ and the denominator as $$D = \cos \theta - \cos 2\theta$$.
<br/><br/>We have to compute $$\frac{d}{d\theta}(\frac{dy}{dx})$$ which is $$\frac{d}{d\theta}(\frac{N}{D})$$.
<br/><br/>We can use the quotient rule for differentiation, which states that if we have a function of the form $$\frac{u}{v}$$, then its derivative is given by $$\frac{vu' - uv'}{v^2}$$.
<br/><br/>So here, $$N' = 2\cos 2\theta - \cos \theta$$ and $$D' = -\sin \theta + 2\sin 2\theta$$.
<br/><br/>Applying the quotient rule :
<br/><br/>$$\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{D N' - N D'}{D^2}$$
<br/><br/>Substituting the expressions for $$N'$$, $$D'$$, $$N$$, and $$D$$ we get :
<br/><br/>$$\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$$
<br/><br/>Now $$
\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \times \frac{d \theta}{d x}
$$
<br/><br/>= $$\frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$$$$
\times \frac{1}{(2 \cos \theta-2 \cos 2 \theta)}
$$
<br/><br/>$$
\begin{aligned}
\left.\therefore \frac{d^2 y}{d x^2}\right|_{\theta=\pi} & =\frac{(-1-1)(2+1)-(0-0)(-0+0)}{2(-1-1)^3} \\\\
& =\frac{-2 \times 3}{-2 \times 8}=\frac{3}{8}
\end{aligned}
$$ | mcq | jee-main-2020-online-9th-january-evening-slot | 1,737 |
1l6nm5311 | maths | differentiation | differentiation-of-parametric-function | <p>Let $$x(t)=2 \sqrt{2} \cos t \sqrt{\sin 2 t}$$ and <br/><br/>$$y(t)=2 \sqrt{2} \sin t \sqrt{\sin 2 t}, t \in\left(0, \frac{\pi}{2}\right)$$. <br/><br/>Then $$\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}$$ at $$t=\frac{\pi}{4}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{-2 \\sqrt{2}}{3}$$"}, {"identifier": "B", "content": "$$\\frac{2}{3}$$"}, {"identifier": "C", "content": "$$\\frac{1}{3}$$"}, {"identifier": "D", "content": "$$ \\frac{-2}{3}$$"}] | ["D"] | null | <p>$$x = 2\sqrt 2 \cos t\sqrt {\sin 2t} ,\,y = 2\sqrt 2 \sin t\sqrt {\sin 2t} $$</p>
<p>$$\therefore$$ $${{dx} \over {dt}} = {{2\sqrt 2 \cos 3t} \over {\sqrt {\sin 2t} }},\,{{dy} \over {dt}} = {{2\sqrt 2 \sin 3t} \over {\sqrt {\sin 2t} }}$$</p>
<p>$$\therefore$$ $${{dy} \over {dx}} = \tan 3t,\,\left( {\mathrm{at}\,t = {\pi \over 4},\,{{dy} \over {dx}} = - 1} \right)$$</p>
<p>and $${{{d^2}y} \over {d{x^2}}} = 3{\sec ^2}3t\,.\,{{dt} \over {dx}} = {{3{{\sec }^2}3t\,.\,\sqrt {\sin 2t} } \over {2\sqrt 2 \cos 3t}}$$</p>
<p>$$\left( {\mathrm{At}\,t = {\pi \over 4},\,{{{d^2}y} \over {d{x^2}}} = - 3} \right)$$</p>
<p>$$\therefore$$ $${{1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \over {{{{d^2}y} \over {d{x^2}}}}} = {2 \over { - 3}} = {{ - 2} \over 3}$$</p> | mcq | jee-main-2022-online-28th-july-evening-shift | 1,738 |
lmUN4HWdFdPr7roD | maths | differentiation | methods-of-differentiation | $${{{d^2}x} \over {d{y^2}}}$$ equals: | [{"identifier": "A", "content": "$$ - {\\left( {{{{d^2}y} \\over {d{x^2}}}} \\right)^{ - 1}}{\\left( {{{dy} \\over {dx}}} \\right)^{ - 3}}$$ "}, {"identifier": "B", "content": "$${\\left( {{{{d^2}y} \\over {d{x^2}}}} \\right)^{}}{\\left( {{{dy} \\over {dx}}} \\right)^{ - 2}}$$ "}, {"identifier": "C", "content": "$$ - \\left( {{{{d^2}y} \\over {d{x^2}}}} \\right){\\left( {{{dy} \\over {dx}}} \\right)^{ - 3}}$$ "}, {"identifier": "D", "content": "$${\\left( {{{{d^2}y} \\over {d{x^2}}}} \\right)^{ - 1}}$$ "}] | ["C"] | null | $${{{d^2}x} \over {d{y^2}}} = {d \over {dy}}\left( {{{dx} \over {dy}}} \right)$$
<br><br>$$ = {d \over {dx}}\left( {{{dx} \over {dy}}} \right){{dx} \over {dy}}$$
<br><br>$$ = {d \over {dx}}\left( {{1 \over {dy/dx}}} \right){{dx} \over {dy}}$$
<br><br>$$ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^2}}}.{{{d^2}y} \over {d{x^2}}}.{1 \over {{{dy} \over {dx}}}}$$
<br><br>$$ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^3}}}{{{d^2}y} \over {d{x^2}}}$$ | mcq | aieee-2011 | 1,739 |
DdF7R8tTzx4Kv40prc7k9k2k5khz7tx | maths | differentiation | methods-of-differentiation | Let ƒ and g be differentiable functions on R
such that fog is the identity function. If for some
a, b $$ \in $$ R, g'(a) = 5 and g(a) = b, then ƒ'(b) is
equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "$${2 \\over 5}$$"}, {"identifier": "D", "content": "$${1 \\over 5}$$"}] | ["D"] | null | Given the function composition f(g(x)) is the identity function, it means f(g(x)) = x for all x.
<br><br>$$ \Rightarrow $$ ƒ'(g(x)) g'(x) = 1
<br><br>put x = a
<br><br>$$ \Rightarrow $$ ƒ'(b) g'(a) = 1
<br><br>$$ \Rightarrow $$ ƒ'(b) = $${1 \over 5}$$ | mcq | jee-main-2020-online-9th-january-evening-slot | 1,740 |
1l54ubqt9 | maths | differentiation | methods-of-differentiation | <p>Let f and g be twice differentiable even functions on ($$-$$2, 2) such that $$f\left( {{1 \over 4}} \right) = 0$$, $$f\left( {{1 \over 2}} \right) = 0$$, $$f(1) = 1$$ and $$g\left( {{3 \over 4}} \right) = 0$$, $$g(1) = 2$$. Then, the minimum number of solutions of $$f(x)g''(x) + f'(x)g'(x) = 0$$ in $$( - 2,2)$$ is equal to ________.</p> | [] | null | 4 | Let $h(x)=f(x) \cdot g^{\prime}(x)$
<br/><br/>
As $f(x)$ is even $f\left(\frac{1}{2}\right)=\left(\frac{1}{4}\right)=0$
<br/><br/>
$\Rightarrow f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{4}\right)=0$
<br/><br/>
and $g(x)$ is even $\Rightarrow g^{\prime}(x)$ is odd <br/><br/>
and $g(1)=2$ ensures one root of $g^{\prime}(x)$ is 0 .
<br/><br/>
So, $h(x)=f(x) \cdot g^{\prime}(x)$ has minimum five zeroes
<br/><br/>
$\therefore h^{\prime}(x)=f^{\prime}(x) \cdot g^{\prime}(x)+f(x) \cdot g^{\prime \prime}(x)=0$,
<br/><br/>
has minimum 4 zeroes
| integer | jee-main-2022-online-29th-june-evening-shift | 1,741 |
1ldswyfk5 | maths | differentiation | methods-of-differentiation | <p>Let $$f:\mathbb{R}\to\mathbb{R}$$ be a differentiable function that satisfies the relation $$f(x+y)=f(x)+f(y)-1,\forall x,y\in\mathbb{R}$$. If $$f'(0)=2$$, then $$|f(-2)|$$ is equal to ___________.</p> | [] | null | 3 | $f(x+y)=f(x)+f(y)-1$
<br/><br/>
$$
\begin{aligned}
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\\\
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=f^{\prime}(0)=2 \\\\
& f^{\prime}(x)=2 \Rightarrow d y=2 d x \\\\
& y=2 x+C \\\\
& \mathrm{x}=0, \mathrm{y}=1, \mathrm{c}=1 \\\\
& \mathrm{y}=2 \mathrm{x}+1 \\\\
& |f(-2)|=|-4+1|=|-3|=3
\end{aligned}
$$ | integer | jee-main-2023-online-29th-january-morning-shift | 1,742 |
1lgq0jgd2 | maths | differentiation | methods-of-differentiation | <p>For the differentiable function $$f: \mathbb{R}-\{0\} \rightarrow \mathbb{R}$$, let $$3 f(x)+2 f\left(\frac{1}{x}\right)=\frac{1}{x}-10$$, then $$\left|f(3)+f^{\prime}\left(\frac{1}{4}\right)\right|$$ is equal to</p> | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "$$\\frac{29}{5}$$"}, {"identifier": "C", "content": "$$\\frac{33}{5}$$"}, {"identifier": "D", "content": "7"}] | ["A"] | null | <ol>
<li><p>Given the equation: $$3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$$</p>
</li>
<li><p>Replace $$x$$ with $$\frac{1}{x}$$ in the original equation:
<br/>$$3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$$</p>
</li>
<li><p>Now, we have two equations:</p>
</li>
</ol>
<p>$$3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$$
<br/><br/>$$3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$$</p>
<ol>
<li>By adding the two equations, we can find $$f(x)$$:</li>
</ol>
<p>$$5f(x) = \frac{3}{x} - 2x - 10$$</p>
<ol>
<li>Now, let's differentiate both sides with respect to $$x$$:</li>
</ol>
<p>$$5f'(x) = -\frac{3}{x^2} - 2$$</p>
<ol>
<li>Now, we can find the values for $$f(3)$$ and $$f'\left(\frac{1}{4}\right)$$:</li>
</ol>
<p>$$f(3) = \frac{1}{5}(1 - 6 - 10) = -3$$
<br/><br/>$$f'\left(\frac{1}{4}\right) = \frac{1}{5}(-48 - 2) = -10$$</p>
<ol>
<li>Finally, calculate the expression we are interested in :</li>
</ol>
<p>$$\left|f(3) + f'\left(\frac{1}{4}\right)\right| = \left|-3 - 10\right| = 13$$</p> | mcq | jee-main-2023-online-13th-april-morning-shift | 1,743 |
lsan2rgn | maths | differentiation | methods-of-differentiation | If $y=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 x$, then $96 y^{\prime}\left(\frac{\pi}{6}\right)$ is equal to : | [] | null | 105 | $\begin{aligned} & y=\frac{(\sqrt{x}+1)\left(x^2-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^2 x-5\right) \cos ^3 x \\\\ & y=\frac{(\sqrt{x}+1)(\sqrt{x})\left((\sqrt{x})^3-1\right)}{(\sqrt{x})\left((\sqrt{x})^2+(\sqrt{x})+1\right)}+\frac{1}{5} \cos ^5 x-\frac{1}{3} \cos ^3 x \\\\ & y=(\sqrt{x}+1)(\sqrt{x}-1)+\frac{1}{5} \cos ^5 x-\frac{1}{3} \cos ^3 x\end{aligned}$
<br/><br/>$\begin{aligned} & y^{\prime}=1-\cos ^4 x \cdot(\sin x)+\cos ^2 x(\sin x \\\\ & y^{\prime}\left(\frac{\pi}{6}\right)=1-\frac{9}{16} \times \frac{1}{2}+\frac{3}{4} \times \frac{1}{2} \\\\ & =\frac{32-9+12}{32}=\frac{35}{32} \\\\ & \therefore 96 y^{\prime}\left(\frac{\pi}{6}\right)=105\end{aligned}$ | integer | jee-main-2024-online-1st-february-evening-shift | 1,744 |
jaoe38c1lsey8r3c | maths | differentiation | methods-of-differentiation | <p>Suppose $$f(x)=\frac{\left(2^x+2^{-x}\right) \tan x \sqrt{\tan ^{-1}\left(x^2-x+1\right)}}{\left(7 x^2+3 x+1\right)^3}$$. Then the value of $$f^{\prime}(0)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\pi$$\n"}, {"identifier": "B", "content": "$$\\sqrt{\\pi}$$\n"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$\\frac{\\pi}{2}$$"}] | ["B"] | null | <p>$$\begin{aligned}
& f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\
& =\lim _{h \rightarrow 0} \frac{\left(2^h+2^{-h}\right) \tan h \sqrt{\tan ^{-1}\left(h^2-h+1\right)}-0}{\left(7 h^2+3 h+1\right)^3 h} \\
& =\sqrt{\pi}
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift | 1,745 |
1lsg3xnzo | maths | differentiation | methods-of-differentiation | <p>Let $$f: \mathbb{R}-\{0\} \rightarrow \mathbb{R}$$ be a function satisfying $$f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$$ for all $$x, y, f(y) \neq 0$$. If $$f^{\prime}(1)=2024$$, then</p> | [{"identifier": "A", "content": "$$x f^{\\prime}(x)+2024 f(x)=0$$\n"}, {"identifier": "B", "content": "$$x f^{\\prime}(x)-2023 f(x)=0$$\n"}, {"identifier": "C", "content": "$$x f^{\\prime}(x)-2024 f(x)=0$$\n"}, {"identifier": "D", "content": "$$x f^{\\prime}(x)+f(x)=2024$$"}] | ["C"] | null | <p>$$f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$$</p>
<p>$$\begin{aligned}
& \mathrm{f}^{\prime}(1)=2024 \\
& \mathrm{f}(1)=1
\end{aligned}$$</p>
<p>Partially differentiating w. r. t. x</p>
<p>$$\begin{aligned}
& \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{\mathrm{y}}\right) \cdot \frac{1}{\mathrm{y}}=\frac{1}{\mathrm{f}(\mathrm{y})} \mathrm{f}^{\prime}(\mathrm{x}) \\
& \mathrm{y} \rightarrow \mathrm{x} \\
& \mathrm{f}^{\prime}(1) \cdot \frac{1}{\mathrm{x}}=\frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \\
& 2024 \mathrm{f}(\mathrm{x})=\mathrm{xf}^{\prime}(\mathrm{x}) \Rightarrow \mathrm{xf}^{\prime}(\mathrm{x})-2024 \mathrm{~f}(\mathrm{x})=0
