2022
Q101
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve $4{x^3} - 3x{y^2} + 6{x^2} - 5xy - 8{y^2} + 9x + 14 = 0$ at the point ($-$2, 3) be A. Then 8A is equal to ______________.
Show Answer
Correct Answer: 170
Explanation:
$
\begin{aligned}
& 4 x^3-3 x y^2+6 x^2-5 x y-8 y^2+9 x+14=0 \text { at } P(-2,3) \\\\
& 12 x^2-3\left(y^2+2 y x y^{\prime}\right)+12 x-5\left(x y^{\prime}+y\right)-16 y y^{\prime} + 9=0 \\\\
& 48-3\left(9-12 y^{\prime}\right)-24-5\left(-2 y^{\prime}+3\right)-48 y^{\prime}+9 =0 \\\\
& y^{\prime}=-9 / 2 \\\\
& \text { Tangent } y-3=-\frac{9}{2}(x+2) \Rightarrow 9 x+2 y=-12 \\\\
& \text { Normal : } y-3=\frac{2}{9}(x+2) \Rightarrow 9 y-2 x=31
\end{aligned}
$
$
\begin{aligned}
& \text { Area }=\frac{1}{2}\left(\frac{31}{2}-4\right) \times 3=\frac{85}{4} \\\\
& 8 \mathrm{~A}=170
\end{aligned}
$
2022
Q102
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If for some $\alpha$ > 0, the area of the region $\{ (x,y):|x + \alpha | \le y \le 2 - |x|\} $ is equal to ${3 \over 2}$, then the area of the region $\{ (x,y):0 \le y \le x + 2\alpha ,\,|x| \le 1\} $ is equal to ____________.
Show Answer
Correct Answer: 4
Explanation:
Point A is the intersection of y = x + $\alpha$ and y = 2 $-$ x lines.
$\therefore$ $y = 2 - y + \alpha $
$ \Rightarrow 2y = 2 + \alpha $
$ \Rightarrow y = {{2 + \alpha } \over 2}$
and $x = 2 - {{2 + \alpha } \over 2} = {{2 - \alpha } \over 2}$
$\therefore$ Point $A = \left( {{{2 - \alpha } \over 2},{{2 + \alpha } \over 2}} \right)$
$\therefore$ AN = y-coordinate of point $A = {{2 + \alpha } \over 2}$
Point B is the intersection of $y = 2 + x$ and $y = - x - \alpha $ lines.
$\therefore$ $y = 2 - y - \alpha $
$ \Rightarrow 2y = 2 - \alpha $
$ \Rightarrow y = {{2 - \alpha } \over 2}$
$\therefore$ $x = {{2 - \alpha } \over 2} - 2 = {{2 - \alpha - 4} \over 2} = {{ - \alpha - 2} \over 2}$
$\therefore$ Point $B = \left( {{{ - \alpha - 2} \over 2},{{2 - \alpha } \over 2}} \right)$
$\therefore$ $BM = y - $coordinate of point $B = {{2 - \alpha } \over 2}$
Area of the common region $BRAE$
$ = \Delta CDE - \left( {\Delta BCR + \Delta ARD} \right)$
$ = {1 \over 2} \times 4 \times 2 - \left( {{1 \over 2}( - \alpha + 2) \times BM + {1 \over 2} \times (2 + \alpha ) \times AN} \right)$
$ = 4 - \left( {{1 \over 2} \times (2 - \alpha ) \times {{(2 - \alpha )} \over 2} + {1 \over 2} \times (2 + \alpha ) \times {{2 + \alpha } \over 2}} \right)$
$ = 4 - \left[ {{{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4}} \right]$
Given, $4 - \left[ {{{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4}} \right] = {3 \over 2}$
$ \Rightarrow {{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4} = {5 \over 2}$
$ \Rightarrow {(2 - \alpha )^2} + {(2 + \alpha )^2} = 10$
$ \Rightarrow 4 + {\alpha ^2} - 4\alpha + 4 + {\alpha ^2} + 4\alpha = 10$
$ \Rightarrow 2{\alpha ^2} + 8 = 10$
$ \Rightarrow 2{\alpha ^2} = 2$
$ \Rightarrow {\alpha ^2} = 1$
$ \Rightarrow \alpha = \, \pm \,1$
Given that $\alpha > 0$ so accepted value of $\alpha = + \,1$.
