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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>$$ \Rightarrow $$ $$\,\,\,$$ b<sup>2</sup> = 2a
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ a<sup>2</sup>(1 $$-$$ e<sup>2</sup>) = 2a [As b<sup>2</sup> = a<sup>2</sup> (1 $$-$$ e<sup>2</sup>)]
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ a (1 $$-$$ e) ( 1 + e) = 2
<br><br>Putting a (1 $$-$$ e) = $${3 \over 2}$$
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ $${3 \over 2}$$ (1 + e) = 2
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ 3 + 3e = 4
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ e = $${1 \over 3}$$ | mcq | jee-main-2018-online-16th-april-morning-slot |
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 eccentricity is $${2 \\over {\\sqrt {r - 1} }},$$ where 0 < r < 1"}, {"identifier": "D", "content": "an ellipse whose eccentricity is $$\\sqrt {{2 \\over {r + 1}}}$$, where r > 1"}] | ["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>$$ = \sqrt {{{\left( {r + 1} \right) - \left( {r - 1} \right)} \over {\left( {r + 1} \right)}}} $$
<br><br>$$ = \sqrt {{2 \over {r + 1}}} = \sqrt {{2 \over {r + 1}}} $$ | mcq | jee-main-2019-online-10th-january-evening-slot |
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 $${{{x^2}} \over {64}} + {{{y^2}} \over {32}} = 1$$ | mcq | jee-main-2019-online-11th-january-evening-slot |
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</sup> = a<sup>2</sup>e<sup>2</sup> . . . . . . (i)
<br><br>$${1 \over 2}$$ S'B.SB = 8
<br><br>S'B.SB = 16
<br><br>a<sup>2</sup>e<sup>2</sup> + b<sup>2</sup> = 16 . . . . .(ii)
<br><br>b<sup>2</sup> = a<sup>2</sup> (1 $$-$$ e<sup>2</sup>) . . . . .(iii)
<br><br>using (i), (ii), (iii) a = 4
<br><br> b = $$2\sqrt 2 $$
<br><br> e = $${1 \over {\sqrt 2 }}$$
<br><br>$$ \therefore $$ $$\ell $$ (L.R) $$=$$ $${{2{b^2}} \over a} = 4$$ | mcq | jee-main-2019-online-12th-january-evening-slot |
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>$$ \Rightarrow $$ b<sup>2</sup> - a<sup>2</sup> = 75 .......(1)
<br><br>Given that,
<br><br>difference of the lengths of major axis and minor
axis is 10.
<br><br>$$ \therefore $$ 2b - 2a = 10
<br><br>$$ \Rightarrow $$ b - a = 5 .......(2)
<br><br>From (1),
<br><br>(b + a)(b - a) = 75
<br><br>$$ \Rightarrow $$ (b + a)5 = 75
<br><br>$$ \Rightarrow $$ (b + a) = 15 .......(3)
<br><br>From (2) and (3) we get,
<br><br>b = 10, a = 5
<br><br>$$ \therefore $$ Length of its latus rectum = $${{{2{a^2}} \over b}}$$
<br><br>= $${{2 \times 25} \over {10}}$$ = 5 | mcq | jee-main-2019-online-8th-april-evening-slot |
iduWy6AjPu6GT3dkEG3rsa0w2w9jxb09lba | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | An ellipse, with foci at (0, 2) and (0, β2) 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</sup> - 4<br><br>
b<sup>2</sup> = 8<br><br>
$$ \because $$ equation of ellipse is $${{{x^2}} \over 4} + {{{y^2}} \over 8} = 1$$<br><br>
then it passes through $$\left( {\sqrt 2 ,2} \right)$$ | mcq | jee-main-2019-online-12th-april-evening-slot |
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<sup>2</sup> = 9
<br><br>$$ \Rightarrow $$ e<sup>2</sup> = $${1 \over 2}$$
<br><br>We know e<sup>2</sup> = 1 - $${{{b^2}} \over {{a^2}}}$$
<br><br>$$ \therefore $$ $${1 \over 2}$$ = 1 - $${{{b^2}} \over {{a^2}}}$$
<br><br>$$ \Rightarrow $$ b<sup>2</sup> = 9
<br><br>$$ \therefore $$ Length of latus rectum = $${{2{b^2}} \over a}$$
<br><br>= $${{2 \times 9} \over {\sqrt {18} }}$$
<br><br>= $$3\sqrt 2 $$ | mcq | jee-main-2020-online-7th-january-morning-slot |
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>$$ \therefore $$ $$\phi {(t)_{\max }} = {8 \over {12}} = {2 \over 3}$$<br><br>$$ \therefore $$ eccentricity (e) = $${2 \over 3}$$<br><br>Also, $${e^2} = 1 - {{{b^2}} \over {{a^2}}}$$<br><br>$$ \Rightarrow {4 \over 9} = 1 - {{{b^2}} \over {{a^2}}}$$<br><br>$$ \Rightarrow {{{b^2}} \over {{a^2}}} = {5 \over 9}$$<br><br>$$ \Rightarrow {{{b^2}} \over a} \times {1 \over a} = {5 \over 9}$$<br><br>$$ \Rightarrow {5 \over a} = {5 \over 9}$$<br><br>$$ \Rightarrow a = 9$$<br><br>$$ \therefore $$ $${b^2} = 5 \times 9 = 45$$<br><br>$$ \therefore $$ $${a^2} + {b^2} = 81 + 45 = 126$$ | mcq | jee-main-2020-online-4th-september-morning-slot |
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 eccentricities, then its value is : | [{"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>$${e^2} = 1 - {{{b^2}} \over {{a^2}}}$$<br><br>$${e^2} = 1 - {{{a^2}} \over {{c^2}}}$$<br><br>$$ \Rightarrow {{{b^2}} \over {{a^2}}} = {{{a^2}} \over {{c^2}}}$$<br><br>$$ \Rightarrow {c^2} = {{{a^4}} \over {{b^2}}} \Rightarrow c = {{{a^2}} \over b}$$<br><br>Also b = ce<br><br>$$ \Rightarrow c = {b \over e}$$<br><br>$${b \over e} = {{{a^2}} \over b}$$<br><br>$$ \Rightarrow e = {{{b^2}} \over {{a^2}}} = 1 - {e^2}$$<br><br>$$ \Rightarrow {e^2} + e - 1 = 0$$<br><br>$$ \Rightarrow e = {{ - 1 + \sqrt 5 } \over 2}$$ | mcq | jee-main-2021-online-22th-july-evening-shift |
1ks0744oi | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | A ray of light through (2, 1) is reflected at a point P on the y-axis and then passes through the point (5, 3). If this reflected ray is the directrix of an ellipse with eccentricity $${1 \over 3}$$ and the distance of the nearer focus from this directrix is $${8 \over {\sqrt {53} }}$$, then the equation of the other directrix can be : | [{"identifier": "A", "content": "11x + 7y + 8 = 0 or 11x + 7y $$-$$ 15 = 0"}, {"identifier": "B", "content": "11x $$-$$ 7y $$-$$ 8 = 0 or 11x + 7y + 15 = 0"}, {"identifier": "C", "content": "2x $$-$$ 7y + 29 = 0 or 2x $$-$$ 7y $$-$$ 7 = 0"}, {"identifier": "D", "content": "2x $$-$$ 7y $$-$$ 39 = 0 or 2x $$-$$ 7y $$-$$ 7 = 0"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266011/exam_images/bvecsk2e3t8vuavydyb4.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264268/exam_images/em5jdqg3jwdghbprqcnp.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263588/exam_images/ffu3bdx7j47o0huvdfzw.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Mathematics - Ellipse Question 40 English Explanation"></picture> <br><br>Equation of reflected Ray<br><br>$$y - 1 = {2 \over 7}(x + 2)$$<br><br>$$7x - 7 = 2x + 4$$<br><br>$$2x - 7y + 11 = 0$$<br><br>Let the equation of other directrix is <br><br>$$2x - 7y + \lambda $$ = 0<br><br>Distance of directrix from focus<br><br>$${a \over e} - e = {8 \over {\sqrt {53} }}$$<br><br>$$3a - {a \over 3} = {8 \over {\sqrt {53} }}$$ or $$a = {3 \over {\sqrt {53} }}$$<br><br>Distance from other focus $${a \over e} + ae$$<br><br>$$3a + {a \over 3} = {{10a} \over 3} = {{10} \over 3} \times {3 \over {\sqrt {53} }} = {{10} \over {\sqrt {53} }}$$<br><br>Distance between two directrix = $$ = {{2a} \over e}$$<br><br>$$ = 2 \times 3 \times {3 \over {\sqrt {53} }} = {{18} \over {\sqrt {53} }}$$<br><br>$$\left| {{{\lambda - 11} \over {\sqrt {53} }}} \right| = {{18} \over {\sqrt {53} }}$$<br><br>$$\lambda - 11 = 18$$ or $$-$$ 18<br><br>$$\lambda = 29$$ or $$-$$7<br><br>$$2x - 7y - 7 = 0$$ or $$2x - 7y + 29 = 0$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
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", "content": "[1, 3] and $$\\left[ { - {1 \\over 3},{1 \\over 3}} \\right]$$"}] | ["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>$${{{{(x - 2)}^2}} \over {{{(1)}^2}}} + {{{y^2}} \over {{{\left( {{1 \over 3}} \right)}^2}}} = 1$$ (equation of an ellipse).<br><br>As it is equation of an ellipse, x & y can vary inside the ellipse.<br><br>So, $$x - 2 \in [ - 1,1]$$ and $$y \in \left[ { - {1 \over 3},{1 \over 3}} \right]$$<br><br>x $$\in$$ [1, 3] $$y \in \left[ { - {1 \over 3},{1 \over 3}} \right]$$ | mcq | jee-main-2021-online-27th-august-morning-shift |
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>$${a^2}\left( {1 - {1 \over {16}}} \right) = {b^2}$$</p>
<p>$$15{a^2} = 16{b^2} \Rightarrow {a^2} = {{16{b^2}} \over {15}}$$</p>
<p>From (i)</p>
<p>$${6 \over {{b^2}}} + {9 \over {{b^2}}} = 1 \Rightarrow {b^2} = 15$$ & $${a^2} = 16$$</p>
<p>$${a^2} + {b^2} = 15 + 16 = 31$$</p> | mcq | jee-main-2022-online-27th-june-morning-shift |
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/1l5v5uvjp/f35361f2-1b79-408c-b8ee-7c34d92e23d8/e7463d60-0905-11ed-a790-b11fa70c8a36/file-1l5v5uvjq.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 24th June Evening Shift Mathematics - Ellipse Question 28 English Explanation"></p>
<p>$$\therefore$$ Let A($$\theta$$) be the area of $$\Delta$$ABB'</p>
<p>Then $$A(\theta ) = {1 \over 2}4\sin \theta (a + a\cos \theta )$$</p>
<p>$$A'(\theta ) = a(2\cos \theta + 2{\cos ^2}\theta )$$</p>
<p>For maxima $$A'(\theta ) = 0$$</p>
<p>$$ \Rightarrow \cos \theta = 1,\,\,\cos \theta = {1 \over 2}$$</p>
<p>But for maximum area $$\cos \theta = {1 \over 2}$$</p>
<p>$$\therefore$$ $$A(\theta ) = 6\sqrt 3 $$</p>
<p>$$ \Rightarrow 2{{\sqrt 3 } \over 2}\left( {a + {a \over 2}} \right) = 6\sqrt 3 $$</p>
<p>$$ \Rightarrow a = 4$$</p>
<p>$$\therefore$$ $$e = \sqrt {1 - {{{b^2}} \over {{a^2}}}} = \sqrt {1 - {4 \over {16}}} = {{\sqrt 3 } \over 2}$$</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
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>$$ \Rightarrow e_1^2 = 1 - {1 \over {{a^2}}}$$</p>
<p>Also, given hyperbola,</p>
<p>$${x^2} - {a^2}{y^2} = 5$$</p>
<p>$$ \Rightarrow {{{x^2}} \over 5} - {{{y^2}} \over {{5 \over {{a^2}}}}} = 1$$</p>
<p>eccentricity $$({e_2}) = \sqrt {1 + {{{b^2}} \over {{a^2}}}} $$</p>
<p>$$ = \sqrt {1 + {5 \over {5{a^2}}}} $$</p>
<p>$$ = \sqrt {1 + {1 \over {{a^2}}}} $$</p>
<p>$$ \Rightarrow e_2^2 = 1 + {1 \over {{a^2}}}$$</p>
<p>Also given,</p>
<p>$${e_1} = b \times {e_2}$$</p>
<p>$$ \Rightarrow e_1^2 = {b^2} \times e_2^2$$</p>
<p>$$ \Rightarrow 1 - {1 \over {{a^2}}} = {b^2}\left( {1 + {1 \over {{a^2}}}} \right)$$</p>
<p>$$ \Rightarrow {{{a^2} - 1} \over {{a^2}}} = {{{b^2}({a^2} + 1)} \over {{a^2}}}$$</p>
<p>$$ \Rightarrow {b^2} = {{{a^2} - 1} \over {{a^2} + 1}}$$</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l60jx84m/7dfa7ac3-4b51-48a5-b102-580ffe2e0760/d3aa7060-0bfc-11ed-8671-c161c1aef925/file-1l60jx84n.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l60jx84m/7dfa7ac3-4b51-48a5-b102-580ffe2e0760/d3aa7060-0bfc-11ed-8671-c161c1aef925/file-1l60jx84n.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Ellipse Question 26 English Explanation"></p>
<p>$$y = \log _e^x$$ is inverse of $$y = {e^x}$$ so it is mirror image of each other with respect to y = x line.</p>
<p>Slope of tangent to y = e<sup>x</sup> curve</p>
<p>$${{dy} \over {dx}} = {e^x}$$</p>
<p>Slope of tangent to $$y = \log _e^x$$ curve,</p>
<p>$${{dy} \over {dx}} = {1 \over x}$$</p>
<p>Both tangents are parallel to y = x line for minimum distance condition.</p>
<p>$$\therefore$$ Slope of y = x line = Slope of both the tangent.</p>
<p>$$\therefore$$ $${{dy} \over {dx}} = {e^x} = 1 \Rightarrow {e^x} = {e^0} = x = 0$$</p>
<p>$$\therefore$$ $$y = {e^x} = {e^0} = 1$$</p>
<p>and $${{dy} \over {dx}} = {1 \over x} = 1 \Rightarrow x = 1$$</p>
<p>$$\therefore$$ $$y = \log _e^1 = 0$$</p>
<p>$$\therefore$$ tangent at (0, 1) point of $$y = {e^x}$$ curve and tangent at (1, 0) point of $$y = \log _e^x$$ curve are parallel.</p>
<p>$$\therefore$$ Minimum distance between point (0, 1) and (1, 0) is $$ = \sqrt {{1^2} + {1^2}} = \sqrt 2 $$</p>
<p>$$\therefore$$ $$a = \sqrt 2 $$</p>
<p>So, $${b^2} = {{{a^2} - 1} \over {{a^2} + 1}} = {{2 - 1} \over {2 + 1}} = {1 \over 3}$$</p>
<p>$$\therefore$$ $${a^2} + {1 \over {{b^2}}} = 2 + {1 \over {1/3}} = 5$$</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
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, $$b = 2\sqrt 6 $$</p>
<p>Therefore, $${e^2} = 1 - {{{b^2}} \over {{a^2}}} = 1 - {{24} \over {49}}$$</p>
