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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": "$$f^{\\prime \\prime}\\left(\\frac{2}{... | ["C"] | null | <p>Given the function:</p>
<p>$ f(x)=\left\{\begin{array}{ll} x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{array}\right. $</p>
<p>we need to find its second derivative at specific points.</p>
<p>First, let’s compute the first derivative $ f^{\prime}(x) $:</p>
<p>$ f^{\prime}(x) = 3x^2 \sin \left( ... | mcq | jee-main-2024-online-6th-april-morning-shift | 6,076 |
1krw1fh1n | maths | ellipse | chord-of-ellipse | Let an ellipse $$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 }}$$, inters... | [{"identifier": "A", "content": "$${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 ... | mcq | jee-main-2021-online-25th-july-morning-shift | 6,077 |
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, $$P(\alpha ,\,\alpha ... | mcq | jee-main-2022-online-25th-june-evening-shift | 6,078 |
1lguvb7is | maths | ellipse | chord-of-ellipse | <p>Consider ellipses $$\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}$$, then ... | [{"identifier": "A", "content": "2870"}, {"identifier": "B", "content": "3210"}, {"identifier": "C", "content": "3320"}, {"identifier": "D", "content": "3080"}] | ["D"] | null | We have, $E_K=K x^2+K^2 y^2=1, K=1,2, \ldots 20$
<br><br>$\Rightarrow \frac{x^2}{\frac{1}{K}}+\frac{y^2}{\frac{1}{K^2}}=1$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ln5ufvyu/8420caac-c5f0-403f-a175-e82f7eb8d93d/8ea00c50-5f75-11ee-8999-67742721d03c/file-6y3zli1ln5ufvyv.png?format=png... | mcq | jee-main-2023-online-11th-april-morning-shift | 6,079 |
lsbkn1ul | maths | ellipse | chord-of-ellipse | The length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid point is $\left(1, \frac{2}{5}\right)$, is equal to : | [{"identifier": "A", "content": "$\\frac{\\sqrt{1691}}{5}$"}, {"identifier": "B", "content": "$\\frac{\\sqrt{2009}}{5}$"}, {"identifier": "C", "content": "$\\frac{\\sqrt{1541}}{5}$"}, {"identifier": "D", "content": "$\\frac{\\sqrt{1741}}{5}$"}] | ["A"] | null | <p>Equation of chord with given middle point.</p>
<p>$$\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$$ (put in original equation)</p... | mcq | jee-main-2024-online-27th-january-morning-shift | 6,080 |
EvZIdx6Rf9PPNIN0 | maths | ellipse | common-tangent | <b>STATEMENT-1 :</b> An equation of a common tangent to the parabola $${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} $$and the el... | [{"identifier": "A", "content": "Statement-1 is false, Statement-2 is true."}, {"identifier": "B", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1."}, {"identifier": "C", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explana... | ["B"] | null | Given equation of ellipse is $$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} \,\,\,\,\,\,\... | mcq | aieee-2012 | 6,081 |
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>$$ \Rightarrow $$ $$2\b... | mcq | jee-main-2019-online-9th-april-evening-slot | 6,082 |
1l58fgd83 | maths | ellipse | common-tangent | <p>If m is the slope of a common tangent to the curves $${{{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} $$</p>
... | mcq | jee-main-2022-online-26th-june-evening-shift | 6,084 |
1lgvqbyet | maths | ellipse | common-tangent | <p>Let a circle of radius 4 be concentric to the ellipse $$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 } ... | mcq | jee-main-2023-online-10th-april-evening-shift | 6,085 |
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": "$$\\le... | ["A"] | null | Given $$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$$ is slope of the tangent
<br><br>So, equation o... | mcq | jee-main-2014-offline | 6,086 |
1ktk74324 | maths | ellipse | locus | The locus of mid-points of the line segments joining ($$-$$3, $$-$$5) and the points on the ellipse $${{{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}$$<br><br>$$ \R... | mcq | jee-main-2021-online-31st-august-evening-shift | 6,087 |
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 ... | [{"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-a9... | mcq | jee-main-2024-online-1st-february-evening-shift | 6,089 |
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 eq... | mcq | jee-main-2017-offline | 6,090 |
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>
$$ \Rightarrow $$ $${2 \over {\sqrt 3 }}\tan \theta = ... | mcq | jee-main-2019-online-12th-april-morning-slot | 6,092 |
7U4w1GA7rQRFV2KmmT7k9k2k5gjiucr | maths | ellipse | normal-to-ellipse | Let the line y = mx and the ellipse 2x<sup>2</sup> + y<sup>2</sup> = 1
intersect at a ponit P in the first quadrant. If the
normal to this ellipse at P meets the co-ordinate axes at $$\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 $$ x<sub... | mcq | jee-main-2020-online-8th-january-morning-slot | 6,093 |
N3TmnUyC4v34yUj1Gpjgy2xukg0cqp9y | maths | ellipse | normal-to-ellipse | If the normal at an end of a latus rectum of an
ellipse passes through an extremity of the
minor axis, then the eccentricity e of the ellipse
satisfies : | [{"identifier": "A", "content": "e<sup>4</sup> + 2e<sup>2</sup> \u2013 1 = 0"}, {"identifier": "B", "content": "e<sup>4</sup> + e<sup>2</sup> \u2013 1 = 0"}, {"identifier": "C", "content": "e<sup>2</sup> + 2e \u2013 1 = 0"}, {"identifier": "D", "content": "e<sup>2</sup> + e \u2013 1 = 0"}] | ["B"] | null | Equation of normal at $$\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 $$ $$... | mcq | jee-main-2020-online-6th-september-evening-slot | 6,095 |
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
... | mcq | jee-main-2023-online-31st-january-morning-shift | 6,096 |
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 ... | [{"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/>$$
\begin{aligned}
& \frac{12 x-36 \sqrt{3}}{\sqrt{3}}=4 y-4 \\\\
& 3 ... | mcq | jee-main-2023-online-13th-april-morning-shift | 6,097 |
2oTqNMY0qPaZCT9gEHjgy2xukfjjqpjo | maths | ellipse | position-of-point-and-chord-joining-of-two-points | If the co-ordinates of two points A and B <br/>are $$\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}$$
<br><br>A and B are foci
<br><br>PA + PB = 2a = 2 × 4 = 8 | mcq | jee-main-2020-online-5th-september-morning-slot | 6,099 |
BcheBfrZmexesZ4Um71klt9cr9j | maths | ellipse | position-of-point-and-chord-joining-of-two-points | If the curve x<sup>2</sup> + 2y<sup>2</sup> = 2 intersects the line x + y = 1 at two points P and Q, then the angle subtended by the line segment PQ at the origin is : | [{"identifier": "A", "content": "$${\\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)$$"}, {"id... | ["D"] | null | Ellipse : $${x \over 2} + {y \over 1} = 1$$<br><br>Line : $$x + y = 1$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267198/exam_images/lmh3h5mktsgvuq0vvk3v.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Eveni... | mcq | jee-main-2021-online-25th-february-evening-slot | 6,100 |
1lgrg5nwu | maths | ellipse | position-of-point-and-chord-joining-of-two-points | <p>Let $$\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}}=\... | [{"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 th... | mcq | jee-main-2023-online-12th-april-morning-shift | 6,101 |
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 $$x^2+9... | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "13"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "10"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwgi30yz/efd8a8c4-b19c-464d-9aab-50da9ed84968/ccf9e7b0-177f-11ef-97dc-2d80937d5077/file-1lwgi30z0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwgi30yz/efd8a8c4-b19c-464d-9aab-50da9ed84968/ccf9e7b0-177f-11ef-97dc-2d80937d5077... | mcq | jee-main-2024-online-5th-april-morning-shift | 6,102 |
fqgR73iD6te7y31I | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | An ellipse has $$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}$$
<br><br>... | mcq | aieee-2005 | 6,104 |
h1jmOU3BvKJGrV5A | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | In the ellipse, the distance between its foci is $$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}$$ | mcq | aieee-2006 | 6,105 |
YPUS6uIIDYQa7Fw1 | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | A focus of an ellipse is at the origin. The directrix is the line $$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$$
<br><br... | mcq | aieee-2008 | 6,106 |
8BGB4L5m1YJ9F01b | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | The ellipse $${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)$$ a... | mcq | aieee-2009 | 6,107 |
nDHwUJAI71QHUK6u | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | An ellipse is drawn by taking a diameter of thec circle $${\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$$ is semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of t... | [{"identifier": "A", "content": "$$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$$... | mcq | aieee-2012 | 6,109 |
