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1l5w0n374 | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Let $$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)$$ and $$\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)$$ where the inverse trigonometric functions take pr... | [{"identifier": "A", "content": "$$15{x^2} - 8x - 7 = 0$$"}, {"identifier": "B", "content": "$$5{x^2} - 12x + 7 = 0$$"}, {"identifier": "C", "content": "$$25{x^2} - 18x - 7 = 0$$"}, {"identifier": "D", "content": "$$25{x^2} - 32x + 7 = 0$$"}] | ["C"] | null | <p>Given,</p>
<p>$$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)$$</p>
<p>We know, $$2{\cos ^{ - 1}}x = {\cos ^{ - 1}}(2{x^2} - 1)$$</p>
<p>$$\therefore$$ $$2{\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right) = {\cos ^1}\left( {2 \times... | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6jduff0 | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>For $$k \in \mathbb{R}$$, let the solutions of the equation $$\cos \left(\sin ^{-1}\left(x \cot \left(\tan ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)\right)\right)\right)=k, 0<|x|<\frac{1}{\sqrt{2}}$$ be $$\alpha$$ and $$\beta$$, where the inverse trigonometric functions take only principal values. If th... | [] | null | 12 | <p>$$\cos \left( {{{\sin }^{ - 1}}\left( {x\cot \left( {{{\tan }^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}} \right)} \right)} \right)} \right)} \right) = k$$</p>
<p>$$ \Rightarrow \cos \left( {{{\sin }^{ - 1}}\left( {x\cot \left( {{{\tan }^{ - 1}}\sqrt {1 - {x^2}} } \right)} \right)} \right) = k$$</p>
<p>$$ \Rightar... | integer | jee-main-2022-online-27th-july-morning-shift |
1l6m56q48 | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Considering only the principal values of the inverse trigonometric functions, the domain of the function $$f(x)=\cos ^{-1}\left(\frac{x^{2}-4 x+2}{x^{2}+3}\right)$$ is :</p> | [{"identifier": "A", "content": "$$\\left(-\\infty, \\frac{1}{4}\\right]$$"}, {"identifier": "B", "content": "$$\\left[-\\frac{1}{4}, \\infty\\right)$$"}, {"identifier": "C", "content": "$$(-1 / 3, \\infty)$$"}, {"identifier": "D", "content": "$$\\left(-\\infty, \\frac{1}{3}\\right]$$"}] | ["B"] | null | <p>$$ - 1 \le {{{x^2} - 4x + 2} \over {{x^2} + 3}} \le 1$$</p>
<p>$$ \Rightarrow - {x^2} - 3 \le {x^2} - 4x + 2 \le {x^2} + 3$$</p>
<p>$$ \Rightarrow 2{x^2} - 4x + 5 \ge 0$$ & $$ - 4x \le 1$$</p>
<p>$$x \in R$$ & $$x \ge - {1 \over 4}$$</p>
<p>So domain is $$\left[ { - {1 \over 4},\infty } \right)$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
1l6m5b57u | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $$\cos ^{-1}(x)-2 \sin ^{-1}(x)=\cos ^{-1}(2 x)$$ is equal to :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$\\frac{1}{2}$$"}, {"identifier": "D", "content": "$$-\\frac{1}{2}$$"}] | ["A"] | null | <p>$${\cos ^{ - 1}}x - 2{\sin ^{ - 1}}x = {\cos ^{ - 1}}2x$$</p>
<p>For Domain : $$x \in \left[ {{{ - 1} \over 2},{1 \over 2}} \right]$$</p>
<p>$${\cos ^{ - 1}}x - 2\left( {{\pi \over 2} - {{\cos }^{ - 1}}x} \right) = {\cos ^{ - 1}}(2x)$$</p>
<p>$$ \Rightarrow {\cos ^{ - 1}}x + 2{\cos ^{ - 1}}x = \pi + {\cos ^{ - 1}}... | mcq | jee-main-2022-online-28th-july-morning-shift |
1ldyay4yl | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>$${\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)$$ is equal to :</p> | [{"identifier": "A", "content": "$${\\pi \\over 2}$$"}, {"identifier": "B", "content": "$${\\pi \\over 3}$$"}, {"identifier": "C", "content": "$${\\pi \\over 6}$$"}, {"identifier": "D", "content": "$${\\pi \\over 4}$$"}] | ["B"] | null | <p>$${\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)$$</p>
<p>$$= {\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}{\left( {{{16 + 8\sqrt 3 } \over {12 + 6\sqrt 3 }}} \right)^{{1 \o... | mcq | jee-main-2023-online-24th-january-morning-shift |
jaoe38c1lscmv27h | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Considering only the principal values of inverse trigonometric functions, the number of positive real values of $$x$$ satisfying $$\tan ^{-1}(x)+\tan ^{-1}(2 x)=\frac{\pi}{4}$$ is :</p> | [{"identifier": "A", "content": "more than 2"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}] | ["D"] | null | <p>$$\begin{aligned}
& \tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4} ; x>0 \\
& \Rightarrow \tan ^{-1} 2 x=\frac{\pi}{4}-\tan ^{-1} x
\end{aligned}$$</p>
<p>Taking tan both sides</p>
<p>$$\begin{aligned}
& \Rightarrow 2 \mathrm{x}=\frac{1-\mathrm{x}}{1+\mathrm{x}} \\
& \Rightarrow 2 \mathrm{x}^2+3 \mathrm{x}-1=0 \\
& \math... | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsd4basi | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>If $$a=\sin ^{-1}(\sin (5))$$ and $$b=\cos ^{-1}(\cos (5))$$, then $$a^2+b^2$$ is equal to</p> | [{"identifier": "A", "content": "25"}, {"identifier": "B", "content": "$$4 \\pi^2+25$$\n"}, {"identifier": "C", "content": "$$8 \\pi^2-40 \\pi+50$$\n"}, {"identifier": "D", "content": "$$4 \\pi^2-20 \\pi+50$$"}] | ["C"] | null | <p>$$\begin{aligned}
& a=\sin ^{-1}(\sin 5)=5-2 \pi \\
& \text { and } b=\cos ^{-1}(\cos 5)=2 \pi-5 \\
& \therefore a^2+b^2=(5-2 \pi)^2+(2 \pi-5)^2 \\
& =8 \pi^2-40 \pi+50
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
luxwcnzi | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $$2 \sin ^{-1} x+3 \cos ^{-1} x=\frac{2 \pi}{5}$$, is __________.</p> | [] | null | 0 | <p>$$\begin{aligned}
& 2 \sin ^{-1} x+3 \cos ^{-1} x=\frac{2 \pi}{5} \\
& \frac{\pi}{2}+\cos ^{-1} x=\frac{2 \pi}{5} \\
& \cos ^{-1} x=\frac{2 \pi}{5}-\frac{\pi}{2} \\
& \cos ^{-1} x=\frac{-\pi}{10}
\end{aligned}$$</p>
<p>Which is not possible as $$\cos ^{-1} x \in[0, \pi]$$</p>
<p>$$\therefore \quad$$ No solution</p> | integer | jee-main-2024-online-9th-april-evening-shift |
lv2erz4o | maths | inverse-trigonometric-functions | principal-value-of-inverse-trigonometric-functions | <p>Given that the inverse trigonometric function assumes principal values only. Let $$x, y$$ be any two real numbers in $$[-1,1]$$ such that $$\cos ^{-1} x-\sin ^{-1} y=\alpha, \frac{-\pi}{2} \leq \alpha \leq \pi$$.
Then, the minimum value of $$x^2+y^2+2 x y \sin \alpha$$ is</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$-$$1"}, {"identifier": "C", "content": "$$\\frac{1}{2}$$"}, {"identifier": "D", "content": "$$\\frac{-1}{2}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \cos ^{-1} x-\frac{\pi}{2}+\cos ^{-1} y=\alpha \\
& \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha \\
& \because \quad \alpha \in\left[\frac{-\pi}{2}, \pi\right] \\
& \text { then } \frac{\pi}{2}+\alpha \in\left(0, \frac{3 \pi}{2}\right) \\
& \cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{1-y^2}\right)=\... | mcq | jee-main-2024-online-4th-april-evening-shift |
KkxLSqZpDRwCLbbj | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | $${\cot ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) - {\tan ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) = x,$$ then sin x is equal to : | [{"identifier": "A", "content": "$${\\tan ^2}\\left( {{\\alpha \\over 2}} \\right)$$ "}, {"identifier": "B", "content": "$${\\cot ^2}\\left( {{\\alpha \\over 2}} \\right)$$"}, {"identifier": "C", "content": "$$\\tan \\alpha $$ "}, {"identifier": "D", "content": "$$cot\\left( {{\\alpha \\over 2}} \\right)$$ "}] | ["A"] | null | Given that,
<br><br>$${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) - {\tan ^{ - 1}}\left( \sqrt{\cos x} \right) = x\,\,\,\,...\left( 1 \right)$$
<br><br>We know,
<br><br>$${\cot ^{ - 1}}x + {\tan ^{ - 1}}x = {\pi \over 2}$$
<br><br>$$\therefore$$ $$\,\,\,$$ $${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) + {\ta... | mcq | aieee-2002 |
VgRPBRM1vhbktxrx | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha ,$$ then $$4{x^2} - 4xy\cos \alpha + {y^2}$$ is equal to : | [{"identifier": "A", "content": "$$2\\sin 2\\alpha $$ "}, {"identifier": "B", "content": "$$4$$ "}, {"identifier": "C", "content": "$$4{\\sin ^2}\\alpha $$ "}, {"identifier": "D", "content": "$$-4{\\sin ^2}\\alpha $$"}] | ["C"] | null | As we know,
<br><br>$${\cos ^{ - 1}}A - {\cos ^{ - 1}}B$$
<br><br>$$ = {\cos ^{ - 1}}\left( {AB + \sqrt {1 - {A^2}} .\sqrt {1 - {B^2}} } \right)$$
<br><br>Given, $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $$
<br><br>$$ \Rightarrow {\cos ^{ - 1}}\left( {x.{y \over 2} + \sqrt {1 - {x^2}} .\sqrt {1 - {{{y^2... | mcq | aieee-2005 |
1TDQHlhEobjkoJLP | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If sin<sup>-1</sup>$$\left( {{x \over 5}} \right)$$ + cosec<sup>-1</sup>$$\left( {{5 \over 4}} \right)$$ = $${\pi \over 2}$$, then the value of x is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3"}] | ["D"] | null | Given sin<sup>-1</sup>$$\left( {{x \over 5}} \right)$$ + cosec<sup>-1</sup>$$\left( {{5 \over 4}} \right)$$ = $${\pi \over 2}$$
<br><br>$$ \Rightarrow $$ sin<sup>-1</sup>$$\left( {{x \over 5}} \right)$$ + sin<sup>-1</sup>$$\left( {{4 \over 5}} \right)$$ = $${\pi \over 2}$$
<br><br>$$ \Rightarrow $$ sin<sup>-1</sup>$$... | mcq | aieee-2007 |
IgvCQgf8GDVXARHa | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | The value of $$cot\left( {\cos e{c^{ - 1}}{5 \over 3} + {{\tan }^{ - 1}}{2 \over 3}} \right)$$ is : | [{"identifier": "A", "content": "$${{6 \\over 17}}$$"}, {"identifier": "B", "content": "$${{3 \\over 17}}$$"}, {"identifier": "C", "content": "$${{4 \\over 17}}$$"}, {"identifier": "D", "content": "$${{5 \\over 17}}$$"}] | ["A"] | null | Given,
<br><br>$$Cot\left( {so{{\sec }^{ - 1}}{5 \over 3} + {{\tan }^{ - 1}}{2 \over 3}} \right)$$