Now, $0 \le y \le x + 2\alpha $ and $|x| \le 1$
$ \Rightarrow 0 \le y \le x + 2$ and $ - 1 \le x \le 1$
Area of $ABCD = {1 \over 2}(1 + 3) \times (1 - ( - 1))$
$ = {1 \over 2} \times 4 \times 2$
$ = 4$ sq. unit
2022
Q103
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
For real numbers a, b (a > b > 0), let
Area $\left\{ {(x,y):{x^2} + {y^2} \le {a^2}\,and\,{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} \ge 1} \right\} = 30\pi $
and
Area $\left\{ {(x,y):{x^2} + {y^2} \le {b^2}\,and\,{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} \le 1} \right\} = 18\pi $
Then, the value of (a $-$ b)2 is equal to ___________.
Show Answer
Correct Answer: 12
Explanation:
$x^{2}+y^{2} \leq a^{2}$ is interior of circle
and $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \geq 1$ is exterior of ellipse
$\therefore$ Area $=\pi a^{2}-\pi a b=30 \pi$
Similarly $x^{2}+y^{2} \geq b^{2}$ and $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1$ gives
$\pi a b-\pi b^{2}=18 \pi$
By (1) and (2), $\frac{a}{b}=\frac{5}{3} \Rightarrow a=\frac{5 b}{3}$
$\Rightarrow \pi \cdot \frac{25 b^{2}}{9}-\pi \cdot \frac{5 b^{2}}{3}=30 \pi$
$\Rightarrow\left(\frac{25}{9}-\frac{5}{3}\right) b^{2}=30$
$\Rightarrow \frac{10}{9} b^{2}=30 \Rightarrow b^{2}=27$
and $a^{2}=\frac{25}{9} \cdot 27=75$
$(a-b)^{2}=(5 \sqrt{3}-3 \sqrt{3})^{2}=3 \cdot 4=12$
2022
Q104
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the area of the region $\left\{ {(x,y):{x^{{2 \over 3}}} + {y^{{2 \over 3}}} \le 1,\,x + y \ge 0,\,y \ge 0} \right\}$ is A, then ${{256A} \over \pi }$ is equal to __________.
Show Answer
Correct Answer: 36
Explanation:
$\therefore$ Area of shaded region
$ = \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {\left( {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}} + x} \right)dx + \int\limits_0^1 {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}}dx} } $
$ = \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}}dx + \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {xdx} } $
Let $x = {\sin ^3}\theta $
$\therefore$ $dx = 3{\sin ^2}\theta \cos \theta d\theta $
$ = \int\limits_{ - {\pi \over 4}}^{{\pi \over 2}} {3{{\sin }^2}\theta {{\cos }^4}\theta d\theta + \left( {0 + {1 \over {16}}} \right)} $
$ = {{9\pi } \over {64}} + {1 \over {16}} - {1 \over {16}} = {{36\pi } \over {256}} = A$
$\therefore$ ${{256A} \over \pi } = 36$
2022
Q105
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let
${A_1} = \left\{ {(x,y):|x| \le {y^2},|x| + 2y \le 8} \right\}$ and
${A_2} = \left\{ {(x,y):|x| + |y| \le k} \right\}$. If 27 (Area A1 ) = 5 (Area A2 ), then k is equal to :
Show Answer
Correct Answer: 6
Explanation:
Required area (above x-axis)
${A_1} = 2\int\limits_0^4 {\left( {{{8 - x} \over 2} - \sqrt x } \right)dx} $
$ = 2\left( {16 - {{16} \over 4} - {8 \over {3/2}}} \right) = {{40} \over 3}$
and ${A_2} = 4\left( {{1 \over 2}\,.\,{k^2}} \right) = 2{k^2}$
$\therefore$ $27\,.\,{{40} \over 3} = 5\,.\,(2{k^2})$
$\Rightarrow$ k = 6
2022
Q106
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the region enclosed between the parabola y2 = 2x and the line x + y = 4 is __________.
Show Answer
Correct Answer: 18
Explanation:
The required area $ = \int_{ - 4}^2 {\left( {4 - y - {{{y^2}} \over 2}} \right)dy} $
$ = \left[ {4y - {{{y^2}} \over 2} - {{{y^3}} \over 6}} \right]_{ - 4}^2$
$ = 18$ square units
2022
Q107
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let S be the region bounded by the curves y = x3 and y2 = x. The curve y = 2|x| divides S into two regions of areas R1 , R2 . If max {R1 , R2 } = R2 , then ${{{R_2}} \over {{R_1}}}$ is equal to ______________.
