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1ktd24qoa | maths | height-and-distance | height-and-distance | A 10 inches long pencil AB with mid point C and a small eraser P are placed on the horizontal top of a table such that PC = $$\sqrt 5 $$ inches and $$\angle$$PCB = tan<sup>-1</sup>(2). The acute angle through which the pencil must be rotated about C so that the perpendicular distance between eraser and pencil becomes e... | [{"identifier": "A", "content": "$${\\tan ^{ - 1}}\\left( {{3 \\over 4}} \\right)$$"}, {"identifier": "B", "content": "tan<sup>$$-$$1</sup>(1)"}, {"identifier": "C", "content": "$${\\tan ^{ - 1}}\\left( {{4 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$${\\tan ^{ - 1}}\\left( {{1 \\over 2}} \\right)$$"}] | ["A"] | null | <br>From figure,<br><br>$$\sin \beta = {1 \over {\sqrt 5 }}$$<br><br>$$\therefore$$ $$\tan \beta = {1 \over 2}$$<br><br>$$\tan (\alpha + \beta ) = 2$$<br><br>$${{\tan \alpha + \tan \beta )} \over {1 - \tan \alpha .\tan \beta }} = 2$$<br><br>$${{\tan \alpha + {1 \over 2}} \over {1 - \tan \alpha \left( {{1 \over 2}}... | mcq | jee-main-2021-online-26th-august-evening-shift | 6,330 |
1ktg351uo | maths | height-and-distance | height-and-distance | Two poles, AB of length a metres and CD of length a + b (b $$\ne$$ a) metres are erected at the same horizontal level with bases at B and D. If BD = x and tan$$\angle$$ACB = $${1 \over 2}$$, then : | [{"identifier": "A", "content": "x<sup>2</sup> + 2(a + 2b)x $$-$$ b(a + b) = 0"}, {"identifier": "B", "content": "x<sup>2</sup> + 2(a + 2b)x + a(a + b) = 0"}, {"identifier": "C", "content": "x<sup>2</sup> $$-$$ 2ax + b(a + b) = 0"}, {"identifier": "D", "content": "x<sup>2</sup> $$-$$ 2ax + a(a + b) = 0"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264596/exam_images/ojvw2ffxtio4f1jffdyo.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Evening Shift Mathematics - Height and Distance Question 12 English Explanation"> <br>... | mcq | jee-main-2021-online-27th-august-evening-shift | 6,331 |
1ktir6j4n | maths | height-and-distance | height-and-distance | A vertical pole fixed to the horizontal ground is divided in the ratio 3 : 7 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole, then the height of the pole (in meters) is : | [{"identifier": "A", "content": "12$$\\sqrt {15} $$"}, {"identifier": "B", "content": "12$$\\sqrt {10} $$"}, {"identifier": "C", "content": "8$$\\sqrt {10} $$"}, {"identifier": "D", "content": "6$$\\sqrt {10} $$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266417/exam_images/igmotqewqjtulhutbk8z.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Morning Shift Mathematics - Height and Distance Question 11 English Explanation"><br><... | mcq | jee-main-2021-online-31st-august-morning-shift | 6,332 |
1l54tdyii | maths | height-and-distance | height-and-distance | <p>From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60$$^\circ$$. The pole subtends an angle 30$$^\circ$$ at the top of the tower. Then the height of the tower is :</p> | [{"identifier": "A", "content": "$$15\\sqrt 3 $$"}, {"identifier": "B", "content": "$$20\\sqrt 3 $$"}, {"identifier": "C", "content": "20 + $$10\\sqrt 3 $$"}, {"identifier": "D", "content": "30"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5flljg6/71a3df8b-f1e7-4fa5-9624-d4e70f01a8df/eb523e50-0076-11ed-bd71-b57b399a7926/file-1l5flljg7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5flljg6/71a3df8b-f1e7-4fa5-9624-d4e70f01a8df/eb523e50-0076-11ed-bd71-b57b399a7926... | mcq | jee-main-2022-online-29th-june-evening-shift | 6,333 |
1l6dwt2de | maths | height-and-distance | height-and-distance | <p>A tower PQ stands on a horizontal ground with base $$Q$$ on the ground. The point $$R$$ divides the tower in two parts such that $$Q R=15 \mathrm{~m}$$. If from a point $$A$$ on the ground the angle of elevation of $$R$$ is $$60^{\circ}$$ and the part $$P R$$ of the tower subtends an angle of $$15^{\circ}$$ at $$A$$... | [{"identifier": "A", "content": "$$5(2 \\sqrt{3}+3) \\,\\mathrm{m}$$"}, {"identifier": "B", "content": "$$5(\\sqrt{3}+3) \\,\\mathrm{m}$$"}, {"identifier": "C", "content": "$$10(\\sqrt{3}+1) \\,\\mathrm{m}$$"}, {"identifier": "D", "content": "$$10(2 \\sqrt{3}+1) \\,\\mathrm{m}$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l79fj3e6/3edc2d44-ec0d-4c06-831f-a466d782744d/eb6330e0-24aa-11ed-8d2e-5f0df5271c2d/file-1l79fj3e7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l79fj3e6/3edc2d44-ec0d-4c06-831f-a466d782744d/eb6330e0-24aa-11ed-8d2e-5f0df5271c2d... | mcq | jee-main-2022-online-25th-july-morning-shift | 6,335 |
1l6jcbdiq | maths | height-and-distance | height-and-distance | <p>Let a vertical tower $$A B$$ of height $$2 h$$ stands on a horizontal ground. Let from a point $$P%$$ on the ground a man can see upto height $$h$$ of the tower with an angle of elevation $$2 \alpha$$. When from $$P$$, he moves a distance $$d$$ in the direction of $$\overrightarrow{A P}$$, he can see the top $$B$$ o... | [{"identifier": "A", "content": "$$\\sqrt{5}-2$$"}, {"identifier": "B", "content": "$$\\sqrt{3}-1$$"}, {"identifier": "C", "content": "$$ \\sqrt{7}-2$$"}, {"identifier": "D", "content": "$$\\sqrt{7}-\\sqrt{3}$$"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7psamcq/756ef252-1665-43d7-bc8f-4dd4bf8b8c96/60d220a0-2da9-11ed-8542-f96181a425b5/file-1l7psamcr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7psamcq/756ef252-1665-43d7-bc8f-4dd4bf8b8c96/60d220a0-2da9-11ed-8542-f96181a425b5... | mcq | jee-main-2022-online-27th-july-morning-shift | 6,336 |
1l6kkz98e | maths | height-and-distance | height-and-distance | <p>The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is $$45^{\circ}$$. Let R be a point on AQ and from a point B, vertically above $$\mathrm{R}$$, the angle of elevation of $$\mathrm{P}$$ is $$60^{\circ}$$. If $$\angle \mathrm{BAQ}=30^{\circ}, \mathrm{AB}=\... | [{"identifier": "A", "content": "$$(10(\\sqrt{3}-1), 25)$$"}, {"identifier": "B", "content": "$$\\left(10(\\sqrt{3}-1), \\frac{25}{2}\\right)$$"}, {"identifier": "C", "content": "$$(10(\\sqrt{3}+1), 25)$$"}, {"identifier": "D", "content": "$$\\left(10(\\sqrt{3}+1), \\frac{25}{2}\\right)$$"}] | ["A"] | null | <p>Let $$BR = x$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7qafj8i/a7a027fb-787f-4a20-838e-deb7a4fd2b6b/4dc88020-2df0-11ed-a744-1fb8f3709cfa/file-1l7qafj8j.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7qafj8i/a7a027fb-787f-4a20-838e-deb7a4fd2b6b/4dc88020-2df0-... | mcq | jee-main-2022-online-27th-july-evening-shift | 6,337 |
1l6nnwt7m | maths | height-and-distance | height-and-distance | <p>A horizontal park is in the shape of a triangle $$\mathrm{OAB}$$ with $$\mathrm{AB}=16$$. A vertical lamp post $$\mathrm{OP}$$ is erected at the point $$\mathrm{O}$$ such that $$\angle \mathrm{PAO}=\angle \mathrm{PBO}=15^{\circ}$$ and $$\angle \mathrm{PCO}=45^{\circ}$$, where $$\mathrm{C}$$ is the midpoint of $$\mat... | [{"identifier": "A", "content": "$$\\frac{32}{\\sqrt{3}}(\\sqrt{3}-1)$$"}, {"identifier": "B", "content": "$$\\frac{32}{\\sqrt{3}}(2-\\sqrt{3})$$"}, {"identifier": "C", "content": "$$\\frac{16}{\\sqrt{3}}(\\sqrt{3}-1)$$"}, {"identifier": "D", "content": "$$\\frac{16}{\\sqrt{3}}(2-\\sqrt{3})$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7rri7rq/1c703de6-5ecf-4544-826f-2ec23ff8478d/dc32c760-2ebf-11ed-b92e-01f1dabc9173/file-1l7rri7rr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7rri7rq/1c703de6-5ecf-4544-826f-2ec23ff8478d/dc32c760-2ebf-11ed-b92e-01f1dabc9173... | mcq | jee-main-2022-online-28th-july-evening-shift | 6,338 |
1l6p25qly | maths | height-and-distance | height-and-distance | <p>The angle of elevation of the top of a tower from a point A due north of it is $$\alpha$$ and from a point B at a distance of 9 units due west of A is $$\cos ^{-1}\left(\frac{3}{\sqrt{13}}\right)$$. If the distance of the point B from the tower is 15 units, then $$\cot \alpha$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{6}{5}$$"}, {"identifier": "B", "content": "$$\\frac{9}{5}$$"}, {"identifier": "C", "content": "$$\\frac{4}{3}$$"}, {"identifier": "D", "content": "$$\\frac{7}{3}$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7ssh0oh/2868d3c3-c2c7-439f-b463-f42683b289b7/6cb0f920-2f50-11ed-88f3-17ddb055f60b/file-1l7ssh0oi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7ssh0oh/2868d3c3-c2c7-439f-b463-f42683b289b7/6cb0f920-2f50-11ed-88f3-17ddb055f60b... | mcq | jee-main-2022-online-29th-july-morning-shift | 6,339 |
1lgsvy5it | maths | height-and-distance | height-and-distance | <p>The angle of elevation of the top $$\mathrm{P}$$ of a tower from the feet of one person standing due South of the tower is $$45^{\circ}$$ and from the feet of another person standing due west of the tower is $$30^{\circ}$$. If the height of the tower is 5 meters, then the distance (in meters) between the two persons... | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "$$\\frac{5}{2} \\sqrt{5}$$"}, {"identifier": "C", "content": "$$5 \\sqrt{5}$$"}, {"identifier": "D", "content": "5"}] | ["A"] | null | Let's denote the person standing due south as S and the one standing due west as W. Also, let the tower be at point T.
<br/><br/>From person S's perspective, we have a right triangle $\triangle SPT$. The height of the tower PT is given as 5m, which is the opposite side for angle S. The angle at S is $45^\circ$. From t... | mcq | jee-main-2023-online-11th-april-evening-shift | 6,340 |
1lh22i34s | maths | height-and-distance | height-and-distance | <p>From the top $$\mathrm{A}$$ of a vertical wall $$\mathrm{AB}$$ of height $$30 \mathrm{~m}$$, the angles of depression of the top $$\mathrm{P}$$ and bottom $$\mathrm{Q}$$ of a vertical tower $$\mathrm{PQ}$$ are $$15^{\circ}$$ and $$60^{\circ}$$ respectively, $$\mathrm{B}$$ and $$\mathrm{Q}$$ are on the same horizonta... | [{"identifier": "A", "content": "$$200(3-\\sqrt{3})$$"}, {"identifier": "B", "content": "$$300(\\sqrt{3}-1)$$"}, {"identifier": "C", "content": "$$300(\\sqrt{3}+1)$$"}, {"identifier": "D", "content": "$$600(\\sqrt{3}-1)$$"}] | ["D"] | null | Given, $A B$ be a vertical wall of height $30 \mathrm{~m}$ and $P Q$ be a vertical tower.