<br><br>$$ = \cot \left( {{{\tan }^{ - 1}}{3 \over 4} + {{\tan }^{ - 1}}{2 \over 3}} \right)$$
<br><br>$$ = cot\left( {{{\tan }^{ - 1}}{{{3 \over 4} + {2 \over 3}} \over {1 - {3 \over 4} - {2 \over 3}}}} \right)$$
<br>... | mcq | aieee-2008 |
uhYKiIhxgTCzlkLr | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $$x, y, z$$ are in A.P. and $${\tan ^{ - 1}}x,{\tan ^{ - 1}}y$$ and $${\tan ^{ - 1}}z$$ are also in A.P., then : | [{"identifier": "A", "content": "$$x=y=z$$ "}, {"identifier": "B", "content": "$$2x=3y=6z$$ "}, {"identifier": "C", "content": "$$6x=3y=2z$$ "}, {"identifier": "D", "content": "$$6x=4y=3z$$"}] | ["A"] | null | Given that, $$x,y,z\,\,$$ are in $$AP$$
<br><br>So, $$\,\,\,$$ $$2y = x + y$$
<br><br>Also given that,
<br><br> $${\tan ^{ - 1}}x,{\tan ^{ - 1}}y\,\,$$ and $$\,\,\,{\tan ^{ - 1}}z\,\,$$ are in $$AP$$
<br><br>So, $$2{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}z$$
<br><br>$$ \Rightarrow {\tan ^{ - 1}}\left( {{{2y... | mcq | jee-main-2013-offline |
tb7Srq7xtkaCBPWe | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | Let $${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right),$$
<br/> where $$\left| x \right| < {1 \over {\sqrt 3 }}.$$ Then a value of $$y$$ is : | [{"identifier": "A", "content": "$${{3x - {x^3}} \\over {1 + 3{x^2}}}$$ "}, {"identifier": "B", "content": "$${{3x + {x^3}} \\over {1 + 3{x^2}}}$$"}, {"identifier": "C", "content": "$${{3x - {x^3}} \\over {1 - 3{x^2}}}$$ "}, {"identifier": "D", "content": "$${{3x + {x^3}} \\over {1 - 3{x^2}}}$$ "}] | ["C"] | null | Given,
<br><br>$${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right)$$
<br><br>$$ \Rightarrow {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{x + {{2x} \over {1 - {x^2}}}} \over {1 - x\left( {{{2x} \over {1 - {x^2}}}} \right)}}} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,... | mcq | jee-main-2015-offline |
wFBhtX1szbvhwRLii253K | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | The value of tan<sup>-1</sup> $$\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right],$$ $$\left| x \right| < {1 \over 2},x \ne 0,$$ is equal to : | [{"identifier": "A", "content": "$${\\pi \\over 4} + {1 \\over 2}{\\cos ^{ - 1}}\\,{x^2}$$"}, {"identifier": "B", "content": "$${\\pi \\over 4} + {\\cos ^{ - 1}}\\,{x^2}$$ "}, {"identifier": "C", "content": "$${\\pi \\over 4} - {1 \\over 2}{\\cos ^{ - 1}}\\,{x^2}$$"}, {"identifier": "D", "content": "$${\\pi \\over ... | ["A"] | null | Given,
<br><br>tan<sup>-1</sup> $$\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]$$
<br><br>Let x<sup>2</sup> = cos $$\theta $$
<br><br>= tan<sup>-1</sup> $$\left[ {{{\sqrt {1 + \cos \theta } + \sqrt {1 - \cos \theta } } \over {\sqrt {1 + \... | mcq | jee-main-2017-online-8th-april-morning-slot |
1AVHKdvqPiPIR7eBWvXeh | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | A value of x satisfying the equation sin[cot<sup>−1</sup> (1+ x)] = cos [tan<sup>−1 </sup>x], is : | [{"identifier": "A", "content": "$$ - {1 \\over 2}$$ "}, {"identifier": "B", "content": "$$-$$ 1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$ {1 \\over 2}$$"}] | ["A"] | null | <p>Let, $${\cot ^{ - 1}}(1 + x) = \alpha $$</p>
<p>$$ \Rightarrow 1 + x = \cot \alpha $$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3b0v2ql/1fa8d111-320f-4de3-8dab-6d23cbe7b0e2/436fe4d0-d65a-11ec-9a06-bd4ec5b93eb4/file-1l3b0v2qm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.... | mcq | jee-main-2017-online-9th-april-morning-slot |
XPZSMXZi1ZlGHmdnVX3rsa0w2w9jx2gxojq | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $$,where –1 $$ \le $$ x $$ \le $$ 1, – 2 $$ \le $$ y $$ \le $$ 2, x $$ \le $$ $${y \over 2}$$
, then for all x, y, 4x<sup>2</sup>
– 4xy cos $$\alpha $$ + y<sup>2</sup>
is equal
to : | [{"identifier": "A", "content": "4 sin<sup>2</sup> $$\\alpha $$"}, {"identifier": "B", "content": "2 sin<sup>2</sup> $$\\alpha $$"}, {"identifier": "C", "content": "4 sin<sup>2</sup> $$\\alpha $$ - 2x<sup>2</sup>y<sup>2</sup>"}, {"identifier": "D", "content": "4 cos<sup>2</sup> $$\\alpha $$ + 2x<sup>2</sup>y<sup>2</sup... | ["A"] | null | $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $$<br><br>
$$ \Rightarrow \cos \left( {{{\cos }^{ - 1}}x - {{\cos }^{ - 1}}\left( {{y \over 2}} \right)} \right) = \cos \alpha $$<br><br>
$$ \Rightarrow x{y \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} = \cos \alpha $$<br><br>
$$\left( {\cos \alpha ... | mcq | jee-main-2019-online-10th-april-evening-slot |
8ggGtIclV7LzC4NVgY3rsa0w2w9jx6ge0vw | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | The value of $${\sin ^{ - 1}}\left( {{{12} \over {13}}} \right) - {\sin ^{ - 1}}\left( {{3 \over 5}} \right)$$ is equal to : | [{"identifier": "A", "content": "$$\\pi - {\\sin ^{ - 1}}\\left( {{{63} \\over {65}}} \\right)$$"}, {"identifier": "B", "content": "$${\\pi \\over 2} - {\\sin ^{ - 1}}\\left( {{{56} \\over {65}}} \\right)$$"}, {"identifier": "C", "content": "$${\\pi \\over 2} - {\\cos ^{ - 1}}\\left( {{9 \\over {65}}} \\right)$$"}, ... | ["B"] | null | $${\sin ^{ - 1}}{{12} \over {13}} - {\sin ^{ - 1}}{3 \over 5} = {\sin ^{ - 1}}\left( {{{12} \over {13}}.{4 \over 5}.{3 \over 5}.{5 \over {13}}} \right)$$<br><br>
$$ \Rightarrow {\sin ^{ - 1}}{{33} \over {65}} = {\pi \over 2} - {\cos ^{ - 1}}{{33} \over {65}}$$<br><br>
$$ \Rightarrow {\pi \over 2} - {\sin ^{ - 1}}{{56... | mcq | jee-main-2019-online-12th-april-morning-slot |
LzROsT3Flxd6M6C9U01L3 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $$\alpha = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$, $$\beta = {\tan ^{ - 1}}\left( {{1 \over 3}} \right)$$ where $$0 < \alpha ,\beta < {\pi \over 2}$$ , then $$\alpha $$ - $$\beta $$ is equal to : | [{"identifier": "A", "content": "$${\\tan ^{ - 1}}\\left( {{9 \\over {14 }}} \\right)$$"}, {"identifier": "B", "content": "$${\\sin ^{ - 1}}\\left( {{9 \\over {5\\sqrt {10} }}} \\right)$$"}, {"identifier": "C", "content": "$${\\cos ^{ - 1}}\\left( {{9 \\over {5\\sqrt {10} }}} \\right)$$"}, {"identifier": "D", "content"... | ["B"] | null | Here $$\cos \alpha = {3 \over 5}$$
<br><br>$$ \therefore $$ $$\tan \alpha = {4 \over 3}$$
<br><br>and $$\tan \beta = {1 \over 3}$$
<br><br>We know,
<br><br>$$\tan \left( {\alpha - \beta } \right) = {{\tan \alpha - \tan \beta } \over {1 + \tan \alpha .\tan \beta }}$$
<br><br>= $${{{4 \over 3} - {1 \over 3}} \over {... | mcq | jee-main-2019-online-8th-april-morning-slot |
0fOD1sLduw0qCWTfaKgYw | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | All x satisfying the inequality (cot<sup>–1</sup>
x)<sup>2</sup>– 7(cot<sup>–1</sup> x) + 10 > 0, lie in the interval : | [{"identifier": "A", "content": "(cot 2, $$\\infty $$)"}, {"identifier": "B", "content": "(\u2013$$\\infty $$, cot 5) $$ \\cup $$ (cot 2, $$\\infty $$)"}, {"identifier": "C", "content": "(cot 5, cot 4)"}, {"identifier": "D", "content": "(\u2013 $$\\infty $$, cot 5) $$ \\cup $$ (cot 4, cot 2)"}] | ["A"] | null | cot<sup>$$-$$1</sup> x > 5, cot<sup>$$-$$1</sup> x < 2
<br><br>$$ \Rightarrow $$ x < cot5, x > cot2 | mcq | jee-main-2019-online-11th-january-evening-slot |
6LQjR16XBRAFABZMYrmun | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If x = sin<sup>$$-$$1</sup>(sin10) and y = cos<sup>$$-$$1</sup>(cos10), then y $$-$$ x is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "7$$\\pi $$"}, {"identifier": "D", "content": "$$\\pi $$"}] | ["D"] | null | x = sin<sup>$$-$$1</sup> sin 10 = 3$$\pi $$ $$-$$ 10
<br><br>y = cos<sup>$$-$$1</sup>cos 10 = 4$$\pi $$ $$-$$ 10
<br><br>y $$-$$ x = (4$$\pi $$ $$-$$ 10) $$-$$ (3$$\pi $$ $$-$$ 10) = $$\pi $$ | mcq | jee-main-2019-online-9th-january-evening-slot |
zmr9GaOF2EgS90jIGdF4s | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $${\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}$$ (x > $$3 \over 4$$), then x is equal to : | [{"identifier": "A", "content": "$${{\\sqrt {145} } \\over {10}}$$"}, {"identifier": "B", "content": "$${{\\sqrt {145} } \\over {11}}$$"}, {"identifier": "C", "content": "$${{\\sqrt {145} } \\over {12}}$$"}, {"identifier": "D", "content": "$${{\\sqrt {146} } \\over {12}}$$"}] | ["C"] | null | Given,
<br><br>$${\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}$$
<br><br>$$ \Rightarrow {\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) = {\pi \over 2} - {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right)$$
<br><br>$$ \Rightarrow \cos \left( {{{\cos }^{ - 1}}\... | mcq | jee-main-2019-online-9th-january-morning-slot |
UKJTY6DgHPShPaIPIOjgy2xukf0y124i | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | 2$$\pi $$ - $$\left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)$$ is equal to : | [{"identifier": "A", "content": "$${{7\\pi } \\over 4}$$"}, {"identifier": "B", "content": "$${{5\\pi } \\over 4}$$"}, {"identifier": "C", "content": "$${{3\\pi } \\over 2}$$"}, {"identifier": "D", "content": "$${\\pi \\over 2}$$"}] | ["C"] | null | $$2\pi - \left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)$$<br><br>$$ = 2\pi - \left( {{{\tan }^{ - 1}}{4 \over 3} + {{\tan }^{ - 1}}{5 \over {12}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)$$<br><br>$$ = 2\pi - \left\{ {{{\tan }^{ - 1}}\left( {{{{4 ... | mcq | jee-main-2020-online-3rd-september-morning-slot |
KKoj2g0io0AWLE2BYi1klrm0vh5 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | A possible value of $$\tan \left( {{1 \over 4}{{\sin }^{ - 1}}{{\sqrt {63} } \over 8}} \right)$$ is : | [{"identifier": "A", "content": "$$\\sqrt 7 - 1$$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 7 }}$$"}, {"identifier": "C", "content": "$$2\\sqrt 2 - 1$$"}, {"identifier": "D", "content": "$${1 \\over {2\\sqrt 2 }}$$"}] | ["B"] | null | $$\tan \left( {{1 \over 4}{{\sin }^{ - 1}}{{\sqrt {63} } \over 8}} \right)$$<br><br>$${\sin ^{ - 1}}\left( {{{\sqrt {63} } \over 8}} \right) = \theta $$ $$\sin \theta = {{\sqrt {63} } \over 8}$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265142/exam_images/vmbo5oe8fbvyyyiih8ml.webp" style=... | mcq | jee-main-2021-online-24th-february-evening-slot |