Show Answer
Correct Answer: 19
Explanation:
$C_{1}: y=x^{3}$
$C_{2}: y^{2}=x$
and $C_{3}=y=2|x|$
$C_{1}$ and $C_{2}$ intersect at $(1,1)$
$C_{2}$ and $C_{3}$ intersect at $\left(\frac{1}{4}, \frac{1}{2}\right)$
Clearly $R_{1}=\int_{0}^{1 / 4}(\sqrt{x}-2 x) d x=\frac{2}{3}\left(\frac{1}{8}\right)-\frac{1}{16}=\frac{1}{48}$
and $R_{1}+R_{2}=\int_{0}^{1}\left(\sqrt{x}-x^{3}\right) d x=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}$
So, $\frac{R_{1}+R_{2}}{R_{1}}=\frac{5 / 12}{1 / 48} \Rightarrow 1+\frac{R_{2}}{R_{1}}=20$
$\Rightarrow \frac{R_{2}}{R_{1}}=19$
2021
Q108
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area, enclosed by the curves $y = \sin x + \cos x$ and $y = \left| {\cos x - \sin x} \right|$ and the lines $x = 0,x = {\pi \over 2}$, is :
A.
$2\sqrt 2 (\sqrt 2 - 1)$
D.
$2\sqrt 2 (\sqrt 2 + 1)$
Show Answer
Correct Answer: A
Explanation:
$A = \int_0^{{\pi \over 2}} {\left( {(\sin x + \cos x) - \left| {\cos x - \sin x} \right|} \right)\,dx} $
2021
Q109
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area of the region bounded by the parabola (y $-$ 2)2 = (x $-$ 1), the tangent to it at the point whose ordinate is 3 and the x-axis is :
Show Answer
Correct Answer: A
Explanation:
y = 3 $\Rightarrow$ x = 2
Point is (2, 3)
Diff. w.r.t x
2 (y $-$ 2) y' = 1
$\Rightarrow$ $y' = {1 \over {2(y - 2)}}$
$ \Rightarrow y{'_{(2,3)}} = {1 \over 2}$
$ \Rightarrow {{y - 3} \over {x - 2}} = {1 \over 2} \Rightarrow x - 2y + 4 = 0$
Area $ = \int\limits_0^3 {\left( {{{(y - 2)}^2} + 1 - (2y - 4)} \right)} \,dy$
= 9 sq. units
2021
Q110
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area of the region bounded by y $-$ x = 2 and x2 = y is equal to :
Show Answer
Correct Answer: C
Explanation:
y $-$ x = 2, x
2 = y
Now, x
2 = 2 + x
$\Rightarrow$ x
2 $-$ x $-$ 2 = 0
$\Rightarrow$ (x + 1)(x $-$ 2) = 0
Area = $\int\limits_{ - 1}^2 {(2 + x - {x^2})} $
$ = \left| {2x + {{{x^2}} \over 2} - {{{x^3}} \over 3}} \right|_{ - 1}^2$
$ = \left( {4 + 2 - {8 \over 3}} \right) - \left( { - 2 + {1 \over 2} + {1 \over 3}} \right)$
$ = 6 - 3 + 2 - {1 \over 2} = {9 \over 2}$
2021
Q111
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the area of the bounded region
Show Answer
Correct Answer: B
Explanation:
$R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\} \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}$
$\int\limits_{{1 \over 2}}^2 {{2^x}dx} - \int\limits_1^2 {\ln xdx} $
$ \Rightarrow \left[ {{{{2^x}} \over {\ln 2}}} \right]_{1/2}^2 - [x\ln x - x]_1^2$
$ \Rightarrow {{({2^2}) - {2^{1/2}}} \over {{{\log }_e}2}} - (2\ln 2 - 1)$
$ \Rightarrow {{\left( {{2^2} - \sqrt 2 } \right)} \over {{{\log }_e}2}} - 2\ln 2 + 1$
$\therefore$ $\alpha = {2^2} - \sqrt 2 $, $\beta = - 2$, $\gamma = 1$
$ \therefore {(\alpha + \beta + 2\gamma )^2}$
$ = {({2^2} - \sqrt 2 - 2 - 2)^2}$
$ = {(\sqrt 2 )^2} = 2$
2021
Q112
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the region, given by the set $\{ (x,y) \in R \times R|x \ge 0,2{x^2} \le y \le 4 - 2x\} $ is :
Show Answer
Correct Answer: D
Explanation:
Required area = $\left. {\int\limits_0^1 {\left( {4 - 2x - 2{x^2}} \right)dx = 4x - {x^2} - {{2{x^3}} \over 3}} } \right|_0^1$
$ = 4 - 1 - {2 \over 3} = {7 \over 3}$
2021
Q113
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area bounded by the curve 4y2 = x2 (4 $-$ x)(x $-$ 2) is equal to :
Show Answer
Correct Answer: C
Explanation:
Given,
4y
2 = x
2 (4 $-$ x)(x $-$ 2) ..... (1)
Here, Left hand side 4y
2 is always positive. So Right hand side should also be positive.