<br><br>Such that $\angle B Q A=\angle T A Q=60^{\circ}$ (Alternate angles)
<br><br>and $\angle C P A=\angle T A P=15^{\circ}$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnoi86a2/300ca189-bd8f-4... | mcq | jee-main-2023-online-6th-april-morning-shift | 6,341 |
oP6I4Fp5P7sBpVKC10jgy2xukfqdqhxe | maths | hyperbola | common-tangent | If the line y = mx + c is a common tangent to
the hyperbola
<br/>$${{{x^2}} \over {100}} - {{{y^2}} \over {64}} = 1$$ and the circle
x<sup>2</sup>
+ y<sup>2</sup>
= 36, then which one of the following is
true? | [{"identifier": "A", "content": "5m = 4"}, {"identifier": "B", "content": "8m + 5 = 0"}, {"identifier": "C", "content": "c<sup>2</sup> = 369"}, {"identifier": "D", "content": "4c<sup>2</sup> = 369"}] | ["D"] | null | $${{{x^2}} \over {100}} - {{{y^2}} \over {64}} = 1$$
<br><br>$$ \therefore $$ c = $$ \pm $$ $$\sqrt {{a^2}{m^2} - {b^2}} $$
<br><br>$$ \Rightarrow $$ c = $$ \pm $$ $$\sqrt {100{m^2} - 64} $$
<br><br>General tangent to hyperbola in slope form is
<br><br>y = mx $$ \pm $$ $$\sqrt {100{m^2} - 64} $$
<br><br>This tangent is... | mcq | jee-main-2020-online-5th-september-evening-slot | 6,343 |
1l6klz0xz | maths | hyperbola | common-tangent | <p>A common tangent $$\mathrm{T}$$ to the curves $$\mathrm{C}_{1}: \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$$ and $$C_{2}: \frac{x^{2}}{42}-\frac{y^{2}}{143}=1$$ does not pass through the fourth quadrant. If $$\mathrm{T}$$ touches $$\mathrm{C}_{1}$$ at $$\left(x_{1}, y_{1}\right)$$ and $$\mathrm{C}_{2}$$ at $$\left(x_{2}, y_{... | [] | null | 20 | <p>Equation of tangent to ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$$ and given slope m is : $$y = mx + \sqrt {4{m^2} + 9} $$ ..... (i)</p>
<p>For slope m equation of tangent to hyperbola is :</p>
<p>$$y = mx + \sqrt {42{m^2} - 143} $$ ....... (ii)</p>
<p>Tangents from (i) and (ii) are identical then</p>
<p>$... | integer | jee-main-2022-online-27th-july-evening-shift | 6,344 |
1l6m6ue5c | maths | hyperbola | common-tangent | <p>For the hyperbola $$\mathrm{H}: x^{2}-y^{2}=1$$ and the ellipse $$\mathrm{E}: \frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1$$, a $$>\mathrm{b}>0$$, let the</p>
<p>(1) eccentricity of $$\mathrm{E}$$ be reciprocal of the eccentricity of $$\mathrm{H}$$, and</p>
<p>(2) the line $$y=\sqrt{\frac{5}{2}... | [] | null | 3 | <p>The equation of tangent to hyperbola $${x^2} - {y^2} = 1$$ within slope $$m$$ is equal to $$y = mx\, \pm \,\sqrt {{m^2} - 1} $$ ...... (i)</p>
<p>And for same slope $$m$$, equation of tangent to ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ is $$y = mx\, \pm \,\sqrt {{a^2}{m^2} + {b^2}} $$ ......... | integer | jee-main-2022-online-28th-july-morning-shift | 6,345 |
Z3V5G1McOharWAOJoJmAb | maths | hyperbola | locus | The locus of the point of intersection of the straight lines,
<br/><br/>tx $$-$$ 2y $$-$$ 3t = 0
<br/><br/>x $$-$$ 2ty + 3 = 0 <b>(t $$ \in $$ R)</b>, is : | [{"identifier": "A", "content": "an ellipse with eccentricity $${2 \\over {\\sqrt 5 }}$$"}, {"identifier": "B", "content": "an ellipse with the length of major axis 6 "}, {"identifier": "C", "content": "a hyperbola with eccentricity $$\\sqrt 5 $$ "}, {"identifier": "D", "content": "a hyperbola with the length of conjug... | ["D"] | null | Here, tx $$-$$ 2y $$-$$ 3t = 0 & x $$-$$ 2ty + 3 = 0
<br><br>On solving, we get;
<br><br>y = $${{6t} \over {2{t^2} - 2}}$$ = $${{3t} \over {{t^2} - 1}}$$ & x = $${{3{t^2} + 3} \over {{t^2} - 1}}$$
<br><br>Put t = tan$$\theta $$
<br><br>$$ \therefore $$ x = $$... | mcq | jee-main-2017-online-8th-april-morning-slot | 6,346 |
Q6AKcTjpG375yPdeFFUqx | maths | hyperbola | locus | If the tangents drawn to the hyperbola 4y<sup>2</sup> = x<sup>2</sup> + 1 intersect the co-ordinate axes at the distinct points A and B then the locus of the mid point of AB is : | [{"identifier": "A", "content": "x<sup>2</sup> $$-$$ 4y<sup>2</sup> + 16x<sup>2</sup>y<sup>2</sup> = 0"}, {"identifier": "B", "content": "x<sup>2</sup> $$-$$ 4y<sup>2</sup> $$-$$ 16x<sup>2</sup>y<sup>2</sup> = 0"}, {"identifier": "C", "content": "4x<sup>2</sup> $$-$$ y<sup>2</sup> + 16x<sup>2</sup>y<sup>2</sup> = 0"}, ... | ["B"] | null | Equation of hyperbola is :
<br><br>4y<sup>2</sup> = x<sup>2</sup> + 1 $$ \Rightarrow $$ $$-$$ x<sup>2</sup> + 4y<sup>2</sup> = 1
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ $$-$$ $${{{x^2}} \over {{1^2}}}$$ + $${{{y^2}} \over {{{\left( {{1 \over 2}} \right)}^2}}}$$ = 1
<br><br>$$ \therefore $$ a = 1, b = $${1 \ove... | mcq | jee-main-2018-online-15th-april-morning-slot | 6,347 |
eAGN1CCOclwn4x0jehdMT | maths | hyperbola | locus | The locus of the point of intersection of the lines, $$\sqrt 2 x - y + 4\sqrt 2 k = 0$$ and $$\sqrt 2 k\,x + k\,y - 4\sqrt 2 = 0$$ (k is any non-zero real parameter), is : | [{"identifier": "A", "content": "an ellipse whose eccentricity is $${1 \\over {\\sqrt 3 }}.$$"}, {"identifier": "B", "content": "an ellipse with length of its major axis $$8\\sqrt 2 .$$"}, {"identifier": "C", "content": "a hyperbola whose eccentricity is $$\\sqrt 3 .$$"}, {"identifier": "D", "content": "a hyperbola wit... | ["D"] | null | Here, lines are :
<br><br>$$\sqrt 2 x$$ $$-$$ y + 4$$\sqrt 2 k$$ = 0
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $$\sqrt 2 x + 4\sqrt 2 k = y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....$$(i)
<br><br>and $$\sqrt 2 kx + ky - 4\sqrt 2 = 0\,\,\,\,\,...\left( {ii} \right)$$
<br><br>Put the value of y from (i) in (ii) we get;
... | mcq | jee-main-2018-online-16th-april-morning-slot | 6,348 |
MiEqnyGdBW0sqgoDHLsL2 | maths | hyperbola | locus | A circle cuts a chord of length 4a on the x-axis and passes through a point on the y-axis, distant 2b from the origin. Then the locus of the centre of this circle, is : | [{"identifier": "A", "content": "an ellipse"}, {"identifier": "B", "content": "a parabola"}, {"identifier": "C", "content": "a hyperbola"}, {"identifier": "D", "content": "a straight line"}] | ["B"] | null | Let equation of circle is
<br><br>x<sup>2</sup><sup></sup> + y<sup>2</sup> + 2fx + 2fy + e = 0, it passes through (0, 2b)
<br><br>$$ \Rightarrow $$ 0 + 4b<sup>2</sup> + 2g $$ \times $$ 0 + 4f + c = 0
<br><br>$$ \Rightarrow $$ 4b<sup>2</sup> + 4f + c = 0 . . . ... | mcq | jee-main-2019-online-11th-january-evening-slot | 6,349 |
mhXzw5ULbGC8X7A7vQ1kls5f8bf | maths | hyperbola | locus | The locus of the point of intersection of the lines $$\left( {\sqrt 3 } \right)kx + ky - 4\sqrt 3 = 0$$ and $$\sqrt 3 x - y - 4\left( {\sqrt 3 } \right)k = 0$$ is a conic, whose eccentricity is _________. | [] | null | 2 | $$\sqrt 3 kx + ky = 4\sqrt 3 $$ ........(1)<br><br>$$\sqrt 3 kx - ky = 4\sqrt 3 {k^2}$$ ....... (2)<br><br>Adding equation (1) & (2)<br><br>$$2\sqrt 3 kx = 4\sqrt 3 ({k^2} + 1)$$<br><br>$$x = 2\left( {k + {1 \over k}} \right)$$ ......... (3)<br><br>Substracting equation (1) & (2)<br><br>$$y = 2\sqrt 3 \left( {{... | integer | jee-main-2021-online-25th-february-morning-slot | 6,350 |
AvQIY4RyhtIXgia6Mp1kmhx9otb | maths | hyperbola | locus | The locus of the midpoints of the chord of the circle, x<sup>2</sup> + y<sup>2</sup> = 25 which is tangent to the hyperbola, $${{{x^2}} \over 9} - {{{y^2}} \over {16}} = 1$$ is : | [{"identifier": "A", "content": "(x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup> $$-$$ 9x<sup>2</sup> + 16y<sup>2</sup> = 0"}, {"identifier": "B", "content": "(x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup> $$-$$ 9x<sup>2</sup> + 144y<sup>2</sup> = 0"}, {"identifier": "C", "content": "(x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup... | ["A"] | null | tangent of hyperbola<br><br>$$y = mx \pm \sqrt {9{m^2} - 16} $$ ..... (i)<br><br>which is a chord of circle with mid-point (h, k)<br><br>so equation of chord T = S<sub>1</sub><br><br>hx + ky = h<sup>2</sup> + k<sup>2</sup><br><br>$$y = - {{hx} \over k} + {{{h^2} + {k^2}} \over k}$$ ..... (ii)<br><br>by (i) and (ii)<br... | mcq | jee-main-2021-online-16th-march-morning-shift | 6,351 |
1krvscm6d | maths | hyperbola | locus | The locus of the centroid of the triangle formed by any point P on the hyperbola $$16{x^2} - 9{y^2} + 32x + 36y - 164 = 0$$, and its foci is : | [{"identifier": "A", "content": "$$16{x^2} - 9{y^2} + 32x + 36y - 36 = 0$$"}, {"identifier": "B", "content": "$$9{x^2} - 16{y^2} + 36x + 32y - 144 = 0$$"}, {"identifier": "C", "content": "$$16{x^2} - 9{y^2} + 32x + 36y - 144 = 0$$"}, {"identifier": "D", "content": "$$9{x^2} - 16{y^2} + 36x + 32y - 36 = 0$$"}] | ["A"] | null | Given hyperbola is <br><br>$$16{(x + 1)^2} - 9{(y - 2)^2} = 164 + 16 - 36 = 144$$<br><br>$$ \Rightarrow {{{{(x + 1)}^2}} \over 9} - {{{{(y - 2)}^2}} \over {16}} = 1$$<br><br>Eccentricity, $$e = \sqrt {1 + {{16} \over 9}} = {5 \over 3}$$<br><br>$$\Rightarrow$$ foci are (4, 2) and ($$-$$6, 2)<br><br><img src="https://re... | mcq | jee-main-2021-online-25th-july-morning-shift | 6,352 |
1ktd0j2lu | maths | hyperbola | locus | The locus of the mid points of the chords of the hyperbola x<sup>2</sup> $$-$$ y<sup>2</sup> = 4, which touch the parabola y<sup>2</sup> = 8x, is : | [{"identifier": "A", "content": "y<sup>3</sup>(x $$-$$ 2) = x<sup>2</sup>"}, {"identifier": "B", "content": "x<sup>3</sup>(x $$-$$ 2) = y<sup>2</sup>"}, {"identifier": "C", "content": "y<sup>2</sup>(x $$-$$ 2) = x<sup>3</sup>"}, {"identifier": "D", "content": "x<sup>2</sup>(x $$-$$ 2) = y<sup>3</sup>"}] | ["C"] | null | T = S<sub>1</sub><br><br>xh $$-$$ yk = h<sup>2</sup> $$-$$ k<sup>2</sup><br><br>$$y = {{xh} \over k} - {{({h^2} - {k^2})} \over k}$$<br><br>this touches y<sup>2</sup> = 8x then $$c = {a \over m}$$<br><br>$$\left( {{{{k^2} - {h^2}} \over k}} \right) = {{2k} \over h}$$<br><br>2y<sup>2</sup> = x(y<sup>2</sup> $$-$$ x<sup>... | mcq | jee-main-2021-online-26th-august-evening-shift | 6,353 |
1lgutulh7 | maths | hyperbola | locus | <p>Let R be a rectangle given by the lines $$x=0, x=2, y=0$$ and $$y=5$$. Let A$$(\alpha,0)$$ and B$$(0,\beta),\alpha\in[0,2]$$ and $$\beta\in[0,5]$$, be such that the line segment AB divides the area of the rectangle R in the ratio 4 : 1. Then, the mid-point of AB lies on a :</p> | [{"identifier": "A", "content": "hyperbola"}, {"identifier": "B", "content": "straight line"}, {"identifier": "C", "content": "parabola"}, {"identifier": "D", "content": "circle"}] | ["A"] | null | We have, $R$ be a rectangle formed by the lines $x=0$, $x=2, y=0$ and $y=5$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ln228t36/fd6163aa-9150-4285-8e02-935227613d64/f02c4220-5d60-11ee-8969-dbcde4c067b7/file-6y3zli1ln228t37.png?format=png" data-orsrc="https://app-content.cdn.examgoal.... | mcq | jee-main-2023-online-11th-april-morning-shift | 6,355 |
ZI8EaNj1ctW4fj8P | maths | hyperbola | normal-to-hyperbola | The normal to a curve at $$P(x,y)$$ meets the $$x$$-axis at $$G$$. If the distance of $$G$$ from the origin is twice the abscissa of $$P$$, then the curve is a : | [{"identifier": "A", "content": "circle "}, {"identifier": "B", "content": "hyperbola "}, {"identifier": "C", "content": "ellipse "}, {"identifier": "D", "content": "parabola"}] | ["B"] | null | Equation of normal at $$P\left( {x,y} \right)$$ is $$Y - y = - {{dx} \over {dy}}\left( {x - x} \right)$$