cBLZQ3EuvzBZDVfyOI1klt7uc51 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | cosec$$\left[ {2{{\cot }^{ - 1}}(5) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right]$$ is equal to : | [{"identifier": "A", "content": "$${{75} \\over {56}}$$"}, {"identifier": "B", "content": "$${{65} \\over {56}}$$"}, {"identifier": "C", "content": "$${{56} \\over {33}}$$"}, {"identifier": "D", "content": "$${{65} \\over {33}}$$"}] | ["B"] | null | $$\cos ec\left( {2{{\cot }^{ - 1}}(5) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)$$<br><br>$$\cos ec\left( {2{{\tan }^{ - 1}}\left( {{1 \over 5}} \right) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)$$<br><br>$$ = \cos ec\left( {{{\tan }^{ - 1}}\left( {{{2\left( {{1 \over 5}} \right)} \over {1 - {... | mcq | jee-main-2021-online-25th-february-evening-slot |
pVJ6lPiMpUPawu7uY81klugzvt7 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $${{{{\sin }^1}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}$$; $$0 < x < 1$$, <br/>then the value of $$\cos \left( {{{\pi c} \over {a + b}}} \right)$$ is : | [{"identifier": "A", "content": "$${{1 - {y^2}} \\over {2y}}$$"}, {"identifier": "B", "content": "$${{1 - {y^2}} \\over {y\\sqrt y }}$$"}, {"identifier": "C", "content": "$$1 - {y^2}$$"}, {"identifier": "D", "content": "$${{1 - {y^2}} \\over {1 + {y^2}}}$$"}] | ["D"] | null | $${{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}$$<br><br>$${{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \over {a + b}} = {\pi \over {2(a + b)}}$$<br><br>Now, $${{{{\tan }^{ - 1}}y} \over c} = {\pi \over {2(a + ... | mcq | jee-main-2021-online-26th-february-morning-slot |
mbltkG0zPddqjhSIul1kluw1xwf | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If 0 < a, b < 1, and tan<sup>$$-$$1</sup>a + tan<sup>$$-$$1</sup>b = $${\pi \over 4}$$, then the value of <br/><br/>$$(a + b) - \left( {{{{a^2} + {b^2}} \over 2}} \right) + \left( {{{{a^3} + {b^3}} \over 3}} \right) - \left( {{{{a^4} + {b^4}} \over 4}} \right) + .....$$ is : | [{"identifier": "A", "content": "$${\\log _e}$$2"}, {"identifier": "B", "content": "e"}, {"identifier": "C", "content": "$${\\log _e}\\left( {{e \\over 2}} \\right)$$"}, {"identifier": "D", "content": "e<sup>2</sup> = 1"}] | ["A"] | null | tan<sup>$$-$$1</sup>a + tan<sup>$$-$$1</sup>b = $${\pi \over 4}$$ 0 < a, b < 1<br><br>$$ \Rightarrow {{a + b} \over {1 - ab}} = 1$$<br><br>a + b = 1 $$-$$ ab<br><br>(a + 1)(b + 1) = 2<br><br>Now $$\left[ {a - {{{a^2}} \over 2} + {{{a^3}} \over 3} + ....} \right] + \left[ {b - {{{b^2}} \over 2} + {{{b^3}} \over 3... | mcq | jee-main-2021-online-26th-february-evening-slot |
1nIy9Pldg3q5vC5Kur1kmjakoz4 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | The sum of possible values of x for <br/><br/>tan<sup>$$-$$1</sup>(x + 1) + cot<sup>$$-$$1</sup>$$\left( {{1 \over {x - 1}}} \right)$$ = tan<sup>$$-$$1</sup>$$\left( {{8 \over {31}}} \right)$$ is : | [{"identifier": "A", "content": "$$-$$$${{{32} \\over 4}}$$"}, {"identifier": "B", "content": "$$-$$$${{{33} \\over 4}}$$"}, {"identifier": "C", "content": "$$-$$$${{{31} \\over 4}}$$"}, {"identifier": "D", "content": "$$-$$$${{{30} \\over 4}}$$"}] | ["A"] | null | tan<sup>$$-$$1</sup>(x + 1) + cot<sup>$$-$$1</sup>$$\left( {{1 \over {x - 1}}} \right)$$ = tan<sup>$$-$$1</sup>$$\left( {{8 \over {31}}} \right)$$
<br><br>$$ \Rightarrow $$ tan<sup>$$-$$1</sup>(x + 1) + tan<sup>$$-$$1</sup>(x - 1) = tan<sup>$$-$$1</sup>$$\left( {{8 \over {31}}} \right)$$
<br><br>$$ \Rightarrow $$ $${\t... | mcq | jee-main-2021-online-17th-march-morning-shift |
1krrolzup | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | The value of $$\tan \left( {2{{\tan }^{ - 1}}\left( {{3 \over 5}} \right) + {{\sin }^{ - 1}}\left( {{5 \over {13}}} \right)} \right)$$ is equal to : | [{"identifier": "A", "content": "$${{ - 181} \\over {69}}$$"}, {"identifier": "B", "content": "$${{220} \\over {21}}$$"}, {"identifier": "C", "content": "$${{ - 291} \\over {76}}$$"}, {"identifier": "D", "content": "$${{151} \\over {63}}$$"}] | ["B"] | null | $$2{\tan ^{ - 1}}\left( {{3 \over 5}} \right) = {\tan ^{ - 1}}\left( {{{6/5} \over {1 - {9 \over {{2^5}}}}}} \right) = {\tan ^{ - 1}}\left( {{{{6 \over 5}} \over {{{16} \over {25}}}}} \right) = {\tan ^{ - 1}}{{15} \over 8}$$<br><br>$$\therefore$$ $$2{\tan ^{ - 1}}\left( {{3 \over 5}} \right) + {\sin ^{ - 1}}\left( {{5 ... | mcq | jee-main-2021-online-20th-july-evening-shift |
1ktei3u10 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | If $${({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a$$; 0 < x < 1, a $$\ne$$ 0, then the value of 2x<sup>2</sup> $$-$$ 1 is : | [{"identifier": "A", "content": "$$\\cos \\left( {{{4a} \\over \\pi }} \\right)$$"}, {"identifier": "B", "content": "$$\\sin \\left( {{{2a} \\over \\pi }} \\right)$$"}, {"identifier": "C", "content": "$$\\cos \\left( {{{2a} \\over \\pi }} \\right)$$"}, {"identifier": "D", "content": "$$\\sin \\left( {{{4a} \\over \\pi ... | ["B"] | null | Given $$a = {({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2}$$<br><br>$$ = ({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)({\sin ^{ - 1}}x - {\cos ^{ - 1}}x)$$<br><br>$$ = {\pi \over 2}\left( {{\pi \over 2} - 2{{\cos }^{ - 1}}x} \right)$$<br><br>$$ \Rightarrow 2{\cos ^{ - 1}}x = {\pi \over 2} - {{2a} \over \pi }$$<br><br>$$ \Right... | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktfwajo8 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | Let M and m respectively be the maximum and minimum values of the function <br/>f(x) = tan<sup>$$-$$1</sup> (sin x + cos x) in $$\left[ {0,{\pi \over 2}} \right]$$, then the value of tan(M $$-$$ m) is equal to : | [{"identifier": "A", "content": "$$2 + \\sqrt 3 $$"}, {"identifier": "B", "content": "$$2 - \\sqrt 3 $$"}, {"identifier": "C", "content": "$$3 + 2\\sqrt 2 $$"}, {"identifier": "D", "content": "$$3 - 2\\sqrt 2 $$"}] | ["D"] | null | Let g(x) = sin x + cos x = $$\sqrt 2 $$ sin$$\left( {x + {\pi \over 4}} \right)$$<br><br>g(x)$$\in$$ $$\left[ {1,\sqrt 2 } \right]$$ for x$$\in$$ [0, $$\pi$$/2]<br><br>f(x) = tan<sup>$$-$$1</sup> (sin x + cos x) $$\in$$ $$\left[ {{\pi \over 4},{{\tan }^{ - 1}}\sqrt 2 } \right]$$<br><br>tan$$({\tan ^{ - 1}}\sqrt 2 - ... | mcq | jee-main-2021-online-27th-august-evening-shift |
1l5464te0 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>$$50\tan \left( {3{{\tan }^{ - 1}}\left( {{1 \over 2}} \right) + 2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right) + 4\sqrt 2 \tan \left( {{1 \over 2}{{\tan }^{ - 1}}(2\sqrt 2 )} \right)$$ is equal to ____________.</p> | [] | null | 29 | $50 \tan \left(\tan ^{-1} \frac{1}{2}+2 \tan ^{-1}\left(\frac{1}{2}\right)+2 \tan ^{-1}(2)\right)$
$$
+4 \sqrt{2} \tan \left(\frac{\tan ^{-1}}{2}(2 \sqrt{2})\right)
$$
<br/><br/>
$\Rightarrow 50 \tan \left(\pi+\tan ^{-1}\left(\frac{1}{2}\right)\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1} 2 \sqrt{2}\right)$
<br... | integer | jee-main-2022-online-29th-june-morning-shift |
1l56rlph0 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>The value of $$\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)$$ is :</p> | [{"identifier": "A", "content": "$${{26} \\over {25}}$$"}, {"identifier": "B", "content": "$${{25} \\over {26}}$$"}, {"identifier": "C", "content": "$${{50} \\over {51}}$$"}, {"identifier": "D", "content": "$${{52} \\over {51}}$$"}] | ["A"] | null | <p>$$\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)$$</p>
<p>$$ = \cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(n + 1) - n} \over {1 + (n + 1)n}}} \right)} } \right)$$</p>
<p>$$ = \cot \left( {\sum\limits_{n = 1}^{50} {({{\tan }^{ - 1}}(n... | mcq | jee-main-2022-online-27th-june-evening-shift |
1l59kzd00 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>The value of $${\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin \left( {{\pi \over 4}} \right)}}} \right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$ - {\\pi \\over 4}$$"}, {"identifier": "B", "content": "$$ - {\\pi \\over 8}$$"}, {"identifier": "C", "content": "$$ - {{5\\pi } \\over {12}}$$"}, {"identifier": "D", "content": "$$ - {{4\\pi } \\over 9}$$"}] | ["B"] | null | <p>$${\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin {\pi \over 4}}}} \right)$$</p>
<p>$$ = {\tan ^{ - 1}}\left( {{{{1 \over {\sqrt 2 }} - 1} \over {{1 \over {\sqrt 2 }}}}} \right)$$</p>
<p>$$ = {\tan ^{ - 1}}(1 - \sqrt 2 ) = - {\tan ^{ - 1}}(\sqrt 2 - 1)$$</p>
<p>$$ = - {\pi \ove... | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5b7rjam | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$x * y = {x^2} + {y^3}$$ and $$(x * 1) * 1 = x * (1 * 1)$$.</p>
<p>Then a value of $$2{\sin ^{ - 1}}\left( {{{{x^4} + {x^2} - 2} \over {{x^4} + {x^2} + 2}}} \right)$$ is :</p> | [{"identifier": "A", "content": "$${\\pi \\over 4}$$"}, {"identifier": "B", "content": "$${\\pi \\over 3}$$"}, {"identifier": "C", "content": "$${\\pi \\over 2}$$"}, {"identifier": "D", "content": "$${\\pi \\over 6}$$"}] | ["B"] | null | The star "*" in this context represents a binary operation, similar to addition (+), subtraction (-), multiplication (×), and division (÷). It is a custom operation defined by the problem statement, and the specific rules of the operation are provided in the problem.