3EgKRF0Qam0Y2DwppaMvV | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | Consider an ellipse, whose center is at the origin and its major axis is along the x-axis. If its eccentricity is $${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 $$ b<sup>2</sup> = 25(1 $$-$$ 9/25)
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265271/e... | mcq | jee-main-2017-online-8th-april-morning-slot | 6,110 |
47Y6lsuqMl34l2lIwhhip | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | If the length of the latus rectum of an ellipse is 4 units and the distance between a focus an its nearest vertex on the major axis is $${3 \over 2}$$ units, then its eccentricity is : | [{"identifier": "A", "content": "$${1 \\over 2}$$ "}, {"identifier": "B", "content": "$${1 \\over 3}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 9}$$"}] | ["B"] | null | If the cordinate of focus and vertex are (ae, 0) and (a, 0) respectively,
<br><br>then distance between focus and vertex,
<br><br>a $$-$$ ae = $${3 \over 2}$$ (given)
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ a (1 $$-$$ e) = $${3 \over 2}$$
<br><br>Length of latus rectum,
<br><br>$${{2{b^2}} \over a} = 4$$
<br><br>$$ \R... | mcq | jee-main-2018-online-16th-april-morning-slot | 6,112 |
KbnTkPZf7qScCM6DN6YHH | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | Let S = $$\left\{ {\left( {x,y} \right) \in {R^2}:{{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}}} \right\};r \ne \pm 1.$$ Then S represents : | [{"identifier": "A", "content": "an ellipse whose eccentricity is $${1 \\over {\\sqrt {r + 1} }},$$ where r > 1"}, {"identifier": "B", "content": "an ellipse whose eccentricity is $${2 \\over {\\sqrt {r + 1} }},$$ where 0 < r < 1"}, {"identifier": "C", "content": "an ellipse whose eccentric... | ["D"] | null | $${{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}} = 1$$
<br><br>for r > 1, $${{{y^2}} \over {1 + r}} + {{{x^2}} \over {1 - r}} = 1$$
<br><br>$$e = \sqrt {1 - \left( {{{r - 1} \over {r + 2}}} \right)} $$
<br><br>$$ = \sqrt {{{\left( {r + 1} \right) - \left( {r - 1} \right)} \over {\left( {r ... | mcq | jee-main-2019-online-10th-january-evening-slot | 6,113 |
N4CHAkYaudXunJ9Heb9Zh | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it? | [{"identifier": "A", "content": "$$\\left( {4\\sqrt 2 ,2\\sqrt 3 } \\right)$$"}, {"identifier": "B", "content": "$$\\left( {4\\sqrt 3 ,2\\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "$$\\left( {4\\sqrt 3 ,2\\sqrt 2 } \\right)$$"}, {"identifier": "D", "content": "$$\\left( {4\\sqrt 2 ,2\\sqrt 2 } \\right)$$"}] | ["C"] | null | $${{2{b^2}} \over a} = 8$$ and 2ae $$=$$ 2b
<br><br>$$ \Rightarrow $$ $${b \over a}$$ = e and 1 $$-$$ e<sup>2</sup> = e<sup>2</sup> $$ \Rightarrow $$ e $$=$$ $${1 \over {\sqrt 2 }}$$
<br><br>$$ \Rightarrow $$ b = 4$$\sqrt 2 $$ and a $$=$$ 8
<br><br>So equation of ellipse is $${{{x^2}} \over {6... | mcq | jee-main-2019-online-11th-january-evening-slot | 6,114 |
VkdvpRdbpHlWvtYoCNZTw | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | Let S and S' be the foci of an ellipse and B be any one of the extremities of its minor axis. If $$\Delta $$S'BS is a right angled triangle with right angle at B and area ($$\Delta $$S'BS) = 8 sq. units, then the length of a latus rectum of the ellipse is :
| [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4$$\\sqrt 2 $$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "2$$\\sqrt 2 $$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263675/exam_images/bt2q58jwtsfdf5ghjlxc.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Evening Slot Mathematics - Ellipse Question 65 English Explanation">
<br>b<sup>2</s... | mcq | jee-main-2019-online-12th-january-evening-slot | 6,115 |
i63vRXUpIdgaQSbeWRu0J | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | In an ellipse, with centre at the origin, if the
difference of the lengths of major axis and minor
axis is 10 and one of the foci is at (0,5$$\sqrt 3$$), then
the length of its latus rectum is : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "6"}] | ["A"] | null | Focus (0, be) = (0, 5$$\sqrt 3$$)
<br><br>$$ \therefore $$ be = 5$$\sqrt 3$$
<br><br>$$ \Rightarrow $$ b<sup>2</sup>e<sup>2</sup> = 75
<br><br>As here b > a
<br><br>so e<sup>2</sup> = $$1 - {{{a^2}} \over {{b^2}}}$$
<br><br>$$ \therefore $$ b<sup>2</sup>$$\left( {1 - {{{a^2}} \over {{b^2}}}} \right)$$ = 75
<br><br>$... | mcq | jee-main-2019-online-8th-april-evening-slot | 6,116 |
iduWy6AjPu6GT3dkEG3rsa0w2w9jxb09lba | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | An ellipse, with foci at (0, 2) and (0, –2) and minor axis of length 4, passes through which of the following points? | [{"identifier": "A", "content": "$$\\left( {2,\\sqrt 2 } \\right)$$"}, {"identifier": "B", "content": "$$\\left( {2,2\\sqrt 2 } \\right)$$"}, {"identifier": "C", "content": "$$\\left( {\\sqrt 2 ,2} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {1,2\\sqrt 2 } \\right)$$"}] | ["C"] | null | Let $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1(a < b)$$ is the equation of ellipse, focii $$(0, \pm 2)$$<br><br>
Given 2a = 4 $$ \Rightarrow $$ a = 2<br><br>
e<sup>2</sup> = 1 - $${{{a^2}} \over {{b^2}}}$$ $$ \Rightarrow $$ b<sup>2</sup>e<sup>2</sup> = b<sup>2</sup> - a<sup>2</sup><br><br>
4 = b<sup>2</... | mcq | jee-main-2019-online-12th-april-evening-slot | 6,117 |
S0LExgr96l5oy2q8S87k9k2k5e2uho9 | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12,
then the length of its latus rectum is : | [{"identifier": "A", "content": "$$\\sqrt 3 $$"}, {"identifier": "B", "content": "$$3\\sqrt 2 $$"}, {"identifier": "C", "content": "$${3 \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$$2\\sqrt 3 $$"}] | ["B"] | null | Distance between foci = 2ae = 6
<br><br>$$ \Rightarrow $$ ae = 3 .....(1)
<br><br>Distance between directrices = $${{2a} \over e}$$ = 12
<br><br>$$ \Rightarrow $$ $${a \over e}$$ = 6 .....(2)
<br><br>from (1) and (2)
<br><br>a<sup>2</sup> = 18
<br><br>also a<sup>2</sup>e<sup>2</sup> = 9
<br><br>$$ \Rightarrow $$ 18e<s... | mcq | jee-main-2020-online-7th-january-morning-slot | 6,118 |
8tth7ZaouyPxK54n7Djgy2xukf7fuiug | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | Let $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ (a > b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, <br/>$$\phi \left( t \right) = {5 \over {12}} + t - {t^2}$$, then a<sup>2</sup> + b<sup>2</sup> is equal to : | [{"identifier": "A", "content": "145"}, {"identifier": "B", "content": "126"}, {"identifier": "C", "content": "135"}, {"identifier": "D", "content": "116"}] | ["B"] | null | Given ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ (a > b)<br><br>Length of latus rectum $$ = {{2{b^2}} \over a} = 10$$<br><br>$$\phi (t) = {5 \over {12}} + t - {t^2}$$<br><br>$$ = {8 \over {12}} - {\left( {t - {1 \over 2}} \right)^2}$$<br><br>$$ \therefore $$ $$\phi {(t)_{\max }} = {8 \over {12... | mcq | jee-main-2020-online-4th-september-morning-slot | 6,119 |
1krua9jnn | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | Let $${E_1}:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1,a > b$$. Let E<sub>2</sub> be another ellipse such that it touches the end points of major axis of E<sub>1</sub> and the foci of E<sub>2</sub> are the end points of minor axis of E<sub>1</sub>. If E<sub>1</sub> and E<sub>2</sub> have same eccentriciti... | [{"identifier": "A", "content": "$${{ - 1 + \\sqrt 5 } \\over 2}$$"}, {"identifier": "B", "content": "$${{ - 1 + \\sqrt 8 } \\over 2}$$"}, {"identifier": "C", "content": "$${{ - 1 + \\sqrt 3 } \\over 2}$$"}, {"identifier": "D", "content": "$${{ - 1 + \\sqrt 6 } \\over 2}$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267797/exam_images/t9q8kuirdplfx1cvehdq.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 22th July Evening Shift Mathematics - Ellipse Question 44 English Explanation"><br>$${e^2} = 1 - {... | mcq | jee-main-2021-online-22th-july-evening-shift | 6,120 |
1ktenjau8 | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | If x<sup>2</sup> + 9y<sup>2</sup> $$-$$ 4x + 3 = 0, x, y $$\in$$ R, then x and y respectively lie in the intervals : | [{"identifier": "A", "content": "$$\\left[ { - {1 \\over 3},{1 \\over 3}} \\right]$$ and $$\\left[ { - {1 \\over 3},{1 \\over 3}} \\right]$$"}, {"identifier": "B", "content": "$$\\left[ { - {1 \\over 3},{1 \\over 3}} \\right]$$ and [1, 3]"}, {"identifier": "C", "content": "[1, 3] and [1, 3]"}, {"identifier": "D", "cont... | ["D"] | null | x<sup>2</sup> + 9y<sup>2</sup> $$-$$ 4x + 3 = 0<br><br>(x<sup>2</sup> $$-$$ 4x) + (9y<sup>2</sup>) + 3 = 0<br><br>(x<sup>2</sup> $$-$$ 4x + 4) + (9y<sup>2</sup>) + 3 $$-$$ 4 = 0<br><br>(x $$-$$ 2)<sup>2</sup> + (3y)<sup>2</sup> = 1<br><br>$${{{{(x - 2)}^2}} \over {{{(1)}^2}}} + {{{y^2}} \over {{{\left( {{1 \over 3}} \r... | mcq | jee-main-2021-online-27th-august-morning-shift | 6,122 |
1l57oi7dz | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | <p>Let the eccentricity of an ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, $$a > b$$, be $${1 \over 4}$$. If this ellipse passes through the point $$\left( { - 4\sqrt {{2 \over 5}} ,3} \right)$$, then $${a^2} + {b^2}$$ is equal to :</p> | [{"identifier": "A", "content": "29"}, {"identifier": "B", "content": "31"}, {"identifier": "C", "content": "32"}, {"identifier": "D", "content": "34"}] | ["B"] | null | <p>$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$</p>