<br><br>Coordinate of $$G$$ at $$X$$ axis is $$\left( {X,0} \right)$$ (let)
<br><br>$$\therefore$$ $$0 - y = - {{dx} \over {dy}}\left( {X - x} \right) \Rightarrow y{{dy} \over {dx}} = X - x$$
<br><br>$$ \Rightar... | mcq | aieee-2007 | 6,356 |
2IuA3TuSpLnwUxoO8Vx2H | maths | hyperbola | normal-to-hyperbola | A normal to the hyperbola, 4x<sup>2</sup> $$-$$ 9y<sup>2</sup> = 36 meets the co-ordinate axes $$x$$ and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the ocus of P is : | [{"identifier": "A", "content": "4x<sup>2</sup> + 9y<sup>2</sup> = 121"}, {"identifier": "B", "content": "9x<sup>2</sup> + 4y<sup>2</sup> = 169"}, {"identifier": "C", "content": "4x<sup>2</sup> $$-$$ 9y<sup>2</sup> = 121"}, {"identifier": "D", "content": "9x<sup>2</sup> $$-$$ 4y<sup>2</sup> = 169"}] | ["D"] | null | Given, 4x<sup>2</sup> $$-$$ 9y<sup>2</sup> = 36
<br><br>After differentiating w.r.t.x, we get
<br><br>4.2x $$-$$ 9.2.y.$${{dy} \over {dx}}$$ = 0
<br><br>$$ \Rightarrow $$ Slope of tangent = $${{dy} \over {dx}}$$ = $${{4x} \over {9y}}$$
<br><br>So, slope of normal = $${{ - 9y} \over {4x}}$$
<br><br>Now, equation of norm... | mcq | jee-main-2018-online-15th-april-evening-slot | 6,357 |
0oVM7UCfTdYqh9poTc18hoxe66ijvwpdyef | maths | hyperbola | normal-to-hyperbola | If the line y = mx + 7$$\sqrt 3 $$ is normal to the
hyperbola
$${{{x^2}} \over {24}} - {{{y^2}} \over {18}} = 1$$ , then a value of m is : | [{"identifier": "A", "content": "$${3 \\over {\\sqrt 5 }}$$"}, {"identifier": "B", "content": "$${{\\sqrt 15 } \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt 5 } \\over 2}$$"}, {"identifier": "D", "content": "$${2 \\over {\\sqrt 5 }}$$"}] | ["D"] | null | Given line y = mx + 7$$\sqrt 3 $$ .....(1)
<br><br>Given hyperbola
<br><br>$${{{x^2}} \over {24}} - {{{y^2}} \over {18}} = 1$$
<br><br>Here $${a^2} = 24$$ and $${b^2} = 18$$
<br><br>We know the equation of normal to the hyperbola is
<br><br>$$y = mx \pm {{m\left( {{a^2} + {b^2}} \right)} \over {\sqrt {{a^2} - {b^2}{m^2... | mcq | jee-main-2019-online-9th-april-morning-slot | 6,358 |
tTDEM7hQQr5lkaZQzQ7k9k2k5hjxu5z | maths | hyperbola | normal-to-hyperbola | If a hyperbola passes through the point
P(10, 16) and it has vertices at (± 6, 0), then the
equation of the normal to it at P is : | [{"identifier": "A", "content": "2x + 5y = 100"}, {"identifier": "B", "content": "x + 3y = 58"}, {"identifier": "C", "content": "x + 2y = 42"}, {"identifier": "D", "content": "3x + 4y = 94"}] | ["A"] | null | Let hyperbola is $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$
<br><br>vertices ($$ \pm $$a, 0 ) = ($$ \pm $$6, 0) $$ \Rightarrow $$ a = 6
<br><br>Hyperbola passes through p(10, 16)
<br><br>$$ \therefore $$ $${{{{10}^2}} \over {{6^2}}} - {{{{16}^2}} \over {{b^2}}} = 1$$
<br><br>$$ \Rightarrow $$ b = 12
<br>... | mcq | jee-main-2020-online-8th-january-evening-slot | 6,359 |
qmELQCfVJYJt8ZkIAejgy2xukf7fotkp | maths | hyperbola | normal-to-hyperbola | Let P(3, 3) be a point on the hyperbola, <br/>$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$. If the normal to it at P intersects the x-axis
at (9, 0) and e is its eccentricity, then the ordered pair (a<sup>2</sup>, e<sup>2</sup>) is equal to : | [{"identifier": "A", "content": "$$\\left( {{9 \\over 2},2} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{3 \\over 2},2} \\right)$$"}, {"identifier": "C", "content": "(9,3) "}, {"identifier": "D", "content": "$$\\left( {{9 \\over 2},3} \\right)$$"}] | ["D"] | null | Given hyperbola, $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$<br><br>Point P (3, 3) is on the parabola<br><br>$$ \therefore $$ $${9 \over {{a^2}}} - {9 \over {{b^2}}} = 1$$ ...(1)<br><br>Equation of normal at (x<sub>1</sub>, y<sub>1</sub>),<br><br>$${{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^... | mcq | jee-main-2020-online-4th-september-morning-slot | 6,360 |
1ktcyu7x8 | maths | hyperbola | normal-to-hyperbola | The point $$P\left( { - 2\sqrt 6 ,\sqrt 3 } \right)$$ lies on the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ having eccentricity $${{\sqrt 5 } \over 2}$$. If the tangent and normal at P to the hyperbola intersect its conjugate axis at the point Q and R respectively, then QR is equal to : | [{"identifier": "A", "content": "$$4\\sqrt 3 $$"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "$$6\\sqrt 3 $$"}, {"identifier": "D", "content": "$$3\\sqrt 6 $$"}] | ["C"] | null | $$P\left( { - 2\sqrt 6 ,\sqrt 3 } \right)$$ lies on hyperbola <br><br>$$ \Rightarrow {{24} \over {{a^2}}} - {3 \over {{b^2}}} = 1$$ ...... (i)<br><br>$$e = {{\sqrt 5 } \over 2} \Rightarrow {b^2} = {a^2}\left( {{5 \over 4} - 1} \right) \Rightarrow 4{b^2} = {a^2}$$<br><br>Put in (i) $$ \Rightarrow {6 \over {{b^2}}} - {3 ... | mcq | jee-main-2021-online-26th-august-evening-shift | 6,361 |
1l58g2o6j | maths | hyperbola | normal-to-hyperbola | <p>The normal to the hyperbola <br/><br/>$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over 9} = 1$$ at the point $$\left( {8,3\sqrt 3 } \right)$$ on it passes through the point :</p> | [{"identifier": "A", "content": "$$\\left( {15, - 2\\sqrt 3 } \\right)$$"}, {"identifier": "B", "content": "$$\\left( {9,2\\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - 1,9\\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - 1,6\\sqrt 3 } \\right)$$"}] | ["C"] | null | <p>Given hyperbola : $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over 9} = 1$$</p>
<p>$$\because$$ It passes through $$(8,3\sqrt 3 )$$</p>
<p>$$\because$$ $${{64} \over {{a^2}}} - {{27} \over 9} = 1 \Rightarrow {a^2} = 16$$</p>
<p>Now, equation of normal to hyperbola</p>
<p>$${{16x} \over 8} + {{9y} \over {3\sqrt 3 }} = 16 +... | mcq | jee-main-2022-online-26th-june-evening-shift | 6,363 |
1l59lgdx4 | maths | hyperbola | normal-to-hyperbola | <p>Let the eccentricity of the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ be $${5 \over 4}$$. If the equation of the normal at the point $$\left( {{8 \over {\sqrt {5} }},{{12} \over {5}}} \right)$$ on the hyperbola is $$8\sqrt 5 x + \beta y = \lambda $$, then $$\lambda$$ $$-$$ $$\beta$$ is equa... | [] | null | 85 | <p>$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1\left( {e = {5 \over 4}} \right)$$</p>
<p>So, $${b^2} = {a^2}\left( {{{25} \over {16}} - 1} \right) \Rightarrow b = {3 \over 4}a$$</p>
<p>Also $$\left( {{8 \over {\sqrt 5 }},{{12} \over 5}} \right)$$ lies on the given hyperbola</p>
<p>So, $${{64} \over {5{a^2}}}... | integer | jee-main-2022-online-25th-june-evening-shift | 6,364 |
1ldv2wg2b | maths | hyperbola | normal-to-hyperbola | <p>The vertices of a hyperbola H are ($$\pm$$ 6, 0) and its eccentricity is $${{\sqrt 5 } \over 2}$$. Let N be the normal to H at a point in the first quadrant and parallel to the line $$\sqrt 2 x + y = 2\sqrt 2 $$. If d is the length of the line segment of N between H and the y-axis then d$$^2$$ is equal to __________... | [] | null | 216 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lebzl8r1/919fa76e-b803-49d8-86c9-4e9df9241622/72ada3d0-b0a8-11ed-98af-57b0199700ba/file-1lebzl8r2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lebzl8r1/919fa76e-b803-49d8-86c9-4e9df9241622/72ada3d0-b0a8-11ed-98af-57b0199700ba/fi... | integer | jee-main-2023-online-25th-january-morning-shift | 6,366 |
p6VqpFyaPpNleFFg | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | The foci of the ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1$$ and the hyperbola $${{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}$$ coincide. Then the value of $${b^2}$$ is : | [{"identifier": "A", "content": "$$9$$"}, {"identifier": "B", "content": "$$1$$ "}, {"identifier": "C", "content": "$$5$$ "}, {"identifier": "D", "content": "$$7$$ "}] | ["D"] | null | $${{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}$$
<br><br>$$a = \sqrt {{{144} \over {25}}} ,b = \sqrt {{{81} \over {25}}} ,\,\,$$
<br><br>$$e = \sqrt {1 + {{81} \over {144}}} = {{15} \over {12}} = {5 \over 4}$$
<br><br>$$\therefore$$ Foci $$ = \left( { \pm 3,0} \right)$$
<br><br>$$\therefore$$ foci of... | mcq | aieee-2003 | 6,368 |
Yyy5G3cAbrt43UGL | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | For the Hyperbola $${{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1$$ , which of the following remains constant when $$\alpha $$ varies$$=$$? | [{"identifier": "A", "content": "abscissae of vertices "}, {"identifier": "B", "content": "abscissae of foci "}, {"identifier": "C", "content": "eccentricity "}, {"identifier": "D", "content": "directrix."}] | ["B"] | null | Given, equation of hyperbola is $${{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1$$
<br><br>We know that the equation of hyperbola is
<br><br>$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ Here, $${a^2} = {\cos ^2}\alpha $$ and $${b^2} = {\sin ^2}\alpha $$
<br><br>We know th... | mcq | aieee-2007 | 6,369 |
LPJEFjsEwH1i9kZv | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | The eccentricity of the hyperbola whose length of the latus rectum is equal to $$8$$ and the length of its conjugate axis is equal to half of the distance between its foci, is : | [{"identifier": "A", "content": "$${2 \\over {\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${\\sqrt 3 }$$ "}, {"identifier": "C", "content": "$${{4 \\over 3}}$$"}, {"identifier": "D", "content": "$${4 \\over {\\sqrt 3 }}$$"}] | ["A"] | null | $${{2{b^2}} \over a} = 8$$ and $$2b = {1 \over 2}\left( {2ae} \right)$$
<br><br>$$ \Rightarrow 4{b^2} = {a^2}{e^2}$$
<br><br>$$ \Rightarrow 4{a^2}\left( {{e^2} - 1} \right) = {a^2}{e^2}$$
<br><br>$$ \Rightarrow 3{e^2} = 4 \Rightarrow e = {2 \over {\sqrt 3 }}$$ | mcq | jee-main-2016-offline | 6,370 |
wTrhrAus7jXGW0OMPsaAA | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | A hyperbola whose transverse axis is along the major axis of the conic, $${{{x^2}} \over 3} + {{{y^2}} \over 4} = 4$$ and has vertices at the foci of this conic. If the eccentricity of the hyperbola is $${3 \over 2},$$ then which of the following points does <b>NOT</b> lie on it? | [{"identifier": "A", "content": "(0, 2)"}, {"identifier": "B", "content": "$$\\left( {\\sqrt 5 ,2\\sqrt 2 } \\right)$$"}, {"identifier": "C", "content": "$$\\left( {\\sqrt {10} ,2\\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$$\\left( {5,2\\sqrt 3 } \\right)$$ "}] | ["D"] | null | $${{{x^2}} \over {12}} + {{{y^2}} \over {16}}$$ = 1
<br><br>e = $$\sqrt {1 - {{12} \over {16}}} $$ = $${1 \over 2}$$
<br><br>Foci (0, 2) & (0, $$-$$ 2)
<br><br>So, transverse axis of hyperbola
<br><br>= 2b = 4 <br><br>$$ \Rightarrow $$ b = 2 & a<sup>2</sup> = 1<sup>2</sup> (e<sup>2</sup... | mcq | jee-main-2016-online-10th-april-morning-slot | 6,372 |
c5COgvW6rHjsd2dmukCP2 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | Let $$0 < \theta < {\pi \over 2}$$. If the eccentricity of the