<br/><br/>
In this case, the operation "*" is defin... | mcq | jee-main-2022-online-24th-june-evening-shift |
1l5c13084 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>The set of all values of k for which <br/><br/>$${({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3},\,x \in R$$, is the interval :</p> | [{"identifier": "A", "content": "$$\\left[ {{1 \\over {32}},{7 \\over 8}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{1 \\over {24}},{{13} \\over {16}}} \\right)$$"}, {"identifier": "C", "content": "$$\\left[ {{1 \\over {48}},{{13} \\over {16}}} \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {{1 ... | ["A"] | null | <p>$${({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3}$$</p>
<p>Let $$f(t) = {t^3} + {\left( {{\pi \over 2} - t} \right)^3}$$</p>
<p>Where $$t = {\tan ^{ - 1}}x$$ ; $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$</p>
<p>$$ = {t^3} + {\left( {{\pi \over 2}} \right)^3} - {{3{\pi ^2}t} \over 4} + {{3... | mcq | jee-main-2022-online-24th-june-morning-shift |
1l5vzrho2 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let m and M respectively be the minimum and the maximum values of $$f(x) = {\sin ^{ - 1}}2x + \sin 2x + {\cos ^{ - 1}}2x + \cos 2x,\,x \in \left[ {0,{\pi \over 8}} \right]$$. Then m + M is equal to :</p> | [{"identifier": "A", "content": "$$1 + \\sqrt 2 + \\pi $$"}, {"identifier": "B", "content": "$$\\left( {1 + \\sqrt 2 } \\right)\\pi $$"}, {"identifier": "C", "content": "$$\\pi + \\sqrt 2 $$"}, {"identifier": "D", "content": "$$1 + \\pi $$"}] | ["A"] | null | <p>$$f(x) = {\sin ^{ - 1}}(2x) + \sin 2x + {\cos ^{ - 1}}(2x) + \cos 2x$$</p>
<p>$$ = {\sin ^{ - 1}}(2x) + {\cos ^{ - 1}}(2x) + \sin 2x + \cos 2x$$</p>
<p>$$ = {\pi \over 2} + \sqrt 2 \left( {{1 \over {\sqrt 2 }}\sin 2x + {1 \over {\sqrt 2 }}\cos 2x} \right)$$</p>
<p>$$ = {\pi \over 2} + \sqrt 2 \left( {\cos {\pi \o... | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6f3v5u9 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$x = \sin (2{\tan ^{ - 1}}\alpha )$$ and $$y = \sin \left( {{1 \over 2}{{\tan }^{ - 1}}{4 \over 3}} \right)$$. If $$S = \{ a \in R:{y^2} = 1 - x\} $$, then $$\sum\limits_{\alpha \in S}^{} {16{\alpha ^3}} $$ is equal to _______________.</p> | [] | null | 130 | <p>$$\because$$ $$x = \sin \left( {2{{\tan }^{ - 1}}\alpha } \right) = {{2\alpha } \over {1 + {\alpha ^2}}}$$ ...... (i)</p>
<p>and $$y = \sin \left( {{1 \over 2}{{\tan }^{ - 1}}{4 \over 3}} \right) = \sin \left( {{{\sin }^{ - 1}}{1 \over {\sqrt 5 }}} \right) = {1 \over {\sqrt 5 }}$$</p>
<p>Now, $${y^2} = 1 - x$$</p>
<... | integer | jee-main-2022-online-25th-july-evening-shift |
1l6gj38zp | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>$$\tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$$\\frac{1}{4}$$"}, {"identifier": "D", "content": "$$\\frac{5}{4}$$"}] | ["B"] | null | <p>$$\tan \left( {2{{\tan }^{ - 1}}{1 \over 5} + {{\sec }^{ - 1}}{{\sqrt 5 } \over 2} + 2{{\tan }^{ - 1}}{1 \over 8}} \right)$$</p>
<p>$$ = \tan \left( {2{{\tan }^{ - 1}}\left( {{{{1 \over 5} + {1 \over 8}} \over {1 - {1 \over 5}\,.\,{1 \over 8}}}} \right) + {{\sec }^{ - 1}}{{\sqrt 5 } \over 2}} \right)$$</p>
<p>$$ = \... | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6hz53dp | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>If $$0 < x < {1 \over {\sqrt 2 }}$$ and $${{{{\sin }^{ - 1}}x} \over \alpha } = {{{{\cos }^{ - 1}}x} \over \beta }$$, then the value of $$\sin \left( {{{2\pi \alpha } \over {\alpha + \beta }}} \right)$$ is :</p> | [{"identifier": "A", "content": "$$4 \\sqrt{\\left(1-x^{2}\\right)}\\left(1-2 x^{2}\\right)$$"}, {"identifier": "B", "content": "$$4 x \\sqrt{\\left(1-x^{2}\\right)}\\left(1-2 x^{2}\\right)$$"}, {"identifier": "C", "content": "$$2 x \\sqrt{\\left(1-x^{2}\\right)}\\left(1-4 x^{2}\\right)$$"}, {"identifier": "D", "conten... | ["B"] | null | <p>Let $${{{{\sin }^{ - 1}}x} \over \alpha } = {{{{\cos }^{ - 1}}x} \over \beta } = k \Rightarrow {\sin ^{ - 1}}x + {\cos ^{ - 1}}x = k(\alpha + \beta )$$</p>
<p>$$ \Rightarrow \alpha + \beta = {\pi \over {2k}}$$</p>
<p>Now, $${{2\pi \,\alpha } \over {\alpha + \beta }} = {{2\pi \,\alpha } \over {{\pi \over {2k}}}... | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6nm2vtd | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>The sum of the absolute maximum and absolute minimum values of the function $$f(x)=\tan ^{-1}(\sin x-\cos x)$$ in the interval $$[0, \pi]$$ is :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$$\\tan ^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right)-\\frac{\\pi}{4}$$"}, {"identifier": "C", "content": "$$\\cos ^{-1}\\left(\\frac{1}{\\sqrt{3}}\\right)-\\frac{\\pi}{4}$$"}, {"identifier": "D", "content": "$$\\frac{-\\pi}{12}$$"}] | ["C"] | null | <p>$$f(x) = {\tan ^{ - 1}}(\sin x - \cos x),\,\,\,\,\,[0,\pi ]$$</p>
<p>Let $$g(x) = \sin x - \cos x$$</p>
<p>$$ = \sqrt 2 \sin \left( {x - {\pi \over 4}} \right)$$ and $$x - {\pi \over 4} \in \left[ {{{ - \pi } \over 4},\,{{3\pi } \over 4}} \right]$$</p>
<p>$$\therefore$$ $$g(x) \in \left[ { - 1,\,\sqrt 2 } \right]$... | mcq | jee-main-2022-online-28th-july-evening-shift |
1ldo6tynu | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$S = \left\{ {x \in R:0 < x < 1\,\mathrm{and}\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)} \right\}$$.</p>
<p>If $$\mathrm{n(S)}$$ denotes the number of elements in $$\mathrm{S}$$ then :</p> | [{"identifier": "A", "content": "$$\\mathrm{n}(\\mathrm{S})=0$$"}, {"identifier": "B", "content": "$$\\mathrm{n}(\\mathrm{S})=1$$ and only one element in $$\\mathrm{S}$$ is less than $$\\frac{1}{2}$$."}, {"identifier": "C", "content": "$$\\mathrm{n}(\\mathrm{S})=1$$ and the elements in $$\\mathrm{S}$$ is more than $$\\... | ["D"] | null | $$ {\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)}$$
<br/><br/>$\begin{aligned} & \text { Put } x=\tan \theta \quad \theta \in\left(0, \frac{\pi}{4}\right) \\\\ & 2 \tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)=\cos ^{-1}\left... | mcq | jee-main-2023-online-1st-february-evening-shift |
ldo7xp9f | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | Let (a, b) $\subset(0,2 \pi)$ be the largest interval for which $\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi)$,
holds. <br/><br/>If $\alpha x^{2}+\beta x+\sin ^{-1}\left(x^{2}-6 x+10\right)+\cos ^{-1}\left(x^{2}-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to : | [{"identifier": "A", "content": "$\\frac{\\pi}{16}$\n"}, {"identifier": "B", "content": "$\\frac{\\pi}{48}$\n"}, {"identifier": "C", "content": "$\\frac{\\pi}{8}$\n"}, {"identifier": "D", "content": "$\\frac{\\pi}{12}$"}] | ["D"] | null | $\sin ^{-1} \sin \theta-\left(\frac{\pi}{2}-\sin ^{-1} \sin \theta\right)>0$
<br/><br/>$\Rightarrow \sin ^{-1} \sin \theta>\frac{\pi}{4}$
<br/><br/>$\Rightarrow \sin \theta>\frac{1}{\sqrt{2}}$
<br/><br/>So, $\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$
<br/><br/>$\theta \in\left(\frac{\pi}{4}, \frac{3 \pi... | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldom652p | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$S$$ be the set of all solutions of the equation $$\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^{2}}\right)=\pi, x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$$. Then $$\sum_\limits{x \in S} 2 \sin ^{-1}\left(x^{2}-1\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\pi-2 \\sin ^{-1}\\left(\\frac{\\sqrt{3}}{4}\\right)$$"}, {"identifier": "B", "content": "$$\\pi-\\sin ^{-1}\\left(\\frac{\\sqrt{3}}{4}\\right)$$"}, {"identifier": "C", "content": "$$\\frac{-2 \\pi}{3}$$"}, {"identifier": "D", "content": "None"}] | ["D"] | null | $$
\begin{aligned}
& \cos ^{-1}(2 \mathrm{x})=\pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2} \\\\
& \text { Since } \cos ^{-1}(2 \mathrm{x}) \in[0, \pi] \\\\
& \text { R.H.S. } \geq \pi \\\\
& \pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2}=\pi \\\\
& \Rightarrow \cos ^{-1} \sqrt{1-\mathrm{x}^2}=0 \\\\
& \Rightarrow \sqrt{1-\mathrm{x}^... | mcq | jee-main-2023-online-1st-february-morning-shift |
1ldprk1i1 | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>If $${\sin ^{ - 1}}{\alpha \over {17}} + {\cos ^{ - 1}}{4 \over 5} - {\tan ^{ - 1}}{{77} \over {36}} = 0,0 < \alpha < 13$$, then $${\sin ^{ - 1}}(\sin \alpha ) + {\cos ^{ - 1}}(\cos \alpha )$$ is equal to :</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "$$\\pi$$"}, {"identifier": "C", "content": "16 $$-$$ 5$$\\pi$$"}, {"identifier": "D", "content": "0"}] | ["B"] | null | $\sin ^{-1}\left(\frac{\alpha}{17}\right)=-\cos ^{4}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{77}{36}\right)$
<br/><br/>Let $\cos ^{-1}\left(\frac{4}{5}\right)=p$ and $\tan ^{-1}\left(\frac{77}{36}\right)=q$
<br/><br/>$\Rightarrow \sin \left(\sin ^{-1} \frac{\alpha}{17}\right)=\sin (q-p)$
<br/><br/>$=\sin q \cd... | mcq | jee-main-2023-online-31st-january-morning-shift |
ldqv4ewv | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | Let $a_{1}=1, a_{2}, a_{3}, a_{4}, \ldots .$. be consecutive natural numbers.
<br/><br/>Then $\tan ^{-1}\left(\frac{1}{1+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{1}{1+a_{2} a_{3}}\right)+\ldots . .+\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right)$ is equal to : | [{"identifier": "A", "content": "$\\frac{\\pi}{4}-\\cot ^{-1}(2022)$"}, {"identifier": "B", "content": "$\\frac{\\pi}{4}-\\tan ^{-1}(2022)$"}, {"identifier": "C", "content": "$\\cot ^{-1}(2022)-\\frac{\\pi}{4}$"}, {"identifier": "D", "content": "$\\tan ^{-1}(2022)-\\frac{\\pi}{4}$"}] | null | null | $a_{1}=1, a_{2}, a_{3}, a_{4}, \ldots .$. be consecutive natural numbers.
<br/><br/>$$
\begin{aligned}
& \therefore \quad a_2=2, a_3=3, \ldots ., a_{2021}=2021, a_{2022}=2022 \\\\
& \tan ^{-1}\left[\frac{1}{1+a_1 a_2}\right]=\tan ^{-1}\left[\frac{1}{1+1 \cdot 2}\right]=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{2}\right) \... | mcqm | jee-main-2023-online-30th-january-evening-shift |
1ldv37zch | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>If the sum of all the solutions of $${\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right) + {\cot ^{ - 1}}\left( {{{1 - {x^2}} \over {2x}}} \right) = {\pi \over 3}, - 1 < x < 1,x \ne 0$$, is $$\alpha - {4 \over {\sqrt 3 }}$$, then $$\alpha$$ is equal to _____________.</p> | [] | null | 2 | <b>Case-I</b>
<br/><br/>
$-1 < x < 0$
<br/><br/>
$\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\pi+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}$
<br/><br/>
$\tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{-\pi}{3}$
<br/><br/>
$2 \tan ^{-1} x=\frac{-\pi}{3}$
<br/><br/>
$\tan ^{-1} x=\frac{-\pi}{6}$
<br/><br/>
$x=\frac{-1... | integer | jee-main-2023-online-25th-january-morning-shift |
1lgoy4b0g | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>For $$x \in(-1,1]$$, the number of solutions of the equation $$\sin ^{-1} x=2 \tan ^{-1} x$$ is equal to __________.</p> | [] | null | 2 | <p>We're given the equation $\sin^{-1}x = 2\tan^{-1}x$ for $x$ in the interval $(-1, 1]$. We want to find the number of solutions.</p>
<p>Step 1: Apply the sine and tangent functions to both sides :</p>
<p>We can rewrite the equation by applying the sine function to both sides :</p>
<p>$$\sin(\sin^{-1}x) = \sin(2\t... | integer | jee-main-2023-online-13th-april-evening-shift |
1lgq0x4gc | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>If $$S=\left\{x \in \mathbb{R}: \sin ^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right)-\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right)=\frac{\pi}{4}\right\}$$, then $$\sum_\limits{x \in s}\left(\sin \left(\left(x^{2}+x+5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^{2}+x+5\right) \pi\right)\right)$$ is equal to ___... | [] | null | 4 | Given equation is
<br/><br/>$$
\sin^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right)-\sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right)=\frac{\pi}{4}.