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-evening-shift | 1,746 |
1lsg8vyuz | maths | differentiation | methods-of-differentiation | <p>Let $$g: \mathbf{R} \rightarrow \mathbf{R}$$ be a non constant twice differentiable function such that $$\mathrm{g}^{\prime}\left(\frac{1}{2}\right)=\mathrm{g}^{\prime}\left(\frac{3}{2}\right)$$. If a real valued function $$f$$ is defined as $$f(x)=\frac{1}{2}[g(x)+g(2-x)]$$, then</p> | [{"identifier": "A", "content": "$$f^{\\prime \\prime}(x)=0$$ for atleast two $$x$$ in $$(0,2)$$\n"}, {"identifier": "B", "content": "$$f^{\\prime}\\left(\\frac{3}{2}\\right)+f^{\\prime}\\left(\\frac{1}{2}\\right)=1$$\n"}, {"identifier": "C", "content": "$$f^{\\prime \\prime}(x)=0$$ for no $$x$$ in $$(0,1)$$\n"}, {"identifier": "D", "content": "$$f^{\\prime \\prime}(x)=0$$ for exactly one $$x$$ in $$(0,1)$$"}] | ["A"] | null | <p>$$f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(2-x)}{2}, f^{\prime}\left(\frac{3}{2}\right)=\frac{g^{\prime}\left(\frac{3}{2}\right)-g^{\prime}\left(\frac{1}{2}\right)}{2}=0$$</p>
<p>Also $$\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\frac{\mathrm{g}^{\prime}\left(\frac{1}{2}\right)-\mathrm{g}^{\prime}\left(\frac{3}{2}\right)}{2}=0, \mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$$</p>
<p>$$\Rightarrow \mathrm{f}^{\prime}\left(\frac{3}{2}\right)=\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$$</p>
<p>$$\Rightarrow \operatorname{rootsin}\left(\frac{1}{2}, 1\right)$$ and $$\left(1, \frac{3}{2}\right)$$</p>
<p>$$\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})$$ is zero at least twice in $$\left(\frac{1}{2}, \frac{3}{2}\right)$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift | 1,747 |
luy6z53a | maths | differentiation | methods-of-differentiation | <p>Let $$f(x)=a x^3+b x^2+c x+41$$ be such that $$f(1)=40, f^{\prime}(1)=2$$ and $$f^{\prime \prime}(1)=4$$. Then $$a^2+b^2+c^2$$ is equal to:</p> | [{"identifier": "A", "content": "54"}, {"identifier": "B", "content": "51"}, {"identifier": "C", "content": "73"}, {"identifier": "D", "content": "62"}] | ["B"] | null | <p>Given the polynomial function:</p>
<p>$$f(x) = ax^3 + bx^2 + cx + 41$$</p>
<p>We are provided the following conditions from the problem:</p>
<p>1. $$f(1) = 40$$</p>
<p>2. $$f^{\prime}(1) = 2$$</p>
<p>3. $$f^{\prime \prime}(1) = 4$$</p>
<p>First, calculate $f(1)$:</p>
<p>$$f(1) = a(1)^3 + b(1)^2 + c(1) + 41 = 40$$</p>
<p>Simplifying, we get:</p>
<p>$$a + b + c + 41 = 40$$</p>
<p>Therefore:</p>
<p>$$a + b + c = -1$$</p>
<p>Next, calculate the first derivative $f^{\prime}(x)$:</p>
<p>$$f^{\prime}(x) = 3ax^2 + 2bx + c$$</p>
<p>Given $f^{\prime}(1) = 2$:</p>
<p>$$f^{\prime}(1) = 3a(1)^2 + 2b(1) + c = 2$$</p>
<p>Simplifying, we get:</p>
<p>$$3a + 2b + c = 2$$</p>
<p>Next, calculate the second derivative $f^{\prime \prime}(x)$:</p>
<p>$$f^{\prime \prime}(x) = 6ax + 2b$$</p>
<p>Given $f^{\prime \prime}(1) = 4$:</p>
<p>$$f^{\prime \prime}(1) = 6a(1) + 2b = 4$$</p>
<p>Simplifying, we get:</p>
<p>$$6a + 2b = 4$$</p>
<p>Dividing the entire equation by 2:</p>
<p>$$3a + b = 2$$</p>
<p>We now have three equations:</p>
<p>1. $$a + b + c = -1$$</p>
<p>2. $$3a + 2b + c = 2$$</p>
<p>3. $$3a + b = 2$$</p>
<p>To solve for $a$, $b$, and $c$, follow these steps:</p>
<p>First, subtract the third equation from the second equation:</p>
<p>$$(3a + 2b + c) - (3a + b) = 2 - 2$$</p>
<p>Which simplifies to:</p>
<p>$$b + c = 0$$</p>
<p>So,</p>
<p>$$c = -b$$</p>
<p>Substitute $c = -b$ into the first equation:</p>
<p>$$(a + b - b = -1)$$</p>
<p>Simplifying, we get:</p>
<p>$$a = -1$$</p>
<p>Now substitute $a = -1$ into the third equation:</p>
<p>$$3(-1) + b = 2$$</p>
<p>Which simplifies to:</p>
<p>$$-3 + b = 2$$</p>
<p>Therefore:</p>
<p>$$b = 5$$</p>
<p>Next, since $c = -b$:</p>
<p>$$c = -5$$</p>
<p>Finally, we need to find $a^2 + b^2 + c^2$:</p>
<p>$$a^2 + b^2 + c^2 = (-1)^2 + 5^2 + (-5)^2$$</p>
<p>Simplifying, we get:</p>
<p>$$1 + 25 + 25 = 51$$</p>
<p>Therefore, the answer is:</p>
<p><strong>Option B: 51</strong></p> | mcq | jee-main-2024-online-9th-april-morning-shift | 1,748 |
aLwt0WnHgsFA2mmz | maths | differentiation | successive-differentiation | If $$f\left( x \right) = {x^n},$$ then the value of
<p>$$f\left( 1 \right) - {{f'\left( 1 \right)} \over {1!}} + {{f''\left( 1 \right)} \over {2!}} - {{f'''\left( 1 \right)} \over {3!}} + ..........{{{{\left( { - 1} \right)}^n}{f^n}\left( 1 \right)} \over {n!}}$$ is</p> | [{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$${{2^n}}$$ "}, {"identifier": "C", "content": "$${{2^n} - 1}$$ "}, {"identifier": "D", "content": "$$0$$"}] | ["D"] | null | $$f\left( x \right) = {x^n} \Rightarrow f\left( 1 \right) = 1$$
<br><br>$$f'\left( x \right) = n{x^{n - 1}} \Rightarrow f'\left( 1 \right) = n$$
<br><br>$$f''\left( x \right) = n\left( {n - 1} \right){x^{n - 2}}$$
<br><br>$$ \Rightarrow f''\left( 1 \right) = n\left( {n - 1} \right)$$
<br><br>$$\therefore$$ $${f^n}\left( x \right) = n!$$
<br><br>$$ \Rightarrow {f^n}\left( 1 \right) = n!$$
<br><br>$$ = 1 - {n \over {1!}} + {{n\left( {n - 1} \right)} \over {2!}}{{n\left( {n - 1} \right)\left( {n - 2} \right)} \over {3!}}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + .... + {\left( { - 1} \right)^n}{{n!} \over {n!}}$$
<br><br>$$ = {}^n\,{C_0} - {}^n\,{C_1} + {}^n\,{C_2} - {}^n\,{C_3}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ...... + {\left( { - 1} \right)^n}\,{}^n{C_n} = 0$$ | mcq | aieee-2003 | 1,749 |
oPx9mrcOVu1a3ZqfY3d42 | maths | differentiation | successive-differentiation | Let f be a polynomial function such that
<br/><br/>f (3x) = f ' (x) . f '' (x), for all x $$ \in $$ <b>R</b>. Then : | [{"identifier": "A", "content": "f (2) + f ' (2) = 28"}, {"identifier": "B", "content": "f '' (2) $$-$$ f ' (2) = 0"}, {"identifier": "C", "content": "f '' (2) $$-$$ f (2) = 4"}, {"identifier": "D", "content": "f (2) $$-$$ f ' (2) + f '' (2) = 10"}] | ["B"] | null | <p>Let $$f(x) = {a_0}{x^n} + {a_1}{x^{n - 1}} + {a_2}{x^{n - 1}} + \,\,....\,\, + {a_{n - 1}}x + {a_n}$$</p>
<p>$$f'(x) = {a_0}n{x^{n - 1}} + {a_1}(n - 1){x^{n - 2}} + \,\,.....\,\, + {a_{n - 1}}$$</p>
<p>$$f''(x) = {a_0}n(n - 1){x^{n - 2}} + {a_1}(n - 1)(n - 2){x^{n - 3}} + \,\,....\,\, + {a_{n - 2}}$$</p>
<p>Now,</p>
<p>$$f(3x) = {3^n}{a_0}{x^n} + {3^{n - 1}}{a_1}{x^{n - 1}} + {3^{n - 2}}{a_2}{x^{n - 2}} + \,\,....\,\, + 3{a_{n - 1}} + {a_n}$$</p>
<p>$$f'(x)\,.\,f''(x) = [{a_0}n{x^{n - 1}} + {a_1}(n - 1){x^{n - 2}} + \,\,....\,\, + {a_{n - 1}}]$$</p>
<p>$$[{a_0}n(n - 1){x^{n - 2}} + {a_1}(n - 1)(n - 2){x^{n - 3}} + \,\,....\,\, + {a_{n - 2}}]$$</p>
<p>Comparing highest powers of x, we get</p>
<p>$${3^n}{a_0}{x_n} = a_0^2(n - 1){x^{n - 1 + n - 2}} = a_0^2{n^2}(n - 1){x^{2n - 3}}$$</p>
<p>Therefore, $$2n - 3 = n$$</p>
<p>$$\Rightarrow$$ n = 3 and $${3^n}{a_0} = a_0^2{n^2}(n - 1)$$</p>
<p>$$ \Rightarrow {a_0} = 27 = {3 \over 2}$$</p>
<p>Therefore, $$f(x) = {3 \over 2}{x^3} + {a_1}{x^2} + {a_2}x + {a_3}$$</p>
<p>$$f'(x) = {9 \over 2}{x^2} + 2{a_1}x + {a_2}$$</p>
<p>$$f''(x) = 9x + 2{a_1}$$</p>
<p>$$f(3x) = {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3}$$</p>
<p>Now, $$f(3x) = f'(x)\,.\,f''(x)$$</p>
<p>$$ \Rightarrow {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3}$$</p>
<p>$$ \Rightarrow \left( {{9 \over 2}{x^2} + 2{a_1}x + {a_2}} \right)(9x + 2{a_1})$$</p>
<p>$$ \Rightarrow {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3} = {{81} \over 2}{x^3} + [9{a_1} + 18{a_1}]{x^2} + [4a_1^2 + 9{a_2}]x + 2{a_1}{a_2}$$</p>
<p>Comparing the coefficients, we get</p>
<p>$$9{a_1} = 27{a_1}$$</p>
<p>$$ \Rightarrow {a_1} = 0,\,3{a_2} = 4a_1^2 + 9{a_2} = 9{a_2}$$</p>
<p>$$ \Rightarrow {a_2} = 0$$</p>
<p>Therefore, $$f(x) = {3 \over 2}{x^3}$$</p>
<p>$$f'(x) = {9 \over 2}{x^2}$$</p>
<p>$$f''(x) = 9x$$</p>
<p>Hence, $$f''(2) - f'(x) = 18 - 18 = 0$$</p> | mcq | jee-main-2017-online-9th-april-morning-slot | 1,750 |
7QKM8xqcNeiQvnGv4PFtQ | maths | differentiation | successive-differentiation | Let f : R $$ \to $$ R be a function such that f(x) = x<sup>3</sup> + x<sup>2</sup>f'(1) + xf''(2) + f'''(3), x $$ \in $$ R. Then f(2) equals - | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "$$-$$ 2"}, {"identifier": "C", "content": "$$-$$ 4"}, {"identifier": "D", "content": "8"}] | ["B"] | null | f(x) = x<sup>3</sup> + x<sup>2</sup>f '(1) + xf ''(2) + f '''(3)
<br><br>$$ \Rightarrow $$ f '(x) = 3x<sup>2</sup> + 2xf '(1) + f ''(x) . . . . . (1)
<br><br>$$ \Rightarrow $$ f ''(x) = 6x + 2f '(1) . . . . . . (2)
<br><br>$$ \Rightarrow $$ f '''(x) = 6 . . . . . .(3)
<br><br>put x = 1 in equation (1) :
<br><br>f '(1) = 3 + 2f '(1) + f ''(2) . . . . .(4)
<br><br>put x = 2 in equation (2) :
<br><br>f ''(2) = 12 + 2f '(1) . . . . .(5)
<br><br>from equation (4) & (5) :
<br><br>$$-$$3 $$-$$ f '(1) = 12 + 2f'(1)
<br><br>$$ \Rightarrow $$ 3f '(1) = $$-$$ 15
<br><br>$$ \Rightarrow $$ f '(1) = $$-$$ 5 $$ \Rightarrow $$ f ''(2) = 2 . . . . .(2)
<br><br>put x = 3 in equation (3) :
<br><br>f ''' (3) = 6
<br><br>$$ \therefore $$ f(x) = x<sup>3</sup> $$-$$ 5x<sup>2</sup> + 2x + 6
<br><br>f(2) = 8 $$-$$ 20 + 4 + 6 = $$-$$ 2 | mcq | jee-main-2019-online-10th-january-morning-slot | 1,751 |
xrX2b9GpaeqoQ7utJtjgy2xukf0qq2t5 | maths | differentiation | successive-differentiation | If y<sup>2</sup> + log<sub>e</sub> (cos<sup>2</sup>x) = y, <br/>$$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$, then : | [{"identifier": "A", "content": "|y''(0)| = 2"}, {"identifier": "B", "content": "|y'(0)| + |y''(0)| = 3"}, {"identifier": "C", "content": "y''(0) = 0"}, {"identifier": "D", "content": "|y'(0)| + |y\"(0)| = 1"}] | ["A"] | null | Given y<sup>2</sup> + log<sub>e</sub> (cos<sup>2</sup>x) = y .....(1)
<br><br>Put x = 0, we get
<br><br>y<sup>2</sup> + log<sub>e</sub> (1) = y
<br><br>$$ \Rightarrow $$ y<sup>2</sup> = y
<br><br>$$ \Rightarrow $$ y = 0, 1
<br><br>Differentiating (1) we get
<br><br>2yy' + $${1 \over {\cos x}}\left( { - \sin x} \right)$$ = y'
<br><br>$$ \Rightarrow $$ 2yy' - 2tanx = y' ....(2)
<br><br>From (2) when x = 0, y = 0 then y'(0) = 0
<br><br>From (2) when x = 0, y = 1 then
<br><br>2y' = y'
<br><br>$$ \Rightarrow $$ y'(0) = 0
<br><br>Again differentiating (2) we get
<br><br>2(y')<sup>2</sup>
+ 2yy'' – 2sec<sup>2</sup>x = y''
<br><br>from (2) when x = 0, y = 0, y’(0) = 0 then
<br><br>y”(0) = -2
<br><br>Also from (2) when x = 0, y = 1, y’(0) = 0 then
<br><br>y”(0) = 2
<br><br>$$ \therefore $$ |y''(0)| = 2 | mcq | jee-main-2020-online-3rd-september-morning-slot | 1,752 |
1l56u56af | maths | differentiation | successive-differentiation | <p>If $$y(x) = {\left( {{x^x}} \right)^x},\,x > 0$$, then $${{{d^2}x} \over {d{y^2}}} + 20$$ at x = 1 is equal to ____________.</p> | [] | null | 16 | <p>$$\because$$ $$y(x) = {\left( {{x^x}} \right)^x}$$</p>
<p>$$\therefore$$ $$y = {x^{{x^2}}}$$</p>