In x$\in$ [2, 4] Right hand side is positive.
By putting y = $-$y in equation (1), equation remains same. So, graph is symmetric about x axis.
Required Area = 2A
1 $ = 2\int_2^4 y dx$
$ = \int_2^4 {x\sqrt {(x - 2)(4 - x)} } dx$
put $x = 4{\sin ^2}\theta + 2{\cos ^2}\theta $
$ \Rightarrow dx = \left[ {4(2\sin \theta \cos \theta - 4\sin \theta \cos \theta )} \right]d\theta $
$ \Rightarrow dx = 4\sin \theta \cos \theta d\theta $
When lower limit = 2 then
$2 = 4{\sin ^2}\theta + 2{\cos ^2}\theta $
$ \Rightarrow 4(1 - {\cos ^2}\theta ) + 2{\cos ^2}\theta = 2$
$ \Rightarrow 4 - 4{\cos ^2}\theta + 2{\cos ^2}\theta = 2$
$ \Rightarrow 2{\cos ^2}\theta = 2$
$ \Rightarrow \cos \theta = \pm 1$
$ \Rightarrow \theta = 0,\pi {} $
When upper limit = 4 then
$4 = 4{\sin ^2}\theta + 2{\cos ^2}\theta $
$ \Rightarrow - 2{\cos ^2}\theta = 0$
$ \Rightarrow \theta = {{\pi {} } \over 2}$
$ \therefore $ Range of $\theta$ = 0 to ${{{\pi {} } \over 2}}$
$ \therefore $ Area = $ \int_0^{{{\pi {} } \over 2}} {(4{{\sin }^2}\theta + 2{{\cos }^2}\theta )\sqrt {(2{{\sin }^2}\theta )(2{{\cos }^2}\theta )} (4\sin \theta \cos \theta )d\theta } $
$ = \int_0^{{{\pi {} } \over 2}} {(4{{\sin }^2}\theta + 2{{\cos }^2}\theta )8{{\sin }^2}\theta {{\cos }^2}\theta d\theta } $
$ = 32\int_0^{{{\pi {} } \over 2}} {{{\sin }^4}\theta {{\cos }^2}\theta d\theta } + 16\int_0^{{{\pi {} } \over 2}} {{{\sin }^2}\theta {{\cos }^2}\theta d\theta } $
Using Wallis formula,
$ = 32.{{3.1.1} \over {6.4.2}}.{{\pi {} } \over 2} + 16.{{1.3.1} \over {6.4.2}}.{{\pi {} } \over 2}$
$ = \pi {} + {{\pi {} } \over 2}$
$ = {{3\pi {} } \over 2}$
2021
Q114
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let A1 be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A2 be the area of the region bounded by the curves y = sinx, y = cosx, x-axis and x = ${\pi \over 2}$ in the first quadrant. Then,
A.
${A_1}:{A_2} = 1:\sqrt 2 $ and ${A_1} + {A_2} = 1$
B.
${A_1} = {A_2}$ and ${A_1} + {A_2} = \sqrt 2 $
C.
$2{A_1} = {A_2}$ and ${A_1} + {A_2} = 1 + \sqrt 2 $
D.
${A_1}:{A_2} = 1:2$ and ${A_1} + {A_2} = 1$
Show Answer
Correct Answer: A
Explanation:
${A_1} + {A_2} = \int\limits_0^{\pi /2} {\cos x.\,dx = \left. {\sin x} \right|_0^{\pi /2}} = 1$
${A_1} = \left. {\int\limits_0^{\pi /4} {(\cos x - \sin x)dx = (\sin x + \cos x)} } \right|_0^{\pi /4} = \sqrt 2 - 1$
$ \therefore $ ${A_2} = 1 - \left( {\sqrt 2 - 1} \right) = 2 - \sqrt 2 $
$ \therefore $ ${{{A_1}} \over {{A_2}}} = {{\sqrt 2 - 1} \over {\sqrt 2 \left( {\sqrt 2 - 1} \right)}} = {1 \over {\sqrt 2 }}$
2021
Q115
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area of the region : $R = \{ (x,y):5{x^2} \le y \le 2{x^2} + 9\} $ is :
A.