<p>$$e = {5 \over 7}$$</p> | mcq | jee-main-2022-online-25th-july-evening-shift |
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 $$ = {{2\,.\,\left( {{{\lambda + 5} \over 4}} \right)} \over {\sqrt {\lambda + 5} }} = 4$$.</p>
<p>$$\therefore$$ $$\lambda = 59$$.</p>
<p>Length of major axis $$ = 2\,.\,\sqrt {\lambda + 5} = 16 = l$$</p>
<p>$$\therefore$$ $$\lambda + l = 75$$.</p> | integer | jee-main-2022-online-27th-july-morning-shift |
1l6p1xv5r | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | <p>Let a line L pass through the point of intersection of the lines $$b x+10 y-8=0$$ and $$2 x-3 y=0, \mathrm{~b} \in \mathbf{R}-\left\{\frac{4}{3}\right\}$$. If the line $$\mathrm{L}$$ also passes through the point $$(1,1)$$ and touches the circle $$17\left(x^{2}+y^{2}\right)=16$$, then the eccentricity of the ellipse $$\frac{x^{2}}{5}+\frac{y^{2}}{\mathrm{~b}^{2}}=1$$ is :</p> | [{"identifier": "A", "content": "$$\n\\frac{2}{\\sqrt{5}}\n$$"}, {"identifier": "B", "content": "$$\\sqrt{\\frac{3}{5}}$$"}, {"identifier": "C", "content": "$$\\frac{1}{\\sqrt{5}}$$"}, {"identifier": "D", "content": "$$\\sqrt{\\frac{2}{5}}$$"}] | ["B"] | null | <p>$${L_1}:bx + 10y - 8 = 0,\,{L_2}:2x - 3y = 0$$</p>
<p>then $$L:(bx + 10y - 8) + \lambda (2x - 3y) = 0$$</p>
<p>$$\because$$ It passes through $$(1,\,1)$$</p>
<p>$$\therefore$$ $$b + 2 - \lambda = 0 \Rightarrow \lambda = b + 2$$</p>
<p>and touches the circle $${x^2} + {y^2} = {{16} \over {17}}$$</p>
<p>$$\left| {{{{8^2}} \over {{{(2\lambda + b)}^2} + {{(10 - 3\lambda )}^2}}}} \right| = {{16} \over {17}}$$</p>
<p>$$ \Rightarrow 4{\lambda ^2} + {b^2} + 4b\lambda + 100 + 9{\lambda ^2} - 60\lambda = 68$$</p>
<p>$$ \Rightarrow 13{(b + 2)^2} + {b^2} + 4b(b + 2) - 60(b + 2) + 32 = 0$$</p>
<p>$$ \Rightarrow 18{b^2} = 36$$</p>
<p>$$\therefore$$ $${b^2} = 2$$</p>
<p>$$\therefore$$ Eccentricity of ellipse : $${{{x^2}} \over 5} + {{{y^2}} \over {{b^2}}} = 1$$ is</p>
<p>$$\therefore$$ $$e = \sqrt {1 - {2 \over 5}} = \sqrt {{3 \over 5}} $$</p> | mcq | jee-main-2022-online-29th-july-morning-shift |
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/>
$$
\begin{aligned}
& 1+\alpha^{2}=\frac{128}{10} \\\\
& \alpha^{2}=\frac{118}{10} \\\\
&=118.00
\end{aligned}
$$ | integer | jee-main-2023-online-24th-january-morning-shift |
lgnyuzc0 | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | Let an ellipse with centre $(1,0)$ and latus rectum of length $\frac{1}{2}$ have its major axis along $\mathrm{x}$-axis. If its minor axis subtends an angle $60^{\circ}$ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to ____________. | [] | null | 9 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgrmkz0q/2a7d38c3-ba55-42ad-a453-25c4fef11776/8da18aa0-e0da-11ed-9ecd-e999028462e7/file-1lgrmkz0r.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgrmkz0q/2a7d38c3-ba55-42ad-a453-25c4fef11776/8da18aa0-e0da-11ed-9ecd-e999028462e7/file-1lgrmkz0r.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Ellipse Question 16 English Explanation"><br>$$
\begin{aligned}
& \frac{2 b^2}{a}=\frac{1}{2}, \quad \tan 30^{\circ}=\frac{b}{a e} \\\\
& b^2=\frac{a}{4}, \frac{1}{3}=\frac{b^2}{a^2-b^2} \Rightarrow a^2-b^2=3 b^2 \Rightarrow b^2=\frac{a^2}{4} \\\\
& \Rightarrow \quad a=1, b^2=\frac{1}{4} \Rightarrow b=\frac{1}{2} \\\\
& \Rightarrow \quad(2 a+2 b)^2=9
\end{aligned}
$$ | integer | jee-main-2023-online-15th-april-morning-shift |
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 $${m \over n}$$, where m and n are coprime, then $$m - n$$ is equal to :</p> | [{"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/6y3zli1lnbcvqbx/b63739e8-c768-420d-a989-3fb42c2b49f4/96ca2bd0-627d-11ee-b4f9-75aabed442fa/file-6y3zli1lnbcvqby.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnbcvqbx/b63739e8-c768-420d-a989-3fb42c2b49f4/96ca2bd0-627d-11ee-b4f9-75aabed442fa/file-6y3zli1lnbcvqby.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Morning Shift Mathematics - Ellipse Question 10 English Explanation">
<br><br>Now, point of intersection of line $x+3 y=3$ and circle
<br><br>$$
\begin{array}{lr}
&x^2+y^2=9 \\\\
&\therefore (3-3 y)^2+y^2=9 \\\\
&\Rightarrow 9-18 y+9 y^2+y^2=9 \\\\
&\Rightarrow -18 y+10 y^2=0 \\\\
&\Rightarrow 18 y=10 y^2 \\\\
&\Rightarrow y=0, \frac{9}{5}
\end{array}
$$
<br><br>Now, from figure, we clearly see that vertices $A, P$ and $O$ makes a $\triangle A P O$, <br><br>whose area is $\frac{1}{2} \times$ Base $\times$ Height
<br><br>$$
\begin{aligned}
& \text { Area }=\frac{1}{2} \times 3 \times \frac{9}{5}=\frac{27}{10}=\frac{m}{n} ~~~~[Given]\\\\
& \therefore m=27, n=10 \\\\
& \Rightarrow m-n=27-10=17
\end{aligned}
$$ | mcq | jee-main-2023-online-10th-april-morning-shift |
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 $$25\left(\beta^{2} x^{2}+\alpha^{2} y^{2}\right)=\alpha^{2} \beta^{2}$$ is :</p> | [{"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-4176-8624-2ce86ca94e2d/d2049270-73f9-11ee-b2c1-dbe2f99c6f1d/file-6y3zli1lo75i5g9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lo75i5g8/e4ea036e-437d-4176-8624-2ce86ca94e2d/d2049270-73f9-11ee-b2c1-dbe2f99c6f1d/file-6y3zli1lo75i5g9.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 6th April Evening Shift Mathematics - Ellipse Question 9 English Explanation">
<br>$\begin{array}{rlrl} &\therefore \alpha+t+\beta =100 ........(i) \\\\ & \alpha+t =75........(i) \\\\ &\text { and }\beta+t =40 ........(iii)\end{array}$
<br><br>From Equations (i) and (ii), $\beta=25$
<br><br>From Equations (i) and (iii), $\alpha=60$ and from Eq. (i), $t=15$
<br><br>We have, equation of ellipse
<br><br>$$
\begin{aligned}
25\left(\beta^2 x^2+\alpha^2 y^2\right) & =\alpha^2 \beta^2 \\\\
\Rightarrow 25\left(\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}\right) & =1
\end{aligned}
$$
<br><br>$\begin{aligned} & \Rightarrow 25\left(\frac{x^2}{3600}+\frac{y^2}{625}\right)=1 \\\\ & \Rightarrow \frac{x^2}{144}+\frac{y^2}{25}=1 \\\\ & \therefore \text { Eccentricity, } e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{25}{144}}=\frac{\sqrt{119}}{12}\end{aligned}$ | mcq | jee-main-2023-online-6th-april-evening-shift |
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:</p>
<p>$ \frac{1}{\sqrt{2}} = \sqrt{1 - \left(\frac{b}{a}\right)^2} $</p>
<p>Squaring both sides to eliminate the square root gives:</p>
<p>$ \frac{1}{2} = 1 - \left(\frac{b}{a}\right)^2 $</p>
<p>$ \left(\frac{b}{a}\right)^2 = 1 - \frac{1}{2} $</p>
<p>$ \left(\frac{b}{a}\right)^2 = \frac{1}{2} $</p>
<p>Taking the square root on both sides:</p>
<p>$ \frac{b}{a} = \frac{1}{\sqrt{2}} $</p>
<p>$ a = b\sqrt{2} $</p>
<p>Now, for the ellipse, the length of the latus rectum is given by the formula:</p>
<p>$ \text{Length of Latus Rectum (L)} = \frac{2b^2}{a} $</p>
<p>It's provided that the length of the latus rectum $ L $ is $ \sqrt{14} $, so substitute the known values to find $ b $:</p>
<p>$ \sqrt{14} = \frac{2b^2}{b\sqrt{2}} = \frac{2b}{\sqrt{2}} $</p>
<p>$ b\sqrt{2} = \sqrt{14} $</p>
<p>$ b^2 = \frac{14}{2} $</p>
<p>$ b^2 = 7 $</p>
<p>And since $ a = b\sqrt{2} $, we can find $ a^2 $:</p>
<p>$ a^2 = (b\sqrt{2})^2 $</p>
<p>$ a^2 = 7 \cdot 2 $</p>
<p>$ a^2 = 14 $</p>
<p>Now we have an ellipse with $ a^2 = 14 $ and $ b^2 = 7 $. The equation of a hyperbola similar to the given ellipse but with the terms subtracted is:</p>
<p>$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $</p>
<p>For the hyperbola, the square of the eccentricity $ e' $ is given by:</p>
<p>$ (e')^2 = 1 + \frac{b^2}{a^2} $</p>
<p>Substitute the values we've found for $ a^2 $ and $ b^2 $ into the formula for the square of the hyperbola's eccentricity:</p>
<p>$ (e')^2 = 1 + \frac{b^2}{a^2} $</p>
<p>$ (e')^2 = 1 + \frac{7}{14} $</p>
<p>$ (e')^2 = 1 + \frac{1}{2} $</p>
<p>$ (e')^2 = \frac{3}{2} $</p>
<p>Therefore, the square of the eccentricity of the hyperbola is $ \frac{3}{2} $, which corresponds to option C.</p>
</p> | mcq | jee-main-2024-online-1st-february-morning-shift |
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-8854-3b5a6c9e9092/file-6y3zli1lsjwesup.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 31st January Evening Shift Mathematics - Ellipse Question 5 English Explanation"></p>
<p>$$\begin{aligned}
& \text { Slope of axis }=\frac{1}{2} \\
& y-3=\frac{1}{2}(x-2) \\
& \Rightarrow 2 y-6=x-2 \\
& \Rightarrow 2 y-x-4=0 \\
& 2 x+y-6=0 \\
& 4 x+2 y-12=0 \\
& \alpha+1.6=4 \Rightarrow \alpha=2.4 \\
& \beta+2.8=6 \Rightarrow \beta=3.2
\end{aligned}$$</p>
<p>Ellipse passes through $$(2.4,3.2)$$</p>
<p>$$\Rightarrow \frac{\left(\frac{24}{10}\right)^2}{\mathrm{a}^2}+\frac{\left(\frac{32}{10}\right)^2}{\mathrm{~b}^2}=1$$ ..... (1)</p>
<p>Also $$1-\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{2}=\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{2}$$</p>
<p>$$\Rightarrow a^2=2 b^2$$</p>
<p>Put in (1) $$\Rightarrow b^2=\frac{328}{25}$$</p>
<p>$$\Rightarrow\left(\frac{2 \mathrm{~b}^2}{\mathrm{a}}\right)^2=\frac{4 \mathrm{~b}^2}{\mathrm{a}^2} \times \mathrm{b}^2=4 \times \frac{1}{2} \times \frac{328}{25}=\frac{656}{25}
$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
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 perpendicular on the $$x$$-axis passes through $$P$$, then the length of the latus rectum of $$E$$ is equal to,</p> | [{"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=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsoxo287/d8da4921-3aa3-467d-a404-6687dd0b31e6/268e3570-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxo288.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 30th January Evening Shift Mathematics - Ellipse Question 4 English Explanation"></p>
<p>$$\begin{aligned}
& \text { ae }=3 \\
& \frac{\mathrm{a}}{\mathrm{e}}=\frac{25}{3} \\
& \mathrm{a}=5 \\
& \mathrm{~b}=4
\end{aligned}$$</p>
<p>Length of $$L R=\frac{2 b^2}{a}=\frac{32}{5}$$</p> | mcq | jee-main-2024-online-30th-january-evening-shift |
1lsgaam1v | maths | ellipse | question-based-on-basic-definition-and-parametric-representation | <p>If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is :</p> | [{"identifier": "A", "content": "$$\\frac{1}{\\sqrt{3}}$$\n"}, {"identifier": "B", "content": "$$\\frac{2}{\\sqrt{5}}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\sqrt{3}}{2}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{5}}{3}$$"}] | ["B"] | null | <p>$$\begin{aligned}
& 2 b=a e \\
& \frac{b}{a}=\frac{e}{2} \\
& e=\sqrt{1-\frac{e^2}{4}} \\
& e=\frac{2}{\sqrt{5}}
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
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>$$
\begin{aligned}
8 e^2+l^2 & =\frac{8 \times 107}{109}+\frac{16}{109} \\
& =8
\end{aligned}
$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
U27dN3hfUPJ4eQBv | maths | ellipse | tangent-to-ellipse | The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 5} = 1$$, is : | [{"identifier": "A", "content": "$${{27 \\over 2}}$$"}, {"identifier": "B", "content": "$$27$$ "}, {"identifier": "C", "content": "$${{27 \\over 4}}$$"}, {"identifier": "D", "content": "$$18$$"}] | ["B"] | null | The end point of latus rectum of ellipse
<br><br>$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ in first quadrant is
<br><br>$$\left( {ae,{{{b^2}} \over a}} \right)$$ and the tangent at this point
<br><br>intersects $$x$$-axis at $$\left( {{a \over e},0} \right)$$ and
<br><br>$$y$$-axis at $$(0,a).$$
<br><br>The given ellipse is $${{x{}^2} \over 9} + {{{y^2}} \over 5} = 1$$
<br><br>Then $${a^2} = 9,{b^2} = 5$$
<br><br>$$ \Rightarrow e = \sqrt {1 - {5 \over 9}} = {2 \over 3}$$
<br><br>$$\therefore$$ end point of latus rectum in first quadrant is
<br><br>$$L\left( {2,\,\,5/3} \right)$$