<p>$$\therefore$$ $${{dy} \over {dx}} = {x^2}\,.\,{x^{{x^2} - 1}} + {x^{{x^2}}}\ln x\,.\,2x$$</p>
<p>$$\therefore$$ $${{dx} \over {dy}} = {1 \over {{x^{{x^2} + 1}}(1 + 2\ln x)}}$$ ..... (i)</p>
<p>Now, $${{{d^2}x} \over {dx^2}} = {d \over {dx}}\left( {{{\left( {{x^{{x^2} + 1}}(1 + 2\ln x)} \right)}^{ - 1}}} \right)\,.\,{{dx} \over {dy}}$$</p>
<p>$$ = {{ - x{{\left( {{x^{{x^2} + 1}}(1 + 2\ln x)} \right)}^{ - 2}}\,.\,{x^{{x^2}}}(1 + 2\ln x)({x^2} + 2{x^2}\ln x + 3)} \over {{x^{{x^2}}}(1 + 2\ln x)}}$$</p>
<p>$$ = {{ - {x^{{x^2}}}(1 + 2\ln x)({x^3} + 3 + 2{x^2}\ln x)} \over {{{\left( {{x^{{x^2}}}(1 + 2\ln x)} \right)}^3}}}$$</p>
<p>$${{{d^2}x} \over {d{y^2}(at\,x = 1)}} = - 4$$</p>
<p>$$\therefore$$ $${{{d^2}x} \over {d{y^2}(at\,x = 1)}} + 20 = 16$$</p> | integer | jee-main-2022-online-27th-june-evening-shift | 1,753 |
1l6klla1m | maths | differentiation | successive-differentiation | <p>For the curve $$C:\left(x^{2}+y^{2}-3\right)+\left(x^{2}-y^{2}-1\right)^{5}=0$$, the value of $$3 y^{\prime}-y^{3} y^{\prime \prime}$$, at the point $$(\alpha, \alpha)$$, $$\alpha>0$$, on C, is equal to ____________.</p> | [] | null | 16 | <p>$$\because$$ $$C:({x^2} + {y^2} - 3) + {({x^2} - {y^2} - 1)^5} = 0$$ for point ($$\alpha$$, $$\alpha$$)</p>
<p>$${\alpha ^2} + {\alpha ^2} - 3 + {({\alpha ^2} - {\alpha ^2} - 1)^5} = 0$$</p>
<p>$$\therefore$$ $$\alpha = \sqrt 2 $$</p>
<p>On differentiating $$({x^2} + {y^2} - 3) + {({x^2} - {y^2} - 1)^5} = 0$$ we get</p>
<p>$$x + yy' + 5{({x^2} - {y^2} - 1)^4}(x - yy') = 0$$ ...... (i)</p>
<p>When $$x = y = \sqrt 2 $$ then $$y' = {3 \over 2}$$</p>
<p>Again on differentiating eq. (i) we get :</p>
<p>$$1 + {(y')^2} + yy'' + 20({x^2} - {y^2} - 1)(2x - 2yy')(x - y'y) + 5{({x^2} - {y^2} - 1)^4}(1 - y{'^2} - yy'') = 0$$</p>
<p>For $$x = y = \sqrt 2 $$ and $$y' = {3 \over 2}$$ we get $$y'' = - {{23} \over {4\sqrt 2 }}$$</p>
<p>$$\therefore$$ $$3y' - {y^3}y'' = 3\,.\,{3 \over 2} - {\left( {\sqrt 2 } \right)^3}\,.\,\left( { - {{23} \over {4\sqrt 2 }}} \right) = 16$$</p> | integer | jee-main-2022-online-27th-july-evening-shift | 1,754 |
1ldon088o | maths | differentiation | successive-differentiation | <p>Let $$f(x) = 2x + {\tan ^{ - 1}}x$$ and $$g(x) = {\log _e}(\sqrt {1 + {x^2}} + x),x \in [0,3]$$. Then</p> | [{"identifier": "A", "content": "there exists $$\\widehat x \\in [0,3]$$ such that $$f'(\\widehat x) < g'(\\widehat x)$$"}, {"identifier": "B", "content": "there exist $$0 < {x_1} < {x_2} < 3$$ such that $$f(x) < g(x),\\forall x \\in ({x_1},{x_2})$$"}, {"identifier": "C", "content": "$$\\min f'(x) = 1 + \\max g'(x)$$"}, {"identifier": "D", "content": "$$\\max f(x) > \\max g(x)$$"}] | ["D"] | null | $$
\begin{aligned}
& f^{\prime}(x)=2+\frac{1}{1+x^2}, g^{\prime}(x)=\frac{1}{\sqrt{x^2+1}} \\\\
& f^{\prime \prime}(x)=-\frac{2 x}{\left(1+x^2\right)^2}<0 \\\\
& g^{\prime \prime}(x)=-\frac{1}{2}\left(x^2+1\right)^{-3 / 2} \cdot 2 x<0 \\\\
& \left.f^{\prime}(x)\right|_{\min }=f^{\prime}(3)=2+\frac{1}{10}=\frac{21}{10} \\\\
& \left.g^{\prime}(x)\right|_{\max }=g^{\prime}(0)=1 \\\\
& \left.f^{\prime}(x)\right|_{\max }=f(3)=2+\tan ^{-1} 3 \\\\
& \left.g(x)\right|_{\max }=g(3)=\ln (3+\sqrt{10})<\ln <7<2
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-morning-shift | 1,755 |
1ldoofa8p | maths | differentiation | successive-differentiation | <p>If $$f(x)=x^{2}+g^{\prime}(1) x+g^{\prime \prime}(2)$$ and $$g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x)$$, then the value of $$f(4)-g(4)$$ is equal to ____________.</p> | [] | null | 14 | Let $g^{\prime}(1)=a$ and $g^{\prime \prime}(2)=b$
<br/><br/>$\Rightarrow f(x)=x^{2}+a x+b$
<br/><br/>Now, $f(1)=1+a+b ; f^{\prime}(x)=2 x+a ; f^{\prime \prime}(x)=2$
<br/><br/>$g(x)=(1+a+b) x^{2}+x(2 x+a)+2$
<br/><br/>$\Rightarrow g(x)=(a+b+3) x^{2}+a x+2$
<br/><br/>$\Rightarrow g^{\prime}(x)=2 x(a+b+3)+a \Rightarrow g^{\prime}(1)=2(a+b+3)$$+a=a$
<br/><br/>$\Rightarrow a+b+3=0$ .......(1)
<br/><br/>$g^{\prime \prime}(x)=2(a+b+3)=b$
<br/><br/>$\Rightarrow 2 a+b+6=0$ ........(2)
<br/><br/>Solving (i) and (ii), we get
<br/><br/>$a=-3$ and $b=0$
<br/><br/>$f(x)=x^{2}-3 x$ and $g(x)=-3 x+2$
<br/><br/>$f(4)=4$ and $g(4)=-12+2=-10$
<br/><br/>$\Rightarrow f(4)-g(4)=16-2=14$ | integer | jee-main-2023-online-1st-february-morning-shift | 1,756 |
1ldsfuib3 | maths | differentiation | successive-differentiation | <p>Let $$f$$ and $$g$$ be the twice differentiable functions on $$\mathbb{R}$$ such that</p>
<p>$$f''(x)=g''(x)+6x$$</p>
<p>$$f'(1)=4g'(1)-3=9$$</p>
<p>$$f(2)=3g(2)=12$$.</p>
<p>Then which of the following is NOT true?</p> | [{"identifier": "A", "content": "$$g(-2)-f(-2)=20$$"}, {"identifier": "B", "content": "There exists $$x_0\\in(1,3/2)$$ such that $$f(x_0)=g(x_0)$$"}, {"identifier": "C", "content": "$$|f'(x)-g'(x)| < 6\\Rightarrow -1 < x < 1$$"}, {"identifier": "D", "content": "If $$-1 < x < 2$$, then $$|f(x)-g(x)| < 8$$"}] | ["D"] | null | <p>$$f''(x) = g''(x) + 6x$$</p>
<p>$$ \Rightarrow f'(x) = g'(x) + 3{x^2} + C$$</p>
<p>$$f'(1) = g'(1) + 3 + C$$</p>
<p>$$ \Rightarrow g = 3 + 3 + C \Rightarrow C = 3$$</p>
<p>$$ \Rightarrow f'(x) = g'(x) + 3{x^2} + 3$$</p>
<p>$$ \Rightarrow f(x) = g(x) + {x^2} + 3x + C'$$</p>
<p>$$x = 2$$</p>
<p>$$f(2) = g(2) + 14 + C'$$</p>
<p>$$12 = 4 + 14 + C'$$</p>
<p>$$ \Rightarrow C' = - 6$$</p>
<p>$$ \Rightarrow f(x) = g(2) + {x^3} + 3x - 6$$</p>
<p>$$f( - 2) = g( - 2) - 8 - 6 - 6$$</p>
<p>$$g( - 2) - f( - 2) = 20$$</p>
<p>$$f'(x) - g'(x) = 3{x^2} + 3$$</p>
<p>$$x \in ( - 1,1)$$</p>
<p>$$3{x^2} + 3 \in (0,6)$$</p>
<p>$$ \Rightarrow f'(x) - g'(x) \in (0,6)$$<?p>
<p>$$f(x) - g(x) = {x^3} + 3x - 6$$</p>
<p>At $$x = - 1$$</p>
<p>$$|f( - 1) - g( - 1)| = 10$$</p>
<p>$$\therefore$$ Option (4) is false.</p> | mcq | jee-main-2023-online-29th-january-evening-shift | 1,757 |
1ldv1t76j | maths | differentiation | successive-differentiation | <p>Let $$y(x) = (1 + x)(1 + {x^2})(1 + {x^4})(1 + {x^8})(1 + {x^{16}})$$. Then $$y' - y''$$ at $$x = - 1$$ is equal to</p> | [{"identifier": "A", "content": "496"}, {"identifier": "B", "content": "976"}, {"identifier": "C", "content": "464"}, {"identifier": "D", "content": "944"}] | ["A"] | null | $$
\begin{aligned}
& y=\frac{1-x^{32}}{1-x}=1+x+x^2+x^3+\ldots+x^{31} \\\\
& y^{\prime}=1+2 x+3 x^2+\ldots+31 x^{30} \\\\
& y^{\prime}(-1)=1-2+3-4+\ldots+31=16 \\\\
& y^{\prime \prime}(x)=2+6 x+12 x^2+\ldots+31.30 x^{29} \\\\
& y^{\prime \prime}(-1)=2-6+12 \ldots 31.30=480 \\\\
& y^{\prime \prime}(-1)-y^{\prime}(-1)=-496
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-morning-shift | 1,758 |
1ldwx6r58 | maths | differentiation | successive-differentiation | <p>If $$f(x) = {x^3} - {x^2}f'(1) + xf''(2) - f'''(3),x \in \mathbb{R}$$, then</p> | [{"identifier": "A", "content": "$$2f(0) - f(1) + f(3) = f(2)$$"}, {"identifier": "B", "content": "$$f(1) + f(2) + f(3) = f(0)$$"}, {"identifier": "C", "content": "$$f(3) - f(2) = f(1)$$"}, {"identifier": "D", "content": "$$3f(1) + f(2) = f(3)$$"}] | ["A"] | null | $$
f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3), x \in R
$$<br/><br/>
Let $\mathrm{f}^{\prime}(1)=\mathrm{a}, \mathrm{f}^{\prime \prime}(2)=\mathrm{b}, \mathrm{f}^{\prime \prime \prime}(3)=\mathrm{c}$<br/><br/>
$$
\begin{aligned}
& f(x)=x^3-a x^2+b x-c \\\\
& f^{\prime}(x)=3 x^2-2 a x+b \\\\
& f^{\prime \prime}(x)=6 x-2 a \\\\
& f^{\prime \prime \prime}(x)=6 \\\\
& c=6, a=3, b=6 \\\\
& f(x)=x^3-3 x^2+6 x-6 \\\\
& f(1)=-2, f(2)=2, f(3)=12, f(0)=-6 \\\\
& 2 f(0)-f(1)+f(3)=2=f(2)
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-evening-shift | 1,759 |
lsble0m7 | maths | differentiation | successive-differentiation | Let $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3), x \in \mathbf{R}$. Then $f^{\prime}(10)$ is equal to ____________. | [] | null | 202 | <p>$$\begin{aligned}
& f(x)=x^3+x^2 \cdot f^{\prime}(1)+x \cdot f^{\prime \prime}(2)+f^{\prime \prime \prime}(3) \\
& f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2) \\
& f^{\prime \prime}(x)=6 x+2 f^{\prime}(1) \\
& f^{\prime \prime \prime}(x)=6 \\
& f^{\prime}(1)=-5, f^{\prime \prime}(2)=2, f^{\prime \prime \prime}(3)=6 \\
& f(x)=x^3+x^2 \cdot(-5)+x \cdot(2)+6 \\
& f^{\prime}(x)=3 x^2-10 x+2 \\
& f^{\prime}(10)=300-100+2=202
\end{aligned}$$</p> | integer | jee-main-2024-online-27th-january-morning-shift | 1,760 |
1lsg94q54 | maths | differentiation | successive-differentiation | <p>If $$f(x)=\left|\begin{array}{ccc}
2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\
3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\
2 \cos ^4 x & 3+2 \sin ^4 x & \sin ^2 2 x
\end{array}\right|,$$ then $$\frac{1}{5} f^{\prime}(0)=$$ is equal to :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "6"}] | ["C"] | null | <p>$$\begin{aligned}
& \left|\begin{array}{ccc}
2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\
3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\
2 \cos ^4 x & 3+2 \sin ^2 4 x & \sin ^2 2 x
\end{array}\right| \\
& \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1, \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \\
& \left|\begin{array}{ccc}
2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\
3 & 0 & -3 \\
0 & 3 & -3
\end{array}\right| \\
& \mathrm{f}(\mathrm{x})=45 \\
& \mathrm{f}^{\prime}(\mathrm{x})=0 \\
&
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift | 1,761 |
lv2er9ry | maths | differentiation | successive-differentiation | <p>Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a thrice differentiable function such that $$f(0)=0, f(1)=1, f(2)=-1, f(3)=2$$ and $$f(4)=-2$$. Then, the minimum number of zeros of $$\left(3 f^{\prime} f^{\prime \prime}+f f^{\prime \prime \prime}\right)(x)$$ is __________.</p> | [] | null | 5 | <p>$$\because f: R \rightarrow R \text { and } f(0)=0, f(1)=1, f(2)=-1 \text {, }$$</p>
<p>$$f(3)=2$$ and $$f(4)=-2$$ then</p>
<p>$$f(x)$$ has atleast 4 real roots.</p>
<p>Then $$f(x)$$ has atleast 3 real roots and $$f^{\prime}(x)$$ has atleast 2 real roots.</p>
<p>Now we know that</p>
<p>$$\begin{aligned}
\frac{d}{d x}\left(f^3 \cdot f^{\prime \prime}\right) & =3 f^2 \cdot f^{\prime} \cdot f^{\prime \prime}+f^3 \cdot f^{\prime \prime \prime} \\
& =f^2\left(3 f^{\prime} \cdot f^{\prime}+f \cdot f^{\prime \prime}\right)
\end{aligned}$$</p>
<p>Here $$f^3 \cdot f'$$ has atleast 6 roots.</p>
<p>Then its differentiation has atleast 5 distinct roots.</p> | integer | jee-main-2024-online-4th-april-evening-shift | 1,762 |
lv9s204y | maths | differentiation | successive-differentiation | <p>If $$y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2}$$, then at $$\theta=\frac{\pi}{2}, y^{\prime \prime}+y^{\prime}+y$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{1}{2}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$\\frac{3}{2}$$"}, {"identifier": "D", "content": "2"}] | ["D"] | null | <p>$$\begin{aligned}