<br/><br/> hyperbola $${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }}$$ = 1 is greater
<br/><br/>than 2, then the length of
its latus rectum lies in the interval :
| [{"identifier": "A", "content": "(3, $$\\infty $$) "}, {"identifier": "B", "content": "$$\\left( {{3 \\over 2},2} \\right]$$"}, {"identifier": "C", "content": "$$\\left( {1,{3 \\over 2}} \\right]$$"}, {"identifier": "D", "content": "$$\\left( {2,3} \\right]$$"}] | ["A"] | null | Given hyperbola,
<br><br>$${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }} = 1$$
<br><br>here a = cos$$\theta $$
<br><br>and b = sin$$\theta $$
<br><br>We know, eccentricity of the hyperbola is,
<br><br>$$\sqrt {1 + {{{b^2}} \over {{a^2}}}} $$
<br><br>$$ \therefore $$ Here eccen... | mcq | jee-main-2019-online-9th-january-morning-slot | 6,373 |
kh3oFVUOUj5Bwo7LP4i2O | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is : | [{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "$$\\sqrt 3 $$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$${2 \\over {\\sqrt 3 }}$$"}] | ["D"] | null | Let the equation of hyperbola
<br><br>$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}}$$ = 1
<br><br>Given 2a = 4
<br><br>$$ \Rightarrow $$ $$a$$ = 2
<br><br>It passes through (4, 2)
<br><br>$$ \therefore $$ $${{16} \over 4} - {4 \over {{b^2}}}$$ = 1
<br><br>$$ \Rightarrow $$ b<sup>2... | mcq | jee-main-2019-online-9th-january-evening-slot | 6,374 |
udGNmB5egVI3TF1a5s5Vu | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the
eccentricity of the hyperbola is : | [{"identifier": "A", "content": "$${{13} \\over 6}$$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$${{13} \\over 12}$$"}, {"identifier": "D", "content": "$${{13} \\over 8}$$"}] | ["C"] | null | 2b = 5 and 2ae = 13
<br><br>b<sup>2</sup> = a<sup>2</sup>(e<sup>2</sup> $$-$$ 1) $$ \Rightarrow $$ $${{25} \over 4}$$ = $${{169} \over 4}$$ $$-$$ a<sup>2</sup>
<br><br>$$ \Rightarrow $$ a $$=$$ 6 $$ \Rightarrow $$ e $$=$$ $${{13} \over {12}}$$ | mcq | jee-main-2019-online-11th-january-evening-slot | 6,375 |
7NBZ9LCV3YKefkvN211Fe | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | If the vertices of a hyperbola be at (–2, 0) and (2, 0) and one of its foci be at (–3, 0), then which one of the following points does not lie on this hyperbola? | [{"identifier": "A", "content": "$$\\left( {6,5\\sqrt 2 } \\right)$$"}, {"identifier": "B", "content": "$$\\left( {2\\sqrt 6 ,5} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - 6,2\\sqrt {10} } \\right)$$"}, {"identifier": "D", "content": "$$\\left( {4,\\sqrt {15} } \\right)$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266288/exam_images/sqc6djvwoxgp7yny26un.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Mathematics - Hyperbola Question 57 English Explanation">
<br>ae = 3, ... | mcq | jee-main-2019-online-12th-january-morning-slot | 6,376 |
sCkkVj9Lsm6OEcTclf3rsa0w2w9jwxs6pli | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | If a directrix of a hyperbola centred at the origin and passing through the point (4, –2$$\sqrt 3 $$ ) is 5x = 4$$\sqrt 5 $$ and
its eccentricity is e, then : | [{"identifier": "A", "content": "4e<sup>4</sup> \u2013 24e<sup>2</sup> + 27 = 0"}, {"identifier": "B", "content": "4e<sup>4</sup> \u2013 24e<sup>2</sup> + 35 = 0"}, {"identifier": "C", "content": "4e<sup>4</sup> \u2013 12e<sup>2</sup> - 27 = 0"}, {"identifier": "D", "content": "4e<sup>4</sup> + 8e<sup>2</sup> - 35 = 0"... | ["B"] | null | 5x = 4$$\sqrt 5 $$
<br><br>$$ \Rightarrow $$ $$x = {4 \over {\sqrt 5 }}$$<br><br>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267388/exam_images/fqdlbbxdjjchifyrvnwu.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image... | mcq | jee-main-2019-online-10th-april-morning-slot | 6,377 |
jnsfPBe0KmkT0ZCwSQ3rsa0w2w9jx24zm31 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | If 5x + 9 = 0 is the directrix of the hyperbola 16x<sup>2</sup>
– 9y<sup>2</sup>
= 144, then its corresponding focus is : | [{"identifier": "A", "content": "$$\\left( {{5 \\over 3},0} \\right)$$"}, {"identifier": "B", "content": "(5, 0)"}, {"identifier": "C", "content": "(- 5, 0)"}, {"identifier": "D", "content": "$$\\left( { - {5 \\over 3},0} \\right)$$"}] | ["C"] | null | $${{{x^2}} \over 9} - {{{y^2}} \over {16}} = 1$$<br><br>
$$ \therefore $$ a = 3 and b = 4<br><br>
$${e^2} = 1 + {{{b^2}} \over {{a^2}}}$$<br><br>
$$ \Rightarrow {e^2} = 1 + {{16} \over 9}$$<br><br>
$$ \Rightarrow $$ e = $$5 \over 3$$<br><br>
$$ \therefore $$ focus is (–ae, 0) = (–5, 0) | mcq | jee-main-2019-online-10th-april-evening-slot | 6,378 |
cKUyQMOGtgxolaNvzyjgy2xukezb2lny | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | For some $$\theta \in \left( {0,{\pi \over 2}} \right)$$, if the eccentricity of the
<br/>hyperbola, x<sup>2</sup>–y<sup>2</sup>sec<sup>2</sup>$$\theta $$ = 10 is
$$\sqrt 5 $$ times the
<br/>eccentricity of the ellipse, x<sup>2</sup>sec<sup>2</sup>$$\theta $$ + y<sup>2</sup> = 5, then
the length of the latus rectum o... | [{"identifier": "A", "content": "$$\\sqrt {30} $$"}, {"identifier": "B", "content": "$$2\\sqrt 6 $$"}, {"identifier": "C", "content": "$${{4\\sqrt 5 } \\over 3}$$"}, {"identifier": "D", "content": "$${{2\\sqrt 5 } \\over 3}$$"}] | ["C"] | null | Given equation of hyperbola $$ \Rightarrow {x^2} - {y^2}{\sec ^2}\theta = 10$$<br><br>
$$ \Rightarrow {{{x^2}} \over {10}} - {{{y^2}} \over {10{{\cos }^2}\theta }} = 1$$<br><br>
Hence eccentricity of hyperbola<br><br>
$$\left( {{e_H}} \right) = \sqrt {1 + {{10{{\cos }^2}\theta } \over {10}}} $$ ...(i)... | mcq | jee-main-2020-online-2nd-september-evening-slot | 6,380 |
iii96ZkKaJJmJnFmw61klt9jx19 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | A hyperbola passes through the foci of the ellipse $${{{x^2}} \over {25}} + {{{y^2}} \over {16}} = 1$$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is : | [{"identifier": "A", "content": "$${{{x^2}} \\over 9} - {{{y^2}} \\over 4} = 1$$"}, {"identifier": "B", "content": "$${{{x^2}} \\over 9} - {{{y^2}} \\over 16} = 1$$"}, {"identifier": "C", "content": "$${{{x^2}} \\over 9} - {{{y^2}} \\over 25} = 1$$"}, {"identifier": "D", "content": "x<sup>2</sup> $$-$$ y<sup>2</sup> = ... | ["B"] | null | $${e_1} = \sqrt {1 - {{16} \over {25}}} = {3 \over 5}$$ foci ($$ \pm $$ae, 0)<br><br>Foci = ($$ \pm $$3, 0)<br><br>Let equation of hyperbola be $${{{x^2}} \over {{A^2}}} - {{{y^2}} \over {{B^2}}} = 1$$<br><br>Passes through ($$ \pm $$3, 0)<br><br>A<sup>2</sup> = 9, A = 3, $${e_2} = {5 \over 3}$$<br><br>$${e_2}^2 = 1 +... | mcq | jee-main-2021-online-25th-february-evening-slot | 6,383 |
1l546bltp | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let $$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$, a > 0, b > 0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $$4(2\sqrt 2 + \sqrt {14} )$$. If the eccentricity H is $${{\sqrt {11} } \over 2}$$, then the value of a<sup>2</sup> + b<sup>2</sup> is equal t... | [] | null | 88 | <p>$$2a + 2b = 4\left( {2\sqrt 2 + \sqrt {14} } \right)$$ ...... (1)</p>
<p>$$1 + {{{b^2}} \over {{a^2}}} = {{11} \over {14}}$$ ....... (2)</p>
<p>$$ \Rightarrow {{{b^2}} \over {{a^2}}} = {7 \over 4}$$ ....... (3)</p>
<p>and $$a + b = 4\sqrt 2 + 2\sqrt {14} $$ ...... (4)</p>
<p>By (3) and (4)</p>
<p>$$ \Rightarrow a ... | integer | jee-main-2022-online-29th-june-morning-shift | 6,384 |
1l55hrsqt | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let a > 0, b > 0. Let e and l respectively be the eccentricity and length of the latus rectum of the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$. Let e' and l' respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If $${e^2} = {{11} \over {14}}l$$ a... | [{"identifier": "A", "content": "100"}, {"identifier": "B", "content": "110"}, {"identifier": "C", "content": "120"}, {"identifier": "D", "content": "130"}] | ["D"] | null | <p>$$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$, then</p>
<p>$${e^2} = {{11} \over {14}}l$$ (l be the length of LR)</p>
<p>$$ \Rightarrow {a^2} + {b^2} = {{11} \over 7}{b^2}a$$ ..... (i)</p>
<p>and $$e{'^2} = {{11} \over 8}l'$$ (l' be the length of LR of conjugate hyperbola)</p>
<p>$$ \Rightarrow {a^2} +... | mcq | jee-main-2022-online-28th-june-evening-shift | 6,385 |
1l6je3gl6 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>An ellipse $$E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ passes through the vertices of the hyperbola $$H: \frac{x^{2}}{49}-\frac{y^{2}}{64}=-1$$. Let the major and minor axes of the ellipse $$E$$ coincide with the transverse and conjugate axes of the hyperbola $$H$$, respectively. Let the product of the eccentr... | [] | null | 1552 | <p>Vertices of hyperbola $$ = (0,\, \pm \,8)$$</p>
<p>As ellipse pass through it i.e.,</p>
<p>$$0 + {{64} \over {{b^2}}} = 1 \Rightarrow {b^2} = 64$$ ...... (1)</p>
<p>As major axis of ellipse coincide with transverse axis of hyperbola we have b > a i.e.</p>
<p>$${e_E} = \sqrt {1 - {{{a^2}} \over {64}}} = {{\sqrt {64 ... | integer | jee-main-2022-online-27th-july-morning-shift | 6,389 |
1l6nni0or | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let the hyperbola $$H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ pass through the point $$(2 \sqrt{2},-2 \sqrt{2})$$. A parabola is drawn whose focus is same as the focus of $$\mathrm{H}$$ with positive abscissa and the directrix of the parabola passes through the other focus of $$\mathrm{H}$$. If the length of th... | [{"identifier": "A", "content": "$$(2 \\sqrt{3}, 3 \\sqrt{2})$$"}, {"identifier": "B", "content": "$$\\mathbf(3 \\sqrt{3},-6 \\sqrt{2})$$"}, {"identifier": "C", "content": "$$(\\sqrt{3},-\\sqrt{6})$$"}, {"identifier": "D", "content": "$$(3 \\sqrt{6}, 6 \\sqrt{2})$$"}] | ["B"] | null | <p>$$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$</p>
<p>Focus of parabola : $$(ae,\,0)$$</p>
<p>Directrix : $$x = - ae$$.</p>
<p>Equation of parabola $$ \equiv {y^2} = 4aex$$</p>
<p>Length of latus rectum of parabola $$ = 4ae$$</p>
<p>Length of latus rectum of hyperbola $$ = {{2.{b^2}} \over a}$$</p>
<p>... | mcq | jee-main-2022-online-28th-july-evening-shift | 6,390 |
ldo9yx82 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | Let $\mathrm{H}$ be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$. Then the length of its latus rectum is : | [{"identifier": "A", "content": "$\\frac{5}{2}$"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$\\frac{3}{2}$"}] | ["C"] | null | $ 2 \mathrm{ae}=|(1+\sqrt{2})-(1+\sqrt{2})|=2 \sqrt{2}$
<br/><br/>$$ \Rightarrow $$ $\mathrm{ae}=\sqrt{2}$
<br/><br/>$$ \Rightarrow $$ $\mathrm{a}=1$
<br/><br/>$\Rightarrow \mathrm{b}=1 \because \mathrm{e}=\sqrt{2} \Rightarrow$ Hyperbola is rectangular
<br/><br/>$\Rightarrow \mathrm{L} . \mathrm{R}=\frac{2 \mathrm{... | mcq | jee-main-2023-online-31st-january-evening-shift | 6,391 |
1lguwg2ap | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let $$\mathrm{H}_{\mathrm{n}}: \frac{x^{2}}{1+n}-\frac{y^{2}}{3+n}=1, n \in N$$. Let $$\mathrm{k}$$ be the smallest even value of $$\mathrm{n}$$ such that the eccentricity of $$\mathrm{H}_{\mathrm{k}}$$ is a rational number. If $$l$$ is the length of the latus rectum of $$\mathrm{H}_{\mathrm{k}}$$, then $$21 l$$ is ... | [] | null | 306 | We have,