<br><br>Equation of tangent at $$L$$ is $${{2x} \over 9} + {y \over 3} - 1$$
<br><br>It meets $$x$$-axis at $$A(9/2, 0)$$
<br><br>and $$y$$-axis at $$B(0,3)$$
<br><br>$$\therefore$$ Area of $$\Delta OAB = {1 \over 2} \times {9 \over 2} \times 3 = {{27} \over 4}$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l91nk9em/51951291-ca13-4e57-8386-abbd403322ab/7e6fcbe0-47fc-11ed-8757-0f869593f41f/file-1l91nk9en.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l91nk9em/51951291-ca13-4e57-8386-abbd403322ab/7e6fcbe0-47fc-11ed-8757-0f869593f41f/file-1l91nk9en.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2015 (Offline) Mathematics - Ellipse Question 74 English Explanation"><br>By symmetry area of quadrilateral
<br><br>$$ = 4 \times \left( {Area\,\,\Delta OAB} \right) = 4 \times {{27} \over 4} = 27\,\,$$ sq. units. | mcq | jee-main-2015-offline |
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>$$\Delta $$ = $${1 \over 2}$$ . $${{\sqrt {27} } \over {\cos \theta }}.{{\sqrt 3 } \over {\sin \theta }}$$ = $${9 \over {\sin 2\theta }}$$
<br><br>$$\Delta $$ = will be minimum when sin 2$$\theta $$ = 1
<br><br>$$\Delta $$<sub>min</sub> = 9 | mcq | jee-main-2016-online-9th-april-morning-slot |
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": "$${{{x^2}} \\over 4} + {{{y^2}} \\over 2} = 1$$"}] | ["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>$$h = {{\sqrt 2 \sec \theta } \over 2} \Rightarrow \cos \theta = {1 \over {\sqrt 2 h}}$$
<br><br>and $$k = {{\cos ec\theta } \over 2} \Rightarrow \sin \theta = {1 \over {2k}}$$
<br><br>$$ \because $$ $${\sin ^2}\theta + {\cos ^2}\theta = 1$$
<br><br>$$ \Rightarrow $$ $${1 \over {2{h^2}}} + {1 \over {4{k^2}}} = 1$$
<br><br>$$ \Rightarrow $$ $${1 \over {2{x^2}}} + {1 \over {4{y^2}}} = 1$$ | mcq | jee-main-2019-online-11th-january-morning-slot |
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>$$ \therefore $$ Slope of this tangent = -2
<br><br>Equation of tangent at point (a, b) is
<br><br>4ax + by = 8
<br><br>Slope of this tangent = $$ - {{4a} \over b}$$
<br><br>As tangent at (1, 2) and (a, b) are perpendicular
<br><br>$$ \therefore $$ $$\left( { - 2} \right)\left( { - {{4a} \over b}} \right) = - 1$$
<br><br>$$ \Rightarrow $$ b = -8$$a$$
<br><br>As point (a, b) lies on the ellipse
<br><br>$$ \therefore $$ $$4{a^2} + {b^2} = 8$$
<br><br>$$ \Rightarrow $$ $$4{a^2} + 64{a^2} = 8$$
<br><br>$$ \Rightarrow $$ $${a^2} = {8 \over {68}}$$ = $${2 \over {17}}$$ | mcq | jee-main-2019-online-8th-april-morning-slot |
elcH9KhH29Dzdmnbd93rsa0w2w9jwy2n90l | maths | ellipse | tangent-to-ellipse | If the line x β 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 β 2y = 12<br><br>
$${3 \over {{a^2}}} = {{ - 9} \over {2{b^2}( - 2)}} = {1 \over {12}}$$ (On comparing)<br><br>
$${a^2} = 3 \times 12$$ and $${b^2} = {{9 \times 12} \over 4}$$<br><br>
$$ \Rightarrow $$ a = 6 and b = $$3\sqrt 3 $$<br><br>
So latus rectum = $${{2{b^2}} \over a} = {{2 \times 27} \over 6} = 9$$ | mcq | jee-main-2019-online-10th-april-morning-slot |
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>$$ \Rightarrow $$ $${\left( {3\sqrt 2 } \right)^2} = {a^2}{\left( { - {3 \over 4}} \right)^2} + 9$$
<br><br>$$ \Rightarrow $$ a<sup>2</sup> = 16
<br><br>Also e = $$\sqrt {1 - {{{b^2}} \over {{a^2}}}} $$
<br><br>= $$\sqrt {1 - {9 \over {16}}} $$ = $${{\sqrt 7 } \over 4}$$
<br><br>Distance between focii = 2ae
<br><br>= $$2 \times 4 \times {{\sqrt 7 } \over 4}$$
<br><br>= $$2\sqrt 7 $$ | mcq | jee-main-2020-online-7th-january-evening-slot |
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 tangent is x + 6y = 8
<br><br>$$ \Rightarrow $$ y = $$ - {1 \over 6}x + {8 \over 6}$$ .....(2)
<br><br>By comparing (1) and (2),
<br><br>m = $$ - {1 \over 6}$$ and $${{a^2}{m^2} + {b^2}}$$ = $${{16} \over 9}$$
<br><br>$$ \Rightarrow $$ $${{a^2}\left( {{1 \over {36}}} \right) + {4 \over 3}}$$ = $${{16} \over 9}$$
<br><br>$$ \Rightarrow $$ $${{a^2} = 16}$$
<br><br>$$ \therefore $$ e = $$\sqrt {1 - {{{b^2}} \over {{a^2}}}} $$
<br><br>= $$\sqrt {1 - {{{4 \over 3}} \over {16}}} $$
<br><br>= $$\sqrt {{{11} \over {12}}} $$ = $${1 \over 2}\sqrt {{{11} \over 3}} $$ | mcq | jee-main-2020-online-9th-january-evening-slot |
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 perpendicular is (h, k)
<br><br>Given $${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$$
<br><br>$$ \therefore $$ a = 2, b = $$\sqrt 2 $$
<br><br>and e = $$\sqrt {1 - {2 \over 4}} $$ = $${1 \over {\sqrt 2 }}$$
<br><br>$$ \therefore $$ Focus(ae, 0) = $$\left( {\sqrt 2 ,0} \right)$$
<br><br>Equation of tangent
<br><br>y = mx + $$\sqrt {{a^2}{m^2} + {b^2}} $$
<br><br>$$ \Rightarrow $$ y = mx + $$\sqrt {4{m^2} + 2} $$
<br><br>Passes through (h, k)
<br><br>(k β mh)<sup>2</sup>
= 4m<sup>2</sup>
+ 2 .....(1)
<br><br>Line perpendicular to tangent will have slope $$ - {1 \over m}$$.
<br><br>y - 0 = $$ - {1 \over m}\left( {x - \sqrt 2 } \right)$$
<br><br>my = -x + $${\sqrt 2 }$$
<br><br>$$ \therefore $$ (h + mk)<sup>2</sup> = 2 .....(2)
<br><br>Add equation (1) and (2)
<br><br>k<sup>2</sup>(1 + m<sup>2</sup>) + h<sup>2</sup>
(1 + m<sup>2</sup>) = 4 (1 + m<sup>2</sup>)
<br><br>h<sup>2</sup>
+ k<sup>2</sup>
= 4
<br><br>x<sup>2</sup>
+ y<sup>2</sup>
= 4 (Auxiliary circle)
<br><br>$$ \therefore $$ $$\left( { - 1,\sqrt 3 } \right)$$ lies on the locus. | mcq | jee-main-2020-online-6th-september-morning-slot |
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>$$ \Rightarrow {b^2} + 48 = 16b$$<br><br>$$ \Rightarrow b = 12$$ | mcq | jee-main-2021-online-16th-march-evening-shift |
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'($$\theta$$) = $${3\sqrt 3 }$$ sec$$\theta$$tan$$\theta$$ $$-$$ cosec$$\theta$$ cot$$\theta$$ = 0<br><br>$$ \Rightarrow $$ $${{3\sqrt 3 \sin \theta } \over {{{\cos }^2}\theta }} = {{\cos \theta } \over {\sin \theta }}$$<br><br>$$ \Rightarrow {\tan ^3}\theta = {\left( {{1 \over {\sqrt 3 }}} \right)^3}$$<br><br>$$ \Rightarrow \tan \theta = {1 \over {\sqrt 3 }}$$<br><br>$$ \Rightarrow $$ $$\theta$$ = $${{\pi \over 6}}$$<br><br>also f'($$\theta$$) changes sign $$-$$ to + hence minimum. | mcq | jee-main-2021-online-18th-march-evening-shift |
1kryftyci | maths | ellipse | tangent-to-ellipse | Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its center at (3, $$-$$4), one focus at (4, $$-$$4) and one vertex at (5, $$-$$4). If mx $$-$$ y = 4, m > 0 is a tangent to the ellipse E, then the value of 5m<sup>2</sup> is equal to _____________. | [] | null | 3 | Given C(3, $$-$$4), S(4, $$-$$4)<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265485/exam_images/w9uwtpcdiyqyzwuecyz5.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift Mathematics - Ellipse Question 42 English Explanation"><br><br>and A(5, $$-$$4)<br><br>Hence, a = 2 & ae = 1<br><br>$$\Rightarrow$$ e = $${1 \over 2}$$<br><br>$$\Rightarrow$$ b<sup>2</sup> = 3<br><br>So, $$E:{{{{(x - 3)}^2}} \over 4} + {{{{(y + 4)}^2}} \over 3} = 1$$<br><br>Intersecting with given tangent.<br><br>$${{{x^2} - 6x + 9} \over 4} + {{{m^2}{x^2}} \over 3} = 1$$<br><br>Now, D = 0 (as it is tngent)<br><br>So, 5m<sup>2</sup> = 3. | integer | jee-main-2021-online-27th-july-evening-shift |
1krznj2rx | maths | ellipse | tangent-to-ellipse | If a tangent to the ellipse x<sup>2</sup> + 4y<sup>2</sup> = 4 meets the tangents at the extremities of it major axis at B and C, then the circle with BC as diameter passes through the point : | [{"identifier": "A", "content": "$$(\\sqrt 3 ,0)$$"}, {"identifier": "B", "content": "$$(\\sqrt 2 ,0)$$"}, {"identifier": "C", "content": "(1, 1)"}, {"identifier": "D", "content": "($$-$$1, 1)"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265844/exam_images/giwajhglcafcfpltohbs.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265389/exam_images/vh731jsttbh5pqpbscn6.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265221/exam_images/re6rjexoww28o7wilr3q.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264404/exam_images/imtsetbtkqgas6ehwtgi.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267751/exam_images/nyqbrtkjlwrjf5y2h1f6.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Evening Shift Mathematics - Ellipse Question 41 English Explanation"></picture> <br><br>$${{{x^2}} \over 4} + {{{y^2}} \over 1} = 1$$<br><br>Equation of tangent i (cos$$\theta$$)x + 2sin$$\theta$$y = 2<br><br>$$B\left( { - 2,{{1 + \cos \theta } \over {\sin \theta }}} \right),C\left( {2,{{1 - \cos \theta } \over {\sin \theta }}} \right)$$<br><br>$$B\left( { - 2,\cot {\theta \over 2}} \right)$$<br><br>$$C\left( {2,\tan {\theta \over 2}} \right)$$<br><br>Equation of circle is <br><br>$$(x + 2)(x - 2) + \left( {y - \cot {\theta \over 2}} \right)\left( {y - \tan {\theta \over 2}} \right) = 0$$<br><br>$${x^2} - 4 + {y^2} - \left( {\tan {\theta \over 2} + \cot {\theta \over 2}} \right)y + 1 = 0$$<br><br>so, $$\left( {\sqrt 3 ,0} \right)$$ satisfying option (1) | mcq | jee-main-2021-online-25th-july-evening-shift |
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<sup>2</sup>). A is : | [{"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 of tangent : y = 2x + 6 at P<br><br>$$\therefore$$ P($$-$$8/3, 2/3)<br><br>$$e = {1 \over {\sqrt 2 }}$$<br><br>S & S' = ($$-$$2, 0) & (2, 0)<br><br>Area of $$\Delta$$SPS' = $${1 \over 2} \times 4 \times {2 \over 3}$$<br><br>A = $${4 \over 3}$$<br><br>$$\therefore$$ (5 $$-$$ e<sup>2</sup>)A = $$\left( {5 - {1 \over 2}} \right)$$$${4 \over 3}$$ = 6 | mcq | jee-main-2021-online-26th-august-morning-shift |
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 August Morning Shift Mathematics - Ellipse Question 38 English Explanation"><br><br>So, area $$(\Delta OAB) = {1 \over 2} \times {b \over {\cos \theta }} \times {{2a} \over {\sin \theta }}$$<br><br>$$ = {{2ab} \over {\sin 2\theta }} \ge 2ab$$<br><br>$$\Rightarrow$$ k = 2 | integer | jee-main-2021-online-27th-august-morning-shift |
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 |
1ktk7nk5w | maths | ellipse | tangent-to-ellipse | An angle of intersection of the curves, $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ and x<sup>2</sup> + y<sup>2</sup> = ab, a > b, is : | [{"identifier": "A", "content": "$${\\tan ^{ - 1}}\\left( {{{a + b} \\over {\\sqrt {ab} }}} \\right)$$"}, {"identifier": "B", "content": "$${\\tan ^{ - 1}}\\left( {{{a - b} \\over {2\\sqrt {ab} }}} \\right)$$"}, {"identifier": "C", "content": "$${\\tan ^{ - 1}}\\left( {{{a - b} \\over {\\sqrt {ab} }}} \\right)$$"}, {"identifier": "D", "content": "$${\\tan ^{ - 1}}\\left( {2\\sqrt {ab} } \\right)$$"}] | ["C"] | null | $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1,{x^2} + {y^2} = ab$$<br><br>$${{2{x_1}} \over {{a^2}}} + {{2{y_1}y'} \over {{b^2}}} = 0$$<br><br>$$ \Rightarrow {y_1}' = {{ - {x_1}} \over {{a^2}}}{{{b^2}} \over {{y_1}}}$$ .... (1)<br><br>$$\therefore$$ $$2{x_1} + 2{y_1}y' = 0$$<br><br>$$ \Rightarrow {y_2}' = {{ - {x_1}} \over {{y_1}}}$$ ..... (2)<br><br>Here (x<sub>1</sub>y<sub>1</sub>) is point of intersection of both curves<br><br>$$\therefore$$ $$x_1^2 = {{{a^2}b} \over {a + b}},y_1^2 = {{a{b^2}} \over {a + b}}$$<br><br>$$\therefore$$ $$\tan \theta = \left| {{{{y_1}' - {y_2}'} \over {1 + {y_1}'{y_2}'}}} \right| = \left| {{{{{ - {x_1}{b^2}} \over {{a^2}{y_1}}} + {{{x_1}} \over {{y_1}}}} \over {1 + {{x_1^2{b^2}} \over {{a^2}y_1^2}}}}} \right|$$<br><br>$$\tan \theta = \left| {{{ - {b^2}{x_1}{y_1} + {a^2}{x_1}{y_1}} \over {{a^2}y_1^2 + {b^2}x_1^2}}} \right|$$<br><br>$$\tan \theta = \left| {{{a - b} \over {\sqrt {ab} }}} \right|$$ | mcq | jee-main-2021-online-31st-august-evening-shift |