& y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2} \\
& =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{4 \cos ^3 \theta+8 \cos ^2 \theta+2 \cos \theta-2} \\
& =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)(2 \cos \theta+2)} \\
& =\frac{1}{2(1+\cos \theta)}=\frac{1}{4 \cos ^2 \theta / 2}=\frac{\sec ^2 \theta / 2}{4} \\
& y^{\prime}(\theta)=\frac{1}{4}\left(2 \sec \frac{\theta}{2} \cdot \sec \frac{\theta}{2} \cdot \tan \frac{\theta}{2} \cdot \frac{1}{2}\right) \\
& =\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2}
\end{aligned}$$</p>
<p>$$y^{\prime \prime}(\theta)=\frac{1}{4}\left(\tan \frac{\theta}{2}\right)\left(\sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2}\right) +\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \sec ^2 \frac{\theta}{2} \cdot \frac{1}{2}$$</p>
<p>$$\begin{aligned}
& \text { at } \theta=\frac{\pi}{2}, y(\theta)=\frac{1}{2}, y^{\prime}(\theta)=\frac{1}{2}, y^{\prime \prime}(\theta)=1 \\
& \therefore \quad y+y^{\prime}+y^{\prime \prime}=2
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift | 1,763 |
lvc57b43 | maths | differentiation | successive-differentiation | <p>$$\text { If } f(x)=\left\{\begin{array}{ll}
x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\
0 & , x=0
\end{array}\right. \text {, then }$$</p> | [{"identifier": "A", "content": "$$f^{\\prime \\prime}(0)=0$$\n"}, {"identifier": "B", "content": "$$f^{\\prime \\prime}(0)=1$$\n"}, {"identifier": "C", "content": "$$f^{\\prime \\prime}\\left(\\frac{2}{\\pi}\\right)=\\frac{24-\\pi^2}{2 \\pi}$$\n"}, {"identifier": "D", "content": "$$f^{\\prime \\prime}\\left(\\frac{2}{\\pi}\\right)=\\frac{12-\\pi^2}{2 \\pi}$$"}] | ["C"] | null | <p>Given the function:</p>
<p>$ f(x)=\left\{\begin{array}{ll} x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{array}\right. $</p>
<p>we need to find its second derivative at specific points.</p>
<p>First, let’s compute the first derivative $ f^{\prime}(x) $:</p>
<p>$ f^{\prime}(x) = 3x^2 \sin \left( \frac{1}{x} \right) - x \cos \left( \frac{1}{x} \right) $</p>
<p>Next, the second derivative $ f^{\prime \prime}(x) $ is:</p>
<p>$ f^{\prime \prime}(x) = 6x \sin \left(\frac{1}{x}\right) - 3x \cos \left(\frac{1}{x}\right) - \cos \left(\frac{1}{x}\right) - \frac{1}{x} \sin \left(\frac{1}{x}\right) $</p>
<p>Therefore, evaluating the second derivative at $ x = \frac{2}{\pi} $:</p>
<p>$ f^{\prime \prime} \left(\frac{2}{\pi}\right) = 6 \left( \frac{2}{\pi} \right) \sin \left(\frac{\pi}{2}\right) - 3 \left( \frac{2}{\pi} \right) \cos \left(\frac{\pi}{2}\right) - \cos \left( \frac{\pi}{2} \right) - \frac{\pi}{2} \sin \left( \frac{\pi}{2} \right) $</p>
<p>Since $\sin(\frac{\pi}{2}) = 1$ and $\cos(\frac{\pi}{2}) = 0$, this simplifies to:</p>
<p>$ f^{\prime \prime} \left( \frac{2}{\pi} \right) = \frac{12}{\pi} - \frac{\pi}{2} = \frac{24 - \pi^2}{2\pi} $</p>
<p>Finally, note that $ f^{\prime}(0) $ is not defined, as it involves terms like $\frac{1}{x}$ when $ x = 0 $.</p> | mcq | jee-main-2024-online-6th-april-morning-shift | 1,764 |
1krw1fh1n | maths | ellipse | chord-of-ellipse | Let an ellipse $$E:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, $${a^2} > {b^2}$$, passes through $$\left( {\sqrt {{3 \over 2}} ,1} \right)$$ and has eccentricity $${1 \over {\sqrt 3 }}$$. If a circle, centered at focus F($$\alpha$$, 0), $$\alpha$$ > 0, of E and radius $${2 \over {\sqrt 3 }}$$, intersects E at two points P and Q, then PQ<sup>2</sup> is equal to : | [{"identifier": "A", "content": "$${8 \\over 3}$$"}, {"identifier": "B", "content": "$${4 \\over 3}$$"}, {"identifier": "C", "content": "$${{16} \\over 3}$$"}, {"identifier": "D", "content": "3"}] | ["C"] | null | $${3 \over {2{a^2}}} + {1 \over {{b^2}}} = 1$$ and $$1 - {{{b^2}} \over {{a^2}}} = {1 \over 3}$$<br><br>$$ \Rightarrow {a^2} = 3{b^2} = 3$$ <br><br>$$ \Rightarrow {{{x^2}} \over 3} + {{{y^2}} \over 2} = 1$$ ...... (i)<br><br>Its focus is (1, 0)<br><br>Now, equation of circle is <br><br>$${(x - 1)^2} + {y^2} = {4 \over 3}$$ ..... (ii)<br><br>Solving (i) and (ii) we get<br><br>$$y = \pm {2 \over {\sqrt 3 }},x = 1$$<br><br>$$ \Rightarrow P{Q^2} = {\left( {{4 \over {\sqrt 3 }}} \right)^2} = {{16} \over 3}$$ | mcq | jee-main-2021-online-25th-july-morning-shift | 1,765 |
1l59l0lop | maths | ellipse | chord-of-ellipse | <p>The line y = x + 1 meets the ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$$ at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)<sup>2</sup> is equal to :</p> | [{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "8"}] | ["A"] | null | <p>Let point (a, a + 1) as the point of intersection of line and ellipse.</p>
<p>So, $${{{a^2}} \over 4} + {{{{(a + 1)}^2}} \over 2} = 1 \Rightarrow {a^2} + 2({a^2} + 2a + 1) = 4$$</p>
<p>$$ \Rightarrow 3{a^2} + 4a - 2 = 0$$</p>
<p>If roots of this equation are $$\alpha$$ and $$\beta$$.</p>
<p>So, $$P(\alpha ,\,\alpha + 1)$$ and $$Q(\beta ,\,\beta + 1)$$</p>
<p>$$PQ = 4{r^2} = {(\alpha - \beta )^2} + {(\alpha - \beta )^2}$$</p>
<p>$$ \Rightarrow 9{r^2} = {9 \over 4}(2{(\alpha - \beta )^2})$$</p>
<p>$$ = {9 \over 2}\left[ {{{(\alpha + \beta )}^2} - 4\alpha \beta } \right]$$</p>
<p>$$ = {9 \over 2}\left[ {{{\left( { - {4 \over 3}} \right)}^2} + {8 \over 3}} \right]$$</p>
<p>$$ = {1 \over 2}[16 + 24] = 20$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift | 1,766 |
1lguvb7is | maths | ellipse | chord-of-ellipse | <p>Consider ellipses $$\mathrm{E}_{k}: k x^{2}+k^{2} y^{2}=1, k=1,2, \ldots, 20$$. Let $$\mathrm{C}_{k}$$ be the circle which touches the four chords joining the end points (one on minor axis and another on major axis) of the ellipse $$\mathrm{E}_{k}$$. If $$r_{k}$$ is the radius of the circle $$\mathrm{C}_{k}$$, then the value of $$\sum_\limits{k=1}^{20} \frac{1}{r_{k}^{2}}$$ is :</p> | [{"identifier": "A", "content": "2870"}, {"identifier": "B", "content": "3210"}, {"identifier": "C", "content": "3320"}, {"identifier": "D", "content": "3080"}] | ["D"] | null | We have, $E_K=K x^2+K^2 y^2=1, K=1,2, \ldots 20$
<br><br>$\Rightarrow \frac{x^2}{\frac{1}{K}}+\frac{y^2}{\frac{1}{K^2}}=1$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ln5ufvyu/8420caac-c5f0-403f-a175-e82f7eb8d93d/8ea00c50-5f75-11ee-8999-67742721d03c/file-6y3zli1ln5ufvyv.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ln5ufvyu/8420caac-c5f0-403f-a175-e82f7eb8d93d/8ea00c50-5f75-11ee-8999-67742721d03c/file-6y3zli1ln5ufvyv.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 11th April Morning Shift Mathematics - Ellipse Question 12 English Explanation">
<br><br>Equation of $A B$ is $\frac{x}{\frac{1}{\sqrt{K}}}+\frac{y}{\frac{1}{K}}=1$
<br><br> or $ \sqrt{K} x+K y=1$
<br><br>$r_K$ is the radius of circle $C_K$,
<br><br>So, $r_K=$ perpendicular distance from $(0,0)$ to $\mathrm{AB}$
<br><br>$$
\begin{aligned}
r_K & =\frac{|0+0-1|}{\sqrt{(\sqrt{K})^2+K^2}} \\\\
& =\frac{1}{\sqrt{K+K^2}} \\\\
\frac{1}{r_K^2} & =K+K^2
\end{aligned}
$$
<br><br>$$
\begin{aligned}
\sum_{K=1}^{20} \frac{1}{r_K^2} & =\sum_{K=1}^{20} K+\sum_{K=1}^{20} K^2 \\\\
& =\frac{20 \times 21}{2}+\frac{20 \times 21 \times 41}{6} \\\\
& =210+2870=3080
\end{aligned}
$$ | mcq | jee-main-2023-online-11th-april-morning-shift | 1,767 |
lsbkn1ul | maths | ellipse | chord-of-ellipse | The length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid point is $\left(1, \frac{2}{5}\right)$, is equal to : | [{"identifier": "A", "content": "$\\frac{\\sqrt{1691}}{5}$"}, {"identifier": "B", "content": "$\\frac{\\sqrt{2009}}{5}$"}, {"identifier": "C", "content": "$\\frac{\\sqrt{1541}}{5}$"}, {"identifier": "D", "content": "$\\frac{\\sqrt{1741}}{5}$"}] | ["A"] | null | <p>Equation of chord with given middle point.</p>
<p>$$\begin{aligned}
& T=S_1 \\
& \frac{x}{25}+\frac{y}{40}=\frac{1}{25}+\frac{1}{100} \\
& \frac{8 x+5 y}{200}=\frac{8+2}{200} \\
& y=\frac{10-8 x}{5} \quad \text{.... (i)}
\end{aligned}$$</p>
<p>$$\frac{x^2}{25}+\frac{(10-8 x)^2}{400}=1$$ (put in original equation)</p>
<p>$$\begin{aligned}
& \frac{16 x^2+100+64 x^2-160 x}{400}=1 \\
& 4 x^2-8 x-15=0 \\
& x=\frac{8 \pm \sqrt{304}}{8} \\
& x_1=\frac{8+\sqrt{304}}{8} ; x_2=\frac{8-\sqrt{304}}{8}
\end{aligned}$$</p>
<p>Similarly, $$y=\frac{10-18 \pm \sqrt{304}}{5}=\frac{2 \pm \sqrt{304}}{5}$$</p>
<p>$$\begin{aligned}
& \mathrm{y}_1=\frac{2-\sqrt{304}}{5} ; \mathrm{y}_2=\frac{2+\sqrt{304}}{5} \\
& \text { Distance }=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2} \\
& =\sqrt{\frac{4 \times 304}{64}+\frac{4 \times 304}{25}}=\frac{\sqrt{1691}}{5} \\
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift | 1,768 |
EvZIdx6Rf9PPNIN0 | maths | ellipse | common-tangent | <b>STATEMENT-1 :</b> An equation of a common tangent to the parabola $${y^2} = 16\sqrt 3 x$$ and the ellipse $$2{x^2} + {y^2} = 4$$ is $$y = 2x + 2\sqrt 3 $$
<p><b>STATEMENT-2 :</b>If line $$y = mx + {{4\sqrt 3 } \over m},\left( {m \ne 0} \right)$$ is a common tangent to the parabola $${y^2} = 16\sqrt {3x} $$and the ellipse $$2{x^2} + {y^2} = 4$$, then $$m$$ satisfies $${m^4} + 2{m^2} = 24$$</p> | [{"identifier": "A", "content": "Statement-1 is false, Statement-2 is true."}, {"identifier": "B", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1."}, {"identifier": "C", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1."}, {"identifier": "D", "content": "Statement-1 is true, Statement-2 is false."}] | ["B"] | null | Given equation of ellipse is $$2{x^2} + {y^2} = 4$$
<br><br>$$ \Rightarrow {{2{x^2}} \over 4} + {{{y^2}} \over 4} = 1 \Rightarrow {{{x_2}} \over 2} + {{{y^2}} \over 4} = 1$$
<br><br>Equation of tangent to the ellipse $${{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$$ is
<br><br>$$y = mx \pm \sqrt {2{m^2} + 4} \,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>( as equation of tangent to the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$
<br><br>is $$y=mx+c$$ where $$c = \pm \sqrt {{a^2}{m^2} + {b^2}} $$ )
<br><br>Now, Equation of tangent to the parabola
<br><br>$${y^2} = 16\sqrt 3 x$$ is $$y = mx + {{4\sqrt 3 } \over m}\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br> ( as equation of tangent to the parabola
<br><br>$${y^2} = 4ax$$ is $$y = mx + {a \over m}$$ )
<br><br>On comparing $$(1)$$ and $$(2),$$ we get
<br><br>$${{4\sqrt 3 } \over m} = \pm \sqrt {2{m^2} + 4} $$
<br><br>Squaring on both the sides, we get
<br><br>$$16\left( 3 \right) = \left( {2{m^2} + 4} \right){m^2}$$
<br><br>$$ \Rightarrow 48 = {m^2}\left( {2{m^2} + 4} \right) \Rightarrow 2{m^4} + 4{m^2} - 48 = 0$$
<br><br>$$ \Rightarrow {m^4} + 2{m^2} - 24 = 0 \Rightarrow \left( {{m^2} + 6} \right)\left( {{m^2} - 4} \right) = 0$$
<br><br>$$ \Rightarrow {m^2} = 4$$ ( as $${m^2} \ne - 6$$ ) $$ \Rightarrow m = \pm 2$$
<br><br>$$ \Rightarrow $$ Equation of common tangents are $$y = \pm 2x \pm 2\sqrt 3 $$
<br><br>Thus, statement - $$1$$ is true.
