Integral Calculus jee mains questions | JEE Main Maths Integral Calculus Previous Year Question

Q 1 :
If the area of the region ${(x, y) : \frac{a}{x^{2}} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1}$ is $(\log_{e} 2) - \frac{1}{7}$ then the value of $7 a - 3$ is equal to : [2024]

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(2)

Area of region ${(x, y) : \frac{a}{x^{2}} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1} = \int_{1}^{2} (\frac{1}{x} - \frac{a}{x^{2}}) d x$

$\Rightarrow (\ln 2) - \frac{1}{7} = {[\ln x + \frac{a}{x}]}_{1}^{2} = \ln 2 + \frac{a}{2} - (\ln (1 + a))$

$= \ln 2 - \frac{a}{2} \Rightarrow \frac{a}{2} = \frac{1}{7} \Rightarrow a = \frac{2}{7}$

$So, 7 a - 3 = \frac{2}{7} \times 7 - 3 = - 1$

Q 2 :
The area (in square units) of the region enclosed by the ellipse $x^{2} + 3 y^{2} = 18$ in the first quadrant below the line $y = x$ is [2024]

$\sqrt{3} π$
$\sqrt{3} π + \frac{3}{4}$
$\sqrt{3} π - \frac{3}{4}$
$\sqrt{3} π + 1$

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(1)

We have, $x^{2} + 3 y^{2} = 18$

$\Rightarrow \frac{x^{2}}{18} + \frac{y^{2}}{6} = 1$

$\Rightarrow \frac{x^{2}}{(3 \sqrt{2})^{2}} + \frac{y^{2}}{(\sqrt{6})^{2}} = 1$

For point of intersection of ellipse and line $y = x,$ we have, $x^{2} + 3 x^{2} = 18$

$\Rightarrow x^{2} = \frac{18}{4} \Rightarrow x^{2} = \frac{9}{2} \Rightarrow x = \pm \frac{3}{\sqrt{2}}$

Required area $= \int_{0}^{3 / \sqrt{2}} x d x + \int_{3 / \sqrt{2}}^{3 \sqrt{2}} \sqrt{\frac{18 - x^{2}}{3}} d x$

$= {[\frac{x^{2}}{2}]}_{0}^{3 / \sqrt{2}} + \frac{1}{\sqrt{3}} {[\frac{x}{2} \sqrt{18 - x^{2}} + 9 \sin^{- 1} \frac{x}{3 \sqrt{2}}]}_{3 / \sqrt{2}}^{3 \sqrt{2}}$

$= \frac{9}{4} + \frac{1}{\sqrt{3}} [9 \sin^{- 1} (1) - \frac{3}{2 \sqrt{2}} \cdot \frac{3 \sqrt{3}}{\sqrt{2}} - 9 \sin^{- 1} (\frac{1}{2})]$

$= \frac{9}{4} + \frac{1}{\sqrt{3}} [\frac{9 π}{2} - \frac{9 \sqrt{3}}{4} - \frac{9 π}{6}] = \frac{1}{\sqrt{3}} 3 π = \sqrt{3} π$

Q 3 :
Three points $O (0, 0), P (a, a^{2}), Q (- b, b^{2}), a > 0, b > 0,$ are on the parabola $y = x^{2} .$ Let $S_{1}$ be the area of the region bounded by the line PQ and the parabola, and $S_{2}$ be the area of the triangle $O P Q .$ If the minimum value of $\frac{S_{1}}{S_{2}}$ is $\frac{m}{n}, g c d (m, n) = 1,$ then $m + n$ is equal to _______ . [2024]

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(7)

Equation of line $P Q$ is, $y - a^{2} = \frac{b^{2} - a^{2}}{- b - a} (x - a)$

$\Rightarrow y = x (a - b) + a b$

$∴ S_{1} = \int_{- b}^{a} (x (a - b) + a b - x^{2}) d x$

$= {[(a - b) \frac{x^{2}}{2} + a b x - \frac{x^{3}}{3}]}_{- b}^{a}$

$= (a - b) \frac{(a^{2} - b^{2})}{2} + a b (a + b) - \frac{(a^{3} + b^{3})}{3}$

$\Rightarrow S_{1} = \frac{1}{6} {(a + b)}^{3}$ ...(i)

Also, $S_{2} = \frac{1}{2}$ $| \begin{matrix} 0 & 0 & 1 \\ a & a^{2} & 1 \\ - b & b^{2} & 1 \end{matrix} |$ $= \frac{1}{2} a b (a + b)$