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 $$\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)$$ is <br><br>$${m_1} = - {1 \over {3\sqrt 3 }}$$<br><br>And slope of tangent to the circle at $$\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)$$ is m<sub>2</sub> $$ = - \sqrt 3 $$<br><br>So, if angle between both curves is $$\theta$$ then<br><br>$$\tan \theta = \left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right| = \left| {{{ - {1 \over {3\sqrt 3 }} + \sqrt 3 } \over {1 + \left( { - {1 \over {3\sqrt 3 }}\left( { - \sqrt 3 } \right)} \right)}}} \right|$$<br><br>$$ = {2 \over {\sqrt 3 }}$$<br><br>Option (b) | mcq | jee-main-2021-online-1st-september-evening-shift |
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 ___________.</p> | [] | 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/>
$$
m_{1}+4 m_{1}=\frac{\beta}{\alpha} \text { and } m_{1} \cdot 4 m_{1}=\frac{1}{\alpha}
$$
<br/><br/>
Gives that $4 \beta^{2}=25 \alpha \quad\quad...(ii)$
<br/><br/>
from (i) and (ii)
<br/><br/>
$25\left(\alpha^{2}+\alpha\right)=1\quad\quad...(iii)$
<br/><br/>
Now, $(10 \alpha+5)^{2}+\left(16 \beta^{2}+50\right)^{2}$
<br/><br/>
$=25(2 \alpha+1)^{2}+2500(2 \alpha+1)^{2}$
<br/><br/>
$=2525\left(4 \alpha^{2}+4 \alpha+1\right)$ from equation (iii)
<br/><br/>
$=2525\left(\frac{4}{25}+1\right)$
<br/><br/>
$=2929$ | integer | jee-main-2022-online-24th-june-morning-shift |
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": "$$\\tan ^{-1}\\left(\\frac{3+8 \\sqrt{5}}{35}\\right)$$"}] | ["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>$${3 \over 2}{m^2} + 6m - {{22} \over 3} = 0$$</p>
<p>$$9{m^2} + 36m - 44 = 0$$</p>
<p>$${m_1} + {m_2} = - 4,\,{m_1}{m_2} = - {{44} \over 9}$$</p>
<p>$${({m_1} - {m_2})^2} = 16 + 4 \times {{44} \over 9} = {{320} \over 9}$$</p>
<p>Acute angle between the tangents is given by</p>
<p>$$\alpha = {\tan ^{ - 1}}\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$$</p>
<p>$$ = {\tan ^{ - 1}}\left| {{{{{8\sqrt 5 } \over 3}} \over {1 - {{44} \over 9}}}} \right|$$</p>
<p>$$ = {\tan ^{ - 1}}\left( {{{24\sqrt 5 } \over {35}}} \right)$$</p>
<p>$$\alpha = {\tan ^{ - 1}}\left( {{{24} \over {7\sqrt 5 }}} \right)$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
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>$$ \Rightarrow \left( {2\sqrt 2 - 2 - m\sqrt 2 } \right) = \pm \,\sqrt {2{m^2} + 4} $$</p>
<p>$$ \Rightarrow 2{m^2} - 2m\sqrt 2 \left( {2\sqrt {2 - 2} } \right) + 4(3 - 2\sqrt 2 ) = 2{m^2} + 4$$</p>
<p>$$ \Rightarrow - 2\sqrt 2 m(2\sqrt 2 - 2) = 4 - 12 + 8\sqrt 2 $$</p>
<p>$$ \Rightarrow - 4\sqrt 2 m(\sqrt 2 - 1) = 8(\sqrt 2 - 1)$$</p>
<p>$$ \Rightarrow m = - \sqrt 2 $$ and $$m \to \infty $$</p>
<p>$$\therefore$$ Tangents are $$x = \sqrt 2 $$ and $$y = - \sqrt 2 x + \sqrt 8 $$</p>
<p>$$\therefore$$ $$P(\sqrt 2 ,0)$$ and $$Q(1,\sqrt 2 )$$</p>
<p>and $$S = (0, - \sqrt 2 )$$</p>
<p>$$\therefore$$ $${(PS)^2} + {(QS)^2} = 4 + 9 = 13$$</p> | integer | jee-main-2022-online-28th-july-evening-shift |
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 $$\mathrm{Q}$$, then $$(3\mathrm{PQ})^{2}$$ is equal to ____________.</p> | [] | null | 39 | $\begin{aligned} & \frac{a}{e}=8 \\\\ & a e=2 .(1) \\\\ & 8 e=\frac{2}{e} \\\\ & e^2=\frac{1}{4} \Rightarrow e=\frac{1}{2} \\\\ & a=4 \\\\ & b^2=a^2\left(1-e^2\right) \\\\ & =16\left(\frac{3}{4}\right)=12 \\\\ & \frac{x \cos \theta}{4}+\frac{y \sin \theta}{2 \sqrt{3}}=1 \\\\ & \sin \theta=\frac{1}{2} \\\\ & \theta=30^{\circ}\end{aligned}$
<br/><br/>$\begin{aligned} & \mathrm{P}(2 \sqrt{3}, \sqrt{3}) \\\\ & \mathrm{Q}\left(\frac{8}{\sqrt{3}}, 0\right) \\\\ & (3 \mathrm{PQ})^2=39\end{aligned}$ | integer | jee-main-2023-online-1st-february-evening-shift |
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>
<p>We know,</p>
<p>Equation of tangent to a given ellipse at its point $$(a\cos\theta ,b\sin \theta )$$ is</p>
<p>$${{x\cos \theta } \over a} + {{y\sin \theta } \over b} = 1$$</p>
<p>$$\therefore$$ Here equation of tangent at point $$(4\cos \theta ,3\sin \theta )$$ is</p>
<p>$${{x\cos \theta } \over 4} + {{y\sin \theta } \over 3} = 1$$</p>
<p>When this tangent cut's x axis then y = 0.</p>
<p>$$\therefore$$ $${{x\cos \theta } \over 4} + 0 = 1$$</p>
<p>$$ \Rightarrow x = 4\sec \theta $$</p>
<p>$$\therefore$$ Point of intersection at x axis is $$A(4\sec \theta ,0)$$.</p>
<p>When this tangent cut's y axis then x = 0.</p>
<p>$$\therefore$$ $$0 + {{y\sin \theta } \over 3} = 1$$</p>
<p>$$ \Rightarrow y = 3\cos ec\theta $$</p>
<p>$$\therefore$$ Point of intersection at y axis is $$B(0,3\cos ec\theta )$$</p>
<p>$$\therefore$$ Length of AB</p>
<p>$$ = \sqrt {{{(4\sec \theta - 0)}^2} + {{(0 - 3\cos ec\theta )}^2}} $$</p>
<p>$$ = \sqrt {16{{\sec }^2}\theta + 9\cos e{c^2}\theta } $$</p>
<p>$$ = \sqrt {16(1 + {{\tan }^2}\theta ) + 9(1 + {{\cot }^2}\theta )} $$</p>
<p>$$ = \sqrt {25 + 16{{\tan }^2}\theta + 9{{\cot }^2}\theta } $$</p>
<p>We know, $$AM \ge GM$$</p>
<p>$$\therefore$$ $${{16{{\tan }^2}\theta + 9{{\cot }^2}\theta } \over 2} \ge \sqrt {(16{{\tan }^2}\theta )(9{{\cot }^2}\theta )} $$</p>
<p>$$ \Rightarrow 16{\tan ^2}\theta + 9{\cot ^2}\theta \ge 2(4\tan \theta )(3\cot \theta )$$</p>
<p>$$ \Rightarrow 16{\tan ^2}\theta + 9{\cot ^2}\theta \ge 2 \times 4 \times 3$$</p>
<p>$$ \Rightarrow 16{\tan ^2}\theta + 9{\cot ^2}\theta \ge 24$$</p>
<p>$$\therefore$$ $$AB = \sqrt {25 + 16{{\tan }^2}\theta + 9{{\cot }^2}\theta } $$</p>
<p>$$ \ge \sqrt {25 + 24} $$</p>
<p>$$ \ge \sqrt {49} $$</p>
<p>$$ \ge 7$$</p>
<p>$$\therefore$$ Minimum length of $$AB = 7$$.</p> | integer | jee-main-2023-online-24th-january-morning-shift |
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/>The equation of a circle with center (2,0) and radius r is :
<br/><br/>$$(x-2)^2 + y^2 = r^2$$
<br/><br/>Substituting y^2 from the ellipse equation into the circle equation gives us :
<br/><br/>$$x^2 - 4x + 4 + \frac{36 - x^2}{4} = r^2$$
<br/><br/>Solving this equation leads to :
<br/><br/>$$3x^2 - 16x + 52 - 4r^2 = 0$$
<br/><br/>For the roots of this quadratic equation to be real (which they must be, since they represent real intersection points), the discriminant (D) must be greater than or equal to zero :
<br/><br/>$$D = b^2 - 4ac = (-16)^2 - 4\times3\times(52 - 4r^2) = 256 - 12\times52 + 48r^2$$
<br/><br/>Setting D = 0 gives the minimum value for r (the radius of the inscribed circle) :
<br/><br/>$$256 - 624 + 48r^2 = 0$$
<br/><br/>$$48r^2 = 368$$
<br/><br/>$$r^2 = \frac{368}{48} = \frac{23}{3}$$
<br/><br/>And we're asked for the value of 12r<sup>2</sup>, so :
<br/><br/>$$12r^2 = 12 \times \frac{23}{3} = 92$$
<br/><br/>So, the correct answer is Option B : 92.
| mcq | jee-main-2023-online-11th-april-evening-shift |
TXeVoSP3SOStAzKB | maths | functions | classification-of-functions | A function $$f$$ from the set of natural numbers to integers defined by
$$$f\left( n \right) = \left\{ {\matrix{
{{{n - 1} \over 2},\,when\,n\,is\,odd} \cr
{ - {n \over 2},\,when\,n\,is\,even} \cr
} } \right.$$$
is | [{"identifier": "A", "content": "neither one -one nor onto"}, {"identifier": "B", "content": "one-one but not onto"}, {"identifier": "C", "content": "onto but not one-one"}, {"identifier": "D", "content": "one-one and onto both"}] | ["D"] | null | We have $$f:N \to I$$
<br><br>If $$x$$ and $$y$$ are two even natural numbers,
<br><br>then $$f\left( x \right) = f\left( y \right) \Rightarrow {{ - x} \over 2} = {{ - y} \over 2} \Rightarrow x = y$$
<br><br>Again if $$x$$ and $$y$$ are two odd natural numbers then
<br><br>$$f\left( x \right) = f\left( y \right) \Rightarrow {{x - 1} \over 2} = {{y - 1} \over 2} \Rightarrow x = y$$
<br><br>$$\therefore$$ $$f$$ is onto.
<br><br>Also each negative integer is an image of even natural number and each positive integer is an image of odd natural number.
<br><br>$$\therefore$$ $$f$$ is onto.
<br><br>Hence $$f$$ is one one and onto both. | mcq | aieee-2003 |
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>$$ - 1 \le 2\sin \left( {x - {\pi \over 3}} \right) + 1 \le 3$$
<br><br>$$\therefore$$ $$f\left( x \right) \in \left[ { - 1,3} \right] = S$$ | mcq | aieee-2004 |
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": "$$\\left( { - {\\pi \\over 2},{\\pi \\over 2}} \\right)$$"}] | ["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>$$ \Rightarrow 2{\tan ^{ - 1}}x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right)$$
<br><br>Clearly, range of $$f\left( x \right) = \left( { - {\pi \over 2},{\pi \over 2}} \right)$$
<br><br>For $$f$$ to be onto, co-domain $$=$$ range
<br><br>$$\therefore$$ Co-domain of function $$ = B = \left( { - {\pi \over 2},{\pi \over 2}} \right).$$ | mcq | aieee-2005 |
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 |
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": "Statement - 1 is true, Statement - 2 is false"}, {"identifier": "D", "content": "Statement - 1 is false, Statement - 2 is true"}] | ["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>$$(x+1)$$ being something $$+ve,$$ $${f^{ - 1}}\left( x \right)$$ will exist where
<br><br>$${\left( {x + 1} \right)^2} - 1 = y$$
<br><br>$$ \Rightarrow x + 1 = \sqrt {y + 1} $$
<br><br>$$\,\,\,\,\,$$ $$\left( { + ve} \right.$$ square root as $$x + 1 \ge 0$$$$\left. {} \right)$$
<br><br>$$ \Rightarrow x = - 1 + \sqrt {y + 1} $$
<br><br>$$ \Rightarrow {f^{ - 1}}\left( x \right) = \sqrt {x + 1} - 1$$
<br><br>Then $$\,\,\,\,\,\,\,\,$$ $$f\left( x \right) = {f^{ - 1}}\left( x \right)$$
<br><br>$$ \Rightarrow {\left( {x + 1} \right)^2} - 1 = \sqrt {x + 1} - 1$$
<br><br>$$ \Rightarrow {\left( {x + 1} \right)^2} = \sqrt {x + 1} $$
<br><br>$$ \Rightarrow {\left( {x + 1} \right)^4} = \left( {x + 1} \right)$$
<br><br>$$ \Rightarrow \left( {x + 1} \right)\left[ {{{\left( {x + 1} \right)}^3} - 1} \right] = 0$$
<br><br>$$ \Rightarrow x = - 1,0$$
<br><br>$$\therefore$$ The statement -$$1$$ is correct but statement- $$2$$ is false. | mcq | aieee-2009 |
PjsG1USbhMNW1Hsh | maths | functions | classification-of-functions | For real x, let f(x) = x<sup>3</sup> + 5x + 1, then | [{"identifier": "A", "content": "f is one-one but not onto R"}, {"identifier": "B", "content": "f is onto R but not one-one"}, {"identifier": "C", "content": "f is one-one and onto R"}, {"identifier": "D", "content": "f is neither one-one nor onto R"}] | ["C"] | null | Given that $$f\left( x \right) = {x^3} + 5x + 1$$
<br><br>$$\therefore$$ $$\,\,\,\,\,$$ $$f'\left( x \right) = 3{x^2} + 5 > 0,\,\,\,\forall x \in R$$
<br><br>$$ \Rightarrow f\left( x \right)\,\,$$ is strictly increasing on $$R$$
<br><br>$$ \Rightarrow f\left( x \right)$$ is one one
<br><br>$$\therefore$$ $$\,\,\,\,\,\,\,$$ Being a polynomial $$f(x)$$ is cont. and inc.
<br><br>on $$R$$ with $$\mathop {\lim }\limits_{x \to \infty } \,f\left( x \right) = - \infty $$
<br><br>and $$\mathop {\lim }\limits_{x \to \infty } \,f\left( x \right) = \infty $$
<br><br>$$\therefore$$ $$\,\,\,\,\,\,\,$$ Range of $$f = \left( { - \infty ,\infty } \right) = R$$
<br><br>Hence $$f$$ is onto also, So, $$f$$ is one and onto $$R.$$ | mcq | aieee-2009 |
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>$$ \Rightarrow $$ y + x<sup>2</sup>y = x
<br><br>$$ \Rightarrow $$ yx - x + y = 0
<br><br>As x $$ \in $$ R
<br><br>$$ \therefore $$ (-1)<sup>2</sup> - 4(y)(y) $$ \ge $$ 0
<br><br>$$ \Rightarrow $$ 1 - 4y<sup>2</sup> $$ \ge $$ 0
<br><br>$$ \Rightarrow $$ y $$ \in $$ $$\left[ { - {1 \over 2},{1 \over 2}} \right]$$
<br><br>$$ \therefore $$ Range = Codomain = $$\left[ { - {1 \over 2},{1 \over 2}} \right]$$
<br><br>So, f(x) is surjective.
<br><br>$$ \therefore $$ f(x) is surjective but not injective
| mcq | jee-main-2017-offline |
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 |
i3Uk81JW5dW81psdZCrdJ | maths | functions | classification-of-functions | Let A = {x $$ \in $$ <b>R</b> : x is not a positive integer}.