$6\sqrt 3 $ square units
B.
$12\sqrt 3 $ square units
C.
$11\sqrt 3 $ square units
D.
$9\sqrt 3 $ square units
Show Answer
Correct Answer: B
Explanation:
Required area
$ = 2\int\limits_0^{\sqrt 3 } {\left( {2{x^2} + 9 - 5{x^2}} \right)} dx$
$ = 2\int\limits_0^{\sqrt 3 } {\left( {9 - 3{x^2}} \right)dx} $
$ = 2|\,9x - {x^3}\,|_0^{\sqrt 3 } = 12\sqrt 3 $
2021
Q116
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the part of the circle x2 + y2 = 36, which is outside the parabola
y2 = 9x, is :
Show Answer
Correct Answer: C
Explanation:
${x^2} + {y^2} = 36$ and ${y^2} = 9x$
$ \therefore $ ${x^2} + 9x - 36 = 0$
$ \Rightarrow x = 3, - 12$
Required Area,
$A = \pi {{{(6)}^2}} - 2\left[ {{A_1} + {A_2}} \right]$
$A = \pi {{{(6)}^2} - 2\left[ {\int_0^3 {\sqrt {9x} dx + \int_3^6 {\sqrt {36 - {x^2}} } dx} } \right]} $
$ = 36\pi - $$2\left[ {\left[ {3 \times {2 \over 3}{x^{{3 \over 2}}}} \right]_0^3 + \left[ {{x \over 2}\sqrt {36 - {x^2}} + {{36} \over 2}{{\sin }^{ - 1}}{x \over 6}} \right]_3^6} \right]$
$ = 36\pi - $$2\left[ {\left[ {2{x^{{3 \over 2}}}} \right]_0^3 + \left[ {{x \over 2}\sqrt {36 - {x^2}} + 18{{\sin }^{ - 1}}{x \over 6}} \right]_3^6} \right]$
$ = 36\pi - $$2\left[ {\left[ {6\sqrt 3 - 0} \right] + \left[ {\left( {0 + 18{{\sin }^{ - 1}}{6 \over 6}} \right) - \left( {{3 \over 2} \times 3\sqrt 3 + 18{{\sin }^{ - 1}}{3 \over 6}} \right)} \right]} \right]$
$ = 36\pi - $$2\left[ {\left[ {6\sqrt 3 } \right] + \left[ {\left( {{{18\pi } \over 2}} \right) - \left( {{3 \over 2} \times 3\sqrt 3 + {{18\pi } \over 6}} \right)} \right]} \right]$
$ = 36\pi - $ $\left( {12\sqrt 3 + 18\pi - 9\sqrt 3 - 6\pi } \right)$
= $24\pi - 3\sqrt 3 $
Note :
(i) $\int {\sqrt {{x^2} + {a^2}} dx = {1 \over 2}\left[ {x\sqrt {{x^2} + {a^2}} + {a^2}\log |x + \sqrt {{x^2} + {a^2}} |} \right]} + C$
(ii) $\int {\sqrt {{a^2} - {x^2}} dx = {1 \over 2}\left[ {x\sqrt {{a^2} - {x^2}} + {a^2}{{\sin }^{ - 1}}\left( {{x \over a}} \right)} \right]} + C$
(iii) $\int {\sqrt {{x^2} - {a^2}} dx = {1 \over 2}\left[ {x\sqrt {{x^2} - {a^2}} - {a^2}\log |x + \sqrt {{x^2} - {a^2}} |} \right]} + C$
2021
Q117
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the line y = mx bisects the area enclosed by the lines x = 0, y = 0, x = ${3 \over 2}$ and the curve y = 1 + 4x $-$ x2 , then 12 m is equal to _____________.
Show Answer
Correct Answer: 26
Explanation:
According to the question,
${1 \over 2}\int_0^{3/2} {(1 + 4x - {x^2})dx = \int_0^{3/2} {mx\,dx} } $
$ \Rightarrow {1 \over 2}\left[ {\left( {x + 2{x^2} - {{{x^3}} \over 3}} \right)} \right]_0^{3/2} = {m \over 2}[x]_0^{3/2} \Rightarrow {3 \over 2} + {9 \over 2} - {9 \over 8} = {{9m} \over 4}$
$\Rightarrow$ m = 39/18 $\Rightarrow$ 12m = 26
2021
Q118
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let a and b respectively be the points of local maximum and local minimum of the function f(x) = 2x3 $-$ 3x2 $-$ 12x. If A is the total area of the region bounded by y = f(x), the x-axis and the lines x = a and x = b, then 4A is equal to ______________.