$Now, \frac{S_{1}}{S_{2}} = \frac{{(a + b)}^{3} / 6}{a b (a + b) / 2} = \frac{1}{3} \frac{{(a + b)}^{2}}{a b}$

$= \frac{a^{2} + b^{2} + 2 a b}{3 a b} = \frac{a}{3 b} + \frac{b}{3 a} + \frac{2}{3} = \frac{1}{3} (\frac{a}{b} + \frac{b}{a} + 2)$

$Now, \frac{a}{b} + \frac{1}{a / b} \geq 2$

$∴ Minimum value of \frac{S_{1}}{S_{2}} = \frac{4}{3}$

$So, m + n = 4 + 3 = 7$

Q 4 :
The sum of squares of all possible values of $k,$ for which area of the region bounded by the parabolas $2 y^{2} = k x$ and $k y^{2} = 2 (y - x)$ is maximum, is equal to _______ . [2024]

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(8)

$k y^{2} = 2 (y - x)$ and $2 y^{2} = k x$

Point of intersection

$y (k y - 2 (1 - \frac{2 y}{k})) = 0$

$\Rightarrow k y^{2} - 2 (y - \frac{2 y^{2}}{k}) = 0$

$\Rightarrow y = 0 and k y = 2 (1 - \frac{2 y}{k}) \Rightarrow k y + \frac{4 y}{k} = 2$

$y = \frac{2}{k + \frac{4}{k}} = \frac{2 k}{k^{2} + 4}$ $∴ A = \int_{0}^{\frac{2 k}{k^{2} + 4}} ((y - \frac{k y^{2}}{2}) - (\frac{2 y^{2}}{k})) \cdot d y$

$= {[\frac{y^{2}}{2} - (\frac{k}{2} + \frac{2}{k}) \cdot \frac{y^{3}}{3}]}_{0}^{\frac{2 k}{k^{2} + 4}}$

$= {(\frac{2 k}{k^{2} + 4})}^{2} [\frac{1}{2} - \frac{k^{2} + 4}{2 k} \times \frac{1}{3} \times \frac{2 k}{k^{2} + 4}] = \frac{1}{6} \times 4 \times {(\frac{1}{k + \frac{4}{k}})}^{2}$

$A.M. \geq G.M. \Rightarrow (\frac{k + \frac{4}{k}}{2}) \geq 2 \Rightarrow k + \frac{4}{k} \geq 4$

$So, area is maximum when k = \frac{4}{k} \Rightarrow k = 2, - 2$

$∴ Sum of squares of all possible values of k$

$= 2^{2} + {(- 2)}^{2} = 8$

Topic Question Set

Q 1 :
If the area of the region ${(x, y) : \frac{a}{x^{2}} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1}$ is $(\log_{e} 2) - \frac{1}{7}$ then the value of $7 a - 3$ is equal to : [2024]

Q 2 :
The area (in square units) of the region enclosed by the ellipse $x^{2} + 3 y^{2} = 18$ in the first quadrant below the line $y = x$ is [2024]

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Q 4 :
The sum of squares of all possible values of $k,$ for which area of the region bounded by the parabolas $2 y^{2} = k x$ and $k y^{2} = 2 (y - x)$ is maximum, is equal to _______ . [2024]

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Topic Question Set

Q 1 : If the area of the region {(x,y):ax2≤y≤1x,1≤x≤2,0<a<1} is (loge2)-17 then the value of 7a-3 is equal to : [2024]

Q 2 : The area (in square units) of the region enclosed by the ellipse x2+3y2=18 in the first quadrant below the line y=x is [2024]

Q 3 : Three points O(0,0),P(a,a2),Q(-b,b2),a>0,b>0, are on the parabola y=x2. Let S1 be the area of the region bounded by the line PQ and the parabola, and S2 be the area of the triangle OPQ. If the minimum value of S1S2 is mn,gcd(m,n)=1, then m+n is equal to _______ . [2024]

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Q 4 : The sum of squares of all possible values of k, for which area of the region bounded by the parabolas 2y2=kx and ky2=2(y-x) is maximum, is equal to _______ . [2024]

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Q 1 :
If the area of the region ${(x, y) : \frac{a}{x^{2}} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1}$ is $(\log_{e} 2) - \frac{1}{7}$ then the value of $7 a - 3$ is equal to : [2024]

Q 2 :
The area (in square units) of the region enclosed by the ellipse $x^{2} + 3 y^{2} = 18$ in the first quadrant below the line $y = x$ is [2024]

Q 4 :
The sum of squares of all possible values of $k,$ for which area of the region bounded by the parabolas $2 y^{2} = k x$ and $k y^{2} = 2 (y - x)$ is maximum, is equal to _______ . [2024]