$$
<br/><br/>Let's denote:
<br/><br/>$$A = \sin^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right),$$
<br/><br/>$$B = \sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right).$$
<br/><... | integer | jee-main-2023-online-13th-april-morning-shift |
jaoe38c1lse4tf9q | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>For $$\alpha, \beta, \gamma \neq 0$$, if $$\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pi$$ and $$(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta$$, then $$\gamma$$ equals</p> | [{"identifier": "A", "content": "$$\\sqrt{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\sqrt{3}}{2}$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{\\sqrt{2}}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{3}-1}{2 \\sqrt{2}}$$"}] | ["B"] | null | <p>Let $$\sin ^{-1} \alpha=A, \sin ^{-1} \beta=B, \sin ^{-1} \gamma=C$$</p>
<p>$$\begin{aligned}
& \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \\
& (\alpha+\beta)^2-\gamma^2=3 \alpha \beta \\
& \alpha^2+\beta^2-\gamma^2=\alpha \beta \\
& \frac{\alpha^2+\beta^2-\gamma^2}{2 \alpha \beta}=\frac{1}{2} \\
& \Rightarrow \cos \mathr... | mcq | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lsfkxule | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>Let $$x=\frac{m}{n}$$ ($$m, n$$ are co-prime natural numbers) be a solution of the equation $$\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$$ and let $$\alpha, \beta(\alpha >\beta)$$ be the roots of the equation $$m x^2-n x-m+ n=0$$. Then the point $$(\alpha, \beta)$$ lies on the line</p> | [{"identifier": "A", "content": "$$3 x-2 y=-2$$\n"}, {"identifier": "B", "content": "$$3 x+2 y=2$$\n"}, {"identifier": "C", "content": "$$5 x+8 y=9$$\n"}, {"identifier": "D", "content": "$$5 x-8 y=-9$$"}] | ["C"] | null | <p>Assume $$\sin ^{-1} x=\theta$$</p>
<p>$$\begin{aligned}
& \cos (2 \theta)=\frac{1}{9} \\
& \sin \theta= \pm \frac{2}{3}
\end{aligned}$$</p>
<p>as $$\mathrm{m}$$ and $$\mathrm{n}$$ are co-prime natural numbers,</p>
<p>$$\mathrm{x}=\frac{2}{3}$$</p>
<p>i.e. $$m=2, n=3$$</p>
<p>So, the quadratic equation becomes $$2 x^... | mcq | jee-main-2024-online-29th-january-evening-shift |
lvc57nxl | maths | inverse-trigonometric-functions | properties-of-inverse-trigonometric-functions | <p>For $$n \in \mathrm{N}$$, if $$\cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5+\cot ^{-1} n=\frac{\pi}{4}$$, then $$n$$ is equal to ________.</p> | [] | null | 47 | <p>For $ n \in \mathbb{N} $, if $ \cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1} n = \frac{\pi}{4} $, then $ n $ is equal to <strong></strong><strong></strong>.</p>
<p>Given the equation:</p>
<p>$ \cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1}(n) = \frac{\pi}{4} $</p>
<p>we can use the identity for the sum... | integer | jee-main-2024-online-6th-april-morning-shift |
wQwDqvp4mARIPRuzLBD6U | maths | inverse-trigonometric-functions | sum-upto-n-terms-and-infinite-terms-of-inverse-series | The value of $$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}} \left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} \right)$$ is : | [{"identifier": "A", "content": "$${{22} \\over {23}}$$"}, {"identifier": "B", "content": "$${{23} \\over {22}}$$"}, {"identifier": "C", "content": "$${{21} \\over {19}}$$"}, {"identifier": "D", "content": "$${{19} \\over {21}}$$"}] | ["C"] | null | $$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + n\left( {n + 1} \right)} \right.} } \right)$$
<br><br>$$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {{n^2} + n + 1} \right)} } \right) = \cot \left( {\sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}{1 \over {1 + n\left( {n + 1} \right)}}} ... | mcq | jee-main-2019-online-10th-january-evening-slot |
P82eExl9uoiR298CyCjgy2xukfg71qct | maths | inverse-trigonometric-functions | sum-upto-n-terms-and-infinite-terms-of-inverse-series | If S is the sum of the first 10 terms of the series <br/><br/>
$${\tan ^{ - 1}}\left( {{1 \over 3}} \right) + {\tan ^{ - 1}}\left( {{1 \over 7}} \right) + {\tan ^{ - 1}}\left( {{1 \over {13}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {21}}} \right) + ....$$<br/><br/>
then tan(S) is equal to : | [{"identifier": "A", "content": "$${10 \\over {11}}$$"}, {"identifier": "B", "content": "$${5 \\over {11}}$$"}, {"identifier": "C", "content": "-$${6 \\over {5}}$$"}, {"identifier": "D", "content": "$${5 \\over {6}}$$"}] | ["D"] | null | S = $${\tan ^{ - 1}}\left( {{1 \over 3}} \right) + {\tan ^{ - 1}}\left( {{1 \over 7}} \right) + {\tan ^{ - 1}}\left( {{1 \over {13}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {21}}} \right) + ....$$
<br><br>= $${\tan ^{ - 1}}\left( {{1 \over {1 + 1 \times 2}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {1 + 2 \times 3}}}... | mcq | jee-main-2020-online-5th-september-morning-slot |
j9QMKExBDBcaXztP8N1kmjbau49 | maths | inverse-trigonometric-functions | sum-upto-n-terms-and-infinite-terms-of-inverse-series | If cot<sup>$$-$$1</sup>($$\alpha$$) = cot<sup>$$-$$1</sup> 2 + cot<sup>$$-$$1</sup> 8 + cot<sup>$$-$$1</sup> 18 + cot<sup>$$-$$1</sup> 32 + ...... upto 100 terms, then $$\alpha$$ is : | [{"identifier": "A", "content": "1.02"}, {"identifier": "B", "content": "1.03"}, {"identifier": "C", "content": "1.01"}, {"identifier": "D", "content": "1.00"}] | ["C"] | null | $${\cot ^{ - 1}}(\alpha ) = co{t^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + ....100$$ terms<br><br>$$ = {\tan ^{ - 1}}{1 \over 2} + {\tan ^{ - 1}}{1 \over 8} + {\tan ^{ - 1}}{1 \over {18}} + {\tan ^{ - 1}}{1 \over {32}} + ....100$$ term<br><br>$$ = \sum\limits_{k = 1}^{100} {{{\tan }^{ - 1}}{1 \... | mcq | jee-main-2021-online-17th-march-morning-shift |
1ktd06pxr | maths | inverse-trigonometric-functions | sum-upto-n-terms-and-infinite-terms-of-inverse-series | If $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}{1 \over {2{r^2}}} = p} $$, then the value of tan p is : | [{"identifier": "A", "content": "$${{101} \\over {102}}$$"}, {"identifier": "B", "content": "$${{50} \\over {51}}$$"}, {"identifier": "C", "content": "100"}, {"identifier": "D", "content": "$${{51} \\over {50}}$$"}] | ["B"] | null | $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{2 \over {4{r^2}}}} \right) = \sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(2r + 1) - (2r - 1)} \over {1 + (2r + 1)(2r - 1)}}} \right)} } $$<br><br>= $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}(2r + 1) - {{\tan }^{ - 1}}(2r - 1)} $$<br><br>= $${\tan ^{ - 1}}(1... | mcq | jee-main-2021-online-26th-august-evening-shift |
qaoJWPlc6NQngtMw | maths | limits-continuity-and-differentiability | continuity | $$f$$ is defined in $$\left[ { - 5,5} \right]$$ as
<br/><br/>$$f\left( x \right) = x$$ if $$x$$ is rational
<br/><br/>$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = - x$$ if $$x$$ is irrational. Then | [{"identifier": "A", "content": "$$f(x)$$ is continuous at every x, except $$x = 0$$"}, {"identifier": "B", "content": "$$f(x)$$ is discontinuous at every $$x,$$ except $$x = 0$$"}, {"identifier": "C", "content": "$$f(x)$$ is continuous everywhere"}, {"identifier": "D", "content": "$$f(x)$$ is discontinuous everywhere"... | ["B"] | null | <b>Let a is a rational number</b> other than $$0,$$ in
<br><br>$$\left[ { - 5,5} \right],$$ then
<br><br>$$f\left( a \right) = a$$ and $$\mathop {\lim }\limits_{x \to a} \,\,f\left( x \right) = - a$$
<br><br>[ As in the immediate neighbourhood of a rational -
<br><br>number, we find irrational numbers ]
<br><br>$$\th... | mcq | aieee-2002 |
0qO7zPDbm3sJgFcz | maths | limits-continuity-and-differentiability | continuity | Let $$f(x) = {{1 - \tan x} \over {4x - \pi }}$$, $$x \ne {\pi \over 4}$$, $$x \in \left[ {0,{\pi \over 2}} \right]$$.
<br/><br/>If $$f(x)$$ is continuous in $$\left[ {0,{\pi \over 2}} \right]$$, then $$f\left( {{\pi \over 4}} \right)$$ is | [{"identifier": "A", "content": "$$-1$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$$-{1 \\over 2}$$"}, {"identifier": "D", "content": "$$1$$"}] | ["C"] | null | $$f\left( x \right) = {{1 - \tan x} \over {4x - \pi }}$$ is continuous in $$\left[ {0,{\pi \over 2}} \right]$$
<br><br>$$\therefore$$ $$f\left( {{\pi \over 4}} \right) = \mathop {\lim }\limits_{x \to {\pi \over 4}} f\left( x \right) = \mathop {\lim }\limits_{x \to {\pi \over 4}} \, + f\left( x \right) = \mathop {\l... | mcq | aieee-2004 |
jUvwcp4tbFocqlD1 | maths | limits-continuity-and-differentiability | continuity | The function $$f:R/\left\{ 0 \right\} \to R$$ given by
<br/><br/>$$f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}}$$
<br/><br/>can be made continuous at $$x$$ = 0 by defining $$f$$(0) as | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$-1$$"}] | ["B"] | null | Given, $$f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}}$$
<br><br>$$ \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} {1 \over x} - {2 \over {{e^{2x}} - 1}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{\left( {{e^{2x}} - 1} \right) - 2x} \over {x\left( {{e^{2x}} - 1} \right)}}$$ $$\left... | mcq | aieee-2007 |
FvET6cmrFqjSbbCw | maths | limits-continuity-and-differentiability | continuity | The value of $$p$$ and $$q$$ for which the function
<br/><br/>$$f\left( x \right) = \left\{ {\matrix{
{{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \cr
q & {,x = 0} \cr
{{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{3/2}}}}} & {,x > 0} \cr
} } \right.$$
<br/><br/>is continuous for all... | [{"identifier": "A", "content": "$$p =$$ $${5 \\over 2}$$, $$q = $$ $${1 \\over 2}$$"}, {"identifier": "B", "content": "$$p =$$ $$-{3 \\over 2}$$, $$q = $$ $${1 \\over 2}$$"}, {"identifier": "C", "content": "$$p =$$ $${1 \\over 2}$$, $$q = $$ $${3 \\over 2}$$"}, {"identifier": "D", "content": "$$p =$$ $${1 \\over 2}$$,... | ["B"] | null | $$L.H.L. = \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{\sin \left\{ {\left( {p + 1} \right)\left( { - h} \right)} \right\} - \sin \left( { - h} \right)} \over { - h}}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {{ - \sin \left( {p + 1} \right)h} \over { - h}} + {{\sin \le... | mcq | aieee-2011 |
kuyx43Dm2TedQW6y | maths | limits-continuity-and-differentiability | continuity | If $$f:R \to R$$ is a function defined by
<br/><br/>$$f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$,
<br/><br/>where [x] denotes the greatest integer function, then $$f$$ is | [{"identifier": "A", "content": "continuous for every real $$x$$"}, {"identifier": "B", "content": "discontinuous only at $$x=0$$"}, {"identifier": "C", "content": "discontinuous only at non-zero integral values of $$x$$"}, {"identifier": "D", "content": "continuous only at $$x=0$$"}] | ["A"] | null | Let $$f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)$$
<br><br>Doubtful points are $$x = n,n \in I$$
<br><br>$$L.H.L$$ $$ = \mathop {\lim }\limits_{x \to {n^ - }} \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$
<br><br>$$ = \left( {n - 1} \right)\cos \left( {{{2n - 1} \over ... | mcq | aieee-2012 |
ktLYuFU6rv2nIELWJagcv | maths | limits-continuity-and-differentiability | continuity | Let a, b $$ \in $$ <b>R</b>, (a $$ \ne $$ 0). If the function <i>f</i> defined as
<br/><br/>$$f\left( x \right) = \left\{ {\matrix{
{{{2{x^2}} \over a}\,\,,} & {0 \le x < 1} \cr
{a\,\,\,,} & {1 \le x < \sqrt 2 } \cr
{{{2{b^2} - 4b} \over {{x^3}}},} & {\sqrt 2 \le x < \infty } \cr
... | [{"identifier": "A", "content": "$$\\left( {\\sqrt 2 ,1 - \\sqrt 3 } \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { - \\sqrt 2 ,1 + \\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "$$\\left( {\\sqrt 2 , - 1 + \\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - \\sqrt 2 ,1 - \\sqrt... | ["A"] | null | f(x) is continuous at x = 1
<br><br>$$ \therefore $$ $${{2{{\left( 1 \right)}^2}} \over a} = a$$
<br><br>$$ \Rightarrow $$ a<sup>2</sup> = 2
<br><br>$$ \Rightarrow $$ a = $$ \pm $$ $$\sqrt 2 $$
<br><br>Also f(x) is continuous at x = $$\sqrt 2 $$
<br><br>$$ \therefore... | mcq | jee-main-2016-online-10th-april-morning-slot |
oKPTJbDBgTv4QOlEtDpJg | maths | limits-continuity-and-differentiability | continuity | The value of k for which the function
<br/><br/>$$f\left( x \right) = \left\{ {\matrix{
{{{\left( {{4 \over 5}} \right)}^{{{\tan \,4x} \over {\tan \,5x}}}}\,\,,} & {0 < x < {\pi \over 2}} \cr
{k + {2 \over 5}\,\,\,,} & {x = {\pi \over 2}} \cr
} } \right.$$
<br/><br/>is continuous at x = $$... | [{"identifier": "A", "content": "$${{17} \\over {20}}$$ "}, {"identifier": "B", "content": "$${{2} \\over {5}}$$"}, {"identifier": "C", "content": "$${{3} \\over {5}}$$"}, {"identifier": "D", "content": "$$-$$ $${{2} \\over {5}}$$"}] | ["C"] | null | $f(x)=\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}$<br/><br/>
$\lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}=k+\frac{2}{5}$<br/><br/>
$\Rightarrow \lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\tan 4 x \cot 5 x}=k+\frac{2}{5}$<br/><br/>
$\Rightarr... | mcq | jee-main-2017-online-9th-april-morning-slot |
RWZ8zTUsm0YzRTvqcemcN | maths | limits-continuity-and-differentiability | continuity | Let f(x) = $$\left\{ {\matrix{
{{{\left( {x - 1} \right)}^{{1 \over {2 - x}}}},} & {x > 1,x \ne 2} \cr
{k\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {,x = 2} \cr
} } \right.$$
<br/><br/>Thevaue of k for which f s continuous at x = 2 is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "e"}, {"identifier": "C", "content": "e<sup>-1</sup>"}, {"identifier": "D", "content": "e<sup>-2</sup>"}] | ["C"] | null | Since f(x) is continuous at x = 2.