<br><br>Statement - $$2$$ is obviously true. | mcq | aieee-2012 | 1,769 |
1uvsN9WuUVTtVaGYvw18hoxe66ijvwq7lwv | maths | ellipse | common-tangent | If the tangent to the parabola y<sup>2</sup> = x at a point
($$\alpha $$, $$\beta $$), ($$\beta $$ > 0) is also a tangent to the ellipse,
x<sup>2</sup> + 2y<sup>2</sup> = 1, then $$\alpha $$ is equal to : | [{"identifier": "A", "content": "$$\\sqrt 2 + 1$$"}, {"identifier": "B", "content": "$$\\sqrt 2 - 1$$"}, {"identifier": "C", "content": "$$2\\sqrt 2 + 1$$"}, {"identifier": "D", "content": "$$2\\sqrt 2 - 1$$"}] | ["A"] | null | Point P($$\alpha $$, $$\beta $$) is on the parabola y<sup>2</sup> = x
<br><br>$$ \therefore $$ $${\beta ^2} = \alpha $$ ...........(1)
<br><br>Equation of tangent to the parabola y<sup>2</sup> = x
<br><br>at ($$\alpha $$, $$\beta $$) is T = 0
<br><br>$$\beta y = {{x + \alpha } \over 2}$$
<br><br>$$ \Rightarrow $$ $$2\beta y = x + \alpha $$
<br><br>$$ \Rightarrow $$ $$y = {1 \over {2\beta }}x + {\alpha \over {2\beta }}$$
<br><br>For this straight line, m = $${1 \over {2\beta }}$$ and c = $${\alpha \over {2\beta }}$$
<br><br>This tangent is also a tangent to ellipse x<sup>2</sup> + 2y<sup>2</sup> = 1.
<br><br>$$ \Rightarrow $$ $${{{x^2}} \over 1} + {{{y^2}} \over {{1 \over 2}}} = 1$$
<br><br>$$ \therefore $$ $${a^2} = 1$$ and $${b^2} = {1 \over 2}$$.
<br><br>We know the condition of tangency on ellipse is
<br><br>$${c^2} = {a^2}{m^2} + {b^2}$$
<br><br>$$ \Rightarrow $$ $${\left( {{\alpha \over {2\beta }}} \right)^2} = 1{\left( {{1 \over {2\beta }}} \right)^2} + {1 \over 2}$$
<br><br>$$ \Rightarrow $$ $${{{\alpha ^2}} \over {4{\beta ^2}}} = {1 \over {4{\beta ^2}}} + {1 \over 2}$$
<br><br>$$ \Rightarrow $$ $${\alpha ^2} = 1 + 2{\beta ^2}$$
<br><br>$$ \Rightarrow $$ $${\alpha ^2} = 1 + 2\alpha $$
<br><br>$$ \Rightarrow $$ $${\alpha ^2} - 2\alpha - 1 = 0$$
<br><br>$$ \Rightarrow $$ $$\alpha = 1 + \sqrt 2 $$ and $$\alpha = 1 - \sqrt 2 $$ | mcq | jee-main-2019-online-9th-april-evening-slot | 1,770 |
xtgmJdXktn5EUKYs6j1kluz2mfj | maths | ellipse | common-tangent | Let L be a common tangent line to the curves <br/><br/>4x<sup>2</sup> + 9y<sup>2</sup> = 36 and (2x)<sup>2</sup> + (2y)<sup>2</sup> = 31. Then the <br/><br/>square of the slope of the line L is __________. | [] | null | 3 | Tangent to the curve $${{{x^2}} \over 9} + {{{y^2}} \over {14}} = 1$$ is <br><br>$$y = mx + \sqrt {9{m^2} + 4} $$<br><br>and equation of tangent to the curve $${x^2} + {y^2} = {{31} \over 4}$$ is<br><br>$$y = mx + \sqrt {{{31} \over 4}{{(1 + m)}^2}} $$<br><br>for common tangent $$9{m^2} + 4 = {{31} \over 4} + {{31} \over 4}{m^2}$$<br><br>$$ \Rightarrow {5 \over 4}{m^2} = {{15} \over 4}$$<br><br>$$ \Rightarrow {m^2} = 3$$ | integer | jee-main-2021-online-26th-february-evening-slot | 1,771 |
1l58fgd83 | maths | ellipse | common-tangent | <p>If m is the slope of a common tangent to the curves $${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$ and $${x^2} + {y^2} = 12$$, then $$12{m^2}$$ is equal to :</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "12"}] | ["B"] | null | <p>$${C_1}:{{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$ and $${C_2}:{x^2} + {y^2} = 12$$</p>
<p>Let $$y = mx \pm \,\sqrt {16{m^2} + 9} $$ be any tangent to C<sub>1</sub> and if this is also tangent to C<sub>2</sub> then</p>
<p>$$\left| {{{\sqrt {16{m^2} + 9} } \over {\sqrt {{m^2} + 1} }}} \right| = \sqrt {12} $$</p>
<p>$$ \Rightarrow 16{m^2} + 9 = 12{m^2} + 12$$</p>
<p>$$ \Rightarrow 4{m^2} = 3 \Rightarrow 12{m^2} = 9$$</p> | mcq | jee-main-2022-online-26th-june-evening-shift | 1,772 |
1lgvqbyet | maths | ellipse | common-tangent | <p>Let a circle of radius 4 be concentric to the ellipse $$15 x^{2}+19 y^{2}=285$$. Then the common tangents are inclined to the minor axis of the ellipse at the angle :</p> | [{"identifier": "A", "content": "$$\\frac{\\pi}{4}$$"}, {"identifier": "B", "content": "$$\\frac{\\pi}{3}$$"}, {"identifier": "C", "content": "$$\\frac{\\pi}{6}$$"}, {"identifier": "D", "content": "$$\\frac{\\pi}{12}$$"}] | ["B"] | null | We have, equation of ellipse : $15 x^2+19 y^2=285$
<br/><br/>or $ \frac{x^2}{19}+\frac{y^2}{15}=1$
<br/><br/>Let the coordinate of center of circle be $(0,0)$.
<br/><br/>Equation of circle is $x^2+y^2=16$
<br/><br/>Equation of tangent of ellipse is
<br/><br/>$$
\begin{gathered}
y=m x \pm \sqrt{19 m^2+15} \text { or } \\\\
m x-y \pm \sqrt{19 m^2+15}=0
\end{gathered}
$$
<br/><br/>It is also tangent to the circle $x^2+y^2=16$
<br/><br/>Perpendicular distance from center of circle to tangent $=4$
<br/><br/>$$
\frac{\left|0-0 \pm \sqrt{19 m^2+15}\right|}{\sqrt{m^2+1}}=4
$$
<br/><br/>On squaring both side, we get
<br/><br/>$$
\begin{gathered}
19 m^2+15=16 m^2+16 \\\\
3 m^2=1 \\\\
m^2=\frac{1}{3} \\\\
m= \pm \frac{1}{\sqrt{3}}
\end{gathered}
$$
<br/><br/>$$
\tan \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=\frac{\pi}{6} \text { with } X \text {-axis or }
$$$\frac{\pi}{3}$ with $Y$-axis
<br/><br/>Hence, the common tangents are inclined to the minor axis of the ellipse at an angle of $\frac{\pi}{3}$. | mcq | jee-main-2023-online-10th-april-evening-shift | 1,773 |
lhi1qU8ZF8dwtl0X | maths | ellipse | locus | The locus of the foot of perpendicular drawn from the centre of the ellipse $${x^2} + 3{y^2} = 6$$ on any tangent to it is : | [{"identifier": "A", "content": "$$\\left( {{x^2} + {y^2}} \\right) ^2 = 6{x^2} + 2{y^2}$$ "}, {"identifier": "B", "content": "$$\\left( {{x^2} + {y^2}} \\right) ^2 = 6{x^2} - 2{y^2}$$"}, {"identifier": "C", "content": "$$\\left( {{x^2} - {y^2}} \\right) ^2 = 6{x^2} + 2{y^2}$$ "}, {"identifier": "D", "content": "$$\\left( {{x^2} - {y^2}} \\right) ^2 = 6{x^2} - 2{y^2}$$ "}] | ["A"] | null | Given $$e{q^n}$$ of ellipse can be written as
<br><br>$${{{x^2}} \over 6} + {{{y^2}} \over 2} = 1 \Rightarrow {a^2} = 6,{b^2} = 2$$
<br><br>Now, equation of any variable tangent is
<br><br>$$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} ....\left( i \right)$$
<br><br>where $$m$$ is slope of the tangent
<br><br>So, equation of perpendicular line drawn-
<br><br>from center to tangent is
<br><br>$$y = {{ - x} \over m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>Eliminating $$m,$$ we get
<br><br>$$\left( {{x^4} + {y^4} + 2{x^2}{y^2}} \right) = {a^2}{x^2} + {b^2}{y^2}$$
<br><br>$$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = {a^2}{x^2} + {b^2}{y^2}$$
<br><br>$$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}$$ | mcq | jee-main-2014-offline | 1,774 |
1ktk74324 | maths | ellipse | locus | The locus of mid-points of the line segments joining ($$-$$3, $$-$$5) and the points on the ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$$ is : | [{"identifier": "A", "content": "$$9{x^2} + 4{y^2} + 18x + 8y + 145 = 0$$"}, {"identifier": "B", "content": "$$36{x^2} + 16{y^2} + 90x + 56y + 145 = 0$$"}, {"identifier": "C", "content": "$$36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$$"}, {"identifier": "D", "content": "$$36{x^2} + 16{y^2} + 72x + 32y + 145 = 0$$"}] | ["C"] | null | General point on $${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$$ is A(2cos$$\theta$$, 3sin$$\theta$$)<br><br>given B($$-$$3, $$-$$5)<br><br>midpoint $$C\left( {{{2\cos \theta - 3} \over 2},{{3\sin \theta - 5} \over 2}} \right)$$<br><br>$$h = {{2\cos \theta - 3} \over 2};k = {{3\sin \theta - 5} \over 2}$$<br><br>$$ \Rightarrow {\left( {{{2h + 3} \over 2}} \right)^2} + {\left( {{{2k + 5} \over 3}} \right)^2} = 1$$<br><br>$$ \Rightarrow 36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$$ | mcq | jee-main-2021-online-31st-august-evening-shift | 1,775 |
1l58g1v33 | maths | ellipse | locus | <p>The locus of the mid point of the line segment joining the point (4, 3) and the points on the ellipse $${x^2} + 2{y^2} = 4$$ is an ellipse with eccentricity :</p> | [{"identifier": "A", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "B", "content": "$${1 \\over {2\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}] | ["C"] | null | <p>Let $$P(2\cos \theta ,\,\sqrt 2 \sin \theta )$$ be any point on ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$$ and Q(4, 3) and let (h, k) be the mid point of PQ</p>
<p>then $$h = {{2\cos \theta + 4} \over 2},\,k = {{\sqrt 2 \sin \theta + 3} \over 2}$$</p>
<p>$$\therefore$$ $$\cos \theta = h - 2,\,\sin \theta = {{2k - 3} \over {\sqrt 2 }}$$</p>
<p>$$\therefore$$ $${(h - 2)^2} + {\left( {{{2k - 3} \over {\sqrt 2 }}} \right)^2} = 1$$</p>
<p>$$ \Rightarrow {{{{(x - 2)}^2}} \over 1} + {{{{\left( {y - {3 \over 2}} \right)}^2}} \over {{1 \over 2}}} = 1$$</p>
<p>$$\therefore$$ $$e = \sqrt {1 - {1 \over 2}} = {1 \over {\sqrt 2 }}$$</p> | mcq | jee-main-2022-online-26th-june-evening-shift | 1,776 |
lsamf0th | maths | ellipse | locus | Let $\mathrm{P}$ be a point on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. Let the line passing through $\mathrm{P}$ and parallel to $y$-axis meet the circle $x^2+y^2=9$ at point $\mathrm{Q}$ such that $\mathrm{P}$ and $\mathrm{Q}$ are on the same side of the $x$-axis. Then, the eccentricity of the locus of the point $R$ on $P Q$ such that $P R: R Q=4: 3$ as $P$ moves on the ellipse, is : | [{"identifier": "A", "content": "$\\frac{13}{21}$"}, {"identifier": "B", "content": "$\\frac{\\sqrt{139}}{23}$"}, {"identifier": "C", "content": "$\\frac{\\sqrt{13}}{7}$"}, {"identifier": "D", "content": "$\\frac{11}{19}$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqdjrqo/772aa650-3e8c-46a8-adce-0608922ed01d/09ef1300-cdbd-11ee-a926-9fabe9a328d8/file-6y3zli1lsqdjrqp.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqdjrqo/772aa650-3e8c-46a8-adce-0608922ed01d/09ef1300-cdbd-11ee-a926-9fabe9a328d8/file-6y3zli1lsqdjrqp.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 1st February Evening Shift Mathematics - Ellipse Question 8 English Explanation 1">