<p>$$ \Rightarrow {{{{\left( { - 4\sqrt {{2 \over 5}} } \right)}^2}} \over {{a^2}}} + {{32} \over {{b^2}}} = 1$$</p>
<p>$$ \Rightarrow {{32} \over {5{a^2}}} + {9 \over {{b^2}}} = 1$$ ..... (i)</p>
<p>$${a^2}(1 - {e^2}) = {b^2}$$</p>
<p>$${a^2}\left( {1 - ... | mcq | jee-main-2022-online-27th-june-morning-shift | 6,123 |
1l5b88py5 | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | <p>Let the maximum area of the triangle that can be inscribed in the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 4} = 1,\,a > 2$$, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be $$6\sqrt 3 $$. Then the eccentricity of the ellipse is :</p> | [{"identifier": "A", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${{\\sqrt 3 } \\over 4}$$"}] | ["A"] | null | <p>Given ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 4} = 1,\,a > 2$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5v5uvjp/f35361f2-1b79-408c-b8ee-7c34d92e23d8/e7463d60-0905-11ed-a790-b11fa70c8a36/file-1l5v5uvjq.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image... | mcq | jee-main-2022-online-24th-june-evening-shift | 6,124 |
1l5w08jcr | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | <p>Let the eccentricity of the ellipse $${x^2} + {a^2}{y^2} = 25{a^2}$$ be b times the eccentricity of the hyperbola $${x^2} - {a^2}{y^2} = 5$$, where a is the minimum distance between the curves y = e<sup>x</sup> and y = log<sub>e</sub>x. Then $${a^2} + {1 \over {{b^2}}}$$ is equal to :</p> | [{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "$${5 \\over 2}$$"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "5"}] | ["D"] | null | <p>Given ellipse $${x^2} + {a^2}{y^2} = 25{a^2} \Rightarrow {{{x^2}} \over {25{a^2}}} + {{{y^2}} \over {25}} = 1$$</p>
<p>eccentricity $$({e_1}) = \sqrt {1 - {{{b^2}} \over {{a^2}}}} $$</p>
<p>$$ = \sqrt {1 - {{25} \over {25{a^2}}}} $$</p>
<p>$$ = \sqrt {1 - {1 \over {{a^2}}}} $$</p>
<p>$$ \Rightarrow e_1^2 = 1 - {1 \o... | mcq | jee-main-2022-online-30th-june-morning-shift | 6,125 |
1l6f1l1fg | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | <p>If the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ meets the line $$\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1$$ on the $$x$$-axis and the line $$\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1$$ on the $$y$$-axis, then the eccentricity of the ellipse is :</p> | [{"identifier": "A", "content": "$$\\frac{5}{7}$$"}, {"identifier": "B", "content": "$$\\frac{2 \\sqrt{6}}{7}$$"}, {"identifier": "C", "content": "$$\\frac{3}{7}$$"}, {"identifier": "D", "content": "$$\\frac{2 \\sqrt{5}}{7}$$"}] | ["A"] | null | <p>$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ meets the line $${x \over 7} + {y \over {2\sqrt 6 }} = 1$$ on the x-axis</p>
<p>So, $$a = 7$$</p>
<p>and $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ meets the line $${x \over 7} - {y \over {2\sqrt 6 }} = 1$$ on the y-axis</p>
<p>So, $$b = 2\sqrt... | mcq | jee-main-2022-online-25th-july-evening-shift | 6,126 |
1l6jem0yd | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | <p>If the length of the latus rectum of the ellipse $$x^{2}+4 y^{2}+2 x+8 y-\lambda=0$$ is 4 , and $$l$$ is the length of its major axis, then $$\lambda+l$$ is equal to ____________.</p> | [] | null | 75 | <p>Equation of ellipse is : $${x^2} + 4{y^2} + 2x + 8y - \lambda = 0$$</p>
<p>$${(x + 1)^2} + 4{(y + 1)^2} = \lambda + 5$$</p>
<p>$${{{{(x + 1)}^2}} \over {\lambda + 5}} + {{{{(y + 1)}^2}} \over {\left( {{{\lambda + 5} \over 4}} \right)}} = 1$$</p>
<p>Length of latus rectum $$ = {{2\,.\,\left( {{{\lambda + 5} \ove... | integer | jee-main-2022-online-27th-july-morning-shift | 6,127 |
1ldyc1pa5 | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | <p>Let C be the largest circle centred at (2, 0) and inscribed in the ellipse $${{{x^2}} \over {36}} + {{{y^2}} \over {16}} = 1$$. If (1, $$\alpha$$) lies on C, then 10 $$\alpha^2$$ is equal to ____________</p> | [] | null | 118 | $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$
<br/><br/>
$r^{2}=(x-2)^{2}+y^{2}$
<br/><br/>
Solving simultaneously
<br/><br/>
$-5 x^{2}+36 x+\left(9 r^{2}-180\right)=0$
<br/><br/>
$D=0$
<br/><br/>
$r^{2}=\frac{128}{10}$
<br/><br/>
Distance between $(1, \alpha)$ and $(2,0)$ should be $r$
<br/><br/>
$$
\begin{aligned}
& 1+\alpha... | integer | jee-main-2023-online-24th-january-morning-shift | 6,129 |
1lgxh2jrz | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | <p>Let the ellipse $$E:{x^2} + 9{y^2} = 9$$ intersect the positive x and y-axes at the points A and B respectively. Let the major axis of E be a diameter of the circle C. Let the line passing through A and B meet the circle C at the point P. If the area of the triangle with vertices A, P and the origin O is $${m \over ... | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "17"}, {"identifier": "D", "content": "18"}] | ["C"] | null | The given equation of the ellipse is
<br><br>$$
\begin{aligned}
& x^2+9 y^2=9 ~..........(i)\\\\
& \Rightarrow \frac{x^2}{9}+\frac{y^2}{1}=1
\end{aligned}
$$
<br><br>Now, equation of line $A B$ is
<br><br>$$
x+3 y=3
~$$...........(ii)
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3z... | mcq | jee-main-2023-online-10th-april-morning-shift | 6,131 |
1lh2yedyt | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | <p>In a group of 100 persons 75 speak English and 40 speak Hindi. Each person speaks at least one of the two languages. If the number of persons, who speak only English is $$\alpha$$ and the number of persons who speak only Hindi is $$\beta$$, then the eccentricity of the ellipse $$25\left(\beta^{2} x^{2}+\alpha^{2} y^... | [{"identifier": "A", "content": "$$\\frac{\\sqrt{129}}{12}$$"}, {"identifier": "B", "content": "$$\\frac{3 \\sqrt{15}}{12}$$"}, {"identifier": "C", "content": "$$\\frac{\\sqrt{119}}{12}$$"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{117}}{12}$$"}] | ["C"] | null | Let $E$ be the person speak, English
<br><br>$\therefore n(E)=75$
<br><br>and $H$ be the person speak Hindi
<br><br>$\therefore n(H)=40$
<br><br>Let number of persons who speak both English and Hindi are $t$.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lo75i5g8/e4ea036e-437d-4176-8624... | mcq | jee-main-2023-online-6th-april-evening-shift | 6,132 |
lsapbol9 | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \mathrm{a}>\mathrm{b}$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latusrectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$${7 \\over 2}$$"}, {"identifier": "C", "content": "$${3 \\over 2}$$"}, {"identifier": "D", "content": "$${5 \\over 2}$$"}] | ["C"] | null | <p>
<p>Given the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a > b$, the eccentricity $ e $ is given by the formula:</p>
<p>$ e = \sqrt{1 - \left(\frac{b}{a}\right)^2} $</p>
<p>It is provided that the eccentricity $ e $ is $ \frac{1}{\sqrt{2}} $ (given), so we can equate the two expressions for eccentricity:</... | mcq | jee-main-2024-online-1st-february-morning-shift | 6,133 |
jaoe38c1lsd4d0vp | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | <p>Let $$P$$ be a parabola with vertex $$(2,3)$$ and directrix $$2 x+y=6$$. Let an ellipse $$E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$$, of eccentricity $$\frac{1}{\sqrt{2}}$$ pass through the focus of the parabola $$P$$. Then, the square of the length of the latus rectum of $$E$$, is</p> | [{"identifier": "A", "content": "$$\\frac{512}{25}$$\n"}, {"identifier": "B", "content": "$$\\frac{656}{25}$$\n"}, {"identifier": "C", "content": "$$\\frac{385}{8}$$\n"}, {"identifier": "D", "content": "$$\\frac{347}{8}$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwesuo/97ad35c7-e930-402f-8ae3-bf6649f70495/50a78900-ca2d-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwesup.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsjwesuo/97ad35c7-e930-402f-8ae3-bf6649f70495/50a78900-ca2d-11ee... | mcq | jee-main-2024-online-31st-january-evening-shift | 6,134 |
1lsg4p9ak | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | <p>Let $$A(\alpha, 0)$$ and $$B(0, \beta)$$ be the points on the line $$5 x+7 y=50$$. Let the point $$P$$ divide the line segment $$A B$$ internally in the ratio $$7:3$$. Let $$3 x-25=0$$ be a directrix of the ellipse $$E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ and the corresponding focus be $$S$$. If from $$S$$, the perp... | [{"identifier": "A", "content": "$$\\frac{25}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{25}{9}$$\n"}, {"identifier": "C", "content": "$$\\frac{32}{5}$$\n"}, {"identifier": "D", "content": "$$\\frac{32}{9}$$"}] | ["C"] | null | <p>$$\left.\begin{array}{l}
\mathrm{A}=(10,0) \\
\mathrm{B}=\left(0, \frac{50}{7}\right)
\end{array}\right\} \mathrm{P}=(3,5)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxo287/d8da4921-3aa3-467d-a404-6687dd0b31e6/268e3570-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxo288.png?format... | mcq | jee-main-2024-online-30th-january-evening-shift | 6,135 |