<br/><br/>Define a function $$f$$ : A $$ \to $$Β Β <b>R</b> Β Β asΒ Β $$f(x)$$ = $${{2x} \over {x - 1}}$$,
<br/><br/>then $$f$$ is : | [{"identifier": "A", "content": "not injective"}, {"identifier": "B", "content": "neither injective nor surjective"}, {"identifier": "C", "content": "surjective but not injective "}, {"identifier": "D", "content": "injective but not surjective"}] | ["D"] | null | f(x) = $${{2x} \over {x - 1}}$$
<br><br>f(x) = 2 + $${2 \over {x - 1}}$$
<br><br>f'(x) = $$-$$ $${2 \over {{{\left( {x - 1} \right)}^2}}}$$ < 0 $$\forall $$ x $$ \in $$ R
<br><br>Hence f(x) is strictly decreasing
<br><br>So, f(x) is one-one
<br><br><b>Range : </b> Let y = $${{2x} \over {x - 1}}$$
<br><br>xy $$-$$ y = 2x
<br><br>$$ \Rightarrow $$ x(y $$-$$ 2) = y
<br><br>$$ \Rightarrow $$ x = $${y \over {y - 2}}$$
<br><br>given that x $$ \in $$ R : x is not a +ve integer
<br><br>$$ \therefore $$ $${y \over {y - 2}} \ne $$ N (N $$ \to $$ Natural number)
<br><br>$$ \Rightarrow $$ y $$ \ne $$ Ny $$-$$ 2N
<br><br>$$ \Rightarrow $$ y $$ \ne $$ $${{2N} \over {N - 1}}$$
<br><br>So range $$ \notin $$ R (in to function) | mcq | jee-main-2019-online-9th-january-evening-slot |
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 5 elements in this subset.
<br/><br/>2. We then consider the subset of the codomain that consists of multiples of 3: {3, 6, 9, 12, 15, 18}. There are 6 elements in this subset.
<br/><br/>3. According to the given condition, each of the 5 multiples of 4 must be mapped to a multiple of 3. This can be done in ${ }^6C_5 \cdot 5! = 6!$ ways, considering that there are 6 options for each of the 5 multiples of 4 (each choice constitutes a combination), and we then consider the permutations of these 5 choices.
<br/><br/>4. The remaining 15 elements in the domain (20 original elements minus the 5 multiples of 4) can be mapped onto the remaining 15 elements in the codomain (20 original elements minus the 6 multiples of 3, plus one multiple of 3 that has been assigned to a multiple of 4). This can be done in $15!$ ways.
<br/><br/>So, combining these two cases, the total number of onto functions $f$ is $6! \times 15!$, which corresponds to option C. | mcq | jee-main-2019-online-11th-january-evening-slot |
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/w8mp3onrybrng94fquzf.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 11th January Evening Slot Mathematics - Functions Question 110 English Explanation">
<br>$$ \Rightarrow $$ f(x) is not injective
<br><br>but range of function is $$\left[ {0,\infty } \right)$$
<br><br><b>Remarks :</b> If co-domain is $$\left[ {0,\infty } \right)$$, then f(x) will be surjective. | mcq | jee-main-2019-online-11th-january-evening-slot |
dfuBeaBmWfo5qPrwGI18hoxe66ijvwpyj7h | maths | functions | classification-of-functions | If the function Ζ : R β {1, β1} $$ \to $$ A defined by <br/>
Ζ(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 Ζ(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 $$ \in $$ $$\left( { - \infty , - 1} \right) \cup \left[ {0,\left. \infty \right)} \right.$$
<br><br>For surjective function co-domain = Range
<br><br>$$ \therefore $$ A is R β [β1, 0). | mcq | jee-main-2019-online-9th-april-morning-slot |
uCRfvNgdGc6yDuxSnqjgy2xukfqfns2t | maths | functions | classification-of-functions | Let A = {a, b, c} and B = {1, 2, 3, 4}. Then the
number of elements in the set <br/>C = {f : A $$ \to $$ B |
2 $$ \in $$ f(A) and f is not one-one} is ______. | [] | null | 19 | The desired functions will contain either one
element or two elements in its codomain of
which '2' always belongs to f(A).
<br><br><b>Case 1 :</b> When 2 is the image of all element of set A.
<br><br>Number of ways this is possible = 1
<br><br><b>Case 2 :</b> When one image is 2 and other one image is one of {1, 3, 4}.
<br><br>Number of ways we can choose one of {1, 3, 4} is = <sup>3</sup>C<sub>1</sub>.
<br><br>Now divide 3 elements {a, b, c} of set A into two parts.
<br>We can do this $${{3!} \over {2!1!}}$$ ways.
<br><br>Now map one part of set A into the element 2 of set B and map other part of set A into one of {1, 3, 4} of set B.
<br>We can do that 2! ways.
<br><br>So number of functions in this case
<br>= <sup>3</sup>C<sub>1</sub> $$ \times $$ $${{3!} \over {2!1!}}$$ $$ \times $$ 2! = 18
<br><br>$$ \therefore $$ Total number of functions = 1 + 18 = 19 | integer | jee-main-2020-online-5th-september-evening-slot |
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 |
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="JEE Main 2021 (Online) 25th February Evening Shift Mathematics - Functions Question 88 English Explanation 1"><br><br>Number of one-one function<br><br>x = 5 $$\times$$ 4 $$\times$$ 3<br><br>x = 60<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266205/exam_images/qm3llf0rmyqukfzhahci.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Evening Shift Mathematics - Functions Question 88 English Explanation 2"><br><br>Number of one-one function<br><br>y = 15 $$\times$$ 14 $$\times$$ 13<br><br>y = 15 $$\times$$ 4 $$\times$$ $${{14} \over 4}$$ $$\times$$ 13<br><br>y = 60 $$\times$$ $${7 \over 2}$$ $$\times$$ 13<br><br>2y = (13)(7x)<br><br>2y = 91x | mcq | jee-main-2021-online-25th-february-evening-slot |
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.cloudinary.com/dckxllbjy/image/upload/v1734264384/exam_images/yrhufchc5jsgfnpsyljo.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Evening Shift Mathematics - Functions Question 87 English Explanation 1"></picture> <br><br>f(1) = 2<br><br>f(2) = 2<br><br>f(3) = 4<br><br>f(4) = 4<br><br>f(5) = 6<br><br>f(6) = 6<br><br>f(7) = 8<br><br>f(8) = 8<br><br>f(9) = 10<br><br>f(10) = 10<br><br>$$ \therefore $$ f(1) = f(2) = 2<br><br>f(3) = f(4) = 4<br><br>f(5) = f(6) = 6<br><br>f(7) = f(8) = 8<br><br>f(9) = f(10) = 10<br><br>Given, g(f(x)) = f(x)<br><br>when x = 1, g(f(1)) = f(1) $$ \Rightarrow $$ g(2) = 2<br><br>when, x = 2, g(f(2)) = f(2) $$ \Rightarrow $$ g(2) = 2<br><br>$$ \therefore $$ x = 1, 2, g(2) = 2<br><br>Similarly, at x = 3, 4, g(4) = 4<br><br>at x = 5, 6, g(6) = 6<br><br>at x = 7, 8, g(8) = 8<br><br>at x = 9, 10, g(10) = 10<br><br> <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266114/exam_images/lpbgi28osadvhp2mzckg.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265162/exam_images/kuhwk8m2iozvbagdkggp.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267806/exam_images/hs7csruvmcfyqi6fwypv.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Evening Shift Mathematics - Functions Question 87 English Explanation 2"></picture> <br><br>Here, you can see for even terms mapping is fixed. But far odd terms 1, 3, 5, 7, 9 we can map to any one of the 10 elements.<br><br>$$ \therefore $$ For 1, number of functions = 10<br><br>For 3, number of functions = 10<br><br>$$\eqalign{
& . \cr
& . \cr
& . \cr
& \cr} $$<br>for 9, number of functions = 10<br><br>$$ \therefore $$ Total number of functions = 10 $$\times$$ 10 $$\times$$ 10 $$\times$$ 10 $$\times$$ 10 = 10<sup>5</sup> | mcq | jee-main-2021-online-26th-february-evening-slot |
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>$$\left| \!{\underline {\,
3 \,}} \right. $$ $$\times$$ $$\left| \!{\underline {\,
5 \,}} \right. $$ = 720 | integer | jee-main-2021-online-22th-july-evening-shift |
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 $$\\to$$ N such that gof = f"}] | ["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>$$g\left( {g(x)} \right) = \left[ {\matrix{
{x + 2} & {x = 3k + 1} \cr
{x - 1} & {x = 3k + 2} \cr
{x - 1} & {x = 3k + 3} \cr
} } \right.$$<br><br>$$g\left( {g\left( {g\left( x \right)} \right)} \right) = \left[ {\matrix{
x & {x = 3k + 1} \cr
x & {x = 3k + 2} \cr
x & {x = 3k + 3} \cr
} } \right.$$<br><br>If f : N $$\to$$ N, if is a one-one function such that f(g(x)) = f(x) $$\Rightarrow$$ g(x) = x, which is not the case<br><br>If f : N $$\to$$ N f is an onto function<br><br>such that f(g(x)) = f(x),<br><br>one possibility is <br><br>$$f(x) = \left[ {\matrix{
x & {x = 3n + 1} \cr
x & {x = 3n + 2} \cr
x & {x = 3n + 3} \cr
} } \right.$$ n$$\in$$N<sub>0</sub><br><br>Here f(x) is onto, also f(g(x)) = f(x) $$\forall$$ x$$\in$$N | mcq | jee-main-2021-online-25th-july-morning-shift |
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 and only element of $A$ has image 1.
<br/><br/>
All elements of set $C$ have image 3 or 4 and all elements of set $D$ have image 2 or 3 or 4 .
<br/><br/>
We will solve this question in two cases.
<br/><br/>
<b>Case I</b>: When no element of set $C$ has image 3.
<br/><br/>
Number of onto functions $=2$ (when elements of set $D$ have images 2 or 3$)$
<br/><br/>
<b>Case II</b>: When atleast one element of set $C$ has image 3.<br/><br/> Number of onto functions $=\left(2^{3}-1\right)(1+2+2)$
<br/><br/>
$$
=35
$$
<br/><br/>
Total number of functions $=37$ | integer | jee-main-2022-online-28th-june-evening-shift |
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 function is one-one and onto.</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l5c1z1fa | maths | functions | classification-of-functions | <p>The number of one-one functions f : {a, b, c, d} $$\to$$ {0, 1, 2, ......, 10} such
<br/><br/>that 2f(a) $$-$$ f(b) + 3f(c) + f(d) = 0 is ___________.</p> | [] | null | 31 | <p>Given one-one function</p>
<p>$$f:\{ a,b,c,d\} \to \{ 0,1,2,\,\,....\,\,10\} $$</p>
<p>and $$2f(a) - f(b) + 3f(c) + f(d) = 0$$</p>
<p>$$ \Rightarrow 3f(c) + 2f(a) + f(d) = f(b)$$</p>
<p>Case I:</p>
<p>(1) Now let $$f(c) = 0$$ and $$f(a) = 1$$ then</p>
<p>$$3 \times 0 + 2 \times 1 + f(d) = f(b)$$</p>
<p>$$ \Rightarrow 2 + f(d) = f(b)$$</p>
<p>Now possible value of $$f(d) = 2,3,4,5,6,7,$$ and $$8$$.</p>
<p>f(d) can't be 9 and 10 as if $$f(d) = 9$$ or 10 then $$f(b) = 2 + 9 = 11$$ or $$f(b) = 2 + 10 = 12$$, which is not possible as here any function's maximum value can be 10.</p>
<p>$$\therefore$$ Total possible functions when $$f(c) = 0$$ and $$f(a) = 1$$ are = 7</p>
<p>(2) When $$f(c) = 0$$ and $$f(a) = 2$$ then</p>
<p>$$3 \times 0 + 2 \times 2 + f(d) = f(b)$$</p>
<p>$$ \Rightarrow 4 + f(d) = f(b)$$</p>
<p>$$\therefore$$ possible value of $$f(d) = 1,3,4,5,6$$</p>
<p>$$\therefore$$ Total possible functions in this case = 5</p>
<p>(3) When $$f(c) = 0$$ and $$f(a) = 3$$ then</p>
<p>$$3 \times 0 + 2 \times 3 + f(d) = f(b)$$</p>
<p>$$ \Rightarrow 6 + f(d) = f(b)$$</p>
<p>$$\therefore$$ Possible value of $$f(d) = 1,2,4$$</p>
<p>$$\therefore$$ Total possible functions in this case = 3</p>
<p>(4) When $$f(c) = 0$$ and $$f(a) = 4$$ then</p>
<p>$$3 \times 0 + 2 \times 4 + f(d) = f(b)$$</p>
<p>$$ \Rightarrow 8 + f(d) = f(b)$$</p>
<p>$$\therefore$$ Possible value of $$f(d) = 1,2$$</p>
<p>$$\therefore$$ Total possible functions in this case = 2</p>
<p>(5) When $$f(c) = 0$$ and $$f(a) = 5$$ then</p>
<p>$$3 \times 0 + 2 \times 5 + f(d) = f(b)$$</p>
<p>$$ \Rightarrow 10 + f(d) = f(b)$$</p>
<p>Possible value of f(d) can be 0 but f(c) is already zero. So, no value to f(d) can satisfy.</p>
<p>$$\therefore$$ No function is possible in this case.</p>