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqdl51c/d5a5c6d5-b0af-43b9-b7e2-7efa69d9eeff/30039c00-cdbd-11ee-a926-9fabe9a328d8/file-6y3zli1lsqdl51d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqdl51c/d5a5c6d5-b0af-43b9-b7e2-7efa69d9eeff/30039c00-cdbd-11ee-a926-9fabe9a328d8/file-6y3zli1lsqdl51d.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 1st February Evening Shift Mathematics - Ellipse Question 8 English Explanation 2">
<br>$\begin{aligned} & \mathrm{h}=3 \cos \theta \\\\ & \mathrm{k}=\frac{18}{7} \sin \theta\end{aligned}$
<br><br>$\begin{aligned} & \therefore \text { locus }=\frac{\mathrm{x}^2}{9}+\frac{49 \mathrm{y}^2}{324}=1 \\\\ & \mathrm{e}=\sqrt{1-\frac{324}{49 \times 9}}=\frac{\sqrt{117}}{21}=\frac{\sqrt{13}}{7}\end{aligned}$ | mcq | jee-main-2024-online-1st-february-evening-shift | 1,777 |
Fdaip3SiqPfH8cxI | maths | ellipse | normal-to-ellipse | The eccentricity of an ellipse whose centre is at the origin is $${1 \over 2}$$. If one of its directrices is x = – 4, then the
equation of the normal to it at $$\left( {1,{3 \over 2}} \right)$$ is : | [{"identifier": "A", "content": "2y \u2013 x = 2"}, {"identifier": "B", "content": "4x \u2013 2y = 1"}, {"identifier": "C", "content": "4x + 2y = 7"}, {"identifier": "D", "content": "x + 2y = 4"}] | ["B"] | null | Given e = $${1 \over 2}$$ and $${a \over e}$$ = 4
<br><br>$$ \therefore $$ $$a$$ = 2
<br><br>We have b<sup>2</sup> = $$a$$<sup>2</sup> (1 – e<sup>2</sup>) = $$4\left( {1 - {1 \over 4}} \right)$$ = 3
<br><br>$$ \therefore $$ Equation of ellipse is
<br><br>$${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$$
<br><br>Now, the equation of normal at $$\left( {1,{3 \over 2}} \right)$$ is
<br><br>$${{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2} - {b^2}$$
<br><br>$$ \Rightarrow $$ $${{4x} \over 1} - {{3y} \over {{3 \over 2}}} = 4 - 3$$
<br><br>$$ \Rightarrow $$ 4x – 2y = 1 | mcq | jee-main-2017-offline | 1,778 |
5Xy9INkIXJfGIM8vGg3rsa0w2w9jx2eobip | maths | ellipse | normal-to-ellipse | The tangent and normal to the ellipse 3x<sup>2</sup>
+ 5y<sup>2</sup>
= 32 at the point P(2, 2) meet the x-axis at Q and R,
respectively. Then the area (in sq. units) of the triangle PQR is : | [{"identifier": "A", "content": "$${{14} \\over 3}$$"}, {"identifier": "B", "content": "$${{16} \\over 3}$$"}, {"identifier": "C", "content": "$${{68} \\over {15}}$$"}, {"identifier": "D", "content": "$${{34} \\over {15}}$$"}] | ["C"] | null | $$3{x^2} + 5{y^2} = 32$$<br><br>
6x + 10yy' = 0<br><br>
$$ \Rightarrow y' = {{ - 3x} \over {5y}}$$<br><br>
$$ \Rightarrow y{'_{(2,2)}} = - {3 \over 5}$$<br><br>
Tangent $$(y - 2) = - {3 \over 5}(x - 2) \Rightarrow Q\left( {{{16} \over 3},0} \right)$$<br><br>
Normal $$(y - 2) = {5 \over 3}(x - 2) \Rightarrow R\left( {{{4} \over 5},0} \right)$$<br><br>
$$ \therefore $$ Area = $${1 \over 2}(QR) \times 2 = QR = {{68} \over {15}}$$
| mcq | jee-main-2019-online-10th-april-evening-slot | 1,779 |
ABobMP93kuwLqACJVx3rsa0w2w9jx65dnxr | maths | ellipse | normal-to-ellipse | If the normal to the ellipse 3x<sup>2</sup>
+ 4y<sup>2</sup>
= 12 at a point P on it is parallel to the line, 2x + y = 4 and the tangent
to the ellipse at P passes through Q(4,4) then PQ is equal to : | [{"identifier": "A", "content": "$${{\\sqrt {61} } \\over 2}$$"}, {"identifier": "B", "content": "$${{\\sqrt {221} } \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt {157} } \\over 2}$$"}, {"identifier": "D", "content": "$${{5\\sqrt 5 } \\over 2}$$"}] | ["D"] | null | Equation of ellipse is $${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$$<br><br>
Normal at P(2 cos $$\theta $$, $$\sqrt 3 \sin \theta $$) is 2x sin$$\theta $$ - $$\sqrt 3 y\,cos\theta $$ = sin $$\theta $$ cos $$\theta $$ as the normal is parallel to 2x + y = 4<br><br>
$$ \Rightarrow $$ $${2 \over {\sqrt 3 }}\tan \theta = - 2$$<br><br>
$$ \Rightarrow \tan \theta = - \sqrt 3 \,\,......\,(i)$$<br><br>
tangent at P(2 cos $$\theta $$, $$\sqrt 3 \sin \theta $$) is <br><br>
$$\sqrt 3 $$ x cos $$\theta $$ + 2y sin $$\theta $$ = 2 $$\sqrt 3 $$<br><br>
Passes through (4, 4)<br><br>
$$ \Rightarrow $$ 4$$\sqrt 3 $$ cos $$\theta $$ + 8 sin $$\theta $$ = 2$$\sqrt 3 $$ ....... (ii)<br><br>
From (i) and (ii) <br><br>
$$ \Rightarrow P\left( { - 1,{3 \over 2}} \right)\,\& \,Q(4,4)$$<br><br>
$$ \Rightarrow PQ = \sqrt {25 + {{25} \over 4}} = {{5\sqrt 5 } \over 2}$$ | mcq | jee-main-2019-online-12th-april-morning-slot | 1,780 |
7U4w1GA7rQRFV2KmmT7k9k2k5gjiucr | maths | ellipse | normal-to-ellipse | Let the line y = mx and the ellipse 2x<sup>2</sup> + y<sup>2</sup> = 1
intersect at a ponit P in the first quadrant. If the
normal to this ellipse at P meets the co-ordinate axes at $$\left( { - {1 \over {3\sqrt 2 }},0} \right)$$ and (0, $$\beta $$), then $$\beta $$ is equal
to : | [{"identifier": "A", "content": "$${{\\sqrt 2 } \\over 3}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${{2\\sqrt 2 } \\over 3}$$"}, {"identifier": "D", "content": "$${2 \\over {\\sqrt 3 }}$$"}] | ["A"] | null | Let P be (x<sub>1</sub>
, y<sub>1</sub>)
<br><br>Equation of normal at P is $${x \over {2{x_1}}} - {y \over {{y_1}}} = {1 \over 2} - 1$$
<br><br>It passes through $$\left( { - {1 \over {3\sqrt 2 }},0} \right)$$
<br><br>$$ \therefore $$ $${{ - 1} \over {6\sqrt 2 {x_1}}} = - {1 \over 2}$$
<br><br>$$ \Rightarrow $$ x<sub>1</sub> = $${1 \over {3\sqrt 2 }}$$
<br><br>Also using 2$$x_1^2$$ + $$y_1^2$$ = 1
<br><br>$$ \Rightarrow $$ $$y_1^2$$ = 1 - $${2 \over {18}}$$
<br><br>$$ \Rightarrow $$ y<sub>1</sub> = $${{2\sqrt 2 } \over 3}$$
<br><br>(0, $$\beta $$) is on the normal.
<br><br>So 0 - $${\beta \over {{{2\sqrt 2 } \over 3}}}$$ = $$ - {1 \over 2}$$
<br><br>$$ \Rightarrow $$ $$\beta $$ = $${{\sqrt 2 } \over 3}$$ | mcq | jee-main-2020-online-8th-january-morning-slot | 1,781 |
1EvusdwISdLMDs07h4jgy2xukfakgvy9 | maths | ellipse | normal-to-ellipse | Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is $${1 \over 2}$$. If P(1, $$\beta $$), $$\beta $$ > 0 is a point on this ellipse, then the equation of the normal to it at P is : | [{"identifier": "A", "content": "4x \u2013 3y = 2\n"}, {"identifier": "B", "content": "8x \u2013 2y = 5"}, {"identifier": "C", "content": "7x \u2013 4y = 1 "}, {"identifier": "D", "content": "4x \u2013 2y = 1"}] | ["D"] | null | $$e = {1 \over 2}$$ <br><br>$$x = {a \over e} = 4$$<br><br>$$ \Rightarrow $$ a = 2<br><br>$${e^2} = 1 - {{{b^2}} \over {{a^2}}} $$
<br><br>$$\Rightarrow {1 \over 4} = 1 - {{{b^2}} \over 4}$$<br><br>$${{{b^2}} \over 4} = {3 \over 4} \Rightarrow {b^2} = 3$$<br><br>$$ \therefore $$ Ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$$<br><br>P(1, $$\beta $$) is on the ellipse<br><br> $${1 \over 4} + {{{\beta ^2}} \over 3} = 1$$<br><br>$${{{\beta ^2}} \over 3} = {3 \over 4} \Rightarrow \beta = {3 \over 2}$$<br><br>$$ \Rightarrow P\left( {1,{3 \over 2}} \right)$$<br><br>Equation of normal $${{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2} - {b^2}$$<br><br>$$ \Rightarrow $$ $${{4x} \over 1} - {{3y} \over {{3 \over 2}}} = 4 - 3$$<br><br>$$ \Rightarrow $$ $$4x - 2y = 1$$ | mcq | jee-main-2020-online-4th-september-evening-slot | 1,782 |
N3TmnUyC4v34yUj1Gpjgy2xukg0cqp9y | maths | ellipse | normal-to-ellipse | If the normal at an end of a latus rectum of an
ellipse passes through an extremity of the
minor axis, then the eccentricity e of the ellipse
satisfies : | [{"identifier": "A", "content": "e<sup>4</sup> + 2e<sup>2</sup> \u2013 1 = 0"}, {"identifier": "B", "content": "e<sup>4</sup> + e<sup>2</sup> \u2013 1 = 0"}, {"identifier": "C", "content": "e<sup>2</sup> + 2e \u2013 1 = 0"}, {"identifier": "D", "content": "e<sup>2</sup> + e \u2013 1 = 0"}] | ["B"] | null | Equation of normal at $$\left( {ae,{{{b^2}} \over a}} \right)$$
<br><br>$${{{a^2}x} \over {ae}} - {{{b^2}y} \over {{{{b^2}} \over a}}} = {a^2} - {b^2}$$
<br><br>It passes through (0,–b)
<br><br>$$ \therefore $$ $$0 - {{{b^2}\left( { - b} \right)} \over {{{{b^2}} \over a}}} = {a^2} - {b^2}$$
<br><br>$$ \Rightarrow $$ $$a$$b = $${a^2} - {b^2}$$
<br><br>$$ \Rightarrow $$ $$a$$b = $${a^2}{e^2}$$ [as b<sup>2</sup> = $${a^2}\left( {1 - {e^2}} \right)$$]
<br><br>$$ \Rightarrow $$ $$a$$<sup>2</sup>b<sup>2</sup> = $${a^4}{e^4}$$
<br><br>$$ \Rightarrow $$ $${{{{b^2}} \over {{a^2}}}}$$ = e<sup>4</sup>
<br><br>$$ \Rightarrow $$ $${1 - {e^2}}$$ = e<sup>4</sup>
<br><br>$$ \Rightarrow $$ e<sup>4</sup> + e<sup>2</sup> – 1 = 0 | mcq | jee-main-2020-online-6th-september-evening-slot | 1,783 |
1ldpt2o9m | maths | ellipse | normal-to-ellipse | <p>If the maximum distance of normal to the ellipse $$\frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1, b < 2$$, from the origin is 1, then the eccentricity of the ellipse is :</p> | [{"identifier": "A", "content": "$$\\frac{\\sqrt{3}}{4}$$"}, {"identifier": "B", "content": "$$\\frac{1}{2}$$"}, {"identifier": "C", "content": "$$\\frac{1}{\\sqrt{2}}$$"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{3}}{2}$$"}] | ["D"] | null | Equation of normal is
<br/><br/>$2 x \sec \theta-b y \operatorname{cosec} \theta=4-b^{2}$
<br/><br/>Distance from $(0,0)=\frac{4-b^{2}}{\sqrt{4 \sec ^{2} \theta+b^{2} \operatorname{cosec}^{2} \theta}}$