luy6z5ey | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | <p>Let $$f(x)=x^2+9, g(x)=\frac{x}{x-9}$$ and $$\mathrm{a}=f \circ g(10), \mathrm{b}=g \circ f(3)$$. If $$\mathrm{e}$$ and $$l$$ denote the eccentricity and the length of the latus rectum of the ellipse $$\frac{x^2}{\mathrm{a}}+\frac{y^2}{\mathrm{~b}}=1$$, then $$8 \mathrm{e}^2+l^2$$ is equal to.</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "16"}] | ["C"] | null | <p>$$\begin{aligned}
& g(10)=10 \\
& a=f(g(10))=f(10)=109 \\
& f(3)=18 \\
& b=g(f(3))=g(18)=2 \\
& \frac{x^2}{109}+\frac{y^2}{2}=1 \\
& e=\sqrt{1-\frac{2}{109}}=\sqrt{\frac{107}{109}} \\
& I=\frac{2 b^2}{a}=\frac{2 \times 2}{\sqrt{109}}
\end{aligned}$$</p>
<p>$$
\begin{aligned}
8 e^2+l^2 & =\frac{8 \times 107}{109}+\fr... | mcq | jee-main-2024-online-9th-april-morning-shift | 6,137 |
iFccnxCwYIpA22oysCVGk | maths | ellipse | tangent-to-ellipse | If the tangent at a point on the ellipse $${{{x^2}} \over {27}} + {{{y^2}} \over 3} = 1$$ meets the coordinate axes at A and B, and O is the origin, then the
minimum area (in sq. units) of the triangle OAB is : | [{"identifier": "A", "content": "$${9 \\over 2}$$ "}, {"identifier": "B", "content": "$$3\\sqrt 3 $$ "}, {"identifier": "C", "content": "$$9\\sqrt 3 $$"}, {"identifier": "D", "content": "9"}] | ["D"] | null | Equation of tangent to ellipse
<br><br>$${x \over {\sqrt {27} }}$$ cos$$\theta $$ + $${y \over {\sqrt 3 }}$$sin$$\theta $$ = 1
<br><br>Area bounded by line and co-ordinate axis
<br><br>$$\Delta $$ = $${1 \over 2}$$ . $${{\sqrt {27} } \over {\cos \theta }}.{{\sqrt 3 } \over {\sin \theta }}$$ = $${9 \over {\sin 2\theta }... | mcq | jee-main-2016-online-9th-april-morning-slot | 6,139 |
l3vZfVNMIwjFeXp51Twd0 | maths | ellipse | tangent-to-ellipse | If tangents are drawn to the ellipse x2<sup></sup> + 2y<sup>2</sup> = 2 at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :
| [{"identifier": "A", "content": "$${{{x^2}} \\over 2} + {{{y^2}} \\over 4} = 1$$"}, {"identifier": "B", "content": "$${1 \\over {2{x^2}}} + {1 \\over {4{y^2}}} = 1$$"}, {"identifier": "C", "content": "$${1 \\over {4{x^2}}} + {1 \\over {2{y^2}}} = 1$$"}, {"identifier": "D", "content": "$${{{x^2}} \\over 4} + {{{y^2}} \\... | ["B"] | null | Equation of general tangent on ellipse
<br><br>$${x \over {a\,\sec \theta }} + {y \over {b\cos ec\theta }} = 1$$
<br><br>$$a = \sqrt 2 ,\,\,b = 1$$
<br><br>$$ \Rightarrow {x \over {\sqrt 2 \sec \theta }} + {y \over {\cos ec\theta }} = 1$$
<br><br>Let the midpoint be (h, k)
<br><br>$$h = {{\sqrt 2 \sec \theta } \over 2... | mcq | jee-main-2019-online-11th-january-morning-slot | 6,140 |
wB61HjUIYDxES0iL8KNoT | maths | ellipse | tangent-to-ellipse | If the tangents on the ellipse 4x<sup>2 </sup>+ y<sup>2</sup> = 8 at the
points (1, 2) and (a, b) are perpendicular to each
other, then a<sup>2</sup> is equal to :
| [{"identifier": "A", "content": "$${{2} \\over {17}}$$"}, {"identifier": "B", "content": "$${{64} \\over {17}}$$"}, {"identifier": "C", "content": "$${{128} \\over {17}}$$"}, {"identifier": "D", "content": "$${{4} \\over {17}}$$"}] | ["A"] | null | Given, Equation of ellipse 4x<sup>2</sup>
+ y<sup>2</sup> = 8
<br><br>We know equation of tangent at any point (x<sub>1</sub>, y<sub>1</sub>) is
<br><br>4xx<sub>1</sub>
+ yy<sub>1</sub> = 8
<br><br>$$ \therefore $$ Equation of tangent at point (1, 2) is
<br><br>4x + 2y = 8
<br><br>$$ \Rightarrow $$ 2x + y = 4
<br><br... | mcq | jee-main-2019-online-8th-april-morning-slot | 6,141 |
elcH9KhH29Dzdmnbd93rsa0w2w9jwy2n90l | maths | ellipse | tangent-to-ellipse | If the line x – 2y = 12 is tangent to the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ at the point $$\left( {3, - {9 \over 2}} \right)$$ , then the length of the latus
rectum of the ellipse is : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "$$8\\sqrt 3 $$"}, {"identifier": "D", "content": "$$12\\sqrt 2 $$"}] | ["B"] | null | Equation of tangent at $$\left( {3, - {9 \over 2}} \right)$$ to $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ is<br><br>
$${{3x} \over {{a^2}}} - {{{y^9}} \over {2{b^2}}} = 1$$ which is equivalent to x – 2y = 12<br><br>
$${3 \over {{a^2}}} = {{ - 9} \over {2{b^2}( - 2)}} = {1 \over {12}}$$ (On c... | mcq | jee-main-2019-online-10th-april-morning-slot | 6,142 |
wb1NBkz0GF74XdzTTH7k9k2k5fisi0x | maths | ellipse | tangent-to-ellipse | If 3x + 4y = 12$$\sqrt 2 $$ is a tangent to the ellipse
<br/>$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9} = 1$$ for some $$a$$ $$ \in $$ R, then the distance
between the foci of the ellipse is : | [{"identifier": "A", "content": "$$2\\sqrt 5 $$"}, {"identifier": "B", "content": "$$2\\sqrt 7 $$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$2\\sqrt 2 $$"}] | ["B"] | null | 3x + 4y = 12$$\sqrt 2 $$
<br><br>$$ \Rightarrow $$ y = $$ - {{3x} \over 4} + 3\sqrt 2 $$
is tangent to
<br><br>$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9} = 1$$
<br><br>$$ \therefore $$ c<sup>2</sup>
= a<sup>2</sup>m<sup>2</sup> + b<sup>2</sup>
<br><br>$$ \Rightarrow $$ $${\left( {3\sqrt 2 } \right)^2} = {a^2}{\l... | mcq | jee-main-2020-online-7th-january-evening-slot | 6,143 |
KQHl8wH8Lo2PgjuhB27k9k2k5k6z7ik | maths | ellipse | tangent-to-ellipse | The length of the minor axis (along y-axis) of
an ellipse in the standard form is $${4 \over {\sqrt 3 }}$$. If this
ellipse touches the line, x + 6y = 8; then its
eccentricity is : | [{"identifier": "A", "content": "$${1 \\over 3}\\sqrt {{{11} \\over 3}} $$"}, {"identifier": "B", "content": "$${1 \\over 2}\\sqrt {{5 \\over 3}} $$"}, {"identifier": "C", "content": "$$\\sqrt {{5 \\over 6}} $$"}, {"identifier": "D", "content": "$${1 \\over 2}\\sqrt {{{11} \\over 3}} $$"}] | ["D"] | null | Let the equation of ellipse
<br><br>$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, ($$a > b$$)
<br><br>Given 2b = $${4 \over {\sqrt 3 }}$$
<br><br>$$ \Rightarrow $$ b = $${2 \over {\sqrt 3 }}$$
<br><br>We know, Equation of tangent y = mx $$ \pm $$ $$\sqrt {{a^2}{m^2} + {b^2}} $$ ....(1)
<br><br>Given tan... | mcq | jee-main-2020-online-9th-january-evening-slot | 6,144 |
FSoSbtqGvvMHrcVzzNjgy2xukfw0csxo | maths | ellipse | tangent-to-ellipse | Which of the following points lies on the locus of the foot of perpedicular drawn upon any tangent
to the ellipse,
<br/>$${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$$
<br/>from any of its foci? | [{"identifier": "A", "content": "$$\\left( { - 1,\\sqrt 3 } \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - 2,\\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - 1,\\sqrt 2 } \\right)$$"}, {"identifier": "D", "content": "$$\\left( {1,2 } \\right)$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264499/exam_images/jkfizacjmqkgdkpzypg0.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Morning Slot Mathematics - Ellipse Question 50 English Explanation">
<br>Let foot of... | mcq | jee-main-2020-online-6th-september-morning-slot | 6,145 |
oU8GEnTCtPB7pxk3bK1kmiwd8ea | maths | ellipse | tangent-to-ellipse | If the points of intersections of the ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1$$ and the <br/>circle x<sup>2</sup> + y<sup>2</sup> = 4b, b > 4 lie on the curve y<sup>2</sup> = 3x<sup>2</sup>, then b is equal to : | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "5"}] | ["A"] | null | $${{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1$$ ... (1)<br><br>$${x^2} + {y^2} = 4b$$ .... (2)<br><br>$${y^2} = 3{x^2}$$ .... (3)<br><br>From eq (2) and (3)
<br><br>x<sup>2</sup> = b and y<sup>2</sup> = 3b<br><br>From equation (1)
<br><br>$${b \over {16}} + {{3b} \over {{b^2}}} = 1$$<br><br>$$ \Rightarrow {b^2... | mcq | jee-main-2021-online-16th-march-evening-shift | 6,146 |
30eir6NIWjeZA7OrQd1kmm3uurg | maths | ellipse | tangent-to-ellipse | Let a tangent be drawn to the ellipse $${{{x^2}} \over {27}} + {y^2} = 1$$ at $$(3\sqrt 3 \cos \theta ,\sin \theta )$$ where $$0 \in \left( {0,{\pi \over 2}} \right)$$. Then the value of $$\theta$$ such that the sum of intercepts on axes made by this tangent is minimum is equal to : | [{"identifier": "A", "content": "$${{\\pi \\over 6}}$$"}, {"identifier": "B", "content": "$${{\\pi \\over 3}}$$"}, {"identifier": "C", "content": "$${{\\pi \\over 8}}$$"}, {"identifier": "D", "content": "$${{\\pi \\over 4}}$$"}] | ["A"] | null | Tangent = $${x \over {3\sqrt 3 }}\cos \theta + y\sin \theta = 1$$<br><br>x-intercept = $${3\sqrt 3 }$$ sec$$\theta$$<br><br>y-intercept = cosec$$\theta$$<br><br>sum = $${3\sqrt 3 }$$ sec$$\theta$$ + cosec$$\theta$$ = f($$\theta$$) $$\theta$$$$\in$$$$\left( {0,{\pi \over 2}} \right)$$<br><br>$$ \Rightarrow $$ f'($$\t... | mcq | jee-main-2021-online-18th-march-evening-shift | 6,147 |
1ktbb0luj | maths | ellipse | tangent-to-ellipse | On the ellipse $${{{x^2}} \over 8} + {{{y^2}} \over 4} = 1$$ let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line x + 2y = 0. Let S and S' be the foci of the ellipse and e be its eccentricity. If A is the area of the triangle SPS' then, the value of (5 $$-$$ e<s... | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "14"}, {"identifier": "D", "content": "24"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263538/exam_images/ufp79zh4grekwjt1hqyc.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th August Morning Shift Mathematics - Ellipse Question 39 English Explanation"><br><br>Equation ... | mcq | jee-main-2021-online-26th-august-morning-shift | 6,150 |