<br/><br/>$$
H_n \Rightarrow \frac{x^2}{1+n}-\frac{y^2}{3+n}=1, n \in N
$$
<br/><br/>Here, $a^2=1+n$ and $b^2=3+n$
<br/><br/>$$
\begin{aligned}
\operatorname{Eccentricity}(e) & =\sqrt{1+\frac{b^2}{a^2}} \\\\
& =\sqrt{1+\left(\frac{3+n}{1+n}\right)}=\sqrt{\frac{2 n+4}{n+1}}=\sqrt{\frac{2(n+2)}{n+1}}
\end{aligne... | integer | jee-main-2023-online-11th-april-morning-shift | 6,392 |
1lh2yqxxv | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let the eccentricity of an ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ is reciprocal to that of the hyperbola $$2 x^{2}-2 y^{2}=1$$. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is ___________.</p> | [] | null | 2 | Equation of hyperbola is $2 x^2-2 y^2=1$
<br/><br/>$\Rightarrow \frac{x^2}{1 / 2}-\frac{y^2}{1 / 2}=1$
<br/><br/>Here, $a=b$
<br/><br/>$\therefore$ Eccentricity of rectangular hyperbola is $\sqrt{2}$
<br/><br/>$\therefore$ Eccentricity of ellipse is $\frac{1}{\sqrt{2}}$
<br/><br/>Since, ellipse intersects the hyperbola... | integer | jee-main-2023-online-6th-april-evening-shift | 6,393 |
lsaosyr4 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | For $0<\theta<\pi / 2$, if the eccentricity of the hyperbola
<br/><br/>$x^2-y^2 \operatorname{cosec}^2 \theta=5$ is $\sqrt{7}$ times eccentricity of the<br/><br/> ellipse $x^2 \operatorname{cosec}^2 \theta+y^2=5$, then the value of $\theta$ is : | [{"identifier": "A", "content": "$\\frac{\\pi}{6}$"}, {"identifier": "B", "content": "$\\frac{5 \\pi}{12}$"}, {"identifier": "C", "content": "$\\frac{\\pi}{3}$"}, {"identifier": "D", "content": "$\\frac{\\pi}{4}$"}] | ["C"] | null | <p>To find the value of $\theta$, we need to determine the relationship between the eccentricities of the given hyperbola and ellipse. Let's start by writing down the standard forms of ellipse and hyperbola and then relate them to the given equations.</p>
<p>The standard form of an ellipse is:
<p>$\frac{x^2}{a^2} +... | mcq | jee-main-2024-online-1st-february-morning-shift | 6,394 |
jaoe38c1lse5xrza | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let the foci and length of the latus rectum of an ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b b e( \pm 5,0)$$ and $$\sqrt{50}$$, respectively. Then, the square of the eccentricity of the hyperbola $$\frac{x^2}{b^2}-\frac{y^2}{a^2 b^2}=1$$ equals</p> | [] | null | 51 | <p>$$\begin{aligned}
& \text { focii } \equiv( \pm 5,0) ; \frac{2 b^2}{a}=\sqrt{50} \\
& a=5 \quad b^2=\frac{5 \sqrt{2} a}{2} \\
& b^2=a^2\left(1-e^2\right)=\frac{5 \sqrt{2} a}{2}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \mathrm{a}\left(1-\mathrm{e}^2\right)=\frac{5 \sqrt{2}}{2} \\
& \Rightarrow \frac{5}{... | integer | jee-main-2024-online-31st-january-morning-shift | 6,396 |
1lsg4ewgx | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let $$P$$ be a point on the hyperbola $$H: \frac{x^2}{9}-\frac{y^2}{4}=1$$, in the first quadrant such that the area of triangle formed by $$P$$ and the two foci of $$H$$ is $$2 \sqrt{13}$$. Then, the square of the distance of $$P$$ from the origin is</p> | [{"identifier": "A", "content": "26"}, {"identifier": "B", "content": "22"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "18"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxrj67/57cc4f4f-ed29-40fc-903d-a6afb2548230/871275f0-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxrj68.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsoxrj67/57cc4f4f-ed29-40fc-903d-a6afb2548230/871275f0-ccf2-11ee... | mcq | jee-main-2024-online-30th-january-evening-shift | 6,397 |
1lsgckxsg | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let the latus rectum of the hyperbola $$\frac{x^2}{9}-\frac{y^2}{b^2}=1$$ subtend an angle of $$\frac{\pi}{3}$$ at the centre of the hyperbola. If $$\mathrm{b}^2$$ is equal to $$\frac{l}{\mathrm{~m}}(1+\sqrt{\mathrm{n}})$$, where $$l$$ and $$\mathrm{m}$$ are co-prime numbers, then $$\mathrm{l}^2+\mathrm{m}^2+\mathrm... | [] | null | 182 | <p>LR subtends $$60^{\circ}$$ at centre</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnnba9/95b74993-a660-473c-9d3c-c4fef0db6ac2/87c00510-cde4-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnnbaa.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqnnba9/95b74993-a6... | integer | jee-main-2024-online-30th-january-morning-shift | 6,398 |
luxwe7n5 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let the foci of a hyperbola $$H$$ coincide with the foci of the ellipse $$E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$$ and the eccentricity of the hyperbola $$H$$ be the reciprocal of the eccentricity of the ellipse $$E$$. If the length of the transverse axis of $$H$$ is $$\alpha$$ and the length of its conjugate a... | [{"identifier": "A", "content": "225"}, {"identifier": "B", "content": "237"}, {"identifier": "C", "content": "242"}, {"identifier": "D", "content": "205"}] | ["A"] | null | <p>$$E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$$</p>
<p>$$\begin{aligned}
\text { Eccentricity of ellipse, } e_E & =\sqrt{1-\frac{b^2}{a^2}} \\
& =\sqrt{1-\frac{75}{100}} \\
& e_E=\frac{1}{2}
\end{aligned}$$</p>
<p>$$\therefore e_H=2$$ [ as eccentricity of hyperbola is reciprocal of eccentricity of ellipse]</p>
<p>Tr... | mcq | jee-main-2024-online-9th-april-evening-shift | 6,399 |
lv3vef1a | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let $$\mathrm{S}$$ be the focus of the hyperbola $$\frac{x^2}{3}-\frac{y^2}{5}=1$$, on the positive $$x$$-axis. Let $$\mathrm{C}$$ be the circle with its centre at $$\mathrm{A}(\sqrt{6}, \sqrt{5})$$ and passing through the point $$\mathrm{S}$$. If $$\mathrm{O}$$ is the origin and $$\mathrm{SAB}$$ is a diameter of $$... | [] | null | 40 | <p>$$\begin{aligned}
& \frac{x^2}{3}-\frac{y^2}{5}=1 \\
& 5=3\left(e^2-1\right) \Rightarrow e=\sqrt{\frac{8}{3}} \\
& S \equiv(2 \sqrt{2}, 0)
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4m7nur/89a66068-e09d-4eb2-bb5b-6e5807946115/8a9cdd30-10f6-11ef-8553-fdfc6347... | integer | jee-main-2024-online-8th-april-evening-shift | 6,400 |
lv5grwkp | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let $$H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$$ be the hyperbola, whose eccentricity is $$\sqrt{3}$$ and the length of the latus rectum is $$4 \sqrt{3}$$. Suppose the point $$(\alpha, 6), \alpha>0$$ lies on $$H$$. If $$\beta$$ is the product of the focal distances of the point $$(\alpha, 6)$$, then $$\alpha^2+\beta... | [{"identifier": "A", "content": "170"}, {"identifier": "B", "content": "171"}, {"identifier": "C", "content": "169"}, {"identifier": "D", "content": "172"}] | ["B"] | null | <p>$$\begin{aligned}
& H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=-1 \\
& e=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{3} \\
& \Rightarrow 1+\frac{a^2}{b^2}=3 \\
& \Rightarrow \frac{a^2}{b^2}=2 \quad \text{.... (1)}\\
& \frac{2 a^2}{b}=4 \sqrt{3}
\end{aligned}$$</p>
<p>Using equation (1)</p>
<p>$$\begin{aligned}
& \frac{4 b^2}{b}=4 \sqrt{... | mcq | jee-main-2024-online-8th-april-morning-shift | 6,401 |
urhXbbS4Mwr9Upfs | maths | hyperbola | tangent-to-hyperbola | The locus of a point $$P\left( {\alpha ,\beta } \right)$$ moving under the condition that the line $$y = \alpha x + \beta $$ is tangent to the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ is : | [{"identifier": "A", "content": "an ellipse "}, {"identifier": "B", "content": "a circle "}, {"identifier": "C", "content": "a parabola "}, {"identifier": "D", "content": "a hyperbola "}] | ["D"] | null | Tangent to the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ is
<br><br>$$y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} $$
<br><br>Given that $$y = \alpha x + \beta $$ is the tangent of hyperbola
<br><br>$$ \Rightarrow m = \alpha $$ and $${a^2}{m^2} - {b^2} = {\beta ^2}$$
<br><br>$$\therefore$$ $${a^2}... | mcq | aieee-2005 | 6,402 |
grnK93LwSd8r9LVV | maths | hyperbola | tangent-to-hyperbola | Tangents are drawn to the hyperbola 4x<sup>2</sup> - y<sup>2</sup> = 36 at the points P and Q.
<br/><br/>If these tangents intersect at the
point T(0, 3) then the area (in sq. units) of $$\Delta $$PTQ is : | [{"identifier": "A", "content": "$$36\\sqrt 5 $$"}, {"identifier": "B", "content": "$$45\\sqrt 5 $$"}, {"identifier": "C", "content": "$$54\\sqrt 3 $$"}, {"identifier": "D", "content": "$$60\\sqrt 3 $$"}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265692/exam_images/dqhksg6rhe7itnkgtxn3.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Hyperbola Question 71 English Explanation">
<br><br>Here PQ is the chord of contact.
<br><br>Equation of PQ is
<br><br>x.0 $$-... | mcq | jee-main-2018-offline | 6,404 |
V1oEoOwIdxchnOnBaDwIG | maths | hyperbola | tangent-to-hyperbola | The equation of a tangent to the hyperbola 4x<sup>2</sup> – 5y<sup>2</sup> = 20 parallel to the line x – y = 2 is : | [{"identifier": "A", "content": "x $$-$$ y + 9 = 0"}, {"identifier": "B", "content": "x $$-$$ y $$-$$ 3 = 0"}, {"identifier": "C", "content": "x $$-$$ y + 1 = 0"}, {"identifier": "D", "content": "x $$-$$ y + 7 = 0"}] | ["C"] | null | Hyperbola $${{{x^2}} \over 5} - {{{y^2}} \over 4} = 1$$
<br><br>slope of tangent = 1
<br><br>equation of tangent y = x $$ \pm $$ $$\sqrt {5 - 4} $$
<br><br>$$ \Rightarrow $$ y = x $$ \pm $$ 1
<br><br>$$ \Rightarrow $$ y = x + 1
<br><br>or
<br><br>$$ \Rightarrow $$ y = x $$... | mcq | jee-main-2019-online-10th-january-morning-slot | 6,405 |
serhntUUiJrREZf1pHxE5 | maths | hyperbola | tangent-to-hyperbola | If the eccentricity of the standard hyperbola
passing through the point (4,6) is 2, then the
equation of the tangent to the hyperbola at (4,6)
is : | [{"identifier": "A", "content": "2x \u2013 y \u2013 2 = 0"}, {"identifier": "B", "content": "3x \u2013 2y = 0"}, {"identifier": "C", "content": "2x \u2013 3y + 10 = 0"}, {"identifier": "D", "content": "x \u2013 2y + 8 = 0"}] | ["A"] | null | Formula for standard hyperbola :
<br><br>$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$
<br><br>It passes through (4, 6)
<br><br>$$ \therefore $$ $${{16} \over {{a^2}}} - {{36} \over {{b^2}}} = 1$$ ........(1)
<br><br>We know, $${e^2} = 1 + {{{b^2}} \over {{a^2}}}$$
<br><br>$$ \Rightarrow $$ 4 = $$1 + {{{b^2... | mcq | jee-main-2019-online-8th-april-evening-slot | 6,406 |
MyQ9Y4KXRg1ZCSnoHLjgy2xukews984x | maths | hyperbola | tangent-to-hyperbola | A line parallel to the straight line 2x – y = 0 is
tangent to the hyperbola
<br/>$${{{x^2}} \over 4} - {{{y^2}} \over 2} = 1$$ at the point
$$\left( {{x_1},{y_1}} \right)$$. Then $$x_1^2 + 5y_1^2$$ is equal to : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "8"}] | ["B"] | null | Tangent of hyperbola $${{{x^2}} \over 4} - {{{y^2}} \over 2} = 1$$ at the point (x<sub>1</sub>, y<sub>1</sub>) is
<br><br>$${{x{x_1}} \over 4} - {{y{y_1}} \over 2} = 1$$ which is parallel to 2x – y = 0