<br/><br/>Distance is maximum if
<br/><br/>$4 \sec ^{2} \theta+b^{2} \operatorname{cosec}^{2} \theta$ is minimum
<br/><br/>$\Rightarrow \tan ^{2} \theta=\frac{\mathrm{b}}{2}$
<br/><br/>$\Rightarrow \frac{4-b^{2}}{\sqrt{4 \cdot \frac{b+2}{2}+b^{2} \cdot \frac{b+2}{b}}}=1$
<br/><br/>$\Rightarrow 4-b^{2}=(b+2) \Rightarrow b^{2}+b-2=0$
<br/><br/>$\Rightarrow(b+2)(b-1)=0$
<br/><br/>$\Rightarrow b=1$
<br/><br/>$\therefore e=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$ | mcq | jee-main-2023-online-31st-january-morning-shift | 1,784 |
1lgpy8jfv | maths | ellipse | normal-to-ellipse | <p>Let the tangent and normal at the point $$(3 \sqrt{3}, 1)$$ on the ellipse $$\frac{x^{2}}{36}+\frac{y^{2}}{4}=1$$ meet the $$y$$-axis at the points $$A$$ and $$B$$ respectively. Let the circle $$C$$ be drawn taking $$A B$$ as a diameter and the line $$x=2 \sqrt{5}$$ intersect $$C$$ at the points $$P$$ and $$Q$$. If the tangents at the points $$P$$ and $$Q$$ on the circle intersect at the point $$(\alpha, \beta)$$, then $$\alpha^{2}-\beta^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "61"}, {"identifier": "B", "content": "$$\\frac{304}{5}\n$$"}, {"identifier": "C", "content": "60"}, {"identifier": "D", "content": "$$\\frac{314}{5}\n$$"}] | ["B"] | null | $$
\begin{aligned}
& \frac{x^2}{36}+\frac{y^2}{4}=1 \\\\
& T: \frac{3 \sqrt{3} x}{36}+\frac{y}{4}=1 \\\\
& T: \frac{\sqrt{3} x}{12}+\frac{y}{4}=1 \\\\
& N: \frac{x-3 \sqrt{3}}{\frac{3 \sqrt{3}}{36}}=\frac{y-1}{\frac{1}{4}}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \frac{12 x-36 \sqrt{3}}{\sqrt{3}}=4 y-4 \\\\
& 3 x-9 \sqrt{3}=\sqrt{3} y-\sqrt{3} \\\\
& N: 3 x-\sqrt{3} y=8 \sqrt{3} \\\\
& A(0,4) \quad B(0,-8) \\\\
& \text { C: } x^2+(y-4)(y+8)=0 \\\\
& \text { Line } x=2 \sqrt{5} \\\\
& 20+y^2+4 y-32=0 \\\\
& y^2+4 y-12=0 \\\\
& (y+6)(y-2)=0
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& P(2 \sqrt{5},-6) \quad Q(2 \sqrt{5}, 2) \\\\
& C: x^2+y^2+4 y-32=0 \\\\
& P(2 \sqrt{5},-6) \quad Q(2 \sqrt{5}, 2) \\\\
& T: x x_1+y y_1+2 y+2 y_1-32=0 \\\\
& T_1: 2 \sqrt{5} x-6 y+2 y-12-32=0 \\\\
& \quad 2 \sqrt{5} x-4 y=44 \\\\
& T_1: \sqrt{5} x-2 y=22 .......(i) \\\\
& T_2: 2 \sqrt{5} x+2 y+2 y+4-32=0 \\\\
& 2 \sqrt{5} x+4 y=28 \\\\
& T_2: \sqrt{5} x+2 y=14 .......(ii)
\end{aligned}
$$
<br/><br/>From (i) and (ii), we get
<br/><br/>$$
\begin{aligned}
& \alpha=\frac{18}{\sqrt{5}} \quad \beta=-2 \\\\
& \alpha^2-\beta^2=\frac{304}{5}
\end{aligned}
$$ | mcq | jee-main-2023-online-13th-april-morning-shift | 1,785 |
rlmIdLEuDAejugVr | maths | ellipse | position-of-point-and-chord-joining-of-two-points | The equation of the circle passing through the foci of the ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$, and having centre at $$(0,3)$$ is : | [{"identifier": "A", "content": "$${x^2} + {y^2} - 6y - 7 = 0$$ "}, {"identifier": "B", "content": "$${x^2} + {y^2} - 6y + 7 = 0$$"}, {"identifier": "C", "content": "$${x^2} + {y^2} - 6y - 5 = 0$$"}, {"identifier": "D", "content": "$${x^2} + {y^2} - 6y + 5 = 0$$"}] | ["A"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263921/exam_images/s76qynlxlel1kkck45r2.webp" loading="lazy" alt="JEE Main 2013 (Offline) Mathematics - Ellipse Question 76 English Explanation">
<br><br>From the given equation of ellipse, we have
<br><br>$$a = 4,b = 3,e = \sqrt {1 - {9 \over {16}}} $$
<br><br>$$ \Rightarrow e = {{\sqrt 7 } \over 4}$$
<br><br>Now, radius of this circle $$ = {a^2} = 16$$
<br><br>$$ \Rightarrow Focii = \left( { \pm \sqrt 7 ,0} \right)$$
<br><br>Now equation of circle is
<br><br>$${\left( {x - 0} \right)^2} + {\left( {y - 3} \right)^2} = 16$$
<br><br>$${x^2} + y{}^2 - 6y - 7 = 0$$ | mcq | jee-main-2013-offline | 1,786 |
2oTqNMY0qPaZCT9gEHjgy2xukfjjqpjo | maths | ellipse | position-of-point-and-chord-joining-of-two-points | If the co-ordinates of two points A and B <br/>are $$\left( {\sqrt 7 ,0} \right)$$ and $$\left( { - \sqrt 7 ,0} \right)$$ respectively and<br/> P is any
point on the conic, 9x<sup>2</sup> + 16y<sup>2</sup> = 144, then PA + PB is equal to : | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "16"}, {"identifier": "D", "content": "6"}] | ["A"] | null | 9x<sup>2</sup> + 16y<sup>2</sup> = 144
<br><br>$$ \Rightarrow $$ $${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$
<br><br>$$ \therefore $$ a = 4; b = 3;
<br><br>Now e = $$\sqrt {1 - {9 \over {16}}} = {{\sqrt 7 } \over 4}$$
<br><br>A and B are foci
<br><br>PA + PB = 2a = 2 × 4 = 8 | mcq | jee-main-2020-online-5th-september-morning-slot | 1,787 |
BcheBfrZmexesZ4Um71klt9cr9j | maths | ellipse | position-of-point-and-chord-joining-of-two-points | If the curve x<sup>2</sup> + 2y<sup>2</sup> = 2 intersects the line x + y = 1 at two points P and Q, then the angle subtended by the line segment PQ at the origin is : | [{"identifier": "A", "content": "$${\\pi \\over 2} - {\\tan ^{ - 1}}\\left( {{1 \\over 4}} \\right)$$"}, {"identifier": "B", "content": "$${\\pi \\over 2} + {\\tan ^{ - 1}}\\left( {{1 \\over 3}} \\right)$$"}, {"identifier": "C", "content": "$${\\pi \\over 2} - {\\tan ^{ - 1}}\\left( {{1 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$${\\pi \\over 2} + {\\tan ^{ - 1}}\\left( {{1 \\over 4}} \\right)$$"}] | ["D"] | null | Ellipse : $${x \over 2} + {y \over 1} = 1$$<br><br>Line : $$x + y = 1$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267198/exam_images/lmh3h5mktsgvuq0vvk3v.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Evening Shift Mathematics - Ellipse Question 48 English Explanation"><br><br>Using homogenisation<br><br>$${x^2} + 2{y^2} = 2{(1)^2}$$<br><br>$${x^2} + 2{y^2} = 2{(x + y)^2}$$<br><br>$${x^2} + 2{y^2} = 2{x^2} + 2{y^2} + 4xy$$<br><br>$${x^2} + 4xy = 0$$<br><br>for $$a{x^2} + 2hxy + b{y^2} = 0$$<br><br>$$\tan \theta = \left| {{{2\sqrt {{h^2} - ab} } \over {a + b}}} \right|$$<br><br>$$\tan \theta = \left| {{{2\sqrt {{{(2)}^2} - 0} } \over {1 + 0}}} \right|$$<br><br>$$\tan \theta = - 4$$<br><br>$$\cot \theta = - {1 \over 4}$$<br><br>$$\theta = {\cot ^{ - 1}}\left( { - {1 \over 4}} \right)$$<br><br>$$\theta = \pi - {\cot ^{ - 1}}\left( {{1 \over 4}} \right)$$<br><br>$$\theta = \pi - \left( {{\pi \over 2} - {{\tan }^{ - 1}}\left( {{1 \over 4}} \right)} \right)$$<br><br>$$\theta = {\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 4}} \right)$$ | mcq | jee-main-2021-online-25th-february-evening-slot | 1,788 |
1lgrg5nwu | maths | ellipse | position-of-point-and-chord-joining-of-two-points | <p>Let $$\mathrm{P}\left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right), \mathrm{Q}, \mathrm{R}$$ and $$\mathrm{S}$$ be four points on the ellipse $$9 x^{2}+4 y^{2}=36$$. Let $$\mathrm{PQ}$$ and $$\mathrm{RS}$$ be mutually perpendicular and pass through the origin. If $$\frac{1}{(P Q)^{2}}+\frac{1}{(R S)^{2}}=\frac{p}{q}$$, where $$p$$ and $$q$$ are coprime, then $$p+q$$ is equal to :</p> | [{"identifier": "A", "content": "143"}, {"identifier": "B", "content": "147"}, {"identifier": "C", "content": "137"}, {"identifier": "D", "content": "157"}] | ["D"] | null | Given, points $P$ and $R$ are on the ellipse defined by $9x^2+4y^2=36$ which simplifies to $\frac{x^2}{4} + \frac{y^2}{9} = 1$.
This is the standard form of the equation of an ellipse centered at the origin, with semi-major axis $a=3$ along the $y$-axis and semi-minor axis $b=2$ along the $x$-axis.
<br/><br/>OP is the distance from origin O to point P, which is given by :
<br/><br/>$$OP =r_1 = \sqrt{\left(\frac{2\sqrt{3}}{\sqrt{7}}\right)^2 + \left(\frac{6}{\sqrt{7}}\right)^2} = \sqrt{\frac{12}{7} + \frac{36}{7}} = \sqrt{\frac{48}{7}} = 2\sqrt{\frac{12}{7}}.$$
<br/><br/>1. Let's represent the given point $P\left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right)$ in polar coordinates. We can write $P$ as $(r_1 \cos \theta, r_1 \sin \theta)$. Since $P$ lies on the ellipse, it must satisfy the equation of the ellipse. Substituting $x=r_1 \cos \theta$ and $y=r_1 \sin \theta$ into the equation of the ellipse gives us :
<br/><br/> $$\frac{r_1^2 \cos ^2 \theta}{4}+\frac{r_1^2 \sin ^2 \theta}{9}=1$$
<br/><br/> Simplifying this, we obtain :
<br/><br/> $$\frac{\cos ^2 \theta}{4}+\frac{\sin ^2 \theta}{9}=\frac{7}{48} \quad \text{--- (equation 1)}$$
<br/><br/>2. Similarly, if we represent the point $R$ as $(-r_2 \sin \theta, r_2 \cos \theta)$, (the negative sign is due to the fact that line RS is perpendicular to line PQ. Since they are perpendicular, the angle between them is 90 degrees or $\pi/2$ radians. In terms of sin and cos, $\sin(\theta + \pi/2) = \cos(\theta)$ and $\cos(\theta + \pi/2) = -\sin(\theta)$.) it too should satisfy the equation of the ellipse. We have :
<br/><br/> $$\frac{r_2^2 \sin ^2 \theta}{4}+\frac{r_2^2 \cos ^2 \theta}{9}=1$$
<br/><br/> Simplifying this, we obtain :
<br/><br/> $$\frac{\sin ^2 \theta}{4}+\frac{\cos ^2 \theta}{9}=\frac{1}{r_2^2} \quad \text{--- (equation 2)}$$
<br/><br/>3. From equations (1) and (2), we have :
<br/><br/> $$\frac{1}{r_2^2}=\frac{1}{4}+\frac{1}{9}-\frac{7}{48}=\frac{31}{144}$$
<br/><br/>4. Now, note that lines $PQ$ and $RS$ are perpendicular and pass through the origin, so $PQ = 2OP$ and $RS = 2OR$. Thus,
<br/><br/> $$\frac{1}{PQ^2} + \frac{1}{RS^2} = \frac{1}{4} \left(\frac{1}{r_1^2} + \frac{1}{r_2^2} \right)$$
<br/><br/>5. Substituting the values of $r_1$ and $r_2$, we obtain :
<br/><br/> $$\frac{1}{PQ^2} + \frac{1}{RS^2} = \frac{1}{4}\left(\frac{7}{48}+\frac{31}{144}\right)=\frac{13}{144} = \frac{p}{q}$$
<br/><br/>6. Hence, $p = 13$ and $q = 144$.