1ktepisx4 | maths | ellipse | tangent-to-ellipse | If the minimum area of the triangle formed by a tangent to the ellipse $${{{x^2}} \over {{b^2}}} + {{{y^2}} \over {4{a^2}}} = 1$$ and the co-ordinate axis is kab, then k is equal to _______________. | [] | null | 2 | Tangent <br><br>$${{x\cos \theta } \over b} + {{y\sin \theta } \over {2a}} = 1$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266918/exam_images/r8q6vmkbouo5l2gusylp.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th Augus... | integer | jee-main-2021-online-27th-august-morning-shift | 6,151 |
1ktirkn2y | maths | ellipse | tangent-to-ellipse | The line $$12x\cos \theta + 5y\sin \theta = 60$$ is tangent to which of the following curves? | [{"identifier": "A", "content": "x<sup>2</sup> + y<sup>2</sup> = 169"}, {"identifier": "B", "content": "144x<sup>2</sup> + 25y<sup>2</sup> = 3600"}, {"identifier": "C", "content": "25x<sup>2</sup> + 12y<sup>2</sup> = 3600"}, {"identifier": "D", "content": "x<sup>2</sup> + y<sup>2</sup> = 60"}] | ["B"] | null | $$12x\cos \theta + 5y\sin \theta = 60$$<br><br>$${{x\cos \theta } \over 5} + {{y\sin \theta } \over {12}} = 1$$<br><br>is tangent to $${{{x^2}} \over {25}} + {{{y^2}} \over {144}} = 1$$<br><br>$$144{x^2} + 25{y^2} = 3600$$ | mcq | jee-main-2021-online-31st-august-morning-shift | 6,152 |
1ktoa2pph | maths | ellipse | tangent-to-ellipse | Let $$\theta$$ be the acute angle between the tangents to the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 1} = 1$$ and the circle $${x^2} + {y^2} = 3$$ at their point of intersection in the first quadrant. Then tan$$\theta$$ is equal to : | [{"identifier": "A", "content": "$${5 \\over {2\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${2 \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$${4 \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "2"}] | ["B"] | null | The point of intersection of the curves $${{{x^2}} \over 9} + {{{y^2}} \over 1} = 1$$ and $${x^2} + {y^2} = 3$$ in the first quadrant is $$\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)$$<br><br>Now slope of tangent to the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 1} = 1$$ at $$\left( {{3 \over 2},{{\sqrt 3 } \ov... | mcq | jee-main-2021-online-1st-september-evening-shift | 6,154 |
1l5c2eqeg | maths | ellipse | tangent-to-ellipse | <p>If two tangents drawn from a point ($$\alpha$$, $$\beta$$) lying on the ellipse 25x<sup>2</sup> + 4y<sup>2</sup> = 1 to the parabola y<sup>2</sup> = 4x are such that the slope of one tangent is four times the other, then the value of (10$$\alpha$$ + 5)<sup>2</sup> + (16$$\beta$$<sup>2</sup> + 50)<sup>2</sup> equals ... | [] | null | 2929 | $\because(\alpha, \beta)$ lies on the given ellipse, $25 \alpha^{2}+4 \beta^{2}=1\quad\quad...(i)$
<br/><br/>
Tangent to the parabola, $y=m x+\frac{1}{m}$ passes through $(\alpha, \beta)$. So, $\alpha m^{2}-\beta m+1=0$ has roots $m_{1}$ and $4 m_{1}$,
<br/><br/>
$$
m_{1}+4 m_{1}=\frac{\beta}{\alpha} \text { and } m_{1... | integer | jee-main-2022-online-24th-june-morning-shift | 6,155 |
1l6hyqio5 | maths | ellipse | tangent-to-ellipse | <p>The acute angle between the pair of tangents drawn to the ellipse $$2 x^{2}+3 y^{2}=5$$ from the point $$(1,3)$$ is :</p> | [{"identifier": "A", "content": "$$\\tan ^{-1}\\left(\\frac{16}{7 \\sqrt{5}}\\right)$$"}, {"identifier": "B", "content": "$$\\tan ^{-1}\\left(\\frac{24}{7 \\sqrt{5}}\\right)$$"}, {"identifier": "C", "content": "$$\\tan ^{-1}\\left(\\frac{32}{7 \\sqrt{5}}\\right)$$"}, {"identifier": "D", "content": "$$\\tan ^{-1}\\left(... | ["B"] | null | <p>$$2{x^2} + 3{y^2} = 5$$</p>
<p>Equation of tangent having slope m.</p>
<p>$$y = mx\, \pm \,\sqrt {{5 \over 2}{m^2} + {5 \over 3}} $$</p>
<p>which passes through $$(1,3)$$</p>
<p>$$3 = m\, \pm \sqrt {{5 \over 2}{m^2} + {5 \over 3}} $$</p>
<p>$${5 \over 2}{m^2} + {5 \over 3} = 9 + {m^2} - 6m$$</p>
<p>$${3 \over 2}{m^2... | mcq | jee-main-2022-online-26th-july-evening-shift | 6,156 |
1l6nozbu0 | maths | ellipse | tangent-to-ellipse | <p>Let the tangents at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$ on the ellipse $$\frac{x^{2}}{2}+\frac{y^{2}}{4}=1$$ meet at the point $$R(\sqrt{2}, 2 \sqrt{2}-2)$$. If $$\mathrm{S}$$ is the focus of the ellipse on its negative major axis, then $$\mathrm{SP}^{2}+\mathrm{SQ}^{2}$$ is equal to ___________.</p> | [] | null | 13 | <p>$$E \equiv {{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$$</p>
<p>$$\eqalign{
& T \equiv y = mx\, \pm \,\sqrt {2{m^2} + 4} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \left( {\sqrt 2 ,2\sqrt 2 - 2} \right) \cr} $$</p>
<p>$$ \Rightarrow \left( {2\sqrt 2 - 2 - m\sqrt 2 } \right) = \pm \... | integer | jee-main-2022-online-28th-july-evening-shift | 6,157 |
1ldo7lgc4 | maths | ellipse | tangent-to-ellipse | <p>The line $$x=8$$ is the directrix of the ellipse $$\mathrm{E}:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ with the corresponding focus $$(2,0)$$. If the tangent to $$\mathrm{E}$$ at the point $$\mathrm{P}$$ in the first quadrant passes through the point $$(0,4\sqrt3)$$ and intersects the $$x$$-axis at $$\mathrm{Q}$$... | [] | null | 39 | $\begin{aligned} & \frac{a}{e}=8 \\\\ & a e=2 .(1) \\\\ & 8 e=\frac{2}{e} \\\\ & e^2=\frac{1}{4} \Rightarrow e=\frac{1}{2} \\\\ & a=4 \\\\ & b^2=a^2\left(1-e^2\right) \\\\ & =16\left(\frac{3}{4}\right)=12 \\\\ & \frac{x \cos \theta}{4}+\frac{y \sin \theta}{2 \sqrt{3}}=1 \\\\ & \sin \theta=\frac{1}{2} \\\\ & \theta=30^{... | integer | jee-main-2023-online-1st-february-evening-shift | 6,158 |
1ldyc4xu3 | maths | ellipse | tangent-to-ellipse | <p>Let a tangent to the curve $$9{x^2} + 16{y^2} = 144$$ intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is ________</p> | [] | null | 7 | <p>Given curve,</p>
<p>$$9{x^2} + 16{y^2} = 144$$</p>
<p>$$ \Rightarrow {{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$</p>
<p>$$ \Rightarrow {{{x^2}} \over {{4^2}}} + {{{y^2}} \over {{3^2}}} = 1$$</p>
<p>$$\therefore$$ a = 4 and b = 3</p>
<p>So, general point on the ellipse is $$ = (4\cos \theta ,3\sin \theta )$$</p>
<... | integer | jee-main-2023-online-24th-january-morning-shift | 6,159 |
1lgsu9jda | maths | ellipse | tangent-to-ellipse | <p>If the radius of the largest circle with centre (2,0) inscribed in the ellipse $$x^2+4y^2=36$$ is r, then 12r$$^2$$ is equal to :</p> | [{"identifier": "A", "content": "72"}, {"identifier": "B", "content": "92"}, {"identifier": "C", "content": "115"}, {"identifier": "D", "content": "69"}] | ["B"] | null | The given ellipse has the equation :
<br/><br/>$$x^2+4y^2=36$$
<br/><br/>We can rewrite this as :
<br/><br/>$$\frac{x^2}{6^2} + \frac{y^2}{(6/2)^2} = 1$$
<br/><br/>This shows that it is an ellipse centered at (0,0) with semi-major axis a = 6 along the x-axis and semi-minor axis b = 3 along the y-axis.
<br/><br/>Th... | mcq | jee-main-2023-online-11th-april-evening-shift | 6,160 |
yihWYw1MdWfWneMT | maths | functions | classification-of-functions | If $$f:R \to S$$, defined by
<br/>$$f\left( x \right) = \sin x - \sqrt 3 \cos x + 1$$,
<br/>is onto, then the interval of $$S$$ is | [{"identifier": "A", "content": "[-1, 3]"}, {"identifier": "B", "content": "[-1, 1]"}, {"identifier": "C", "content": "[0, 1]"}, {"identifier": "D", "content": "[0, 3]"}] | ["A"] | null | $$f\left( x \right)$$ is onto
<br><br>$$\therefore$$ $$S=$$ range of $$f(x)$$
<br><br>Now $$f\left( x \right) = \sin \,x - \sqrt 3 \,\cos \,x + 1$$
<br><br>$$ = 2\sin \left( {x - {\pi \over 3}} \right) + 1$$
<br><br>As $$1 - \le \sin \left( {x - {\pi \over 3}} \right) \le 1$$
<br><br>$$ - 1 \le 2\sin \left( {x - {\... | mcq | aieee-2004 | 6,162 |
TmvD2W4ixwMufq70 | maths | functions | classification-of-functions | Let $$f:( - 1,1) \to B$$, be a function defined by
<br/>$$f\left( x \right) = {\tan ^{ - 1}}{{2x} \over {1 - {x^2}}}$$,
<br/>then $$f$$ is both one-one and onto when B is the interval | [{"identifier": "A", "content": "$$\\left( {0,{\\pi \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$\\left[ {0,{\\pi \\over 2}} \\right)$$"}, {"identifier": "C", "content": "$$\\left[ { - {\\pi \\over 2},{\\pi \\over 2}} \\right]$$"}, {"identifier": "D", "content": "$$\\left( { - {\\pi \\over 2},{\\pi ... | ["D"] | null | Given $$\,\,f\left( x \right) = {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right) = 2{\tan ^{ - 1}}x$$
<br><br>for $$x \in \left( { - 1,1} \right)$$
<br><br>If$$\,\,x \in \left( { - 1,1} \right) \Rightarrow {\tan ^{ - 1}}x \in \left( {{{ - \pi } \over 4},{\pi \over 4}} \right)$$
<br><br>$$ \Rightarrow 2{\tan ^{... | mcq | aieee-2005 | 6,163 |
Jh1LdFJkemGj6fH0 | maths | functions | classification-of-functions | Let $$f:N \to Y$$ be a function defined as f(x) = 4x + 3 where
<br/>Y = { y $$ \in $$ N, y = 4x + 3 for some x $$ \in $$ N }.