<p>$$\therefore$$ Total possible functions when $$f(c) = 0$$ and $$f(a) = 1,2,3$$ and $$4$$ are $$ = 7 + 5 + 3 + 2 = 17$$</p>
<p>Case II:</p>
<p>(1) When $$f(c) = 1$$ and $$f(a) = 0$$ then</p>
<p>$$3 \times 1 + 2 \times 0 + f(d) = f(b)$$</p>
<p>$$ \Rightarrow 3 + f(d) = f(b)$$</p>
<p>$$\therefore$$ Possible value of $$f(d) = 2,3,4,5,6,7$$</p>
<p>$$\therefore$$ Total possible functions in this case = 6</p>
<p>(2) When $$f(c) = 1$$ and $$f(a) = 2$$ then</p>
<p>$$3 \times 1 + 2 \times 2 + f(d) = f(b)$$</p>
<p>$$ \Rightarrow 7 + f(d) = f(b)$$</p>
<p>$$\therefore$$ Possible value of $$f(d) = 0,3$$</p>
<p>$$\therefore$$ Total possible functions in this case = 2</p>
<p>(3) When $$f(c) = 1$$ and $$f(a) = 3$$ then</p>
<p>$$3 \times 1 + 2 \times 3 + f(d) = f(b)$$</p>
<p>$$ \Rightarrow 9 + f(d) = f(b)$$</p>
<p>$$\therefore$$ Possible value of $$f(d) = 0$$</p>
<p>$$\therefore$$ Total possible functions in this case = 1</p>
<p>$$\therefore$$ Total possible functions when $$f(c) = 1$$ and $$f(a) = 0,2$$ and $$3$$ are</p>
<p>$$ = 6 + 2 + 1 = 9$$</p>
<p>Case III:</p>
<p>(1) When $$f(c) = 2$$ and $$f(a) = 0$$ then</p>
<p>$$3 \times 2 + 2 \times 0 + f(d) = f(b)$$</p>
<p>$$ \Rightarrow 6 + f(d) = f(b)$$</p>
<p>$$\therefore$$ Possible values of $$f(d) = 1,3,4$$</p>
<p>$$\therefore$$ Total possible functions in this case = 3</p>
<p>(2) When $$f(c) = 2$$ and $$f(a) = 1$$ then,</p>
<p>$$3 \times 2 + 2 \times 1 + f(d) = f(b)$$</p>
<p>$$ \Rightarrow 8 + f(d) = f(b)$$</p>
<p>$$\therefore$$ Possible values of $$f(d) = 0$$</p>
<p>$$\therefore$$ Total possible function in this case = 1</p>
<p>$$\therefore$$ Total possible functions when $$f(c) = 2$$ and $$f(a) = 0,1$$ are</p>
<p>$$ = 3 + 1 = 4$$</p>
<p>Case IV:</p>
<p>(1) When $$f(c) = 3$$ and $$f(a) = 0$$ then</p>
<p>$$3 \times 3 + 2 \times 0 + f(d) = f(b)$$</p>
<p>$$ \Rightarrow 9 + f(d) = f(b)$$</p>
<p>$$\therefore$$ Possible values of $$f(d) = 1$$</p>
<p>$$\therefore$$ Total one-one functions from four cases</p>
<p>$$ = 17 + 9 + 4 + 1 = 31$$</p> | integer | jee-main-2022-online-24th-june-morning-shift |
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>
<p>Case 1 :</p>
<p>When $$f(3) = 2$$ then possible values of $$f(1)$$ and $$f(2)$$ which satisfy $$f(1) + f(2) = f(3)$$ is $$f(1) = 1$$ and $$f(2) = 1$$.</p>
<p>And $$f(4)$$ can be = 1, 2, 3, 4, 5, 6</p>
<p>$$\therefore$$ Total possible functions $$=1\times6=6$$</p>
<p>Case 2 :</p>
<p>When $$f(3) = 3$$ then possible values</p>
<p>(1) $$f(1) = 1$$ and $$f(2) = 2$$</p>
<p>(2) $$f(1) = 2$$ and $$f(2) = 1$$</p>
<p>And $$f(4)$$ can be = 1, 2, 3, 4, 5, 6.</p>
<p>$$\therefore$$ Total functions $$ = 2 \times 6 = 12$$</p>
<p>Case 3 :</p>
<p>When $$f(3) = 4$$ then</p>
<p>(1) $$f(1) = 1$$ and $$f(2) = 3$$</p>
<p>(2) $$f(1) = 2$$ and $$f(2) = 2$$</p>
<p>(3) $$f(1) = 3$$ and $$f(2) = 1$$</p>
<p>And $$f(4)$$ can be = 1, 2, 3, 4, 5, 6</p>
<p>$$\therefore$$ Total functions $$ = 3 \times 6 = 18$$</p>
<p>Case 4 :</p>
<p>When $$f(3) = 5$$ then</p>
<p>(1) $$f(1) = 1$$ and $$f(4) = 4$$</p>
<p>(2) $$f(1) = 2$$ and $$f(4) = 3$$</p>
<p>(3) $$f(1) = 3$$ and $$f(4) = 2$$</p>
<p>(4) $$f(1) = 4$$ and $$f(4) = 1$$</p>
<p>And $$f(4)$$ can be = 1, 2, 3, 4, 5 and 6</p>
<p>$$\therefore$$ Total functions $$ = 4 \times 6 = 24$$</p>
<p>Case 5 :</p>
<p>When $$f(3) = 6$$ then</p>
<p>(1) $$f(1) = 1$$ and $$f(2) = 5$$</p>
<p>(2) $$f(1) = 2$$ and $$f(2) = 4$$</p>
<p>(3) $$f(1) = 3$$ and $$f(2) = 3$$</p>
<p>(4) $$f(1) = 4$$ and $$f(2) = 2$$</p>
<p>(5) $$f(1) = 5$$ and $$f(2) = 1$$</p>
<p>And $$f(4)$$ can be = 1, 2, 3, 4, 5 and 6</p>
<p>$$\therefore$$ Total possible functions $$ = 5 \times 6 = 30$$</p>
<p>$$\therefore$$ Total functions from those 5 cases we get</p>
<p>$$ = 6 + 12 + 18 + 24 + 30$$</p>
<p>$$ = 90$$</p> | mcq | jee-main-2022-online-25th-july-morning-shift |
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 |
1l6klddyv | maths | functions | classification-of-functions | <p>The number of functions $$f$$, from the set $$\mathrm{A}=\left\{x \in \mathbf{N}: x^{2}-10 x+9 \leq 0\right\}$$ to the set $$\mathrm{B}=\left\{\mathrm{n}^{2}: \mathrm{n} \in \mathbf{N}\right\}$$ such that $$f(x) \leq(x-3)^{2}+1$$, for every $$x \in \mathrm{A}$$, is ___________.</p> | [] | null | 1440 | <p>$$A = \left\{ {\matrix{
{x \in N,} & {{x^2} - 10x + 9 \le 0} \cr
} } \right\}$$</p>
<p>$$ = \{ 1,2,3,\,....,\,9\} $$</p>
<p>$$B = \{ 1,4,9,16,\,.....\} $$</p>
<p>$$f(x) \le {(x - 3)^2} + 1$$</p>
<p>$$f(1) \le 5,\,f(2) \le 2,\,\,..........\,f(9) \le 37$$</p>
<p>$$x = 1$$ has 2 choices</p>
<p>$$x = 2$$ has 1 choice</p>
<p>$$x = 3$$ has 1 choice</p>
<p>$$x = 4$$ has 1 choice</p>
<p>$$x = 5$$ has 2 choices</p>
<p>$$x = 6$$ has 3 choices</p>
<p>$$x = 7$$ has 4 choices</p>
<p>$$x = 8$$ has 5 choices</p>
<p>$$x = 9$$ has 6 choices</p>
<p>$$\therefore$$ Total functions = $$2\times1\times1\times1\times2\times3\times4\times5\times6=1440$$</p> | integer | jee-main-2022-online-27th-july-evening-shift |
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 then f(2) has 3, f(3) has 3 and so on and f(6) has 6 elements. Total number of possible functions = 6! = 720 .... (1)
<br><br>If f(1) has no elements (i.e. null set $$\phi$$) then</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leq0kxw4/6ceb3c53-46ad-4822-99ec-6ae534f92aed/4c82d540-b85f-11ed-9fed-b1659a6c339b/file-1leq0kxw5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leq0kxw4/6ceb3c53-46ad-4822-99ec-6ae534f92aed/4c82d540-b85f-11ed-9fed-b1659a6c339b/file-1leq0kxw5.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 30th January Morning Shift Mathematics - Functions Question 42 English Explanation"></p>
<p>Each index number represents the number of elements in respective rows</p>
<p>Taking every series of arrow and counting number of such possible functions (sets)</p>
<p>$$ = {}^6{C_2} \times {}^4{C_1} \times {}^3{C_1} \times {}^2{C_1} + {}^6{C_1} \times {}^5{C_2} \times {}^3{C_1} \times {}^2{C_1} + {}^6{C_1} \times {}^5{C_1} \times {}^4{C_2} \times {}^2{C_1} + {}^6{C_1} \times {}^5{C_1} \times {}^4{C_1} \times {}^3{C_2} + {}^6{C_1} \times {}^5{C_1} \times {}^4{C_1} \times {}^3{C_1} \times {}^2{C_2} + {}^6{C_1} \times {}^5{C_1} \times {}^4{C_1} \times {}^3{C_1} \times {}^2{C_1}$$</p>
<p>$$ = 2520$$ ..........(2)</p>
<p>From (1) and (2) : Total number of functions</p>
<p>= 2520 + 720 = 3240</p> | integer | jee-main-2023-online-30th-january-morning-shift |
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": "$$f(x)$$ is many-one in $$(1,\\infty )$$"}] | ["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>$$ = {{2\left( {1 + x} \right)\left( {1 - x} \right)} \over {{{\left( {{x^2} + 1} \right)}^2}}}$$</p>
<p>$$ \therefore $$ $$f'(x) = 0$$ at x = 1 and x = -1. So, x = 1 and x = -1 are point of maxima/minima.</p>
<p><b>Option A :</b> $$f(x)$$ is many-one in $$( - \infty , - 1)$$.
<br/><br/>As x = -1 is a point of maxima/minima. So it is a boundary point in the range $$( - \infty , - 1)$$. So, $$f(x)$$ is one-one in $$( - \infty , - 1)$$.
<br/><br/>$$ \therefore $$ Option A is incorrect.</p>
<p><b>Option B :</b> $$f(x)$$ is one-one in $$( - \infty ,\infty )$$.
<br/><br/>As, x = 1 and x = -1 are point of maxima/minima which is present inside the range $$( - \infty ,\infty )$$. So, $$f(x)$$ is many-one function in $$( - \infty ,\infty )$$.
<br/><br/>$$ \therefore $$ Option B is incorrect.</p>
<p><b>Option C :</b> $$f(x)$$ is one-one in $$[1,\infty )$$ but not in $$( - \infty ,\infty )$$.
<br/><br/>As x = 1 is a point of maxima/minima. So it is a boundary point in the range $$[1,\infty )$$. So, $$f(x)$$ is one-one in $$[1,\infty )$$.
<br/><br/>As, x = 1 and x = -1 are point of maxima/minima which is present inside the range $$( - \infty ,\infty )$$. So, $$f(x)$$ is many-one function in $$( - \infty ,\infty )$$ .
<br/><br/>$$ \therefore $$ Option C is correct.</p>
<p><b>Option D :</b> $$f(x)$$ is many-one in $$(1,\infty )$$.
<br/><br/>As x = 1 is a point of maxima/minima. So it is a boundary point in the range $$[1,\infty )$$. So, $$f(x)$$ is one-one in $$[1,\infty )$$.
<br/><br/>$$ \therefore $$ Option D is incorrect.</p>
<p><b>Note :</b></p>
<p><b>Methods to Check One-One Function :</b>
<br/><br/>(i) If a function is one-one, any line parallel to $x$-axis cuts the graph of the function maximum at one point.
<br/><br/>(ii) Any continuous function which is entirely increasing or decreasing (no maxima/minima is present) in whole domain will always be one-one function.
</p> | mcq | jee-main-2023-online-29th-january-morning-shift |
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, $f(2) \in[-8,8] \Rightarrow f(1) \in[-7,8] \quad\ldots(\mathrm{A})$
<br/><br/>
and $2 f(2)+f(3)=2 \Rightarrow f(3)=2 f(1)$
<br/><br/>
$\therefore 2 f(1) \in[-8,8] \Rightarrow f(1) \in[-4,4] \quad\ldots(\mathrm{B})$
<br/><br/>
and $3 f(3)+f(4)=3 \Rightarrow f(4)=3-6 f(1)$
<br/><br/>
$\therefore f(1) \in\left[-\frac{5}{6}, \frac{11}{6}\right]\quad...(C)$
<br/><br/>
From (A), (B) and (C) : $f(1)=0$ or 1
<br/><br/>
$\therefore$ Only two functions are possible. | mcq | jee-main-2023-online-25th-january-evening-shift |
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 be rewritten as $$f(1) + f(2) + 1 = f(4)$$. So, the sum of $$f(1), f(2),$$ and 1 is equal to $$f(4)$$. Since $$f(4)$$ is a value in set B, it can take values from 1 to 6.
<br/><br/>The maximum value of $$f(1) + f(2) + 1$$ can be $$6 + 6 + 1 = 13$$, but this is more than 6 (the maximum value of $$f(4)$$), so it's not possible. Thus, the maximum value of $$f(4)$$ in this case can be 6.
<br/><br/>Let's now analyze the number of functions for each value of $$f(4)$$ from 3 to 6 (we start from 3 because $$f(1)$$ and $$f(2)$$ take values from set B and their minimum sum plus 1 is 3):
<br/><br/>1. When $$f(4) = 6$$, then $$f(1) + f(2) = 5$$. The pairs $$(f(1), f(2))$$ that satisfy this equation are $$(1, 4), (2, 3), (3, 2), (4, 1)$$. For each of these 4 cases, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$4 \times 6 \times 6 = 144$$ functions.
<br/><br/>2. When $$f(4) = 5$$, then $$f(1) + f(2) = 4$$. The pairs $$(f(1), f(2))$$ that satisfy this equation are $$(1, 3), (2, 2), (3, 1)$$. For each of these 3 cases, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$3 \times 6 \times 6 = 108$$ functions.
<br/><br/>3. When $$f(4) = 4$$, then $$f(1) + f(2) = 3$$. The pairs $$(f(1), f(2))$$ that satisfy this equation are $$(1, 2), (2, 1)$$. For each of these 2 cases, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$2 \times 6 \times 6 = 72$$ functions.
<br/><br/>4. When $$f(4) = 3$$, then $$f(1) + f(2) = 2$$. The only pair $$(f(1), f(2))$$ that satisfies this equation is $$(1, 1)$$. For this case, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$1 \times 6 \times 6 = 36$$ functions.
<br/><br/>Adding the numbers of functions from all these cases, we get a total of $$144 + 108 + 72 + 36 = 360$$ functions from $$A$$ to $$B$$ that satisfy the given condition.
<br/><br/>Therefore, the number of functions $$f : A \rightarrow B$$ satisfying the condition $$f(1) + f(2) = f(4) - 1$$ is 360.