<br><br>$$ \therefore $$ Slope of tangent $${{x{x_1}} \over 4} - {{y{y_1}} \over 2} = 1$$ = Slope of 2x – y = 0
<br><b... | mcq | jee-main-2020-online-2nd-september-morning-slot | 6,407 |
jH56590JjDMSL9PSa01kmm3kmwv | maths | hyperbola | tangent-to-hyperbola | Consider a hyperbola H : x<sup>2</sup> $$-$$ 2y<sup>2</sup> = 4. Let the tangent at a <br/>point P(4, $${\sqrt 6 }$$) meet the x-axis at Q and latus rectum at R(x<sub>1</sub>, y<sub>1</sub>), x<sub>1</sub> > 0. If F is a focus of H which is nearer to the point P, then the area of $$\Delta$$QFR is equal to : | [{"identifier": "A", "content": "$${\\sqrt 6 }$$ $$-$$ 1"}, {"identifier": "B", "content": "$${7 \\over {\\sqrt 6 }}$$ $$-$$ 2"}, {"identifier": "C", "content": "$${4\\sqrt 6 }$$ $$-$$ 1"}, {"identifier": "D", "content": "$${4\\sqrt 6 }$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3b0ez3v/5fa06fb6-2e73-4b95-aa16-dacb000be7c4/83af90b0-d658-11ec-9a06-bd4ec5b93eb4/file-1l3b0ez3w.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3b0ez3v/5fa06fb6-2e73-4b95-aa16-dacb000be7c4/83af90b0-d658-11ec-9a06-bd4ec5b93eb4... | mcq | jee-main-2021-online-18th-march-evening-shift | 6,408 |
1krua3nf6 | maths | hyperbola | tangent-to-hyperbola | Let a line L : 2x + y = k, k > 0 be a tangent to the hyperbola x<sup>2</sup> $$-$$ y<sup>2</sup> = 3. If L is also a tangent to the parabola y<sup>2</sup> = $$\alpha$$x, then $$\alpha$$ is equal to : | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "$$-$$12"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "$$-$$24"}] | ["D"] | null | Tangent to hyperbola of <br><br>Slope m = $$-$$2 (given)<br><br>y = $$-$$2x $$\pm$$ $$\sqrt {3(3)} $$<br><br>$$\left( {y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} } \right)$$<br><br>$$\Rightarrow$$ y + 2x = $$\pm$$ 3 $$\Rightarrow$$ 2x + y = 3 (k > 0)<br><br>For parabola y<sup>2</sup> = $$\alpha$$x<br><br>$$y = mx + {\alp... | mcq | jee-main-2021-online-22th-july-evening-shift | 6,409 |
1l566s5yb | maths | hyperbola | tangent-to-hyperbola | <p>Let the eccentricity of the hyperbola $$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ be $$\sqrt {{5 \over 2}} $$ and length of its latus rectum be $$6\sqrt 2 $$. If $$y = 2x + c$$ is a tangent to the hyperbola H, then the value of c<sup>2</sup> is equal to :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "20"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "32"}] | ["B"] | null | <p>$$1 + {{{b^2}} \over {{a^2}}} = {5 \over 2} \Rightarrow {{{b^2}} \over {{a^2}}} = {3 \over 2}$$</p>
<p>$${{2{b^2}} \over a} = 6\sqrt 2 \Rightarrow 2.\,{3 \over 2}.\,a = 6\sqrt 2 $$</p>
<p>$$ \Rightarrow a = 2\sqrt 2 ,\,{b^2} = 12$$</p>
<p>$${c^2} = {a^2}{m^2} - {b^2} = 8.4 - 12 = 20$$</p> | mcq | jee-main-2022-online-28th-june-morning-shift | 6,410 |
1l6dxf7ak | maths | hyperbola | tangent-to-hyperbola | <p>Let the equation of two diameters of a circle $$x^{2}+y^{2}-2 x+2 f y+1=0$$ be $$2 p x-y=1$$ and $$2 x+p y=4 p$$. Then the slope m $$ \in $$ $$(0, \infty)$$ of the tangent to the hyperbola $$3 x^{2}-y^{2}=3$$ passing through the centre of the circle is equal to _______________.</p> | [] | null | 2 | $$
\begin{aligned}
&2 p+f-1=0 \quad\dots(1)\\\\
&2-p f-4 p=0 \quad\dots(2)\\\\
&2=p(f+4) \\\\
&p=\frac{2}{f+4} \\\\
&2 p=1-f \\\\
&\frac{4}{f+4}=1-f \\\\
&f^2+3 f=0 \\\\
&f=0 \text { or }-3
\end{aligned}
$$<br/><br/>
Hyperbola $3 x^2-y^2=3, x^2-\frac{y^2}{3}=1$<br/><br/>
$$
\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{m}^2... | integer | jee-main-2022-online-25th-july-morning-shift | 6,412 |
1ldo4xfyk | maths | hyperbola | tangent-to-hyperbola | <p>Let $$\mathrm{P}\left(x_{0}, y_{0}\right)$$ be the point on the hyperbola $$3 x^{2}-4 y^{2}=36$$, which is nearest to the line $$3 x+2 y=1$$. Then $$\sqrt{2}\left(y_{0}-x_{0}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$$-$$9"}, {"identifier": "C", "content": "$$-$$3"}, {"identifier": "D", "content": "9"}] | ["B"] | null | If $\left(x_0, y_0\right)$ is point on hyperbola then tangent at $\left(x_0, y_0\right)$ is parallel to $3 x+2 y=1$
<br/><br/>Equation of tangent $= \frac{x x_0}{12}-\frac{y y_0}{9}=2$
<br/><br/>Slope of tangent $=\frac{-3}{2}$
<br/><br/>Equation of tangent in slope form
<br/><br/>$y=\frac{-3}{2} x \pm \sqrt{12 \cdot \... | mcq | jee-main-2023-online-1st-february-evening-shift | 6,413 |
1lgq0o0y8 | maths | hyperbola | tangent-to-hyperbola | <p>Let $$m_{1}$$ and $$m_{2}$$ be the slopes of the tangents drawn from the point $$\mathrm{P}(4,1)$$ to the hyperbola $$H: \frac{y^{2}}{25}-\frac{x^{2}}{16}=1$$. If $$\mathrm{Q}$$ is the point from which the tangents drawn to $$\mathrm{H}$$ have slopes $$\left|m_{1}\right|$$ and $$\left|m_{2}\right|$$ and they make po... | [] | null | 8 | Equation of tangent to the hyperbola $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
<br/><br/>$$
y=m x \pm \sqrt{a^2-b^2 m^2}
$$
<br/><br/>Given the hyperbola $H: \frac{y^2}{25} - \frac{x^2}{16} = 1$, the equation of the tangent to this hyperbola can be written as :
<br/><br/>$$y=mx \pm \sqrt{25 - 16m^2}$$
<br/><br/>We know tha... | integer | jee-main-2023-online-13th-april-morning-shift | 6,415 |
lv2er3wn | maths | hyperbola | tangent-to-hyperbola | <p>Consider a hyperbola $$\mathrm{H}$$ having centre at the origin and foci on the $$\mathrm{x}$$-axis. Let $$\mathrm{C}_1$$ be the circle touching the hyperbola $$\mathrm{H}$$ and having the centre at the origin. Let $$\mathrm{C}_2$$ be the circle touching the hyperbola $$\mathrm{H}$$ at its vertex and having the cent... | [{"identifier": "A", "content": "$$\\frac{28}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{11}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{14}{3}$$\n"}, {"identifier": "D", "content": "$$\\frac{10}{3}$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhfif9k/1a3e3d85-d2e7-4099-8558-58b863f1c6ee/866ac480-1802-11ef-b156-f754785ad3ce/file-1lwhfif9l.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwhfif9k/1a3e3d85-d2e7-4099-8558-58b863f1c6ee/866ac480-1802-11ef-b156-f754785ad3ce... | mcq | jee-main-2024-online-4th-april-evening-shift | 6,416 |
1ktke5n0m | maths | indefinite-integrals | integration-by-partial-fraction | If $$\int {{{\sin x} \over {{{\sin }^3}x + {{\cos }^3}x}}dx = } $$
<br/><br/>$$\alpha {\log _e}|1 + \tan x| + \beta {\log _e}|1 - \tan x + {\tan ^2}x| + \gamma {\tan ^{ - 1}}\left( {{{2\tan x - 1} \over {\sqrt 3 }}} \right) + C$$, when C is constant of integration, then the value of $$18(\alpha + \beta + {\gamma ^2}... | [] | null | 3 | $$ = \int {{{{{\sin x} \over {{{\cos }^3}x}}} \over {1 + {{\tan }^3}x}}dx = \int {{{\tan x.{{\sec }^2}x} \over {(\tan x + 1)(1 + {{\tan }^2}x - \tan x)}}dx} } $$<br><br>Let $$\tan x = t \Rightarrow {\sec ^2}x.\,dx = dt$$<br><br>$$ = \int {{t \over {(t + 1)({t^2} - t + 1)}}dt} $$<br><br>$$ = \int {\left( {{A \over {t + ... | integer | jee-main-2021-online-31st-august-evening-shift | 6,419 |
Uw47dOEvkjyfJj7r | maths | indefinite-integrals | integration-by-parts | The integral $$\int {\left( {1 + x - {1 \over x}} \right){e^{x + {1 \over x}}}dx} $$ is equal to | [{"identifier": "A", "content": "<img class=\"question-image\" src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264772/exam_images/je7jltiyhhqoek98uddy.webp\" loading=\"lazy\" alt=\"JEE Main 2014 (Offline) Mathematics - Indefinite Integrals Question 68 English Option 1\"> "}, {"identifier": "B", "content": ... | ["D"] | null | Let $$I = \int {\left( {1 + x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle x$}}}}dx$$
<br><br>$$ = \int {{e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle x$}}}}} dx + \int {\left( {x - {1 ... | mcq | jee-main-2014-offline | 6,420 |
k0vrHx9KC2v0TCKPU1nFZ | maths | indefinite-integrals | integration-by-parts | The integral $$\int \, $$cos(log<sub>e</sub> x) dx is equal to : (where C is a constant of integration) | [{"identifier": "A", "content": "$${x \\over 2}$$[sin(log<sub>e</sub> x) $$-$$ cos(log<sub>e</sub> x)] + C"}, {"identifier": "B", "content": "x[cos(log<sub>e</sub> x) + sin(log<sub>e</sub> x)] + C"}, {"identifier": "C", "content": "$${x \\over 2}$$[cos(log<sub>e</sub> x) + sin(log<sub>e</sub> x)] + C"}, {"identifier": ... | ["C"] | null | $${\rm I} = \int {\cos \left( {\ell nx} \right)} dx$$
<br><br>$${\rm I} = \cos (\ln x).x + \int {\sin \left( {\ell nx} \right)dx} $$
<br><br>$${\rm I} = \cos \left( {\ell nx} \right)x + \left[ {\sin \left( {\ell nx} \right).x - \int {\cos \left( {\ell nx} \right)dx} } \right]$$
<br><br>$${\rm I} = {x \over 2}\left[ {\s... | mcq | jee-main-2019-online-12th-january-morning-slot | 6,421 |
1l6re7z1v | maths | indefinite-integrals | integration-by-parts | <p>For $$I(x)=\int \frac{\sec ^{2} x-2022}{\sin ^{2022} x} d x$$, if $$I\left(\frac{\pi}{4}\right)=2^{1011}$$, then</p> | [{"identifier": "A", "content": "$$3^{1010} I\\left(\\frac{\\pi}{3}\\right)-I\\left(\\frac{\\pi}{6}\\right)=0$$"}, {"identifier": "B", "content": "$$3^{1010} I\\left(\\frac{\\pi}{6}\\right)-I\\left(\\frac{\\pi}{3}\\right)=0$$"}, {"identifier": "C", "content": "$$3^{1011} I\\left(\\frac{\\pi}{3}\\right)-I\\left(\\frac{\... | ["A"] | null | <p>Given,</p>
<p>$$I(x) = \int {{{{{\sec }^2}x - 2022} \over {{{\sin }^{2022}}x}}dx} $$</p>
<p>$$ = \int {{{{{\sec }^2}x} \over {{{\sin }^{2022}}x}}dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} } $$</p>
<p>$$ = \int {{1 \over {{{\sin }^{2022}}x}}\,.\,{{\sec }^2}x\,dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} } $... | mcq | jee-main-2022-online-29th-july-evening-shift | 6,422 |
1lgxgz9mo | maths | indefinite-integrals | integration-by-parts | <p>If $$I(x) = \int {{e^{{{\sin }^2}x}}(\cos x\sin 2x - \sin x)dx} $$ and $$I(0) = 1$$, then $$I\left( {{\pi \over 3}} \right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$ - {e^{{3 \\over 4}}}$$"}, {"identifier": "B", "content": "$$ - {1 \\over 2}{e^{{3 \\over 4}}}$$"}, {"identifier": "C", "content": "$${e^{{3 \\over 4}}}$$"}, {"identifier": "D", "content": "$${1 \\over 2}{e^{{3 \\over 4}}}$$"}] | ["D"] | null | $$
\begin{aligned}
& \text { Given, } I(x)=\int e^{\sin ^2 x}(\cos x \sin 2 x-\sin x) d x \\\\
& =\int e^{\sin ^2 x} \cdot \cos x \cdot \sin 2 x d x-\int \sin x e^{\sin ^2 x} d x \\\\
& =\int \frac{\cos x}{\mathrm{I}} \cdot \frac{e^{\sin ^2 x} \cdot \sin 2 x}{\mathrm{II}} d x-\int \sin x \cdot e^{\sin ^2 x} d x \\\\
& ... | mcq | jee-main-2023-online-10th-april-morning-shift | 6,423 |
1lh21fmey | maths | indefinite-integrals | integration-by-parts | <p>Let $$I(x)=\int \frac{x^{2}\left(x \sec ^{2} x+\tan x\right)}{(x \tan x+1)^{2}} d x$$. If $$I(0)=0$$, then $$I\left(\frac{\pi}{4}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\log _{e} \\frac{(\\pi+4)^{2}}{32}-\\frac{\\pi^{2}}{4(\\pi+4)}$$"}, {"identifier": "B", "content": "$$\\log _{e} \\frac{(\\pi+4)^{2}}{16}-\\frac{\\pi^{2}}{4(\\pi+4)}$$"}, {"identifier": "C", "content": "$$\\log _{e} \\frac{(\\pi+4)^{2}}{16}+\\frac{\\pi^{2}}{4(\\pi+4)}$$"}, {"identifi... | ["A"] | null | We have,
<br/><br/>$$
\begin{aligned}
I(x)= & \int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x \\\\