<br/>Show that f is invertible and its inverse is | [{"identifier": "A", "content": "$$g\\left( y \\right) = {{3y + 4} \\over 4}$$"}, {"identifier": "B", "content": "$$g\\left( y \\right) = 4 + {{y + 3} \\over 4}$$"}, {"identifier": "C", "content": "$$g\\left( y \\right) = {{y + 3} \\over 4}$$"}, {"identifier": "D", "content": "$$g\\left( y \\right) = {{y - 3} \\over 4}... | ["D"] | null | Clearly $$f$$ is one one and onto, so invertible
<br><br>Also $$f\left( x \right) = 4x + 3 = y \Rightarrow x = {{y - 3} \over 4}$$
<br><br>$$\therefore$$ $$\,\,\,\,g\left( y \right) = {{y - 3} \over 4}$$ | mcq | aieee-2008 | 6,164 |
tP4DY9Co6Tlcreny | maths | functions | classification-of-functions | Let $$f\left( x \right) = {\left( {x + 1} \right)^2} - 1,x \ge - 1$$
<br/><br/><b>Statement - 1 :</b> The set $$\left\{ {x:f\left( x \right) = {f^{ - 1}}\left( x \right)} \right\} = \left\{ {0, - 1} \right\}$$.
<br/><br/><b>Statement - 2 :</b> $$f$$ is a bijection. | [{"identifier": "A", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1"}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is <b>not</b> a correct explanation for Statement - 1"}, {"identifier": "C", "content": "... | ["C"] | null | Given that $$f\left( x \right) = {\left( {x + 1} \right)^2} - 1,\,x \ge - 1$$
<br><br>Clearly $${D_f} = \left[ { -1 ,\infty } \right)$$ but co-domain is not given
<br><br>$$\therefore$$ $$f(x)$$ need not be necessarily onto.
<br><br>But if $$f(x)$$ is onto then as $$f\left( x \right)$$ is one one also,
<br><br>$$(x+1... | mcq | aieee-2009 | 6,165 |
KGE8ikCoX9SUExIo | maths | functions | classification-of-functions | The function $$f:R \to \left[ { - {1 \over 2},{1 \over 2}} \right]$$ defined as
<br/><br/>$$f\left( x \right) = {x \over {1 + {x^2}}}$$, is | [{"identifier": "A", "content": "invertible"}, {"identifier": "B", "content": "injective but not surjective. "}, {"identifier": "C", "content": "surjective but not injective"}, {"identifier": "D", "content": "neither injective nor surjective."}] | ["C"] | null | $$f\left( x \right) = {x \over {1 + {x^2}}}$$
<br><br>$$ \therefore $$ $$f\left( {{1 \over x}} \right) = {{{1 \over x}} \over {1 + {1 \over {{x^2}}}}} = {x \over {1 + {x^2}}} = f\left( x \right)$$
<br><br>$$ \therefore $$ f(x) is many-one function.
<br><br>Now let y = f(x) = $${x \over {1 + {x^2}}}$$
<br><br>$$ \Righta... | mcq | jee-main-2017-offline | 6,167 |
igQxD7UV54aBTe7EncVNQ | maths | functions | classification-of-functions | The function f : <b>N</b> $$ \to $$ <b>N</b> defined by f (x) = x $$-$$ 5 $$\left[ {{x \over 5}} \right],$$ Where <b>N</b> is the set of natural numbers and [x] denotes the greatest integer less than or equal to x, is : | [{"identifier": "A", "content": "one-one and onto"}, {"identifier": "B", "content": "one-one but not onto."}, {"identifier": "C", "content": "onto but not one-one."}, {"identifier": "D", "content": "neither one-one nor onto."}] | ["D"] | null | f(1) = 1 - 5$$\left[ {{1 \over 5}} \right]$$ = 1
<br><br>f(6) = 6 - 5$$\left[ {{6 \over 5}} \right]$$ = 1
<br><br>So, this function is many to one.
<br><br>f(10) = 10 - 5$$\left[ {{10 \over 5}} \right]$$ = 0 which is not present in the set of natural numbers.
<br><br>So this function is neither one-one nor onto. | mcq | jee-main-2017-online-9th-april-morning-slot | 6,168 |
0WUFs0GDVAFUAg9aXVS3o | maths | functions | classification-of-functions | The number of functions f from {1, 2, 3, ...., 20} onto {1, 2, 3, ...., 20} such that f(k) is a multiple of 3,
whenever k is a multiple of 4, is : | [{"identifier": "A", "content": "6<sup>5</sup> $$ \\times $$ (15)!"}, {"identifier": "B", "content": "5<sup>6</sup> $$ \\times $$ 15"}, {"identifier": "C", "content": "(15)! $$ \\times $$ 6!"}, {"identifier": "D", "content": "5! $$ \\times $$ 6!"}] | ["C"] | null | Given that $f(k)$ is a multiple of 3 whenever $k$ is a multiple of 4, we need to consider how to map elements from the domain {1, 2, 3, ..., 20} to the codomain {1, 2, 3, ..., 20} following this rule.
<br/><br/>1. We first consider the subset of the domain that consists of multiples of 4: {4, 8, 12, 16, 20}. There are... | mcq | jee-main-2019-online-11th-january-evening-slot | 6,170 |
pBqVHNcxP55bNHm9oo93G | maths | functions | classification-of-functions | Let a function f : (0, $$\infty $$) $$ \to $$ (0, $$\infty $$) be defined by f(x) = $$\left| {1 - {1 \over x}} \right|$$. Then f is : | [{"identifier": "A", "content": "not injective but it is surjective"}, {"identifier": "B", "content": "neiter injective nor surjective"}, {"identifier": "C", "content": "injective only"}, {"identifier": "D", "content": "both injective as well as surjective"}] | ["B"] | null | $$f\left( x \right) = \left| {1 - {1 \over x}} \right| = {{\left| {x - 1} \right|} \over x} = \left\{ {\matrix{
{{{1 - x} \over x}} & {0 < x \le 1} \cr
{{{x - 1} \over x}} & {x \ge 1} \cr
} } \right.$$
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266122/exam_images/w8m... | mcq | jee-main-2019-online-11th-january-evening-slot | 6,171 |
dfuBeaBmWfo5qPrwGI18hoxe66ijvwpyj7h | maths | functions | classification-of-functions | If the function ƒ : R – {1, –1} $$ \to $$ A defined by <br/>
ƒ(x) = $${{{x^2}} \over {1 - {x^2}}}$$ , is surjective, then A is equal to | [{"identifier": "A", "content": "R \u2013 (\u20131, 0)"}, {"identifier": "B", "content": "R \u2013 {\u20131}"}, {"identifier": "C", "content": "R \u2013 [\u20131, 0)"}, {"identifier": "D", "content": "[0, $$\\infty $$)"}] | ["C"] | null | Let ƒ(x) = $${{{x^2}} \over {1 - {x^2}}}$$ = y
<br><br>$$ \Rightarrow $$ $$y\left( {1 - {x^2}} \right) = {x^2}$$
<br><br>$$ \Rightarrow $$ $${x^2} = {y \over {1 + y}}$$
<br><br>As $${x^2}$$ is always $$ \ge $$ 0.
<br><br>$$ \therefore $$ $${y \over {1 + y}}$$ $$ \ge $$ 0
<br><br> y $$ \in $$ $$\left( { - \infty , - 1} ... | mcq | jee-main-2019-online-9th-april-morning-slot | 6,172 |
tXk6WTp3s2gOZF5efd1kls4q62e | maths | functions | classification-of-functions | Let f, g : N $$ \to $$ N such that f(n + 1) = f(n) + f(1) $$\forall $$ n$$\in$$N and g be any arbitrary function. Which of the following statements is NOT true? | [{"identifier": "A", "content": "If g is onto, then fog is one-one"}, {"identifier": "B", "content": "f is one-one"}, {"identifier": "C", "content": "If f is onto, then f(n) = n $$\\forall $$n$$\\in$$N"}, {"identifier": "D", "content": "If fog is one-one, then g is one-one"}] | ["A"] | null | $$f(n + 1) = f(n) + 1$$<br><br>$$f(2) = 2f(1)$$<br><br>$$f(3) = 3f(1)$$<br><br>$$f(4) = 4f(1)$$<br><br>.....<br><br>$$f(n) = nf(1)$$<br><br>$$f(x)$$ is one-one | mcq | jee-main-2021-online-25th-february-morning-slot | 6,174 |
G6M026PNBglYe91OIP1klt7gk5q | maths | functions | classification-of-functions | Let x denote the total number of one-one functions from a set A with 3 elements to a set B with 5 elements and y denote the total number of one-one functions form the set A to the set A $$\times$$ B. Then : | [{"identifier": "A", "content": "2y = 273x"}, {"identifier": "B", "content": "y = 91x"}, {"identifier": "C", "content": "2y = 91x"}, {"identifier": "D", "content": "y = 273x"}] | ["C"] | null | Number of elements in A = 3<br><br>Number of elements in B = 5<br><br>Number of elements in A $$\times$$ B = 15<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264770/exam_images/jzoaua9vt8av712nmhev.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JE... | mcq | jee-main-2021-online-25th-february-evening-slot | 6,175 |
ogppq1ohbkBSL5dMSx1kluwzw9x | maths | functions | classification-of-functions | Let $$A = \{ 1,2,3,....,10\} $$ and $$f:A \to A$$ be defined as<br/><br/>$$f(k) = \left\{ {\matrix{
{k + 1} & {if\,k\,is\,odd} \cr
k & {if\,k\,is\,even} \cr
} } \right.$$<br/><br/>Then the number of possible functions $$g:A \to A$$ such that $$gof = f$$ is : | [{"identifier": "A", "content": "5<sup>5</sup>"}, {"identifier": "B", "content": "10<sup>5</sup>"}, {"identifier": "C", "content": "5!"}, {"identifier": "D", "content": "<sup>10</sup>C<sub>5</sub>"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265048/exam_images/wt80rncxso0qezuj3tvv.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266951/exam_images/y5zg7fnmudxcxu7h9gyf.webp"><img src="https://res.c... | mcq | jee-main-2021-online-26th-february-evening-slot | 6,176 |
1kruaeehj | maths | functions | classification-of-functions | Let A = {0, 1, 2, 3, 4, 5, 6, 7}. Then the number of bijective functions f : A $$\to$$ A such that f(1) + f(2) = 3 $$-$$ f(3) is equal to | [] | null | 720 | f(1) + f(2) = 3 $$-$$ f(3)<br><br>$$\Rightarrow$$ f(1) + f(2) = 3 + f(3) = 3<br><br>The only possibility is : 0 + 1 + 2 = 3<br><br>$$\Rightarrow$$ Elements 1, 2, 3 in the domain can be mapped with 0, 1, 2 only.<br><br>So number of bijective functions.<br><br>$$\left| \!{\underline {\,
3 \,}} \right. $$ $$\times$$ $$\... | integer | jee-main-2021-online-22th-july-evening-shift | 6,177 |
1krvzt6at | maths | functions | classification-of-functions | Let g : N $$\to$$ N be defined as<br/><br/>g(3n + 1) = 3n + 2,<br/><br/>g(3n + 2) = 3n + 3,<br/><br/>g(3n + 3) = 3n + 1, for all n $$\ge$$ 0. <br/><br/>Then which of the following statements is true? | [{"identifier": "A", "content": "There exists an onto function f : N $$\\to$$ N such that fog = f"}, {"identifier": "B", "content": "There exists a one-one function f : N $$\\to$$ N such that fog = f"}, {"identifier": "C", "content": "gogog = g"}, {"identifier": "D", "content": "There exists a function : f : N $$\\to$$... | ["A"] | null | g : N $$\to$$ N <br><br>g(3n + 1) = 3n + 2,<br><br>g(3n + 2) = 3n + 3,<br><br>g(3n + 3) = 3n + 1<br><br>$$g(x) = \left[ {\matrix{
{x + 1} & {x = 3k + 1} \cr
{x + 1} & {x = 3k + 2} \cr
{x - 2} & {x = 3k + 3} \cr
} } \right.$$<br><br>$$g\left( {g(x)} \right) = \left[ {\matrix{
{x + 2} &... | mcq | jee-main-2021-online-25th-july-morning-shift | 6,178 |
1l55j9tyq | maths | functions | classification-of-functions | <p>Let S = {1, 2, 3, 4}. Then the number of elements in the set { f : S $$\times$$ S $$\to$$ S : f is onto and f (a, b) = f (b, a) $$\ge$$ a $$\forall$$ (a, b) $$\in$$ S $$\times$$ S } is ______________.</p> | [] | null | 37 | There are 16 ordered pairs in $S \times S$. We write all these ordered pairs in 4 sets as follows.