| integer | jee-main-2023-online-11th-april-evening-shift |
1lgyoqd24 | maths | functions | classification-of-functions | <p>Let $$\mathrm{R}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}\}$$ and $$\mathrm{S}=\{1,2,3,4\}$$. Total number of onto functions $$f: \mathrm{R} \rightarrow \mathrm{S}$$
such that $$f(\mathrm{a}) \neq 1$$, is equal to ______________.</p> | [] | null | 180 | Total number of onto functions
<br/><br/>$$
\begin{aligned}
& =\frac{5 !}{3 ! 2 !} \times 4 ! \\\\
& =\frac{5 \times 4}{2} \times 24=240
\end{aligned}
$$
<br/><br/>When $f(a)=1$, number of onto functions
<br/><br/>$$
\begin{aligned}
& =4 !+\frac{4 !}{2 ! 2 !} \times 3 ! \\\\
& =24+36=60
\end{aligned}
$$
<br/><br/>So, required number of onto functions <br/><br/>$=240-60=180$ | integer | jee-main-2023-online-8th-april-evening-shift |
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>$$\begin{aligned}
& \mathrm{f}: \mathrm{N}-\{1\} \rightarrow \mathrm{N} \\
& \mathrm{f}(\mathrm{n})=\text { The highest prime factor of } \mathrm{n} . \\
& \mathrm{f}(2)=2 \\
& \mathrm{f}(4)=2 \\
& \Rightarrow \text { many one } \\
& 4 \text { is not image of any element } \\
& \Rightarrow \text { into }
\end{aligned}$$</p>
<p>Hence many one and into</p>
<p>Neither one-one nor onto.</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
1lsgcodrp | maths | functions | classification-of-functions | <p>Let $$\mathrm{A}=\{1,2,3, \ldots, 7\}$$ and let $$\mathrm{P}(\mathrm{A})$$ denote the power set of $$\mathrm{A}$$. If the number of functions $$f: \mathrm{A} \rightarrow \mathrm{P}(\mathrm{A})$$ such that $$\mathrm{a} \in f(\mathrm{a}), \forall \mathrm{a} \in \mathrm{A}$$ is $$\mathrm{m}^{\mathrm{n}}, \mathrm{m}$$ and $$\mathrm{n} \in \mathrm{N}$$ and $$\mathrm{m}$$ is least, then $$\mathrm{m}+\mathrm{n}$$ is equal to _________.</p> | [] | null | 44 | <p>$$\begin{aligned}
& f: A \rightarrow P(A) \\
& a \in f(a)
\end{aligned}$$</p>
<p>That means '$$a$$' will connect with subset which contain element '$$a$$'.</p>
<p>Total options for 1 will be $$2^6$$. (Because $$2^6$$ subsets contains 1)</p>
<p>Similarly, for every other element</p>
<p>Hence, total is $$2^6 \times 2^6 \times 2^6 \times 2^6 \times 2^6 \times 2^6 \times 2^6=2^{42}$$</p>
<p>Ans. $$2+42=44$$</p> | integer | jee-main-2024-online-30th-january-morning-shift |
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 |
lv7v4o17 | maths | functions | classification-of-functions | <p>Let $$A=\{1,3,7,9,11\}$$ and $$B=\{2,4,5,7,8,10,12\}$$. Then the total number of one-one maps $$f: A \rightarrow B$$, such that $$f(1)+f(3)=14$$, is :</p> | [{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "180"}, {"identifier": "C", "content": "240"}, {"identifier": "D", "content": "480"}] | ["C"] | null | <p>$$f(1)+f(3)=14$$</p>
<p>Case I</p>
<p>$$\begin{aligned}
& f(1)=2, f(3)=12 \\
& f(1)=12, f(3)=2
\end{aligned}$$</p>
<p>Total one-one function</p>
<p>$$\begin{aligned}
& =2 \times 5 \times 4 \times 3 \\
& =120
\end{aligned}$$</p>
<p>Case II</p>
<p>$$\begin{aligned}
& f(1)=4, f(3)=10 \\
& f(1)=10, f(3)=4
\end{aligned}$$</p>
<p>Total one-one function</p>
<p>$$\begin{aligned}
& =2 \times 5 \times 4 \times 3 \\
& =120 \\
& \text { Total cases }=120+120=240
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift |
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] $, indicating that $ f(x) $ does not cover all possible real values. Consequently, $ f(x) $ is not onto.</p>
<p>Thus, the function is neither one-one nor onto.</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
q6wg32x7mKUhAZRYfi7cS | maths | functions | composite-functions | For x $$ \in $$ <b>R</b>, x $$ \ne $$ 0, Let f<sub>0</sub>(x) = $${1 \over {1 - x}}$$ and
<br/>f<sub>n+1</sub> (x) = f<sub>0</sub>(f<sub>n</sub>(x)), n = 0, 1, 2, . . . .
<br/><br/>Then the value of f<sub>100</sub><sub></sub>(3) + f<sub>1</sub>$$\left( {{2 \over 3}} \right)$$ + f<sub>2</sub>$$\left( {{3 \over 2}} \right)$$ is equal to : | [{"identifier": "A", "content": "$${8 \\over 3}$$"}, {"identifier": "B", "content": "$${5 \\over 3}$$"}, {"identifier": "C", "content": "$${4 \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["B"] | null | As f<sub>n+1</sub>(x) = f<sub>0</sub>(f<sub>n</sub>(x))
<br><br>$$ \therefore $$ f<sub>1</sub>(x) = f<sub>0+1</sub>(x) = f<sub>0</sub>(f<sub>0</sub>(x)) = $${1 \over {1 - {1 \over {1 - x}}}}$$ = $${{x - 1} \over x}$$
<br><br>f<sub>2</sub>(x) = f<sub>1+1</sub>(x) = f<sub>0</sub>(f<sub>1</sub>(x)) = $${1 \over {1 - {{x - 1} \over x}}}$$ = x
<br><br>f<sub>3</sub>(x) = f<sub>2+1</sub>(x) = f<sub>0</sub>(f<sub>2</sub>(x)) = $${1 \over {1 - x}}$$
<br><br>f<sub>4</sub>(x) = f<sub>3+1</sub>(x) = f<sub>0</sub>(f<sub>3</sub>(x)) = $${1 \over {1 - {1 \over {1 - x}}}}$$ = $${{x - 1} \over x}$$
<br><br>$$ \therefore $$ f<sub>0</sub>(x) = f<sub>3</sub>(x) = f<sub>6</sub>(x) = . . . . . = $${1 \over {1 - x}}$$
<br><br>f<sub>1</sub>(x) = f<sub>4</sub>(x) = f<sub>7</sub>(x) = . . . . . .= $${{x - 1} \over x}$$
<br><br>f<sub>2</sub>(x) = f<sub>5</sub>(x) = f<sub>8</sub>(x) = . . . . . . = x
<br><br>So, f<sub>100</sub>(3) = $${{3 - 1} \over 3}$$ = $${2 \over 3}$$
<br><br>f<sub>1</sub>$$\left( {{2 \over 3}} \right)$$ = $${{{2 \over 3} - 1} \over {{2 \over 3}}}$$ = $$-$$ $${1 \over 2}$$
<br><br>f<sub>2</sub>$$\left( {{3 \over 2}} \right)$$ = $${{3 \over 2}}$$
<br><br>$$ \therefore $$ f<sub>100</sub>(3) + f<sub>1</sub>$$\left( {{2 \over 3}} \right)$$ + f<sub>2</sub>$$\left( {{3 \over 2}} \right)$$
<br><br>= $${2 \over 3}$$ $$-$$ $${1 \over 2}$$ + $${3 \over 2}$$
<br><br>= $${5 \over 3}$$ | mcq | jee-main-2016-online-9th-april-morning-slot |
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": "$${{1 - {2^{ - 10}}} \\over {{3^{10}} - {2^{ - 10}}}}$$"}] | ["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> . (3<sup>10</sup> . x $$-$$ 1) + 1 = x
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 3<sup>10</sup> . x $$-$$ 1 + 2<sup>$$-$$10</sup> = x . 2<sup>$$-$$10</sup> [dividing by 2<sup>10</sup>]
<br><br>$$ \Rightarrow $$$$\,\,\,$$3<sup>10</sup> . x $$-$$ 2<sup>$$-$$10</sup> . x = 1 $$-$$ 2<sup>$$-$$10</sup>
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x (3<sup>10</sup> $$-$$ 2<sup>$$-$$ 10</sup>) = 1$$-$$ 2<sup>$$-$$10</sup>
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ x = $${{1 - {2^{ - 10}}} \over {{3^{10}} - {2^{ - 10}}}}$$ | mcq | jee-main-2017-online-8th-april-morning-slot |
ZRBU7jlXwR6K9rsXplsf6 | maths | functions | composite-functions | For $$x \in R - \left\{ {0,1} \right\}$$, Let f<sub>1</sub>(x) = $$1\over x$$, f<sub>2</sub> (x) = 1 β x <br/><br/>and f<sub>3</sub> (x) = $$1 \over {1 - x}$$
be three given
<br/><br/> functions. If a function, J(x) satisfies
<br/><br/>(f<sub>2</sub> o J o f<sub>1</sub>) (x) = f<sub>3</sub> (x) then J(x) is equal to : | [{"identifier": "A", "content": "f<sub>1</sub> (x)"}, {"identifier": "B", "content": "$$1 \\over x$$ f<sub>3</sub> (x)"}, {"identifier": "C", "content": "f<sub>2</sub> (x)"}, {"identifier": "D", "content": "f<sub>3</sub> (x)"}] | ["D"] | null | Given,
<br><br>f<sub>1</sub>(x) = $${1 \over x}$$
<br><br>f<sub>2</sub>(x) = 1 $$-$$ x
<br><br>f<sub>3</sub>(x) = $${1 \over {1 - x}}$$
<br><br>(f<sub>2</sub> $$ \cdot $$ J $$ \cdot $$ f<sub>1</sub>) (x) = f<sub>3</sub>(x)
<br><br>$$ \Rightarrow $$ f<sub>2</sub> {J(f<sub>1</sub>(x))} = f<sub>3</sub>(x)
<br><br>$$ \Rightarrow $$ f<sub>2</sub>{J ($${1 \over x}$$)} = $${1 \over {1 - x}}$$
<br><br>$$ \Rightarrow $$ 1 $$-$$ J($${1 \over x}$$) = $${1 \over {1 - x}}$$
<br><br>$$ \Rightarrow $$ J($${{1 \over x}}$$) = 1 $$-$$ $${{1 \over {1 - x}}}$$
<br><br>$$ \Rightarrow $$ J ($${{1 \over { x}}}$$) = $${{ - x} \over {1 - x}}$$ = $${x \over {x - 1}}$$
<br><br>Put x inplace of $${1 \over x}$$
<br><br>$$ \therefore $$ J(x) = $${{{1 \over x}} \over {{1 \over x} - 1}}$$
<br><br>= $${1 \over {1 - x}} = {f_3}\left( x \right)$$ | mcq | jee-main-2019-online-9th-january-morning-slot |
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/>Β Β Β Β Β Β and g(n) = n $$-$$($$-$$ 1)<sup>n</sup>.
<br/><br/>Then fog is - | [{"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> $$\left\{ {\matrix{
{n + 1;\,\,\,\,n\,\,is\,\,odd} \cr
{n - 1;\,\,\,\,n\,\,is\,\,even} \cr
} } \right.$$
<br><br>f(g(n)) = $$\left\{ {\matrix{
{{n \over 2};\,\,\,\,n\,\,is\,\,even} \cr
{{{n + 1} \over 2};\,\,\,\,n\,\,is\,\,odd} \cr
} } \right.$$
<br><br>$$ \therefore $$ many one but onto | mcq | jee-main-2019-online-10th-january-evening-slot |
9RGTj3z7ViLdFImDSD3rsa0w2w9jwxrfvb9 | maths | functions | composite-functions | Let f(x) = x<sup>2</sup>
, x $$ \in $$ R. For any A $$ \subseteq $$ R, define g (A) = { x $$ \in $$ R : f(x) $$ \in $$ A}. If S = [0,4], then which one of the
following statements is not true ? | [{"identifier": "A", "content": "g(f(S)) $$ \\ne $$ S "}, {"identifier": "B", "content": "f(g(S)) = S"}, {"identifier": "C", "content": "f(g(S)) $$ \\ne $$ f(S)"}, {"identifier": "D", "content": "g(f(S)) = g(S)"}] | ["D"] | null | f(x) = x<sup>2</sup> x $$ \in $$ R<br><br>
g(A) = {x $$ \in $$ R : f(x) $$ \in $$ A} S $$ \equiv $$ [0, 4]<br><br>
g(S) = {x $$ \in $$ R : f(x) $$ \in $$ S}<br><br>
= {x $$ \in $$ R : 0 $$ \le $$ x<sup>2</sup> $$ \le $$ 4}<br><br>
= {x $$ \in $$ R : β2 $$ \le $$ x $$ \le $$ 2}<br><br>
$$ \therefore $$ g(S) $$ \ne $$ S<br><br>
$$ \therefore $$ f(g(S)) $$ \ne $$ f(S)<br><br>
g(f(S)) = {x $$ \in $$ R : f(x) $$ \in $$ f(S)}<br><br>
= {x $$ \in $$ R : x<sup>2</sup> $$ \in $$ S<sup>2</sup>}<br><br>
= {x $$ \in $$ R : 0 $$ \le $$ x<sup>2</sup> $$ \le $$ 16}<br><br>
= {x $$ \in $$ R : β4 $$ \le $$ x $$ \le $$ 4}<br><br>
$$ \therefore $$ g(f(S)) $$ \ne $$ g(S)<br><br>
$$ \therefore $$ g(f(S)) = g(S) is incorrect | mcq | jee-main-2019-online-10th-april-morning-slot |
YES7TxcXPLt3WXthh83rsa0w2w9jwy29l0i | maths | functions | composite-functions | Let f(x) = e<sup>x</sup> β x and g(x) = x<sup>2</sup> β x, $$\forall $$ x $$ \in $$ R. Then the set of all x $$ \in $$ R, where the function h(x) = (fog) (x) is increasing, is : | [{"identifier": "A", "content": "[0, $$\\infty $$)"}, {"identifier": "B", "content": "$$\\left[ { - 1, - {1 \\over 2}} \\right] \\cup \\left[ {{1 \\over 2},\\infty } \\right)$$"}, {"identifier": "C", "content": "$$\\left[ { - {1 \\over 2},0} \\right] \\cup \\left[ {1,\\infty } \\right)$$"}, {"identifier": "D", "content": "$$\\left[ {0,{1 \\over 2}} \\right] \\cup \\left[ {1,\\infty } \\right)$$"}] | ["D"] | null | f(x) = ex β x, g(x) = x<sup>2</sup> β x<br><br>
f(g(x)) = e<sup>(x<sup>2</sup> - x)</sup> - (x<sup>2</sup> - x)<br><br>
If f(g(x)) is increasing function<br><br>
(f (g(x)))<sup>x</sup> = $${e^{\left( {{x^2} - x} \right)}} \times (2x - 1) - 2x + 1$$<br><br>
$$ \Rightarrow \mathop {(2x - 1)}\limits_A \mathop {[{e^{\left( {{x^2} - x} \right)}} - 1]}\limits_B $$<br><br>
A & B are either both positive or negative<br><br>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265461/exam_images/ptsqro56x6slhf2pox2l.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263348/exam_images/zagaiqzxpahbwihv2stm.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266416/exam_images/zvq4sh4mjcxsga2twdtf.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th April Morning Slot Mathematics - Functions Question 103 English Explanation"></picture>
for (f (g(x)))<sup>'</sup> $$ \ge $$ 0,<br><br>
$$x \in \left[ {0,{1 \over 2}} \right] \cup \left[ {1,\infty } \right)$$ | mcq | jee-main-2019-online-10th-april-morning-slot |
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>
$$ \therefore $$ $$\phi \left( {{\pi \over 3}} \right) = \tan \left( {{\pi \over 4} - {\pi \over 3}} \right)$$<br><br>
$$ \Rightarrow \tan \left( { - {\pi \over {12}}} \right) = \tan {{11\pi } \over {12}}$$
| mcq | jee-main-2019-online-12th-april-morning-slot |
mH9qMoWcAAQdKfLocU7k9k2k5e36adj | maths | functions | composite-functions | If g(x) = x<sup>2</sup> + x - 1 and<br/> (goΖ) (x) = 4x<sup>2</sup> - 10x + 5, then Ζ$$\left( {{5 \over 4}} \right)$$ is equal to: | [{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "-$${1 \\over 2}$$"}, {"identifier": "D", "content": "-$${3 \\over 2}$$"}] | ["C"] | null | Given, (goΖ) (x) = 4x<sup>2</sup> - 10x + 5
<br><br>$$ \Rightarrow $$ g(f(x)) = 4x<sup>2</sup> - 10x + 5
<br><br>$$ \therefore $$ g(f($${5 \over 4}$$)) = $$4 \times {{25} \over {16}} - {{50} \over 4} + 5$$ = $$ - {5 \over 4}$$ ...(1)
<br><br>Also given, g(x) = x<sup>2</sup> + x - 1
<br><br>$$ \therefore $$ g(f(x)) = f<sup>2</sup>(x) + f(x) β1
<br><br>$$ \Rightarrow $$ g(f($${5 \over 4}$$)) = f<sup>2</sup>($${5 \over 4}$$) + f($${5 \over 4}$$) β1 ....(2)
<br><br>from (1) & (2)
<br><br>f<sup>2</sup>($${5 \over 4}$$) + f($${5 \over 4}$$) β1 = $$ - {5 \over 4}$$
<br><br>$$ \Rightarrow $$ f<sup>2</sup>($${5 \over 4}$$) + f($${5 \over 4}$$) + $${1 \over 4}$$ = 0
<br><br>$$ \Rightarrow $$ (f($${5 \over 4}$$) + $${1 \over 2}$$)<sup>2</sup> = 0
<br><br>$$ \Rightarrow $$ f($${5 \over 4}$$) = -$${1 \over 2}$$ | mcq | jee-main-2020-online-7th-january-morning-slot |
CRBJuTU7aMZV61IoHjjgy2xukg390lo1 | maths | functions | composite-functions | For a suitably chosen real constant a, let a<br/><br/>
function, $$f:R - \left\{ { - a} \right\} \to R$$ be defined by<br/><br/>
$$f(x) = {{a - x} \over {a + x}}$$. Further suppose that for any real
number $$x \ne - a$$ and $$f(x) \ne - a$$,<br/><br/>
(fof)(x) = x. Then $$f\left( { - {1 \over 2}} \right)$$ is equal to :
| [{"identifier": "A", "content": "$$ {1 \\over 3}$$"}, {"identifier": "B", "content": "\u20133"}, {"identifier": "C", "content": "$$ - {1 \\over 3}$$"}, {"identifier": "D", "content": "3"}] | ["D"] | null | Given, $$f(x) = {{a - x} \over {a + x}}$$
<br><br>and (fof)(x) = x
<br><br>$$ \Rightarrow $$ f(f(x)) = $${{a - f\left( x \right)} \over {a + f\left( x \right)}}$$ = x<br><br>$$ \Rightarrow $$ $${{a - \left( {{{a - x} \over {a + x}}} \right)} \over {a + \left( {{{a - x} \over {a + x}}} \right)}}$$ = x
<br><br>$$ \Rightarrow $$ $${{{a^2} + ax - a + x} \over {{a^2} + ax + a + x}}$$ = x
<br><br>$$ \Rightarrow $$ (a<sup>2</sup>
β a) + x(a + 1) = (a<sup>2</sup>
+ a)x + x<sup>2</sup>(a β 1)
<br><br>$$ \Rightarrow $$ a(a β 1) + x(1 β a<sup>2</sup>) β x<sup>2</sup>(a β 1) = 0
<br><br>$$ \Rightarrow $$ a = 1
<br><br>$$ \therefore $$ f(x) = $${{1 - x} \over {1 + x}}$$
<br><br>So, $$f\left( { - {1 \over 2}} \right)$$ = $${{1 + {1 \over 2}} \over {1 - {1 \over 2}}}$$ = 3 | mcq | jee-main-2020-online-6th-september-evening-slot |
CctLvYFl1Ev2YXQG8W1klrexqta | maths | functions | composite-functions | Let f : R β R be defined as f (x) = 2x β 1 and g : R - {1} β R be defined as g(x) =
$${{x - {1 \over 2}} \over {x - 1}}$$.