<br><br>$$\therefore\,\,\,$$$$\mathop {\lim }\limits_{x \to 2} $$ f(x) = f(2)
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 2} $$ $${\left( {x - 1} \right)^{{1 \over {2 - x}}}}$$ = k ($${{1^\infty }}$$ form)
<br><br>$$ \therefore $$&... | mcq | jee-main-2018-online-15th-april-evening-slot |
6hRA7o4FY7sWQFr59ThCp | maths | limits-continuity-and-differentiability | continuity | If the function f defined as
<br/><br/>$$f\left( x \right) = {1 \over x} - {{k - 1} \over {{e^{2x}} - 1}},x \ne 0,$$ is continuous at
<br/><br/> x = 0, then the ordered pair (k, f(0)) is equal to : | [{"identifier": "A", "content": "(3, 2)"}, {"identifier": "B", "content": "(3, 1)"}, {"identifier": "C", "content": "(2, 1)"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 3},\\,2} \\right)$$"}] | ["B"] | null | If the function is continuous at x = 0, then
<br>$$\mathop {\lim }\limits_{x \to 0} $$ f(x) will exist and f(0) = $$\mathop {\lim }\limits_{x \to 0} $$ f(x)
<br><br>Now, $$\mathop {\lim }\limits_{x \to 0} $$ f(x) = $$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} - {{k - 1} \over {{e^{2x}} - 1}}} \right)$$
<br><... | mcq | jee-main-2018-online-16th-april-morning-slot |
m9GFj6XbKauU5vPk0j9WR | maths | limits-continuity-and-differentiability | continuity | Let f : R $$ \to $$ R be a function defined as
<br/>$$f(x) = \left\{ {\matrix{
5 & ; & {x \le 1} \cr
{a + bx} & ; & {1 < x < 3} \cr
{b + 5x} & ; & {3 \le x < 5} \cr
{30} & ; & {x \ge 5} \cr
} } \right.$$
<br/><br/>Then, f is | [{"identifier": "A", "content": "continuous if a = 0 and b = 5"}, {"identifier": "B", "content": "continuous if a = \u20135 and b = 10"}, {"identifier": "C", "content": "continuous if a = 5 and b = 5 "}, {"identifier": "D", "content": "not continuous for any values of a and b"}] | ["D"] | null | Checking
<br><br>if f(x) is continuous at x = 1 :
<br><br>f(1<sup>$$-$$</sup>) = 5
<br><br>f(1) = 5
<br><br>f(1<sup>+</sup>) = a + b
<br><br>if f(x) is continuous at x = 1,
<br><br>then
<br><br>f(1<sup>$$-$$</sup>) = f(1) = f(1<sup>+</sup>)
<br><br>$$ \Rightarrow $$ 5 = 5 = a + b
<br><br>$$ \therefore $$&... | mcq | jee-main-2019-online-9th-january-morning-slot |
yrQAhYb5CYLwaNHm7fLB7 | maths | limits-continuity-and-differentiability | continuity | Let ƒ : [–1,3] $$ \to $$ R be defined as<br/><br/>
$$f(x) = \left\{ {\matrix{
{\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr
{x + \left| x \right|} & , & {1 \le x < 2} \cr
{x + \left[ x \right]} & , & {2 \le x \le 3} \cr
} } \right.$$<br/><br/>
where [t] ... | [{"identifier": "A", "content": "only three points"}, {"identifier": "B", "content": "four or more points"}, {"identifier": "C", "content": "only two points"}, {"identifier": "D", "content": "only one point"}] | ["A"] | null | $$f(x) = \left\{ {\matrix{
{\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr
{x + \left| x \right|} & , & {1 \le x < 2} \cr
{x + \left[ x \right]} & , & {2 \le x \le 3} \cr
} } \right.$$
<br><br>= $$ = \left\{ {\matrix{
{ - x - 1,} & { - 1 \le x < ... | mcq | jee-main-2019-online-8th-april-evening-slot |
8fazrTwsz2FrcJd12a18hoxe66ijvwpqk6h | maths | limits-continuity-and-differentiability | continuity | If the function ƒ defined on , $$\left( {{\pi \over 6},{\pi \over 3}} \right)$$ by
$$$f(x) = \left\{ {\matrix{
{{{\sqrt 2 {\mathop{\rm cosx}\nolimits} - 1} \over {\cot x - 1}},} & {x \ne {\pi \over 4}} \cr
{k,} & {x = {\pi \over 4}} \cr
} } \right.$$$
is continuous, then
k is equal to | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "1 / $$\\sqrt 2$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "2"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}$$ = f($${{\pi \over 4}}$$) = k
<br><br>$$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}$$ ($${0 \over 0}$$ form) = k
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to {\pi \over 4... | mcq | jee-main-2019-online-9th-april-morning-slot |
jhCiGFJptUGSuZbsqt18hoxe66ijvwvnf7c | maths | limits-continuity-and-differentiability | continuity | If $$f(x) = [x] - \left[ {{x \over 4}} \right]$$ ,x $$ \in $$
4
, where [x] denotes the
greatest integer function, then | [{"identifier": "A", "content": "Both $$\\mathop {\\lim }\\limits_{x \\to 4 - } f(x)$$ and $$\\mathop {\\lim }\\limits_{x \\to 4 + } f(x)$$ exist but are not\nequal"}, {"identifier": "B", "content": "f is continuous at x = 4"}, {"identifier": "C", "content": "$$\\mathop {\\lim }\\limits_{x \\to 4 + } f(x)$$ exists but ... | ["B"] | null | $$f(x) = [x] - \left[ {{x \over 4}} \right]$$
<br><br>Here check continuty at x = 4
<br><br>LHL = $$\mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to {4^ - }} \left[ x \right] - \left[ {{x \over 4}} \right]$$
<br><br>= $$\mathop {\lim }\limits_{h \to 0} \left[ {4 - h} ... | mcq | jee-main-2019-online-9th-april-evening-slot |
TCbtDz0RxmdD0SVTI218hoxe66ijvwvjoqb | maths | limits-continuity-and-differentiability | continuity | If the function $$f(x) = \left\{ {\matrix{
{a|\pi - x| + 1,x \le 5} \cr
{b|x - \pi | + 3,x > 5} \cr
} } \right.$$<br/>
is
continuous at x = 5, then the value of a – b is :- | [{"identifier": "A", "content": "$${2 \\over {\\pi - 5 }}$$"}, {"identifier": "B", "content": "$${2 \\over {5 - \\pi }}$$"}, {"identifier": "C", "content": "$${-2 \\over {\\pi + 5 }}$$"}, {"identifier": "D", "content": "$${2 \\over {\\pi + 5 }}$$"}] | ["B"] | null | As f(x) is continuous at x = 5 then
<br><br>$$\mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right) = f\left( 5 \right) = \mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right)$$
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{h \to 0} f\left( {5 - h} \right) = f\left( 5 \right) = \mathop {\lim }\limits_{h \to... | mcq | jee-main-2019-online-9th-april-evening-slot |
b2zSpcAT0sllT3pXyX3rsa0w2w9jwxsw3fk | maths | limits-continuity-and-differentiability | continuity | If$$f(x) = \left\{ {\matrix{
{{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \cr
q & {,x = 0} \cr
{{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{{\raise0.5ex\hbox{$\scriptstyle 3$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}}}} & {,x > 0} \cr
} } \right.$$
<br/>is co... | [{"identifier": "A", "content": "$$\\left( { - {3 \\over 2}, - {1 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - {1 \\over 2},{3 \\over 2}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - {3 \\over 2}, {1 \\over 2}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { {5 \\ove... | ["C"] | null | $$f(x) = \left\{ {\matrix{
{{{\sin (p + 1)x + \sin x} \over x}} & {x < 0} \cr
q & {x = 0} \cr
{{{\sqrt {{x^2} + x} - \sqrt x } \over {{x^{{3 \over 2}}}}}} & {x > 0} \cr
} } \right.$$<br><br>
is continuous at x = 0<br><br>
So f(0<sup>–</sup>) = f(0) = f (0<sup>+</sup>) ... (1)<br><br>... | mcq | jee-main-2019-online-10th-april-morning-slot |
LCqQPEx1rXZyyJNIHG7k9k2k5fp3lto | maths | limits-continuity-and-differentiability | continuity | If the function ƒ defined on $$\left( { - {1 \over 3},{1 \over 3}} \right)$$ by
<br/><br/>f(x) = $$\left\{ {\matrix{
{{1 \over x}{{\log }_e}\left( {{{1 + 3x} \over {1 - 2x}}} \right),} & {when\,x \ne 0} \cr
{k,} & {when\,x = 0} \cr
} } \right.$$
<br/><br/>is continuous, then
k is equal to_______. | [] | null | 5 | $$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left( {{{\ln \left( {1 + 3x} \right)} \over x} - {{\ln \left( {1 - 2x} \right)} \over x}} \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left( {3{{\ln \left( {1 + 3x} \right)} \over {3x}} - \left( { - 2} \right... | integer | jee-main-2020-online-7th-january-evening-slot |
9ZFCv3vcgoXxoGMIksjgy2xukfjjuex1 | maths | limits-continuity-and-differentiability | continuity | Let $$f(x) = x.\left[ {{x \over 2}} \right]$$, for -10< x < 10, where [t] denotes the greatest integer function. Then the number of points of discontinuity of f is equal to _____. | [] | null | 8 | $$x \in ( - 10,10)$$<br><br>$$ \Rightarrow $$ $${x \over 2} \in ( - 5,5) \to 9$$ integers<br><br>check continuity at x = 0<br><br>$$\left. {\matrix{
f & {(0) = } & 0 \cr
f & {({0^ + }) = } & 0 \cr
f & {({0^ - }) = } & 0 \cr
} } \right\}continuous\,at\,x = 0$$<br><br>function wil... | integer | jee-main-2020-online-5th-september-morning-slot |
IDBHWW16iY3z0Pmutmjgy2xukewms5il | maths | limits-continuity-and-differentiability | continuity | If a function f(x) defined by
<br/><br/>$$f\left( x \right) = \left\{ {\matrix{
{a{e^x} + b{e^{ - x}},} & { - 1 \le x < 1} \cr
{c{x^2},} & {1 \le x \le 3} \cr
{a{x^2} + 2cx,} & {3 < x \le 4} \cr
} } \right.$$
<br/><br/>be continuous for some $$a$$, b, c $$ \in $$ R and f'(0) + f'(2) =... | [{"identifier": "A", "content": "$${e \\over {{e^2} - 3e - 13}}$$"}, {"identifier": "B", "content": "$${1 \\over {{e^2} - 3e + 13}}$$"}, {"identifier": "C", "content": "$${e \\over {{e^2} - 3e + 13}}$$"}, {"identifier": "D", "content": "$${e \\over {{e^2} + 3e + 13}}$$"}] | ["C"] | null | Given function,
<br>$$f\left( x \right) = \left\{ {\matrix{
{a{e^x} + b{e^{ - x}},} & { - 1 \le x < 1} \cr
{c{x^2},} & {1 \le x \le 3} \cr
{a{x^2} + 2cx,} & {3 < x \le 4} \cr
} } \right.$$
<br><br>For continuity at x = 1
<br><br>$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \rig... | mcq | jee-main-2020-online-2nd-september-morning-slot |
605rQcjA9lhdt5uZLM7k9k2k5k6lz22 | maths | limits-continuity-and-differentiability | continuity | Let [t] denote the greatest integer $$ \le $$ t
and $$\mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right] = A$$. <br/>Then the function,
f(x) = [x<sup>2</sup>]sin($$\pi $$x) is discontinuous, when x is
equal to : | [{"identifier": "A", "content": "$$\\sqrt {A + 1} $$"}, {"identifier": "B", "content": "$$\\sqrt {A + 5} $$"}, {"identifier": "C", "content": "$$\\sqrt {A + 21} $$"}, {"identifier": "D", "content": "$$\\sqrt {A} $$"}] | ["A"] | null | A = $$\mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right]$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} x\left( {{4 \over x} - \left\{ {{4 \over x}} \right\}} \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left( {4 - \left\{ {{4 \over x}} \right\}} \right)$$
<br><br>= 4
<br><br>Now, when x = $$\sq... | mcq | jee-main-2020-online-9th-january-evening-slot |