Show Answer
Correct Answer: 114
Explanation:
f'(x) = 6x
2 $-$ 6x $-$ 12 = 6(x $-$ 2) (x + 1)
Point = (2, $-$20) & ($-$1, 7)
$A = \int\limits_{ - 1}^0 {(2{x^3} - 3{x^2} - 12x)dx + \int\limits_0^2 {(12x + 3{x^2} - 2{x^3})\,dx} } $
$A = \left( {{{{x^4}} \over 2} - {x^3} - 6{x^2}} \right)_{ - 1}^0 + \left( {6{x^2} + {x^3} - {{{x^4}} \over 2}} \right)_0^2$
4A = 114
2021
Q119
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area of the region $S = \{ (x,y):3{x^2} \le 4y \le 6x + 24\} $ is ____________.
Show Answer
Correct Answer: 27
Explanation:
For A & B
3x
2 = 6x + 24 $\Rightarrow$ x
2 $-$ 2x $-$ 8 = 0
$\Rightarrow$ x = $-$2, 4
Area $ = \int\limits_{ - 2}^4 {\left( {{3 \over 2}x + 6 - {3 \over 4}{x^2}} \right)dx} $
$ = \left[ {{{3{x^2}} \over 4} + 6x - {{{x^3}} \over 4}} \right]_{ - 2}^4 = 27$
2021
Q120
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the region bounded by the curves x2 + 2y $-$ 1 = 0, y2 + 4x $-$ 4 = 0 and y2 $-$ 4x $-$ 4 = 0, in the upper half plane is _______________.
Show Answer
Correct Answer: 2
Explanation:
Required area (shaded)
$ = 2\left[ {\int\limits_0^2 {\left( {{{4 - {y^2}} \over 4}} \right)dy - \int\limits_0^1 {\left( {{{1 - {x^2}} \over 2}} \right)dx} } } \right]$
$ = 2\left[ {{4 \over 3} - {1 \over 3}} \right] = (2)$
2021
Q121
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let T be the tangent to the ellipse E : x2 + 4y2 = 5 at the point P(1, 1). If the area of the region bounded by the tangent T, ellipse E, lines x = 1 and x = $\sqrt 5 $ is $\alpha$$\sqrt 5 $ + $\beta$ + $\gamma$ cos$-$1 $\left( {{1 \over {\sqrt 5 }}} \right)$, then |$\alpha$ + $\beta$ + $\gamma$| is equal to ______________.
Show Answer
Correct Answer: 1.25
Explanation:
E : x
2 + 4y
2 = 5
Tangent at P : x + 4y = 5
Required area
$ = \int\limits_1^{\sqrt 5 } {\left( {{{5 - x} \over 4} - {{\sqrt {5 - {x^2}} } \over 2}} \right)dx} $
$ = \left[ {{{5x} \over 4} - {{{x^2}} \over 8} - {x \over 4}\sqrt {5 - {x^2}} - {5 \over 2}{{\sin }^{ - 1}}{x \over {\sqrt 5 }}} \right]_1^{\sqrt 5 }$
$ = {5 \over 4}\sqrt 5 - {5 \over 4} - {5 \over 4}{\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)$
If we assume $\alpha$, $\beta$, $\gamma$, $\in$ Q (Not given in question) then $\alpha$ = ${5 \over 4}$, $\beta$ = $-$${5 \over 4}$ & $\gamma$ = $-$${5 \over 4}$
|$\alpha$ + $\beta$ + $\gamma$| = 1.25
2021
Q122
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let f : [$-$3, 1] $ \to $ R be given as
Show Answer
Correct Answer: 41
Explanation:
Area is $\int\limits_{ - 3}^{ - 2} {(x + 6)dx + \int\limits_{ - 2}^0 {{x^2}dx + \int\limits_0^1 {\sqrt {x}dx = A} } } $
$ = {7 \over 2} + \left[ {{{{x^3}} \over 3}} \right]_{ - 2}^0 + \left[ {{2 \over 3}{x^{3/2}}} \right]_0^1$
$ = {7 \over 2} + {8 \over 3} + {2 \over 3} = {{41} \over 6}$
So, 6A = 41
2021
Q123
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area bounded by the lines y = || x $-$ 1 | $-$ 2 | is ___________.