<br/><br/>7. So, the final answer $p+q = 13 + 144 = 157$. | mcq | jee-main-2023-online-12th-april-morning-shift | 1,789 |
lv7v4g1l | maths | ellipse | position-of-point-and-chord-joining-of-two-points | <p>Let the line $$2 x+3 y-\mathrm{k}=0, \mathrm{k}>0$$, intersect the $$x$$-axis and $$y$$-axis at the points $$\mathrm{A}$$ and $$\mathrm{B}$$, respectively. If the equation of the circle having the line segment $$A B$$ as a diameter is $$x^2+y^2-3 x-2 y=0$$ and the length of the latus rectum of the ellipse $$x^2+9 y^2=k^2$$ is $$\frac{m}{n}$$, where $$m$$ and $$n$$ are coprime, then $$2 \mathrm{~m}+\mathrm{n}$$ is equal to</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "13"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "10"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwgi30yz/efd8a8c4-b19c-464d-9aab-50da9ed84968/ccf9e7b0-177f-11ef-97dc-2d80937d5077/file-1lwgi30z0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwgi30yz/efd8a8c4-b19c-464d-9aab-50da9ed84968/ccf9e7b0-177f-11ef-97dc-2d80937d5077/file-1lwgi30z0.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Morning Shift Mathematics - Ellipse Question 1 English Explanation"></p>
<p>Equation of circle with $$A B$$ as diameter</p>
<p>$$\begin{aligned}
& \left(x-\frac{k}{2}\right) x+y\left(y-\frac{k}{3}\right)=0 \\
& \Rightarrow x^2+y^2-\frac{k x}{2}-\frac{k y}{3}=0
\end{aligned}$$</p>
<p>Comparing, $$k=6$$</p>
<p>Latus rectum of ellipse</p>
<p>$$\begin{aligned}
& x^2+9 y^2=k^2=6^2 \\
& \Rightarrow \frac{x^2}{6^2}+\frac{y^2}{2^2}=1 \\
& \text { L.R }=\frac{2 b^2}{a}=\frac{2 \times 4}{6}=\frac{4}{3} \\
& m=4 \\
& n=3 \\
& 2 m+n=8+3=11
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift | 1,790 |
UpdYGwyWWDYrcvXF | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | The eccentricity of an ellipse, with its centre at the origin, is $${1 \over 2}$$. If one of the directrices is $$x=4$$, then the equation of the ellipse is : | [{"identifier": "A", "content": "$$4{x^2} + 3{y^2} = 1$$ "}, {"identifier": "B", "content": "$$3{x^2} + 4{y^2} = 12$$"}, {"identifier": "C", "content": "$$4{x^2} + 3{y^2} = 12$$"}, {"identifier": "D", "content": "$$3{x^2} + 4{y^2} = 1$$"}] | ["B"] | null | $$e = {1 \over 2}.\,\,$$ Directrix, $$x = {a \over e} = 4$$
<br><br>$$\therefore$$ $$a = 4 \times {1 \over 2} = 2$$
<br><br>$$\therefore$$ $$b = 2\sqrt {1 - {1 \over 4}} = \sqrt 3 $$
<br><br>Equation of elhipe is
<br><br>$${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1 \Rightarrow 3{x^2} + 4{y^2} = 12$$ | mcq | aieee-2004 | 1,791 |
fqgR73iD6te7y31I | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | An ellipse has $$OB$$ as semi minor axis, $$F$$ and $$F$$' its focii and theangle $$FBF$$' is a right angle. Then the eccentricity of the ellipse is : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 4}$$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 3 }}$$"}] | ["A"] | null | as $$\angle FBF' = {90^ \circ }$$
<br><br>$$ \Rightarrow F{B^2} + F'{B^2} = FF{'^2}$$
<br><br>$$\therefore$$ $${\left( {\sqrt {{a^2}{e^2} + {b^2}} } \right)^2} + \left( {\sqrt {{a^2}{e^2} + {b^2}} } \right) = {\left( {2ae} \right)^2}$$
<br><br>$$ \Rightarrow 2\left( {{a^2}{e^2} + {b^2}} \right) = 4{a^2}{e^2}$$
<br><br>$$ \Rightarrow {e^2} = {{{b^2}} \over {{a^2}}}$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263637/exam_images/evoxpwc1lpvv4diyvenq.webp" loading="lazy" alt="AIEEE 2005 Mathematics - Ellipse Question 84 English Explanation">
<br><br>Also $${e^2} = 1 - {b^2}/{a^2} = 1 - {e^2}$$
<br><br>$$ \Rightarrow 2{e^2} = 1,\,\,e = {1 \over {\sqrt 2 }}$$ | mcq | aieee-2005 | 1,792 |
h1jmOU3BvKJGrV5A | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | In the ellipse, the distance between its foci is $$6$$ and minor axis is $$8$$. Then its eccentricity is : | [{"identifier": "A", "content": "$${3 \\over 5}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${4 \\over 5}$$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 5 }}$$"}] | ["A"] | null | $$2ae = 6 \Rightarrow ae = 3;\,\,2b = 8 \Rightarrow b = 4$$
<br><br>$${b^2} = {a^2}\left( {1 - {e^2}} \right);16 = {a^2} - {a^2}{e^2}$$
<br><br>$$ \Rightarrow a{}^2 = 16 + 9 = 25$$
<br><br>$$ \Rightarrow a = 5$$
<br><br>$$\therefore$$ $$e = {3 \over a} = {3 \over 5}$$ | mcq | aieee-2006 | 1,793 |
YPUS6uIIDYQa7Fw1 | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | A focus of an ellipse is at the origin. The directrix is the line $$x=4$$ and the eccentricity is $${{1 \over 2}}$$. Then the length of the semi-major axis is : | [{"identifier": "A", "content": "$${{8 \\over 3}}$$"}, {"identifier": "B", "content": "$${{2 \\over 3}}$$"}, {"identifier": "C", "content": "$${{4 \\over 3}}$$"}, {"identifier": "D", "content": "$${{5 \\over 3}}$$"}] | ["A"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264328/exam_images/fxzt3slso48o3sotippm.webp" loading="lazy" alt="AIEEE 2008 Mathematics - Ellipse Question 81 English Explanation">
<br><br>Perpendicular distance of directrix from focus
<br><br>$$ = {a \over e} - ae = 4$$
<br><br>$$ \Rightarrow a\left( {2 - {1 \over 2}} \right) = 4$$
<br><br>$$ \Rightarrow a = {8 \over 3}$$
<br><br>$$\therefore$$ Semi major axis $$=8/3$$ | mcq | aieee-2008 | 1,794 |
8BGB4L5m1YJ9F01b | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | The ellipse $${x^2} + 4{y^2} = 4$$ is inscribed in a rectangle aligned with the coordinate axex, which in turn is inscribed in another ellipse that passes through the point $$(4,0)$$. Then the equation of the ellipse is : | [{"identifier": "A", "content": "$${x^2} + 12{y^2} = 16$$ "}, {"identifier": "B", "content": "$$4{x^2} + 48{y^2} = 48$$ "}, {"identifier": "C", "content": "$$4{x^2} + 64{y^2} = 48$$ "}, {"identifier": "D", "content": "$${x^2} + 16{y^2} = 16$$ "}] | ["A"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265178/exam_images/ymnjbdlge7ihvtgsvwzr.webp" loading="lazy" alt="AIEEE 2009 Mathematics - Ellipse Question 80 English Explanation">
<br><br>The given ellipse is $${{{x^2}} \over 4} + {{{y^2}} \over 1} = 1$$
<br><br>So $$A=(2,0)$$ and $$B = \left( {0,1} \right)$$
<br><br>If $$PQRS$$ is the rectangular in which it is inscribed, then
<br><br>$$P = \left( {2,1} \right).$$
<br><br>Let $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$
<br><br>be the ellipse circumscribing the rectangular $$PQRS$$.
<br><br>Then it passes through $$P\,\,(2,1)$$
<br><br>$$\therefore$$ $${4 \over {a{}^2}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( a \right)$$
<br><br>Also, given that, it passes through $$(4,0)$$
<br><br>$$\therefore$$ $${{16} \over {{a^2}}} + 0 = 1 \Rightarrow {a^2} = 16$$
<br><br>$$ \Rightarrow {b^2} = 4/3$$ $$\left[ {\,\,} \right.$$ substituting $${{a^2} = 16\,\,}$$ in $$\left. {e{q^n}\left( a \right)\,\,} \right]$$
<br><br>$$\therefore$$ The required ellipse is $${{{x^2}} \over {16}} + {{{y^2}} \over {4/3}} = 1$$
<br><br>or $${x^2} + 12y{}^2 = 16$$ | mcq | aieee-2009 | 1,795 |
e0Okjyna0slrDuAf | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | Equation of the ellipse whose axes of coordinates and which passes through the point $$(-3,1)$$ and has eccentricity $$\sqrt {{2 \over 5}} $$ is : | [{"identifier": "A", "content": "$$5{x^2} + 3{y^2} - 48 = 0$$ "}, {"identifier": "B", "content": "$$3{x^2} + 5{y^2} - 15 = 0$$"}, {"identifier": "C", "content": "$$5{x^2} + 3{y^2} - 32 = 0$$"}, {"identifier": "D", "content": "$$3{x^2} + 5{y^2} - 32 = 0$$"}] | ["D"] | null | Let the ellipse be $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$
<br><br>It press through $$(-3, 1)$$ so $${9 \over {{a^2}}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( i \right)$$
<br><br>Also, $${b^2} = {a^2}\left( {1 - 2/5} \right)$$
<br><br>$$ \Rightarrow 5{b^2} = 3{a^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>Solving $$(i)$$ and $$(ii)$$ we get $${a^2} = {{32} \over 3},{b^2} = {{32} \over 5}$$
<br><br>So, the equation of the ellipse is $$3{x^2} + 5{y^2} = 32$$ | mcq | aieee-2011 | 1,796 |
nDHwUJAI71QHUK6u | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | An ellipse is drawn by taking a diameter of thec circle $${\left( {x - 1} \right)^2} + {y^2} = 1$$ as its semi-minor axis and a diameter of the circle $${x^2} + {\left( {y - 2} \right)^2} = 4$$ is semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is : | [{"identifier": "A", "content": "$$4{x^2} + {y^2} = 4$$ "}, {"identifier": "B", "content": "$${x^2} + 4{y^2} = 8$$"}, {"identifier": "C", "content": "$$4{x^2} + {y^2} = 8$$"}, {"identifier": "D", "content": "$${x^2} + 4{y^2} = 16$$"}] | ["D"] | null | Equation of circle is $${\left( {x - 1} \right)^2} + {y^2} = 1$$
<br><br>$$ \Rightarrow $$ radius $$=1$$ and diameter $$=2$$
<br><br>$$\therefore$$ Length of semi-minor axis is $$2.$$
<br><br>Equation of circle is $${x^2} + {\left( {y - 2} \right)^2} = 4 = {\left( 2 \right)^2}$$
<br><br>$$ \Rightarrow $$ radius $$=2$$ and diameter $$=4$$
<br><br>$$\therefore$$ Length of semi major axis is $$4$$
<br><br>We know, equation of ellipse is given by
<br><br>$${{{x^2}} \over {\left( {Major\,\,\,axi{s^{\,\,2}}} \right)}} + {{{y^2}} \over {\left( {Minor\,\,\,axi{s^{\,\,2}}} \right)}} = 1$$
<br><br>$$ \Rightarrow {{{x^2}} \over {{{\left( 4 \right)}^2}}} + {{{y^2}} \over {{{\left( 2 \right)}^2}}} = 1$$
<br><br>$$ \Rightarrow {{{x^2}} \over {16}} + {{{y^2}} \over 4} = 1$$
<br><br>$$ \Rightarrow {x^2} + 4{y^2} = 16$$ | mcq | aieee-2012 | 1,797 |
3EgKRF0Qam0Y2DwppaMvV | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | Consider an ellipse, whose center is at the origin and its major axis is along the x-axis. If its eccentricity is $${3 \over 5}$$ and the distance between its foci is 6, then the area (in sq. units) of the quadrilatateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is : | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "32"}, {"identifier": "C", "content": "80"}, {"identifier": "D", "content": "40"}] | ["D"] | null | e = 3/5 & 2ae = 6 $$ \Rightarrow $$ a = 5
<br><br>$$ \because $$ b<sup>2</sup> = a<sup>2</sup> (1 $$-$$ e<sup>2</sup>)
<br><br>$$ \Rightarrow $$ b<sup>2</sup> = 25(1 $$-$$ 9/25)
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265271/exam_images/qlwwjazgwqnl9oqbsmlw.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Online) 8th April Morning Slot Mathematics - Ellipse Question 71 English Explanation">
<br>$$ \Rightarrow $$ b = 4
<br><br>$$ \therefore $$ Area of required quadrilateral
<br><br>= 4(1/2 ab) = 2ab = 40 | mcq | jee-main-2017-online-8th-april-morning-slot | 1,798 |
pH7BEcGlUt4jdIi97WFap | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate
axes and passing through the points (4, −1) and (−2, 2) is :
| [{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$${2 \\over {\\sqrt 5 }}$$ "}, {"identifier": "C", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "D", "content": "$${{\\sqrt 3 } \\over 4}$$ "}] | ["C"] | null | Centre at (0, 0)
<br><br>$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}}$$ = 1
<br><br>at point (4, $$-$$ 1)
<br><br>$${{16} \over {{a^2}}} + {1 \over {{b^2}}}$$ = 1
<br><br>$$ \Rightarrow $$ 16b<sup>2</sup> + a<sup>2</sup> = a<sup>2</sup>b<sup>2</sup> . . . .(i)
<br><br>at point ($$-$$ 2, 2)
<br><br>$${4 \over {{a^2}}} + {4 \over {{b^2}}} = 1$$
<br><br>$$ \Rightarrow $$ 4b<sup>2</sup> + 4a<sup>2</sup> = a<sup>2</sup>b<sup>2</sup> . . . .(ii)
<br><br>$$ \Rightarrow $$ 16b<sup>2</sup> + a<sup>2</sup> = 4a<sup>2</sup> + 4b<sup>2</sup>
<br><br>From equations (i) and (ii)
<br><br>$$ \Rightarrow $$ 3a<sup>2</sup> = 12b<sup>2</sup>
<br><br>$$ \Rightarrow $$ <b>a<sup>2</sup> = 4b<sup>2</sup></b>
<br><br>b<sup>2</sup> = a<sup>2</sup>(1 $$-$$ e<sup>2</sup>)
<br><br>$$ \Rightarrow $$ e<sup>2</sup> = $${3 \over 4}$$
<br><br>$$ \Rightarrow $$ e = $${{\sqrt 3 } \over 2}$$ | mcq | jee-main-2017-online-9th-april-morning-slot | 1,799 |
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