zx9Jvq6OgUw6bi7xYi7k9k2k5itocz7 | maths | limits-continuity-and-differentiability | continuity | If $$f(x) = \left\{ {\matrix{
{{{\sin (a + 2)x + \sin x} \over x};} & {x < 0} \cr
{b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,;} & {x = 0} \cr
{{{{{\left( {x + 3{x^2}} \right)}^{{1 \over 3}}} - {x^{ {1 \over 3}}}} \over {{x^{{4 \over 3}}}}};} & {x > 0} \cr
} } \r... | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "-1"}, {"identifier": "C", "content": "-2"}, {"identifier": "D", "content": "1"}] | ["A"] | null | f(0<sup>-</sup>) = $$\mathop {\lim }\limits_{x \to {0^ - }} {{\sin \left( {a + 2} \right)x + \sin x} \over x}$$
<br><br>= $$\mathop {\lim }\limits_{x \to {0^ - }} {{\sin \left( {a + 2} \right)x} \over {\left( {a + 2} \right)x}} \times \left( {a + 2} \right)$$ + $$\mathop {\lim }\limits_{x \to {0^ - }} {{\sin x} \over x... | mcq | jee-main-2020-online-9th-january-morning-slot |
ha7ang0gyEpK5LIsYn1klrekjnm | maths | limits-continuity-and-differentiability | continuity | If f : R $$ \to $$ R is a function defined by f(x)= [x - 1] $$\cos \left( {{{2x - 1} \over 2}} \right)\pi $$, where [.] denotes the greatest
integer function, then f is : | [{"identifier": "A", "content": "continuous for every real x"}, {"identifier": "B", "content": "discontinuous at all integral values of x except at x = 1"}, {"identifier": "C", "content": "discontinuous only at x = 1"}, {"identifier": "D", "content": "continuous only at x = 1"}] | ["A"] | null | Given, $$f(x) = [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$ where [ . ] is greatest integer function and f : R $$\to$$ R<br/><br/>$$\because$$ It is a greatest integer function then we need to check its continuity at x $$\in$$ I except these it is continuous.<br/><br/>Let, x = n where n $$\in$$ I<br/><br/>The... | mcq | jee-main-2021-online-24th-february-morning-slot |
0uHZpf1JbjDDpEOZKe1kluxgk8f | maths | limits-continuity-and-differentiability | continuity | Let f : R $$ \to $$ R be defined as <br/><br/>$$f(x) = \left\{ \matrix{
2\sin \left( { - {{\pi x} \over 2}} \right),if\,x < - 1 \hfill \cr
|a{x^2} + x + b|,\,if - 1 \le x \le 1 \hfill \cr
\sin (\pi x),\,if\,x > 1 \hfill \cr} \right.$$ If f(x) is continuous on R, then a + b equals : | [{"identifier": "A", "content": "$$-$$3"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "$$-$$1"}, {"identifier": "D", "content": "1"}] | ["C"] | null | $$f( - {1^ - }) = 2$$<br><br>$$f( - {1^ + }) = |a + b - 1|$$<br><br>$$|a + b - 1|\, = 2$$ ... (i)<br><br>$$f({1^ - }) = |a + b + 1|$$<br><br>$$f({1^ + }) = 0$$<br><br>$$|a + b + 1| = 0 \Rightarrow a + b + 1 = 0$$<br><br>$$ \Rightarrow a + b = - 1$$ .... (ii) | mcq | jee-main-2021-online-26th-february-evening-slot |
OthbyCL4SlClGBaQfD1kmix3tnd | maths | limits-continuity-and-differentiability | continuity | Let $$\alpha$$ $$\in$$ R be such that the function $$f(x) = \left\{ {\matrix{
{{{{{\cos }^{ - 1}}(1 - {{\{ x\} }^2}){{\sin }^{ - 1}}(1 - \{ x\} )} \over {\{ x\} - {{\{ x\} }^3}}},} & {x \ne 0} \cr
{\alpha ,} & {x = 0} \cr
} } \right.$$ is continuous at x = 0, where {x} = x $$-$$ [ x ] is the greate... | [{"identifier": "A", "content": "no such $$\\alpha$$ exists"}, {"identifier": "B", "content": "$$\\alpha$$ = 0"}, {"identifier": "C", "content": "$$\\alpha$$ = $${\\pi \\over 4}$$"}, {"identifier": "D", "content": "$$\\alpha$$ = $${\\pi \\over {\\sqrt 2 }}$$"}] | ["A"] | null | $$RHL = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(1 - {x^2}){{\sin }^{ - 1}}(1 - x)} \over {x(1 - {x^2})}} $$
<br><br>$$= {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(1 - {x^2})} \over x}$$<br><br>$$ = {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ + }} {{ - 1} \over {\sqrt {... | mcq | jee-main-2021-online-16th-march-evening-shift |
dTxhqnR9k40ZyMN5n31kmizqg2q | maths | limits-continuity-and-differentiability | continuity | Let f : R $$ \to $$ R and g : R $$ \to $$ R be defined as <br/><br/>$$f(x) = \left\{ {\matrix{
{x + a,} & {x < 0} \cr
{|x - 1|,} & {x \ge 0} \cr
} } \right.$$ and <br/><br/>$$g(x) = \left\{ {\matrix{
{x + 1,} & {x < 0} \cr
{{{(x - 1)}^2} + b,} & {x \ge 0} \cr
} } \right.$$,... | [] | null | 1 | $$g[f(x)] = \left[ {\matrix{
{f(x) + 1} & {f(x) < 0} \cr
{{{(f(x) - 1)}^2} + b} & {f(x) \ge 0} \cr
} } \right.$$<br><br>$$g[f(x)] = \left[ {\matrix{
{x + a + 1} & {x + a < 0\& x < 0} \cr
{|x - 1| + 1} & {|x - 1| < 0\& x \ge 0} \cr
{{{(x + a - 1)}^2} + b} &... | integer | jee-main-2021-online-16th-march-evening-shift |
IiJVI2bm82owjrynl21kmjcls4g | maths | limits-continuity-and-differentiability | continuity | If the function $$f(x) = {{\cos (\sin x) - \cos x} \over {{x^4}}}$$ is continuous at each point in its domain and $$f(0) = {1 \over k}$$, then k is ____________. | [] | null | 6 | $$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} {{\cos \left( {\sin x} \right) - \cos x} \over {{x^4}}}$$
<br><br>$$ \Rightarrow $$ $${1 \over k} = \mathop {\lim }\limits_{x \to 0} {{2\sin \left( {{{\sin x + x} \over 2}} \right)\sin \left( {{{x - \sin x} \over 2}} \right)} \over ... | integer | jee-main-2021-online-17th-march-morning-shift |
h98WMUthi0WrOatQRp1kmm32325 | maths | limits-continuity-and-differentiability | continuity | Let f : R $$ \to $$ R be a function defined as<br/><br/>$$f(x) = \left\{ \matrix{
{{\sin (a + 1)x + \sin 2x} \over {2x}},if\,x < 0 \hfill \cr
b,\,if\,x\, = 0 \hfill \cr
{{\sqrt {x + b{x^3}} - \sqrt x } \over {b{x^{5/2}}}},\,if\,x > 0 \hfill \cr} \right.$$<br/><br/>If f is continuous at x = 0, then the v... | [{"identifier": "A", "content": "$$-$$3"}, {"identifier": "B", "content": "$$-$$2"}, {"identifier": "C", "content": "$$ - {5 \\over 2}$$"}, {"identifier": "D", "content": "$$ - {3 \\over 2}$$<br/>"}] | ["D"] | null | Given, $f(x)=\left\{\begin{array}{cl}\frac{\sin (a+1) x+\sin 2 x}{2 x}, & x<0 \\ b, & x=0 \\ \frac{\sqrt{x+b x^3}-\sqrt{x}}{b x^{5 / 2}}, & x>0\end{array}\right.$<br/><br/>
$$
\begin{array}{ll}
\because & f(x) \text { is continuous at } x=0 . \\\\
\therefore & \lim _\limits{x \rightarrow 0^{-}} f(x)=\lim _\limits{x \ri... | mcq | jee-main-2021-online-18th-march-evening-shift |
1krpzg126 | maths | limits-continuity-and-differentiability | continuity | Let a function f : R $$\to$$ R be defined as $$f(x) = \left\{ {\matrix{
{\sin x - {e^x}} & {if} & {x \le 0} \cr
{a + [ - x]} & {if} & {0 < x < 1} \cr
{2x - b} & {if} & {x \ge 1} \cr
} } \right.$$ <br/><br/>where [ x ] is the greatest integer less than or equal to x. If f i... | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "5"}] | ["B"] | null | Continuous x = 0<br><br>f(0<sup>+</sup>) = f(0<sup>$$-$$</sup>) $$\Rightarrow$$ a $$-$$ 1 = 0 $$-$$ e<sup>0</sup><br><br>$$\Rightarrow$$ a = 0<br><br>Continuous at x = 1<br><br>f(1<sup>+</sup>) = f(1<sup>$$-$$</sup>)<br><br>$$\Rightarrow$$ 2(1) $$-$$ b = a + ($$-$$1)<br><br>$$\Rightarrow$$ b = 2 $$-$$ a + 1 $$\Rightar... | mcq | jee-main-2021-online-20th-july-morning-shift |
1kru9vuuq | maths | limits-continuity-and-differentiability | continuity | Let f : R $$\to$$ R be defined as $$f(x) = \left\{ {\matrix{
{{{{x^3}} \over {{{(1 - \cos 2x)}^2}}}{{\log }_e}\left( {{{1 + 2x{e^{ - 2x}}} \over {{{(1 - x{e^{ - x}})}^2}}}} \right),} & {x \ne 0} \cr
{\alpha ,} & {x = 0} \cr
} } \right.$$<br/><br/>If f is continuous at x = 0, then $$\alpha$$ is equal... | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "2"}] | ["A"] | null | For continuity <br><br>$$\mathop {\lim }\limits_{x \to 0} {{{x^3}} \over {4{{\sin }^4}x}}(\ln (1 + 2x{e^{ - 2x}}) - 2\ln (1 - x{e^{ - x}})) = \alpha $$<br><br>$$\mathop {\lim }\limits_{x \to 0} {1 \over {4x}}[2x{e^{ - 2x}} + 2x{e^{ - x}}] = \alpha $$<br><br>$$ = {1 \over 4}(4) = \alpha = 1$$ | mcq | jee-main-2021-online-22th-july-evening-shift |
1krvv39l9 | maths | limits-continuity-and-differentiability | continuity | Let f : R $$\to$$ R be defined as<br/><br/>$$f(x) = \left\{ {\matrix{
{{{\lambda \left| {{x^2} - 5x + 6} \right|} \over {\mu (5x - {x^2} - 6)}},} & {x < 2} \cr
{{e^{{{\tan (x - 2)} \over {x - [x]}}}},} & {x > 2} \cr
{\mu ,} & {x = 2} \cr
} } \right.$$<br/><br/>where [x] is the greates... | [{"identifier": "A", "content": "e($$-$$e + 1)"}, {"identifier": "B", "content": "e(e $$-$$ 2)"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2e $$-$$ 1"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} {e^{{{\tan (x - 2)} \over {x - 2}}}} = {e^1}$$<br><br>$$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} {{ - \lambda (x - 2)(x - 3)} \over {\mu (x - 2)(x - 3)}} = - {\lambda \over \mu }$$<br><br... | mcq | jee-main-2021-online-25th-july-morning-shift |
1krzqujtw | maths | limits-continuity-and-differentiability | continuity | Consider the function<br/><br/><br/>where P(x) is a polynomial such that P'' (x) is always a constant and P(3) = 9. If f(x) is continuous at x = 2, then P(5) is equal to _____________.<img src="data:image/png;base64,UklGRsAKAABXRUJQVlA4ILQKAACQPACdASqQAZ8APm00l0ikIqIhI5JpsIANiWlu4XNBG/Nj8bf0rsz/1fQPencvjHH01alnxz7T/qv7... | [] | null | 39 | $$f(x) = \left\{ {\matrix{
{{{P(x)} \over {\sin (x - 2)}},} & {x \ne 2} \cr
{7,} & {x = 2} \cr
} } \right.$$<br><br>P''(x) = const. $$\Rightarrow$$ P(x) is a 2 degree polynomial<br><br>f(x) is cont. at x = 2<br><br>f(2<sup>+</sup>) = f(2<sup>$$-$$</sup>)<br><br>$$\mathop {\lim }\limits_{x \to {2^ + ... | integer | jee-main-2021-online-25th-july-evening-shift |
1ks08g1ia | maths | limits-continuity-and-differentiability | continuity | Let $$f:\left( { - {\pi \over 4},{\pi \over 4}} \right) \to R$$ be defined as $$f(x) = \left\{ {\matrix{
{{{(1 + |\sin x|)}^{{{3a} \over {|\sin x|}}}}} & , & { - {\pi \over 4} < x < 0} \cr
b & , & {x = 0} \cr
{{e^{\cot 4x/\cot 2x}}} & , & {0 < x < {\pi \over 4}} \cr... | [{"identifier": "A", "content": "1 $$-$$ e"}, {"identifier": "B", "content": "e $$-$$ 1"}, {"identifier": "C", "content": "1 + e"}, {"identifier": "D", "content": "e"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to 0} f(x) = b$$<br><br>$$\mathop {\lim }\limits_{x \to {0^ + }} x{e^{{{\cot 4x} \over {\cot 2x}}}} = {e^{{1 \over 2}}} = b$$<br><br>$$\mathop {\lim }\limits_{x \to {0^ - }} {(1 + |\sin x|)^{{{3a} \over {|\sin x|}}}} = {e^{3a}} = {e^{{1 \over 2}}}$$<br><br>$$a = {1 \over 6} \Rightarrow 6a =... | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktbj2omr | maths | limits-continuity-and-differentiability | continuity | Let a, b $$\in$$ R, b $$\in$$ 0, Define a function <br/><br/>$$f(x) = \left\{ {\matrix{