<br/><br/>
$A=\{(1,1)\}$
<br/><br/>
$B=\{(1,4),(2,4),(3,4)(4,4),(4,3),(4,2),(4,1)\}$
<br/><br/>
$C=\{(1,3),(2,3),(3,3),(3,2),(3,1)\}$
<br/><br/>
$D=\{(1,2),(2,2),(2,1)\}$
<br/><br/>
All elements of set $B$ have image 4 an... | integer | jee-main-2022-online-28th-june-evening-shift | 6,179 |
1l56640do | maths | functions | classification-of-functions | <p>Let a function f : N $$\to$$ N be defined by</p>
<p>$$f(n) = \left[ {\matrix{
{2n,} & {n = 2,4,6,8,......} \cr
{n - 1,} & {n = 3,7,11,15,......} \cr
{{{n + 1} \over 2},} & {n = 1,5,9,13,......} \cr
} } \right.$$</p>
<p>then, f is</p> | [{"identifier": "A", "content": "one-one but not onto"}, {"identifier": "B", "content": "onto but not one-one"}, {"identifier": "C", "content": "neither one-one nor onto"}, {"identifier": "D", "content": "one-one and onto"}] | ["D"] | null | <p>When n = 1, 5, 9, 13 then $${{n + 1} \over 2}$$ will give all odd numbers.</p>
<p>When n = 3, 7, 11, 15 .....</p>
<p>n $$-$$ 1 will be even but not divisible by 4</p>
<p>When n = 2, 4, 6, 8 .....</p>
<p>Then 2n will give all multiples of 4</p>
<p>So range will be N.</p>
<p>And no two values of n give same y, so func... | mcq | jee-main-2022-online-28th-june-morning-shift | 6,180 |
1l6dusstk | maths | functions | classification-of-functions | <p>The total number of functions,</p>
<p>$$
f:\{1,2,3,4\} \rightarrow\{1,2,3,4,5,6\}
$$
such that $$f(1)+f(2)=f(3)$$, is equal to :
</p> | [{"identifier": "A", "content": "60"}, {"identifier": "B", "content": "90"}, {"identifier": "C", "content": "108"}, {"identifier": "D", "content": "126"}] | ["B"] | null | <p>Given, $$f(1) + f(2) = f(3)$$</p>
<p>It means $$f(1),f(2)$$ and $$f(3)$$ are dependent on each other. But there is no condition on $$f(4)$$, so $$f(4)$$ can be $$f(4) = 1,2,3,4,5,6$$.</p>
<p>For $$f(1),f(2)$$ and we have to find how many functions possible which will satisfy the condition $$f(1) + f(2) = f(3)$$</p>
... | mcq | jee-main-2022-online-25th-july-morning-shift | 6,182 |
1l6f0troy | maths | functions | classification-of-functions | <p>The number of bijective functions $$f:\{1,3,5,7, \ldots, 99\} \rightarrow\{2,4,6,8, \ldots .100\}$$, such that $$f(3) \geq f(9) \geq f(15) \geq f(21) \geq \ldots . . f(99)$$, is ____________.</p> | [{"identifier": "A", "content": "$${ }^{50} P_{17}$$"}, {"identifier": "B", "content": "$${ }^{50} P_{33}$$"}, {"identifier": "C", "content": "$$33 ! \\times 17$$!"}, {"identifier": "D", "content": "$$\\frac{50!}{2}$$"}] | ["B"] | null | <p>As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction $$f(3) > f(9) > f(15)\,.......\, > f(99)$$</p>
<p>So number of ways $$ = {}^{50}{C_{17}}\,.\,1\,.\,33!$$</p>
<p>$$ = {}^{50}{P_{33}}$$</p> | mcq | jee-main-2022-online-25th-july-evening-shift | 6,183 |
1ldr7zkxa | maths | functions | classification-of-functions | <p>Let $$S=\{1,2,3,4,5,6\}$$. Then the number of one-one functions $$f: \mathrm{S} \rightarrow \mathrm{P}(\mathrm{S})$$, where $$\mathrm{P}(\mathrm{S})$$ denote the power set of $$\mathrm{S}$$, such that $$f(n) \subset f(\mathrm{~m})$$ where $$n < m$$ is ____________.</p> | [] | null | 3240 | <p>$$\because S={1,2,3,4,5,6}$$ and $$P(S) = \{ \phi ,\{ 1\} ,\{ 2\} ,....,\{ 1,2,3,4,5,6\} \} $$</p>
<p>$$f(n)$$ corresponding a set having m elements which belongs to P(S), should be a subset of $$f(n+1)$$, so $$f(n+1)$$ should be a subset of P(S) having at least $$m+1$$ elements.</p>
<p>Now, if f(1) has one element ... | integer | jee-main-2023-online-30th-january-morning-shift | 6,185 |
1ldsvim1d | maths | functions | classification-of-functions | <p>Let $$f:R \to R$$ be a function such that $$f(x) = {{{x^2} + 2x + 1} \over {{x^2} + 1}}$$. Then</p> | [{"identifier": "A", "content": "$$f(x)$$ is many-one in $$( - \\infty , - 1)$$"}, {"identifier": "B", "content": "$$f(x)$$ is one-one in $$( - \\infty ,\\infty )$$"}, {"identifier": "C", "content": "$$f(x)$$ is one-one in $$[1,\\infty )$$ but not in $$( - \\infty ,\\infty )$$"}, {"identifier": "D", "content": "$$f(x)$... | ["C"] | null | <p>$$f(x) = {{{x^2} + 2x + 1} \over {({x^2} + 1)}}$$</p>
<p>$$ \Rightarrow f'(x) = {{({x^2} + 1)(2x + 2) - ({x^2} + 2x + 1)(2x)} \over {{{({x^2} + 1)}^2}}}$$</p>
<p>$$ \Rightarrow f'(x) = {{2 - 2{x^2}} \over {{{({x^2} + 1)}^2}}}$$</p>
<p>$$ = {{2\left( {1 + x} \right)\left( {1 - x} \right)} \over {{{\left( {{x^2} + 1} ... | mcq | jee-main-2023-online-29th-january-morning-shift | 6,186 |
1ldu4e543 | maths | functions | classification-of-functions | <p>The number of functions</p>
<p>$$f:\{ 1,2,3,4\} \to \{ a \in Z|a| \le 8\} $$</p>
<p>satisfying $$f(n) + {1 \over n}f(n + 1) = 1,\forall n \in \{ 1,2,3\} $$ is</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "4"}] | ["A"] | null | $\because f:\{1,2,3,4\} \rightarrow\{a \in \mathbb{Z}:|9| \leq 8\}$
<br/><br/>
and $f(n)+\frac{1}{n} f(n+1)=1$
<br/><br/>
$\Rightarrow n f(n)+f(n+1)=n \quad \ldots$ (i)
<br/><br/>
$\therefore f(1)+f(2)=1 \Rightarrow f(2)=1-f(1)$
<br/><br/>
But $f(1) \in[-8,8]$
<br/><br/>
Hence, $f(2) \in[-8,8] \Rightarrow f(1) \in[-7,8... | mcq | jee-main-2023-online-25th-january-evening-shift | 6,187 |
1lgsw4p1x | maths | functions | classification-of-functions | <p>Let $$\mathrm{A}=\{1,2,3,4,5\}$$ and $$\mathrm{B}=\{1,2,3,4,5,6\}$$. Then the number of functions $$f: \mathrm{A} \rightarrow \mathrm{B}$$ satisfying $$f(1)+f(2)=f(4)-1$$ is equal to __________.</p> | [] | null | 360 | Given that the function $$f : A \rightarrow B$$ satisfies the condition $$f(1) + f(2) = f(4) - 1$$, where the set $$A = \{1, 2, 3, 4, 5\}$$ and the set $$B = \{1, 2, 3, 4, 5, 6\}$$.