= & x^2 \int \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2} d x \\\\
& \quad-\int\left\{\frac{d}{d x}\left(x^2\right) \int \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2} d x\right\} d x \text { (integration by pa... | mcq | jee-main-2023-online-6th-april-morning-shift | 6,424 |
lv2eqxr8 | maths | indefinite-integrals | integration-by-parts | <p>If $$\int \operatorname{cosec}^5 x d x=\alpha \cot x \operatorname{cosec} x\left(\operatorname{cosec}^2 x+\frac{3}{2}\right)+\beta \log _x\left|\tan \frac{x}{2}\right|+\mathrm{C}$$
where $$\alpha, \beta \in \mathbb{R}$$ and $$\mathrm{C}$$ is the constant of integration, then the value of $$8(\alpha+\beta)$$ equals _... | [] | null | 1 | <p>$$\begin{aligned}
& I=\int(\operatorname{cosec} x)^5 d x=\int(\operatorname{cosec} x)^3(\operatorname{cosec} x)^2 d x \\
& =(\operatorname{cosec} x)^3 \int \operatorname{cosec}^2 x d x- \\
& \int\left(\frac{d}{d x}(\operatorname{cosec} x)^3 \int \operatorname{cosec}^2 x d x\right) d x
\end{aligned}$$</p>
<p>$$\begin... | integer | jee-main-2024-online-4th-april-evening-shift | 6,425 |
hKe3lNduHiApUmZ8 | maths | indefinite-integrals | integration-by-substitution | $$\int {{{\left\{ {{{\left( {\log x - 1} \right)} \over {1 + {{\left( {\log x} \right)}^2}}}} \right\}}^2}\,\,dx} $$ is equal to | [{"identifier": "A", "content": "$${{\\log x} \\over {{{\\left( {\\log x} \\right)}^2} + 1}} + C$$ "}, {"identifier": "B", "content": "$${x \\over {{x^2} + 1}} + C$$ "}, {"identifier": "C", "content": "$${{x{e^x}} \\over {1 + {x^2}}} + C$$ "}, {"identifier": "D", "content": "$${x \\over {{{\\left( {\\log x} \\right)}^2... | ["D"] | null | $$\int {{{{{\left( {\log x - 1} \right)}^2}} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}}} dx$$
<br><br>$$ = \int {{{1 + {{\left( {\log x} \right)}^2} - 2\log x} \over {{{\left[ {1 + {{\left( {\log x} \right)}^2}} \right]}^2}}}} $$
<br><br>$$ = \int {\left[ {{1 \over {\left( {1 + {{\left( {\log x} ... | mcq | aieee-2005 | 6,426 |
2ACvJSdqk8b5slgg | maths | indefinite-integrals | integration-by-substitution | If the $$\int {{{5\tan x} \over {\tan x - 2}}dx = x + a\,\ln \,\left| {\sin x - 2\cos x} \right| + k,} $$ then $$a$$ is
<br/>equal to : | [{"identifier": "A", "content": "$$-1$$ "}, {"identifier": "B", "content": "$$-2$$ "}, {"identifier": "C", "content": "$$1$$ "}, {"identifier": "D", "content": "$$2$$ "}] | ["D"] | null | $$\int {{{5\tan x} \over {\tan x - 2}}} dx$$
<br><br>$$ = \int {{{5{{\sin x} \over {\cos x}}} \over {{{\sin x} \over {\cos x}} - 2}}} \,dx$$
<br><br>$$ = \int {\left( {{{5\sin x} \over {\cos x}} \times {{\cos x} \over {\sin x - 2\cos x}}} \right)} \,dx$$
<br><br>$$ = \int {{{5\,\sin \,x\,dx} \over {\sin x - 2\,\cos x}}... | mcq | aieee-2012 | 6,428 |
nZ3cEdvpgAwa9myY | maths | indefinite-integrals | integration-by-substitution | The integral $$\int {{{dx} \over {{x^2}{{\left( {{x^4} + 1} \right)}^{3/4}}}}} $$ equals : | [{"identifier": "A", "content": "$$ - {\\left( {{x^4} + 1} \\right)^{{1 \\over 4}}} + c$$"}, {"identifier": "B", "content": "$$ - {\\left( {{{{x^4} + 1} \\over {{x^4}}}} \\right)^{{1 \\over 4}}} + c$$ "}, {"identifier": "C", "content": "$$ {\\left( {{{{x^4} + 1} \\over {{x^4}}}} \\right)^{{1 \\over 4}}} + c$$ "}, {"id... | ["B"] | null | $$1 = \int {{{dx} \over {{x^2}{{\left( {{x^4} + 1} \right)}^{3/4}}}}} $$
<br><br>$$ = \int {{{dx} \over {{x^3}{{\left( {1 + {x^{ - 4}}} \right)}^{3/4}}}}} $$
<br><br>Let $${x^{ - 4}} = y$$
<br><br>$$ \Rightarrow - 4{x^{ - 3}}\,dx = dy$$
<br><br>$$ \Rightarrow dx = {{ - 1} \over 4}{x^3}dy$$
<br><br>$$\therefore$$ $$I ... | mcq | jee-main-2015-offline | 6,430 |
LQe1HRHuBYA8dRTJJfiHH | maths | indefinite-integrals | integration-by-substitution | The integral $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}} $$ is equal to :
<br/><br/>(where C is a constant of integration.) | [{"identifier": "A", "content": "$$ - 2\\sqrt {{{1 + \\sqrt x } \\over {1 - \\sqrt x }}} + C$$ "}, {"identifier": "B", "content": "$$ - 2\\sqrt {{{1 - \\sqrt x } \\over {1 + \\sqrt x }}} + C$$"}, {"identifier": "C", "content": "$$ - \\sqrt {{{1 - \\sqrt x } \\over {1 + \\sqrt x }}} + C$$ "}, {"identifier": "D", "con... | ["B"] | null | I = $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}} $$
<br><br>= $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt x \sqrt {1 - x} }}} $$
<br><br>Let 1 + $$\sqrt x $$ = t
<br><br>$$ \Rightarrow $$$$\,\,\,$$$${1 \over {2\sqrt x }}\,dx$$ = dt
<br... | mcq | jee-main-2016-online-10th-april-morning-slot | 6,432 |
7Gnzjnba9fRxVw93NEa9p | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}} = {\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k,$$
<br/><br/>where k is a constant of integration, then A + B +C equals : | [{"identifier": "A", "content": "$${{21} \\over 5}$$ "}, {"identifier": "B", "content": "$${{16} \\over 5}$$"}, {"identifier": "C", "content": "$${{7} \\over 10}$$"}, {"identifier": "D", "content": "$${{27} \\over 10}$$"}] | ["B"] | null | $$\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}} $$
<br><br>= $$\int {{{dx} \over {{{\cos }^3}x\sqrt {4\sin x\cos x} }}} $$
<br><br>= $$\int {{{dx} \over {2{{\cos }^4}x\sqrt {\tan x} }}} $$
<br><br>Let tan x = t<sup>2</sup>
<br><br>$$ \Rightarrow $$$$\,\,\,$$ sec<sup>2... | mcq | jee-main-2016-online-9th-april-morning-slot | 6,433 |
5rgUmlOIq0Q0L7jm | maths | indefinite-integrals | integration-by-substitution | Let $${I_n} = \int {{{\tan }^n}x\,dx} ,\,\left( {n > 1} \right).$$
<br/><br/>If $${I_4} + {I_6}$$ = $$a{\tan ^5}x + b{x^5} + C$$, where C is a constant of integration,
<br/><br/>then the ordered pair $$\left( {a,b} \right)$$ is equal to | [{"identifier": "A", "content": "$$\\left( {{1 \\over 5},0} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{1 \\over 5}, - 1} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - {1 \\over 5},0} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - {1 \\over 5},1} \\right)$$"}] | ["A"] | null | Given,
<br><br>In = $$\int {{{\tan }^n}x\,dx,\,\,\,n > 1} $$
<br><br>$$\therefore\,\,\,$$ I<sub>4</sub> = $$\int {{{\tan }^4}x\,dx} $$
<br><br>and I<sub>6</sub> = $$\int {{{\tan }^6}} x\,dx$$
<br><br>$$\therefore\,\,\,$$ I = I<sub>4</sub> + I<sub>6</sub>
<br><br>= $$\int {\left( {{{\tan }^4}x + {{\tan }^6}x} \right... | mcq | jee-main-2017-offline | 6,434 |
4FTEIQa2Co5vKRg6Ns20r | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}dx = x - {K \over {\sqrt A }}{{\tan }^{ - 1}}} $$ $$\left( {{{K\,\tan x + 1} \over {\sqrt A }}} \right) + C,(C\,\,$$ is a constant of integration) then the ordered pair (K, A) is equal to : | [{"identifier": "A", "content": "(2, 1)"}, {"identifier": "B", "content": "($$-$$2, 3)"}, {"identifier": "C", "content": "(2, 3)"}, {"identifier": "D", "content": "($$-$$2, 1)"}] | ["C"] | null | Given,
<br><br>$$\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}} \,\,dx$$
<br><br>Let tanx = t
<br><br>$$ \Rightarrow $$$$\,\,\,$$ sec<sup>2</sup>x dx = dt
<br><br>$$ \Rightarrow $$$$\,\,\,$$ dx = $${{dt} \over {{{\sec }^2}x}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $$dx = {{dt} \over {1 + {{\tan }^2}x}}$$
<br><br>$... | mcq | jee-main-2018-online-16th-april-morning-slot | 6,435 |
sWW8X3j3VBEBvcDd | maths | indefinite-integrals | integration-by-substitution | The integral
<br/><br/>$$\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}} dx$$
<br/><br/>is equal to | [{"identifier": "A", "content": "$${{ - 1} \\over {1 + {{\\cot }^3}x}} + C$$"}, {"identifier": "B", "content": "$${1 \\over {3\\left( {1 + {{\\tan }^3}x} \\right)}} + C$$"}, {"identifier": "C", "content": "$${{ - 1} \\over {3\\left( {1 + {{\\tan }^3}x} \\right)}} + C$$"}, {"identifier": "D", "content": "$${1 \\over {1 ... | ["C"] | null | Given,
<br><br>$$\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}}\,dx $$
<br><br>$$ = \int {{{{{\sin }^2}x\,{{\cos }^2}x} \over {{{\left[ {{{\sin }^3}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + {{\cos }^3}x\left( {... | mcq | jee-main-2018-offline | 6,436 |
H8ln4uLN3lq8lQeMmZ3rsa0w2w9jxacphio | maths | indefinite-integrals | integration-by-substitution | Let $$a \in \left( {0,{\pi \over 2}} \right)$$ be fixed. If the integral
<br/><br>$$\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx$$ = A(x) cos 2$$\alpha $$ + B(x) sin 2$$\alpha $$ + C, where C is a
<br/><br/>constant of integration, then the functions A(x) and B(x) are respectively : </br> | [{"identifier": "A", "content": "$$x - \\alpha $$ and $${\\log _e}\\left| {\\cos \\left( {x - \\alpha } \\right)} \\right|$$"}, {"identifier": "B", "content": "$$x + \\alpha $$ and $${\\log _e}\\left| {\\sin \\left( {x - \\alpha } \\right)} \\right|$$"}, {"identifier": "C", "content": "$$x + \\alpha $$ and $${\\log _e}... | ["D"] | null | <p>To solve the given integral, first we simplify the expression in the integral as follows :</p>
<p>$$\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx = \int {{{\sin x \over \cos x} + {\sin \alpha \over \cos \alpha }} \over {{\sin x \over \cos x} - {\sin \alpha \over \cos \alpha }}} dx$$ </p>
<p>Simpl... | mcq | jee-main-2019-online-12th-april-evening-slot | 6,438 |
2TA2SAVPabICEfCIXG3rsa0w2w9jx2b6x3x | maths | indefinite-integrals | integration-by-substitution | If $$\int {{x^5}} {e^{ - {x^2}}}dx = g\left( x \right){e^{ - {x^2}}} + c$$, where c is a constant of integration, then $$g$$(–1) is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "- 1"}, {"identifier": "C", "content": "$$ - {5 \\over 2}$$"}, {"identifier": "D", "content": "$$ - {1 \\over 2}$$"}] | ["C"] | null | Let x<sup>2</sup> = t<br><br>
$$ \Rightarrow {1 \over 2}\int {{t^2}{e^{ - t}}dt} $$<br><br>
$$ \Rightarrow {1 \over 2}\left[ { - {t^2}{e^{ - t}} + \int {2t{e^{ - t}}dt} } \right]$$<br><br>
$$ \Rightarrow {{ - {t^2}{e^{ - t}}} \over 2} - t{e^{ - t}} - {e^{ - t}}$$<br><br>
$$ \Rightarrow \left( { - {{{x^4}} \over 2} - {x... | mcq | jee-main-2019-online-10th-april-evening-slot | 6,439 |
NGp83ukLX0F1lubeJ33rsa0w2w9jwxvclk0 | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{dx} \over {{{\left( {{x^2} - 2x + 10} \right)}^2}}}} = A\left( {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {{f\left( x \right)} \over {{x^2} - 2x + 10}}} \right) + C$$
<br/><br/> where C is a constant of integration then : | [{"identifier": "A", "content": "A =$${1 \\over {54}}$$ and f(x) = 9(x\u20131)<sup>2</sup>"}, {"identifier": "B", "content": "A =$${1 \\over {54}}$$ and f(x) = 3(x\u20131)"}, {"identifier": "C", "content": "A =$${1 \\over {81}}$$ and f(x) = 3(x\u20131)"}, {"identifier": "D", "content": "A =$${1 \\over {27}}$$ and f(x) ... | ["B"] | null | $$\int {{{dx} \over {{{({x^2} - 2x + 10)}^2}}} = \int {{{dx} \over {{{({{(x - 1)}^2} + 9)}^2}}}} } $$<br><br>
$$Let{\rm{ }}{\left( {x{\rm{ }}-{\rm{ }}1} \right)^2}{\rm{ }} = {\rm{ }}9ta{n^2}\theta \,\,\,\,...\left( i \right)$$<br><br>
$$ \Rightarrow \tan \theta = {{x - 1} \over 3}$$<br><br>
On Differentiating ...(i)<b... | mcq | jee-main-2019-online-10th-april-morning-slot | 6,440 |