Show Answer
Correct Answer: 8
Explanation:
Question is incomplete it should be area bounded
by y = || x $-$ 1 | $-$ 2 | and y = 2.
2021
Q124
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area A. Then A4 is equal to __________.
Show Answer
Correct Answer: 64
Explanation:
$A = \int\limits_{{\pi \over 4}}^{{{5\pi } \over 4}} {(\sin x - \cos x)dx} $
$= [ - \cos x - \sin x]_{\pi /4}^{5\pi /4}$
$ = - \left[ {\left( {\cos {{5\pi } \over 4} + \sin {5\pi \over 4}} \right) - \left( {\cos {\pi \over 4} + \sin {\pi \over 4}} \right)} \right]$
$ = - \left[ {\left( { - {1 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }}} \right) - \left( {{1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }}} \right)} \right]$
$ = {4 \over {\sqrt 2 }} = 2\sqrt 2 $
$ \Rightarrow {A^4} = {\left( {2\sqrt 2 } \right)^4} = 64$
2020
Q125
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the region enclosed
2 – 1 and y = 1 – x2 is equal to :
Show Answer
Correct Answer: A
Explanation:
A = $\int\limits_{ - 1}^1 {\left( {\left( {1 - {x^2}} \right) - \left( {{x^2} - 1} \right)} \right)dx} $
= $\int\limits_{ - 1}^1 {\left( {2 - 2{x^2}} \right)dx} $
= $4\int\limits_0^1 {\left( {1 - {x^2}} \right)dx} $
= $4\left( {x - {{{x^3}} \over 3}} \right)_0^1$
= $4\left( {{2 \over 3}} \right)$ = ${8 \over 3}$
2020
Q126
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the region 2 $ \ge $ |x|}
Show Answer
Correct Answer: B
Explanation:
For point of intersection
x + y = 1 $ \Rightarrow $ x = 1 – y
y
2 = ${x \over 2}$ $ \Rightarrow $ 2y
2
= x
2y
2
= 1 – y $ \Rightarrow $ 2y
2
+ y – 1 = 0
$ \Rightarrow $ (2y – 1) (y + 1) = 0
$ \Rightarrow $ y = ${1 \over 2}$ or -1
Total area = $4\int\limits_0^{{1 \over 2}} {\left[ {\left( {1 - x} \right) - \left( {\sqrt {{x \over 2}} } \right)} \right]} dx$
= $4\left[ {x - {{{x^2}} \over 2} - {1 \over {\sqrt 2 }}{{{x^{3/2}}} \over {3/2}}} \right]_0^{{1 \over 2}}$
= $4\left[ {{1 \over 2} - {1 \over 8} - {{\sqrt 2 } \over 3}{{\left( {{1 \over 2}} \right)}^{3/2}}} \right]$
= 4 $ \times $ ${5 \over {24}}$ = ${5 \over 6}$
2020
Q127
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the region
A.
${8 \over 3}\sqrt 2 - 1$
B.
${4 \over 3}\sqrt 2 + 1$
C.
${8 \over 3}\sqrt 2 - {1 \over 2}$
D.
${4 \over 3}\sqrt 2 - {1 \over 2}$
Show Answer
Correct Answer: C
Explanation:
y = (x – 1)[x] = $\left\{ {\matrix{
{0,} & {0 \le x < 1} \cr
{x - 1,} & {1 \le x < 2} \cr
{2,} & {x = 2} \cr
} } \right.$
A = $\int\limits_0^2 {2\sqrt x } dx - {1 \over 2}.1.1$
= 2$\left[ {{{{x^{3/2}}} \over {{3 \over 2}}}} \right]_0^2$ - ${1 \over 2}$
= ${{8\sqrt 2 } \over 3} - {1 \over 2}$
2020
Q128
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the region
2 + 1, 0 $ \le $ y $ \le $ x + 1,
Show Answer
Correct Answer: B
Explanation:
$A = \int\limits_{{1 \over 2}}^1 {({x^2} + 1)dx + } $$\int\limits_1^2 {} $${(x + 1)dx}$
= ${\left[ {{{{x^3}} \over 3} + x} \right]_{{1 \over 2}}^1 + \left[ {{{{x^2}} \over 2} + x} \right]_1^2}$
${ = \left( {{4 \over 3} - {{13} \over {24}}} \right) + \left( {4 - {3 \over 2}} \right)}$
${ = {{19} \over {24}} + {5 \over 2}}$
${ = {{79} \over {24}}}$
2020
Q129
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Consider a region R = {(x, y) $ \in $ R : x2 $ \le $ y $ \le $ 2x}.
if a line y = $\alpha $ divides the area of region R into
two equal parts, then which of the following is
true?