<br/><br/>We want to find out how many such functions exist.
<br/><br/>First, observe that the condition $$f(1) + f(2) = f(4) - 1$$ can... | integer | jee-main-2023-online-11th-april-evening-shift | 6,188 |
lsbl9xjw | maths | functions | classification-of-functions | The function $f: \mathbf{N}-\{1\} \rightarrow \mathbf{N}$; defined by $f(\mathrm{n})=$ the highest prime factor of $\mathrm{n}$, is : | [{"identifier": "A", "content": "one-one only"}, {"identifier": "B", "content": "neither one-one nor onto"}, {"identifier": "C", "content": "onto only"}, {"identifier": "D", "content": "both one-one and onto"}] | ["B"] | null | <p>$$\begin{aligned}
& \mathrm{f}: \mathrm{N}-\{1\} \rightarrow \mathrm{N} \\
& \mathrm{f}(\mathrm{n})=\text { The highest prime factor of } \mathrm{n} . \\
& \mathrm{f}(2)=2 \\
& \mathrm{f}(4)=2 \\
& \Rightarrow \text { many one } \\
& 4 \text { is not image of any element } \\
& \Rightarrow \text { into }
\end{aligne... | mcq | jee-main-2024-online-27th-january-morning-shift | 6,190 |
luxwdqq9 | maths | functions | classification-of-functions | <p>Let $$A=\{(x, y): 2 x+3 y=23, x, y \in \mathbb{N}\}$$ and $$B=\{x:(x, y) \in A\}$$. Then the number of one-one functions from $$A$$ to $$B$$ is equal to _________.</p> | [] | null | 24 | <p>$$\begin{aligned}
& A=\{(x, y) ; 2 x+3 y=23, x, y \in N\} \\
& A=\{(1,7),(4,5),(7,3),(10,1)\} \\
& B=\{x:(x, y) \in A\} \\
& B=\{1,4,7,10\}
\end{aligned}$$</p>
<p>So, total number of one-one functions from A to B is $$4!=24$$</p> | integer | jee-main-2024-online-9th-april-evening-shift | 6,192 |
lvc57nwe | maths | functions | classification-of-functions | <p>The function $$f(x)=\frac{x^2+2 x-15}{x^2-4 x+9}, x \in \mathbb{R}$$ is</p> | [{"identifier": "A", "content": "both one-one and onto.\n"}, {"identifier": "B", "content": "onto but not one-one.\n"}, {"identifier": "C", "content": "neither one-one nor onto.\n"}, {"identifier": "D", "content": "one-one but not onto."}] | ["C"] | null | <p>The function $ f(x)=\frac{x^2+2x-15}{x^2-4x+9}, x \in \mathbb{R} $ can be simplified to $ f(x)=\frac{(x-3)(x+5)}{x^2-4x+9} $.</p>
<p>For $ x=3 $ and $ x=-5 $, $ f(x) $ equals 0. Therefore, $ f(x) $ is not one-one as it yields the same output for different input values.</p>
<p>The range of $ f(x) $ is $ [-2, 1.6] $... | mcq | jee-main-2024-online-6th-april-morning-shift | 6,194 |
g5GINXf1pcMuUqCs3v0PL | maths | functions | composite-functions | Let f(x) = 2<sup>10</sup>.x + 1 and g(x)=3<sup>10</sup>.x $$-$$ 1. If (fog) (x) = x, then x is equal to : | [{"identifier": "A", "content": "$${{{3^{10}} - 1} \\over {{3^{10}} - {2^{ - 10}}}}$$"}, {"identifier": "B", "content": "$${{{2^{10}} - 1} \\over {{2^{10}} - {3^{ - 10}}}}$$"}, {"identifier": "C", "content": "$${{1 - {3^{ - 10}}} \\over {{2^{10}} - {3^{ - 10}}}}$$"}, {"identifier": "D", "content": "$${{1 - {2^{ - 10}}}... | ["D"] | null | (fog) (x) = x
<br><br>$$ \Rightarrow $$$$\,\,\,$$ f (g(x)) = x
<br><br>$$ \Rightarrow $$$$\,\,\,$$ f (3<sup>10</sup>. x $$-$$ 1) = x [ as g(x) = 3<sup>10</sup>. x $$-$$ 1]
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 2<sup>10</sup... | mcq | jee-main-2017-online-8th-april-morning-slot | 6,196 |
obRwtmqr5ZJkly6xfb5WH | maths | functions | composite-functions | Let N be the set of natural numbers and two functions f and g be defined as f, g : N $$ \to $$ N such that
<br/><br/>f(n) = $$\left\{ {\matrix{
{{{n + 1} \over 2};} & {if\,\,n\,\,is\,\,odd} \cr
{{n \over 2};} & {if\,\,n\,\,is\,\,even} \cr
} \,\,} \right.$$;
<br/><br/> and g(n) = n $$-$$($$-$$ ... | [{"identifier": "A", "content": "neither one-one nor onto"}, {"identifier": "B", "content": "onto but not one-one"}, {"identifier": "C", "content": "both one-one and onto"}, {"identifier": "D", "content": "one-one but not onto"}] | ["B"] | null | f(x) = $$\left\{ {\matrix{
{{{n + 1} \over 2};} & {if\,\,n\,\,is\,\,odd} \cr
{{n \over 2};} & {if\,\,n\,\,is\,\,even} \cr
} \,\,} \right.$$;
<br><br>g(x) = n $$-$$ ($$-$$ 1)<sup>n</sup> $$\left\{ {\matrix{
{n + 1;\,\,\,\,n\,\,is\,\,odd} \cr
{n - 1;\,\,\,\,n\,\,is\,\,even} \cr
} } \right... | mcq | jee-main-2019-online-10th-january-evening-slot | 6,198 |
l6ARTWKmS9no4QtJ6Q3rsa0w2w9jx5c382k | maths | functions | composite-functions | For x $$ \in $$ (0, 3/2), let f(x) = $$\sqrt x $$ , g(x) = tan x and h(x) = $${{1 - {x^2}} \over {1 + {x^2}}}$$. If $$\phi $$ (x) = ((hof)og)(x), then $$\phi \left( {{\pi \over 3}} \right)$$
is equal to : | [{"identifier": "A", "content": "$$\\tan {{7\\pi } \\over {12}}$$"}, {"identifier": "B", "content": "$$\\tan {{11\\pi } \\over {12}}$$"}, {"identifier": "C", "content": "$$\\tan {\\pi \\over {12}}$$"}, {"identifier": "D", "content": "$$\\tan {{5\\pi } \\over {12}}$$"}] | ["B"] | null | $$\phi \left( x \right) = \left( {\left( {hof} \right)og} \right)(x) = h\left( {\sqrt {\tan x} } \right)$$<br><br>
$$ \Rightarrow \phi (x) = {{1 - \tan x} \over {1 + \tan x}} = \tan \left( {{\pi \over 4} - 4} \right)$$<br><br>
$$ \therefore $$ $$\phi \left( {{\pi \over 3}} \right) = \tan \left( {{\pi \over 4} - {\pi... | mcq | jee-main-2019-online-12th-april-morning-slot | 6,201 |
pDG9X0KeEGgr6xvj4v1kmlj1fg6 | maths | functions | composite-functions | If the functions are defined as $$f(x) = \sqrt x $$ and $$g(x) = \sqrt {1 - x} $$, then what is the common domain of the following functions :<br/><br/>f + g, f $$-$$ g, f/g, g/f, g $$-$$ f where $$(f \pm g)(x) = f(x) \pm g(x),(f/g)x = {{f(x)} \over {g(x)}}$$ | [{"identifier": "A", "content": "$$0 \\le x \\le 1$$"}, {"identifier": "B", "content": "$$0 \\le x < 1$$"}, {"identifier": "C", "content": "$$0 < x < 1$$"}, {"identifier": "D", "content": "$$0 < x \\le 1$$"}] | ["C"] | null | $$f + g = \sqrt x + \sqrt {1 - x} $$<br><br>$$ \Rightarrow x \ge 0$$ & $$1 - x \ge 0 \Rightarrow x \in [0,1]$$<br><br>$$f - g = \sqrt x - \sqrt {1 - x} $$<br><br>$$ \Rightarrow x \ge 0$$ & $$1 - x \ge 0 \Rightarrow x \in [0,1]$$<br><br>$$f/g = {{\sqrt x } \over {\sqrt {1 - x} }}$$<br><br>$$ \Rightarrow x \ge ... | mcq | jee-main-2021-online-18th-march-morning-shift | 6,205 |
1krrqrepo | maths | functions | composite-functions | Let $$f:R - \left\{ {{\alpha \over 6}} \right\} \to R$$ be defined by $$f(x) = {{5x + 3} \over {6x - \alpha }}$$. Then the value of $$\alpha$$ for which (fof)(x) = x, for all $$x \in R - \left\{ {{\alpha \over 6}} \right\}$$, is : | [{"identifier": "A", "content": "No such $$\\alpha$$ exists"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "6"}] | ["B"] | null | $$f(x) = {{5x + 3} \over {6x - \alpha }} = y$$ ..... (i)<br><br>$$5x + 3 = 6xy - \alpha y$$<br><br>$$x(6y - 5) = \alpha y + 3$$<br><br>$$x = {{\alpha y + 3} \over {6y - 5}}$$<br><br>$${f^{ - 1}}(x) = {{\alpha x + 3} \over {6x - 5}}$$ ...... (ii)<br><br>fo $$f(x) = x$$<br><br>$$f(x) = {f^{ - 1}}(x)$$<br><br>From eq<sup>... | mcq | jee-main-2021-online-20th-july-evening-shift | 6,206 |
1krzn2o7b | maths | functions | composite-functions | Consider function f : A $$\to$$ B and g : B $$\to$$ C (A, B, C $$ \subseteq $$ R) such that (gof)<sup>$$-$$1</sup> exists, then : | [{"identifier": "A", "content": "f and g both are one-one"}, {"identifier": "B", "content": "f and g both are onto"}, {"identifier": "C", "content": "f is one-one and g is onto"}, {"identifier": "D", "content": "f is onto and g is one-one"}] | ["C"] | null | $$\therefore$$ (gof)<sup>$$-$$1</sup> exist $$\Rightarrow$$ gof is bijective<br><br>$$\Rightarrow$$ 'f' must be one-one and 'g' must be ONTO. | mcq | jee-main-2021-online-25th-july-evening-shift | 6,207 |
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