QtWndUPN5ICUBfhiUWRfe | maths | indefinite-integrals | integration-by-substitution | The integral $$\int {{\rm{se}}{{\rm{c}}^{{\rm{2/ 3}}}}\,{\rm{x }}\,{\rm{cose}}{{\rm{c}}^{{\rm{4 / 3}}}}{\rm{x \,dx}}} $$ is equal to
(Hence C is a constant of integration) | [{"identifier": "A", "content": "-3/4 tan <sup>- 4 / 3</sup> x + C"}, {"identifier": "B", "content": "3tan<sup>\u20131/3</sup>x + C"}, {"identifier": "C", "content": "\u20133cot<sup>\u20131/3</sup>x+ C"}, {"identifier": "D", "content": "- 3tan<sup>\u20131/3</sup>x + C"}] | ["D"] | null | $$\int {{{\sec }^{{2 \over 3}}}} x\cos e{c^{{4 \over 3}}}xdx$$
<br><br>= $$\int {{{{{\sec }^{{2 \over 3}}}x} \over {\cos e{c^{{2 \over 3}}}x}}\cos e{c^2}xdx} $$
<br><br>= $$\int {{1 \over {{{\cot }^{{2 \over 3}}}x}}\cos e{c^2}xdx} $$
<br><br>Let cot x = t<sup>3</sup>
<br><br>$$ \Rightarrow $$ - cosec<sup>2</sup>x dx = ... | mcq | jee-main-2019-online-9th-april-morning-slot | 6,441 |
AQn2qq06q2fjAXy7QknRR | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{dx} \over {{x^3}{{(1 + {x^6})}^{2/3}}}} = xf(x){{(1 + {x^6})}^{{1 \over 3}}} + C} $$ <br/>
where C is a constant of integration, then the
function ƒ(x) is equal to | [{"identifier": "A", "content": "$${3 \\over {{x^2}}}$$"}, {"identifier": "B", "content": "$$ - {1 \\over {6{x^3}}}$$"}, {"identifier": "C", "content": "$$ - {1 \\over {2{x^3}}}$$"}, {"identifier": "D", "content": "$$ - {1 \\over {2{x^2}}}$$"}] | ["C"] | null | I = $$\int {{{dx} \over {{x^3}{{\left( {1 + {x^6}} \right)}^{{2 \over 3}}}}}} $$
<br><br>= $$\int {{{dx} \over {{x^7}{{\left( {{1 \over {{x^6}}} + 1} \right)}^{{2 \over 3}}}}}} $$
<br><br>Let $${{1 \over {{x^6}}} + 1}$$ = t
<br><br>$$ \Rightarrow $$ $${{ - 6} \over {{x^7}}}dx = dt$$
<br><br>$$ \Rightarrow $$ $${{dx} \o... | mcq | jee-main-2019-online-8th-april-evening-slot | 6,442 |
ye4tbM04B4WyItNxoFvXn | maths | indefinite-integrals | integration-by-substitution | The integral $$\int {{{3{x^{13}} + 2{x^{11}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^4}}}} \,dx$$ is equal to : (where C is a constant of integration)
| [{"identifier": "A", "content": "$${{{x^{12}}} \\over {6{{\\left( {2{x^4} + 3{x^2} + 1} \\right)}^3}}}$$ + $$C$$"}, {"identifier": "B", "content": "$${{{x^4}} \\over {6{{\\left( {2{x^4} + 3{x^2} + 1} \\right)}^3}}} + C$$"}, {"identifier": "C", "content": "$${{{x^{12}}} \\over {{{\\left( {2{x^4} + 3{x^2} + 1} \\right)}^... | ["A"] | null | $$\int {{{3{x^{13}} + 2{x^{11}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^4}}}} dx$$
<br><br>$$\int {{{\left( {{3 \over {{x^3}}} + {2 \over {{x^5}}}} \right)dx} \over {{{\left( {2 + {3 \over {{x^2}}} + {1 \over {{x^4}}}} \right)}^4}}}} $$
<br><br>Let $$\left( {2 + {3 \over {{x^2}}} + {1 \over {{x^4}}}}... | mcq | jee-main-2019-online-12th-january-evening-slot | 6,443 |
wH0XiFnWxUutIMcwlOwvx | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} $$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))<sup>m</sup> equals : | [{"identifier": "A", "content": "$${1 \\over {27{x^6}}}$$"}, {"identifier": "B", "content": "$${{ - 1} \\over {27{x^9}}}$$"}, {"identifier": "C", "content": "$${1 \\over {9{x^4}}}$$"}, {"identifier": "D", "content": "$${1 \\over {3{x^3}}}$$"}] | ["B"] | null | $$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} $$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C
<br><br>$$\int {{{\left| x \right|\sqrt {{1 \over {{x^2}}} - 1} } \over {{x^4}}}} \,dx,$$
<br><br>Put $${1 \over {{x^2}}} - 1 = t \Rightarrow {{dt} \over {dx}} = {{ - 2} \over {{x^3}}}$$
<br><br><b>Case-... | mcq | jee-main-2019-online-11th-january-morning-slot | 6,445 |
GNwlKDvLfwVDVMJ3J4cOC | maths | indefinite-integrals | integration-by-substitution | If $$\int \, $$x<sup>5</sup>.e<sup>$$-$$4x<sup>3</sup></sup> dx = $${1 \over {48}}$$e<sup>$$-$$4x<sup>3</sup></sup> f(x) + C, where C is a constant of inegration, then f(x) is equal to - | [{"identifier": "A", "content": "$$-$$2x<sup>3</sup> $$-$$ 1"}, {"identifier": "B", "content": "$$-$$ 2x<sup>3</sup> + 1 "}, {"identifier": "C", "content": "4x<sup>3</sup> + 1"}, {"identifier": "D", "content": "$$-$$4x<sup>3</sup> $$-$$ 1"}] | ["D"] | null | $$\int {{x^5}} .{e^{ - 4{x^3}}}\,dx = {1 \over {48}}{e^{ - 4{x^3}}}f\left( x \right) + c$$
<br><br>Put $${x^3} = t$$
<br><br>$$3{x^2}\,dx = dt$$
<br><br>$$\int {{x^3}.{e^{ - 4{x^3}}}.\,{x^2}} dx$$
<br><br>$${1 \over 3}\int {t.{e^{ - 4t}}dt} $$
<br><br>$${1 \over 3}\left[ {t.{{{e^{ - 4t}}} \over { - 4}} - \in... | mcq | jee-main-2019-online-10th-january-evening-slot | 6,446 |
XbsoEUqQUgLdlc6T8Qlx5 | maths | indefinite-integrals | integration-by-substitution | If $$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx,\,\left( {x \ge 0} \right),$$
<br/><br/>$$f\left( 0 \right) = 0,$$ then the value of $$f(1)$$ is : | [{"identifier": "A", "content": "$$ - $$ $${1 \\over 2}$$"}, {"identifier": "B", "content": "$$ - $$ $${1 \\over 4}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over 4}$$"}] | ["D"] | null | $$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx$$
<br><br>$$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{x^{14}}{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx$$
<br><br>$$f\left( x \right) = \int {{{5{x^{ - 6}} + 7{x^{ - 8}}} \over {{{\left( {{... | mcq | jee-main-2019-online-9th-january-evening-slot | 6,447 |
Hj0hWZ3LNJm6lC7Dh5vpM | maths | indefinite-integrals | integration-by-substitution | For x<sup>2</sup> $$ \ne $$ n$$\pi $$ + 1, n $$ \in $$ N (the set of natural numbers), the integral <br/><br/>$$\int {x\sqrt {{{2\sin ({x^2} - 1) - \sin 2({x^2} - 1)} \over {2\sin ({x^2} - 1) + \sin 2({x^2} - 1)}}} dx} $$ is equal to :
<br/><br/>(where c is a constant of integration) | [{"identifier": "A", "content": "$${\\log _e}\\left| {{1 \\over 2}{{\\sec }^2}\\left( {{x^2} - 1} \\right)} \\right| + c$$"}, {"identifier": "B", "content": "$${1 \\over 2}{\\log _e}\\left| {\\sec \\left( {{x^2} - 1} \\right)} \\right| + c$$"}, {"identifier": "C", "content": "$${1 \\over 2}{\\log _e}\\left| {{{\\sec }^... | ["D"] | null | $$\int {x\sqrt {{{2\sin \left( {{x^2} - } \right) - \sin 2\left( {{x^2} - 1} \right)} \over {2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} } \,\,dx$$
<br><br>$$ = \int {x\sqrt {{{2\sin \left( {{x^2} - 1} \right) - 2sin\left( {{x^2} - 1} \right)\cos \left( {{x^2} - 1} \right)} \over {2\sin \left... | mcq | jee-main-2019-online-9th-january-morning-slot | 6,448 |
H0FHSffmm2l7UtVzAnjgy2xukfqemvyt | maths | indefinite-integrals | integration-by-substitution | If
<br/>$$\int {{{\cos \theta } \over {5 + 7\sin \theta - 2{{\cos }^2}\theta }}} d\theta $$ = A$${\log _e}\left| {B\left( \theta \right)} \right| + C$$,
<br/><br/>where C is a constant of integration, then $${{{B\left( \theta \right)} \over A}}$$
<br/>can be : | [{"identifier": "A", "content": "$${{2\\sin \\theta + 1} \\over {5\\left( {\\sin \\theta + 3} \\right)}}$$"}, {"identifier": "B", "content": "$${{2\\sin \\theta + 1} \\over {\\sin \\theta + 3}}$$"}, {"identifier": "C", "content": "$${{5\\left( {2\\sin \\theta + 1} \\right)} \\over {\\sin \\theta + 3}}$$"}, {"iden... | ["C"] | null | $$\int {{{\cos \theta } \over {5 + 7\sin \theta - 2{{\cos }^2}\theta }}} d\theta $$
<br><br>= $$\int {{{\cos \theta d\theta } \over {5 + 7\sin \theta - 2\left( {1 - {{\sin }^2}\theta } \right)}}} $$
<br><br>= $$\int {{{\cos \theta d\theta } \over {3 + 7\sin \theta + 2{{\sin }^2}\theta }}} $$
<br><br>Let sin $$\theta... | mcq | jee-main-2020-online-5th-september-evening-slot | 6,450 |
aEHNBpLgZlUpf8cadcjgy2xukfjjgzep | maths | indefinite-integrals | integration-by-substitution | If <br/>$$\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $$ = $$g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$$<br/><br/>
where c is a constant of integration,
then g(0) is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "e"}, {"identifier": "D", "content": "e<sup>2</sup>"}] | ["B"] | null | I = $$\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $$
<br><br>= $$\int {\left( {\left( {{e^{2x}} + {e^x} - 1} \right) + \left( {{e^x} - {e^{ - x}}} \right)} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx$$
<br><br>= $$\int {\left( {{e^{2x}} + {e^x} - 1} \ri... | mcq | jee-main-2020-online-5th-september-morning-slot | 6,451 |
E6ZDVN1s3vLFiRj1o5jgy2xukf7g2hr5 | maths | indefinite-integrals | integration-by-substitution | Let $$f\left( x \right) = \int {{{\sqrt x } \over {{{\left( {1 + x} \right)}^2}}}dx\left( {x \ge 0} \right)} $$. Then f(3) – f(1) is eqaul to : | [{"identifier": "A", "content": "$$ - {\\pi \\over {12}} + {1 \\over 2} + {{\\sqrt 3 } \\over 4}$$"}, {"identifier": "B", "content": "$$ {\\pi \\over {12}} + {1 \\over 2} - {{\\sqrt 3 } \\over 4}$$"}, {"identifier": "C", "content": "$$ - {\\pi \\over 6} + {1 \\over 2} + {{\\sqrt 3 } \\over 4}$$"}, {"identifier": "D"... | ["B"] | null | $$\int {{{\sqrt x } \over {{{(1 + x)}^2}}}} dx(x > 0)$$<br><br>Put x = tan<sup>2</sup>$$\theta $$ $$ \Rightarrow $$ 2xdx = 2tan$$\theta $$sec<sup>2</sup>$$\theta $$d$$\theta $$<br><br>$$I = \int {{{2{{\tan }^2}\theta .{{\sec }^2}\theta } \over 2}} d\theta = \int {2{{\sin }^2}\theta d\theta = \int {(1 - \cos 2\thet... | mcq | jee-main-2020-online-4th-september-morning-slot | 6,452 |
m2nLubDEJo1qVTMWIO7k9k2k5ith9ii | maths | indefinite-integrals | integration-by-substitution | The integral $$\int {{{dx} \over {{{(x + 4)}^{{8 \over 7}}}{{(x - 3)}^{{6 \over 7}}}}}} $$ is equal to :<br/>
(where C is a constant of integration) | [{"identifier": "A", "content": "$${1 \\over 2}{\\left( {{{x - 3} \\over {x + 4}}} \\right)^{{3 \\over 7}}} + C$$"}, {"identifier": "B", "content": "$${\\left( {{{x - 3} \\over {x + 4}}} \\right)^{{1 \\over 7}}} + C$$"}, {"identifier": "C", "content": "$$ - {1 \\over {13}}{\\left( {{{x - 3} \\over {x + 4}}} \\right)^{{... | ["B"] | null | $$\int {{{dx} \over {{{(x + 4)}^{{8 \over 7}}}{{(x - 3)}^{{6 \over 7}}}}}} $$
<br><br>= $$\int {{{dx} \over {{{\left( {x + 4} \right)}^2}{{\left( {{{x - 3} \over {x + 4}}} \right)}^{{6 \over 7}}}}}} $$
<br><br>Put $${{{x - 3} \over {x + 4}}}$$ = t
<br><br>$$ \Rightarrow $$ $$\left\{ {{{\left( {x + 4} \right) - \left( {... | mcq | jee-main-2020-online-9th-january-morning-slot | 6,453 |
nGPDGf6uHUWzEnzMjQ1klreafmf | maths | indefinite-integrals | integration-by-substitution | If $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx = a{\sin ^{ - 1}}\left( {{{\sin x + \cos x} \over b}} \right) + c$$, where c is a constant of integration, then
the ordered pair (a, b) is equal to : | [{"identifier": "A", "content": "(-1, 3)"}, {"identifier": "B", "content": "(1, 3)"}, {"identifier": "C", "content": "(1, -3)"}, {"identifier": "D", "content": "(3, 1)"}] | ["B"] | null | Given $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx$$
<br><br>Write sin2x = 1 + sin2x - 1
<br><br>= $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {1 + \sin 2x - 1} \right]} }}} dx$$
<br><br>= $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x - 1} \righ... | mcq | jee-main-2021-online-24th-february-morning-slot | 6,456 |
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