{a\sin {\pi \over 2}(x - 1),} & {for\,x \le 0} \cr
{{{\tan 2x - \sin 2x} \over {b{x^3}}},} & {for\,x > 0} \cr
} } \right.$$. <br/><br/>If f is continuous at x = 0, then 10 $$-$$ ab is equal to _______________... | [] | null | 14 | $$f(x) = \left\{ {\matrix{
{a\sin {\pi \over 2}(x - 1),} & {for\,x \le 0} \cr
{{{\tan 2x - \sin 2x} \over {b{x^3}}},} & {for\,x > 0} \cr
} } \right.$$<br><br>For continuity at '0'<br><br>$$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$$<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to {0... | integer | jee-main-2021-online-26th-august-morning-shift |
1ktiqpn6h | maths | limits-continuity-and-differentiability | continuity | If the function <br/>$$f(x) = \left\{ {\matrix{
{{1 \over x}{{\log }_e}\left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)} & , & {x < 0} \cr
k & , & {x = 0} \cr
{{{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}} & , & {x > 0} \cr
} } \right.$$ i... | [{"identifier": "A", "content": "$$-$$5"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "$$-$$4"}, {"identifier": "D", "content": "4"}] | ["A"] | null | If f(x) is continuous at x = 0, RHL = LHL = f(0)<br><br>$$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}.{{\sqrt {{x^2} + 1} + 1} \over {\sqrt {{x^2} + 1} + 1}}$$ (Rationalisation)<br><br>$$\mathop {\lim }\limits_... | mcq | jee-main-2021-online-31st-august-morning-shift |
1ktoc7wu0 | maths | limits-continuity-and-differentiability | continuity | Let [t] denote the greatest integer $$\le$$ t. The number of points where the function $$f(x) = [x]\left| {{x^2} - 1} \right| + \sin \left( {{\pi \over {[x] + 3}}} \right) - [x + 1],x \in ( - 2,2)$$ is not continuous is _____________. | [] | null | 2 | $$f(x) = [x]\left| {{x^2} - 1} \right| + \sin \left( {{\pi \over {[x] + 3}}} \right) - [x + 1]$$<br><br>$$f\left( x \right) = \left\{ {\matrix{
{ - 2\left| {{x^2} - 1} \right| + 1,} & { - 2 < x < - 1} \cr
{ - \left| {{x^2} - 1} \right| + 1,} & { - 1 \le x < 0} \cr
{\sin {\pi \over 3} + ... | integer | jee-main-2021-online-1st-september-evening-shift |
1l55hacch | maths | limits-continuity-and-differentiability | continuity | <p>Let f, g : R $$\to$$ R be functions defined by</p>
<p>$$f(x) = \left\{ {\matrix{
{[x]} & , & {x < 0} \cr
{|1 - x|} & , & {x \ge 0} \cr
} } \right.$$ and $$g(x) = \left\{ {\matrix{
{{e^x} - x} & , & {x < 0} \cr
{{{(x - 1)}^2} - 1} & , & {x \ge 0} \cr
} } \... | [{"identifier": "A", "content": "one point"}, {"identifier": "B", "content": "two points"}, {"identifier": "C", "content": "three points"}, {"identifier": "D", "content": "four points"}] | ["B"] | null | $f(x)=\left\{\begin{array}{ll}{[x],} & x<0 \\ |1-x|, & x \geq 0\end{array}\right.$ and $g(x)= \begin{cases}e^{x}-x, & x<0 \\ (x-1)^{2}-1, & x \geq 0\end{cases}$
<br><br>
$$
f \circ g(x)= \begin{cases}{[g(x)],} & g(x)<0 \\ |1-g(x)|, & g(x) \geq 0\end{cases}
$$<br><br>
<img src="https://a... | mcq | jee-main-2022-online-28th-june-evening-shift |
1l566fwdl | maths | limits-continuity-and-differentiability | continuity | <p>Let f : R $$\to$$ R be defined as</p>
<p>$$f(x) = \left[ {\matrix{
{[{e^x}],} & {x < 0} \cr
{a{e^x} + [x - 1],} & {0 \le x < 1} \cr
{b + [\sin (\pi x)],} & {1 \le x < 2} \cr
{[{e^{ - x}}] - c,} & {x \ge 2} \cr
} } \right.$$</p>
<p>where a, b, c $$\in$$ R and [t] denotes... | [{"identifier": "A", "content": "There exists a, b, c $$\\in$$ R such that f is continuous on R."}, {"identifier": "B", "content": "If f is discontinuous at exactly one point, then a + b + c = 1"}, {"identifier": "C", "content": "If f is discontinuous at exactly one point, then a + b + c $$\\ne$$ 1"}, {"identifier": "D... | ["C"] | null | <p>$$f(x) = \left\{ {\matrix{
0 & {x < 0} \cr
{a{e^x} - 1} & {0 \le x < 1} \cr
b & {x = 1} \cr
{b - 1} & {1 < x < 2} \cr
{ - c} & {x \ge 2} \cr
} } \right.$$</p>
<p>To be continuous at x = 0</p>
<p>a $$-$$ 1 = 0</p>
<p>to be continuous at x = 1</p>
<p>ae $$-$$ 1 = b = b $$-$$ 1 $$\Rightarrow$... | mcq | jee-main-2022-online-28th-june-morning-shift |
1l59l625z | maths | limits-continuity-and-differentiability | continuity | <p>Let $$f(x) = \left[ {2{x^2} + 1} \right]$$ and $$g(x) = \left\{ {\matrix{
{2x - 3,} & {x < 0} \cr
{2x + 3,} & {x \ge 0} \cr
} } \right.$$, where [t] is the greatest integer $$\le$$ t. Then, in the open interval ($$-$$1, 1), the number of points where fog is discontinuous is equal to __________... | [] | null | 62 | $$
\mathrm{f}(\mathrm{g}(\mathrm{x}))=\left[2 \mathrm{~g}^2(\mathrm{x})\right]+1
$$<br/><br/>
$$
=\left\{\begin{array}{l}
{\left[2(2 x-3)^2\right]+1 ; x<0} \\
{\left[2(2 x+3)^2\right]+1 ; x \geq 0}
\end{array}\right.
$$<br/><br/>
$\therefore$ fog is discontinuous whenever $2(2 x-3)^2$ or $2(2 x+3)^2$ belongs to integer... | integer | jee-main-2022-online-25th-june-evening-shift |
1l5c2bcte | maths | limits-continuity-and-differentiability | continuity | <p>The number of points where the function</p>
<p>$$f(x) = \left\{ {\matrix{
{|2{x^2} - 3x - 7|} & {if} & {x \le - 1} \cr
{[4{x^2} - 1]} & {if} & { - 1 < x < 1} \cr
{|x + 1| + |x - 2|} & {if} & {x \ge 1} \cr
} } \right.$$</p>
<p>[t] denotes the greatest integer $$\le$$ t,... | [] | null | 7 | $\because f(-1)=2$ and $f(1)=3$
<br/><br/>
For $x \in(-1,1),\left(4 x^{2}-1\right) \in[-1,3)$
<br/><br/>
hence $f(x)$ will be discontinuous at $x=1$ and also
<br/><br/>
whenever $4 x^{2}-1=0,1$ or 2
<br/><br/>
$$
\Rightarrow x=\pm \frac{1}{2}, \pm \frac{1}{\sqrt{2}} \text { and } \pm \frac{\sqrt{3}}{2}
$$
<br/><br/>
So... | integer | jee-main-2022-online-24th-june-morning-shift |
1l6ggcf4g | maths | limits-continuity-and-differentiability | continuity | <p>Let f : R $$\to$$ R be a continuous function such that $$f(3x) - f(x) = x$$. If $$f(8) = 7$$, then $$f(14)$$ is equal to :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "16"}] | ["B"] | null | <p>$$f(3x) - f(x) = x$$ ...... (1)</p>
<p>$$x \to {x \over 3}$$</p>
<p>$$f(x) - f\left( {{x \over 3}} \right) = {x \over 3}$$ ....... (2)</p>
<p>Again $$x \to {x \over 3}$$</p>
<p>$$f\left( {{x \over 3}} \right) - f\left( {{x \over 9}} \right) = {x \over {{3^2}}}$$ ...... (3)</p>
<p>Similarly</p>
<p>$$f\left( {{x \over... | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6ggrtzq | maths | limits-continuity-and-differentiability | continuity | <p>If $$f(x) = \left\{ {\matrix{
{x + a} & , & {x \le 0} \cr
{|x - 4|} & , & {x > 0} \cr
} } \right.$$ and $$g(x) = \left\{ {\matrix{
{x + 1} & , & {x < 0} \cr
{{{(x - 4)}^2} + b} & , & {x \ge 0} \cr
} } \right.$$ are continuous on R, then $$(gof)(2) + (fog)... | [{"identifier": "A", "content": "$$-$$10"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "$$-$$8"}] | ["D"] | null | <p>$$f(x) = \left\{ {\matrix{
{x + a} & , & {x \le 0} \cr
{|x - 4|} & , & {x > 0} \cr
} } \right.$$ and $$g(x) = \left\{ {\matrix{
{x + 1} & , & {x < 0} \cr
{{{(x - 4)}^2} + b} & , & {x \ge 0} \cr
} } \right.$$</p>
<p>$$\because$$ $$f(x)$$ and $$g(x)$$ are continuous on R</p>
<p>$$\therefore$$ $... | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6gh1jfg | maths | limits-continuity-and-differentiability | continuity | <p>Let $$f(x) = \left\{ {\matrix{
{{x^3} - {x^2} + 10x - 7,} & {x \le 1} \cr
{ - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \cr
} } \right.$$.</p>
<p>Then the set of all values of b, for which f(x) has maximum value at x = 1, is :</p> | [{"identifier": "A", "content": "($$-$$6, $$-$$2)"}, {"identifier": "B", "content": "(2, 6)"}, {"identifier": "C", "content": "$$[ - 6, - 2) \\cup (2,6]$$"}, {"identifier": "D", "content": "$$\\left[ {-\\sqrt 6 , - 2} \\right) \\cup \\left( {2,\\sqrt 6 } \\right]$$"}] | ["C"] | null | <p>$$f(x) = \left\{ {\matrix{
{{x^3} - {x^2} + 10x - 7,} & {x \le 1} \cr
{ - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \cr
} } \right.$$</p>
<p>If $$f(x)$$ has maximum value at $$x = 1$$ then $$f(1 + ) \le f(1)$$</p>
<p>$$ - 2 + {\log _2}({b^2} - 4) \le 1 - 1 + 10 - 7$$</p>
<p>$${\log _2}({b^2} - 4) \le 5$$</... | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6kimxb4 | maths | limits-continuity-and-differentiability | continuity | <p>If for $$\mathrm{p} \neq \mathrm{q} \neq 0$$, the function $$f(x)=\frac{\sqrt[7]{\mathrm{p}(729+x)}-3}{\sqrt[3]{729+\mathrm{q} x}-9}$$ is continuous at $$x=0$$, then :</p> | [{"identifier": "A", "content": "$$7 p q \\,f(0)-1=0$$"}, {"identifier": "B", "content": "$$63 q \\,f(0)-\\mathrm{p}^{2}=0$$"}, {"identifier": "C", "content": "$$21 q \\,f(0)-\\mathrm{p}^{2}=0$$"}, {"identifier": "D", "content": "$$7 p q \\,f(0)-9=0$$"}] | ["B"] | null | <p>$$f(x) = {{\root 7 \of {p(729 + x)} - 3} \over {\root 3 \of {729 + qx} - 9}}$$</p>
<p>for continuity at $$x = 0$$, $$\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$$</p>
<p>Now, $$\therefore$$ $$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {{\root 7 \of {p(729 + x)} - 3} \over {\root 3 \... | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6nlhfp1 | maths | limits-continuity-and-differentiability | continuity | <p>The function $$f: \mathbb{R} \rightarrow \mathbb{R}$$ defined by <br/><br/>$$f(x)=\lim\limits_{n \rightarrow \infty} \frac{\cos (2 \pi x)-x^{2 n} \sin (x-1)}{1+x^{2 n+1}-x^{2 n}}$$ is continuous for all x in :</p> | [{"identifier": "A", "content": "$$R-\\{-1\\}$$"}, {"identifier": "B", "content": "$$ \\mathbb{R}-\\{-1,1\\}$$"}, {"identifier": "C", "content": "$$R-\\{1\\}$$"}, {"identifier": "D", "content": "$$R-\\{0\\}$$"}] | ["B"] | null | <p>$$f(x) = \mathop {\lim }\limits_{n \to \infty } {{\cos (2\pi x) - {x^{2n}}\sin (x - 1)} \over {1 + {x^{2n + 1}} - {x^{2n}}}}$$</p>
<p>For $$|x| < 1,\,f(x) = \cos 2\pi x$$, continuous function</p>
<p>$$|x| > 1,\,f(x) = \mathop {\lim }\limits_{n \to \infty } {{{1 \over {{x^{2n}}}}\cos 2\pi x - \sin (x - 1)} \over {{1 ... | mcq | jee-main-2022-online-28th-july-evening-shift |
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