Then the composition function f(g(x)) is : | [{"identifier": "A", "content": "one-one but not onto"}, {"identifier": "B", "content": "onto but not one-one"}, {"identifier": "C", "content": "both one-one and onto"}, {"identifier": "D", "content": "neither one-one nor onto"}] | ["A"] | null | Given, f(x) = 2x $$-$$ 1; f : R $$\to$$ R<br><br>$$g(x) = {{x - 1/2} \over {x - 1}};g:R - \{ 1) \to R$$<br><br>$$f[g(x)] = 2g(x) - 1$$<br><br>$$ = 2 \times \left( {{{x - {1 \over 2}} \over {x - 1}}} \right) - 1 = 2 \times \left( {{{2x - 1} \over {2(x - 1)}}} \right) - 1$$<br><br>$$ = {{2x - 1} \over {x - 1}} - 1 = {{2x - 1 - x + 1} \over {x - 1}} = {x \over {x - 1}}$$<br><br>$$\therefore$$ $$f[g(x)] = 1 + {1 \over {x - 1}}$$<br><br>Now, draw the graph of $$1 + {1 \over {x - 1'}}$$<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxodylbv/0599df3d-1cd0-46d8-af57-05e82c944997/6012ada0-66ea-11ec-ad63-8320eb85293b/file-1kxodylbw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxodylbv/0599df3d-1cd0-46d8-af57-05e82c944997/6012ada0-66ea-11ec-ad63-8320eb85293b/file-1kxodylbw.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 80vh" alt="JEE Main 2021 (Online) 24th February Morning Shift Mathematics - Functions Question 92 English Explanation"><br>$$\because$$ Any horizontal line does not cut the graph at more than one points, so it is one-one and here, co-domain and range are not equal, so it is into.<br><br>Hence, the required function is one-one into. | mcq | jee-main-2021-online-24th-february-morning-slot |
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>$$ \Rightarrow x \ge 0$$ & $$1 - x > 0 \Rightarrow x \in [0,1)$$<br><br>$$g/f = {{\sqrt {1 - x} } \over {\sqrt x }}$$<br><br>$$ \Rightarrow 1 - x \ge 0$$ & $$x > 0 \Rightarrow x \in (0,1]$$<br><br>$$g - f = \sqrt {1 - x} - \sqrt x $$<br><br>$$ \Rightarrow 1 - x \ge 0$$ & $$x \ge 0 \Rightarrow x \in [0,1]$$<br><br>$$ \Rightarrow $$ $$x \in (0,1)$$ | mcq | jee-main-2021-online-18th-march-morning-shift |
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>n</sup> (i) & (ii)<br><br>Clearly $$(\alpha = 5)$$ | mcq | jee-main-2021-online-20th-july-evening-shift |
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 |
1l54ug8v8 | maths | functions | composite-functions | <p>Let f(x) and g(x) be two real polynomials of degree 2 and 1 respectively. If $$f(g(x)) = 8{x^2} - 2x$$ and $$g(f(x)) = 4{x^2} + 6x + 1$$, then the value of $$f(2) + g(2)$$ is _________.</p> | [] | null | 18 | $f(g(x))=8 x^{2}-2 x$
$$
g(f(x))=4 x^{2}+6 x+1
$$
<br/><br/>
let $f(x)=c x^{2}+d x+e$
<br/><br/>
$g(x)=a x+b$
<br/><br/>
$f(g(x))=c(a x+b)^{2}+d(a x+b)+e \equiv 8 x^{2}-2 x$
<br/><br/>
$g(f(x))=a\left(c x^{2}+d x+e\right)+b \equiv 4 x^{2}+6 x+1$
<br/><br/>
$\therefore \quad a c=4 \quad a d=6 \quad a e+b=1$
<br/><br/>
$a^{2} c=8 \quad 2 a b c+a d=-2 \quad c b^{2}+b d+e=0$
<br/><br/>
By solving
<br/><br/>
$a=2 \quad b=-1$
<br/><br/>
$c=2 \quad d=3 \quad e=1$
<br/><br/>
$\therefore \quad f(x)=2 x^{2}+3 x+1$
<br/><br/>
$g(x)=2 x-1$
<br/><br/>
$f(2)+g(2)=2(2)^{2}+3(2)+1+2(2)-1$
<br/><br/>
$=18$ | integer | jee-main-2022-online-29th-june-evening-shift |
1l56rupn7 | maths | functions | composite-functions | <p>Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Define f : S $$\to$$ S as</p>
<p>$$f(n) = \left\{ {\matrix{
{2n} & , & {if\,n = 1,2,3,4,5} \cr
{2n - 11} & , & {if\,n = 6,7,8,9,10} \cr
} } \right.$$.</p>
<p>Let g : S $$\to$$ S be a function such that $$fog(n) = \left\{ {\matrix{
{n + 1} & , & {if\,n\,\,is\,odd} \cr
{n - 1} & , & {if\,n\,\,is\,even} \cr
} } \right.$$.</p>
<p>Then $$g(10)g(1) + g(2) + g(3) + g(4) + g(5))$$ is equal to _____________.<p></p></p> | [] | null | 190 | <p>$$\because$$ $$f(n) = \left\{ {\matrix{
{2n,} & {n = 1,2,3,4,5} \cr
{2n - 11,} & {n = 6,7,8,9,10} \cr} } \right.$$</p>
<p>$$\therefore$$ f(1) = 2, f(2) = 4, ......, f(5) = 10</p>
<p>and f(6) = 1, f(7) = 3, f(8) = 5, ......, f(10) = 9</p>
<p>Now, $$f(g(n)) = \left\{ {\matrix{
{n + 1,} & {if\,n\,is\,odd} \cr
{n - 1,} & {if\,n\,is\,even} \cr
} } \right.$$</p>
<p>$$\therefore$$ $$\matrix{
{f(g(10)) = 9} & { \Rightarrow g(10) = 10} \cr
{f(g(1)) = 2} & { \Rightarrow g(1) = 1} \cr
{f(g(2)) = 1} & { \Rightarrow g(2) = 6} \cr
{f(g(3)) = 4} & { \Rightarrow g(3) = 2} \cr
{f(g(4)) = 3} & { \Rightarrow g(4) = 7} \cr
{f(g(5)) = 6} & { \Rightarrow g(5) = 3} \cr
} $$</p>
<p>$$\therefore$$ $$g(10)g(1) + g(2) + g(3) + g(4) + g(5)) = 190$$</p>
| integer | jee-main-2022-online-27th-june-evening-shift |
1l587b7lt | maths | functions | composite-functions | <p>Let $$f(x) = {{x - 1} \over {x + 1}},\,x \in R - \{ 0, - 1,1\} $$. If $${f^{n + 1}}(x) = f({f^n}(x))$$ for all n $$\in$$ N, then $${f^6}(6) + {f^7}(7)$$ is equal to :</p> | [{"identifier": "A", "content": "$${7 \\over 6}$$"}, {"identifier": "B", "content": "$$ - {3 \\over 2}$$"}, {"identifier": "C", "content": "$${7 \\over {12}}$$"}, {"identifier": "D", "content": "$$ - {{11} \\over {12}}$$"}] | ["B"] | null | <p>Given,</p>
<p>$$f(x) = {{x - 1} \over {x + 1}}$$</p>
<p>Also given,</p>
<p>$${f^{n + 1}}(x) = f({f^n}(x))$$ ..... (1)</p>
<p>$$\therefore$$ For $$n = 1$$</p>
<p>$${f^{1 + 1}}(x) = f({f^1}(x))$$</p>
<p>$$ \Rightarrow {f^2}(x) = f(f(x))$$</p>
<p>$$ = f\left( {{{x - 1} \over {x + 1}}} \right)$$</p>
<p>$$ = {{{{x - 1} \over {x + 1}} - 1} \over {{{x - 1} \over {x + 1}} + 1}}$$</p>
<p>$$ = {{{{x - 1 - x - 1} \over {x + 1}}} \over {{{x - 1 + x + 1} \over {x + 1}}}}$$</p>
<p>$$ = {{ - 2} \over {2x}}$$</p>
<p>$$ = - {1 \over x}$$</p>
<p>From equation (1), when n = 2</p>
<p>$${f^{2 + 1}}(x) = f({f^2}(x))$$</p>
<p>$$ \Rightarrow {f^3}(x) = f({f^2}(x))$$</p>
<p>$$ = f\left( { - {1 \over x}} \right)$$</p>
<p>$$ = {{ - {1 \over x} - 1} \over { - {1 \over x} + 1}}$$</p>
<p>$$ = {{{{ - 1 - x} \over x}} \over {{{ - 1 + x} \over x}}}$$</p>
<p>$$ = {{ - 1 - x} \over { - 1 + x}} = {{ - (x + 1)} \over {x - 1}}$$</p>
<p>Similarly,</p>
<p>$${f^4}(x) = f({f^3}(x))$$</p>
<p>$$ = f\left( {{{ - x + 1} \over {x - 1}}} \right)$$</p>
<p>$$ = {{{{ - (x + 1)} \over {x - 1}} - 1} \over {{{ - (x + 1)} \over {x - 1}} + 1}}$$</p>
<p>$$ = {{{{ - x - 1 - x + 1} \over {x - 1}}} \over {{{ - x - 1 + x - 1} \over {x - 1}}}}$$</p>
<p>$$ = {{ - 2x} \over { - 2}} = x$$</p>
<p>$$\therefore$$ $${f^5}(x) = f({f^4}(x))$$</p>
<p>$$ = f(x)$$</p>
<p>$$ = {{x - 1} \over {x + 1}}$$</p>
<p>$${f^6}(x) = f({f^5}(x))$$</p>
<p>$$ = f\left( {{{x - 1} \over {x + 1}}} \right)$$</p>
<p>$$ = - {1 \over x}$$ (Already calculated earlier)</p>
<p>$${f^7}(x) = f({f^6}(x))$$</p>
<p>$$ = f\left( { - {1 \over x}} \right)$$</p>
<p>$$ = {{ - {1 \over x} - 1} \over { - {1 \over x} + 1}}$$</p>
<p>$$ = {{ - (x + 1)} \over {x - 1}}$$</p>
<p>$$\therefore$$ $${f^6}(6) = - {1 \over 6}$$</p>
<p>and $${f^7}(7) = {{ - (7 + 1)} \over {7 - 1}} = - {8 \over 6}$$</p>
<p>So, $${f^6}(6) + {f^7}(7)$$</p>
<p>$$ = - {1 \over 6} - {8 \over 6}$$</p>
<p>$$ = - {3 \over 2}$$</p> | mcq | jee-main-2022-online-26th-june-morning-shift |
1l58ex9ng | maths | functions | composite-functions | <p>Let f : R $$\to$$ R be defined as f (x) = x $$-$$ 1 and g : R $$-$$ {1, $$-$$1} $$\to$$ R be defined as $$g(x) = {{{x^2}} \over {{x^2} - 1}}$$.</p>
<p>Then the function fog is :</p> | [{"identifier": "A", "content": "one-one but not onto"}, {"identifier": "B", "content": "onto but not one-one"}, {"identifier": "C", "content": "both one-one and onto"}, {"identifier": "D", "content": "neither one-one nor onto"}] | ["D"] | null | <p>$$f:R \to R$$ defined as</p>
<p>$$f(x) = x - 1$$ and $$g:R \to \{ 1, - 1\} \to R,\,g(x) = {{{x^2}} \over {{x^2} - 1}}$$</p>
<p>Now $$fog(x) = {{{x^2}} \over {{x^2} - 1}} - 1 = {1 \over {{x^2} - 1}}$$</p>
<p>$$\therefore$$ Domain of $$fog(x) = R - \{ - 1,1\} $$</p>
<p>And range of $$fog(x) = ( - \infty , - 1] \cup (0,\infty )$$</p>
<p>Now, $${d \over {dx}}(fog(x)) = {{ - 1} \over {{x^2} - 1}}\,.\,2x = {{2x} \over {1 - {x^2}}}$$</p>
<p>$$\therefore$$ $${d \over {dx}}(fog(x)) > 0$$ for $${{2x} \over {(1 - x)(1 + x)}} > 0$$</p>
<p>$$ \Rightarrow {x \over {(x - 1)(x + 1)}} < 0$$</p>
<p>$$\therefore$$ $$x \in ( - \infty , - 1) \cup (0,1)$$</p>
<p>and $${d \over {dx}}(fog(x)) < 0$$ for $$x \in ( - 1,0) \cup (1,\infty )$$</p>
<p>$$\therefore$$ $$fog(x)$$ is neither one-one nor onto.</p> | mcq | jee-main-2022-online-26th-june-evening-shift |
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