A.
3$\alpha $2 - 8$\alpha $ + 8 = 0
B.
$\alpha $3 - 6$\alpha $3/2 - 16 = 0
C.
3$\alpha $2 - 8$\alpha $3/2 + 8 = 0
D.
$\alpha $3 - 6$\alpha $2 + 16 = 0
Show Answer
Correct Answer: C
Explanation:
y $ \ge $ x
2 $ \Rightarrow $ upper region of y = x
2
y $ \le $ 2x $ \Rightarrow $ lower region of y = 2x
According to question, area of OABC = 2 $ \times $ area of OAC
$ \Rightarrow $ $\int\limits^{4}_{0} \left( \sqrt{y} -\frac{y}{2} \right) dy$ = 2$\int\limits^{\alpha }_{0} \left( \sqrt{y} -\frac{y}{2} \right) dy$
$\Rightarrow \left[ {{2 \over 3}{y^{{3 \over 2}}} - {{{y^2}} \over 4}} \right]_0^4 = 2\left[ {{2 \over 3}{y^{{3 \over 2}}} - {{{y^2}} \over 4}} \right]_0^\alpha $
$ \Rightarrow {{16} \over 3} - 4 = 2\left[ {{2 \over 3}{{\left( \alpha \right)}^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}} \right]$
$ \Rightarrow {4 \over 3} = 2\left[ {{2 \over 3}{\alpha ^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}} \right]$
$ \Rightarrow {2 \over 3} = {2 \over 3}{\alpha ^{{3 \over 2}}} - {{{\alpha ^2}} \over 4}$
$ \Rightarrow 8 = 8{\alpha ^{{3 \over 2}}} - 3{\alpha ^2}$
$ \Rightarrow 3{\alpha ^2} - 8{\alpha ^{{3 \over 2}}} + 8 = 0$
2020
Q130
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Area (in sq. units) of the region outside
A.
$6\left( {4 - \pi } \right)$
B.
$3\left( {4 - \pi } \right)$
C.
$6\left( {\pi - 2} \right)$
D.
$3\left( {\pi - 2} \right)$
Show Answer
Correct Answer: C
Explanation:
Area of Ellipse = $\pi $ab = 6$\pi $
$ \therefore $ Required area = Area of ellipse
– 4 (Area of triangle OAB)
= 6$\pi $ - $4\left( {{1 \over 2} \times 2 \times 3} \right)$
= 6$\pi $ - 12
= $6\left( {\pi - 2} \right)$ sq.units
2020
Q131
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Given : $f(x) = \left\{ {\matrix{
{x\,\,\,\,\,,} & {0 \le x < {1 \over 2}} \cr
{{1 \over 2}\,\,\,\,,} & {x = {1 \over 2}} \cr
{1 - x\,\,\,,} & {{1 \over 2} < x \le 1} \cr
} } \right.$
A.
${1 \over 2} + {{\sqrt 3 } \over 4}$
B.
${1 \over 2} - {{\sqrt 3 } \over 4}$
C.
${1 \over 3} + {{\sqrt 3 } \over 4}$
D.
${{\sqrt 3 } \over 4} - {1 \over 3}$
Show Answer
Correct Answer: D
Explanation:
Required area = Area of trepezium ABCD –
Area of parabola between x = ${1 \over 2}$ and x = ${{\sqrt 3 } \over 2}$
= Area of trepezium ABCD - $\int\limits_{{1 \over 2}}^{{{\sqrt 3 } \over 2}} {{{\left( {x - {1 \over 2}} \right)}^2}dx} $
= ${1 \over 2}\left( {{{\sqrt 3 } \over 2} - {1 \over 2}} \right)\left( {{1 \over 2} + 1 - {{\sqrt 3 } \over 2}} \right)$ - ${1 \over 3}\left[ {{{\left( {x - {1 \over 2}} \right)}^3}} \right]_{{1 \over 2}}^{{{\sqrt 3 } \over 2}}$
= ${1 \over 2}\left( {{{\sqrt 3 - 1} \over 2}} \right)\left( {{{3 - \sqrt 3 } \over 2}} \right)$ $ - {1 \over 3}\left[ {{{\left( {{{\sqrt 3 - 1} \over 2}} \right)}^3} - 0} \right]$
= ${{\sqrt 3 } \over 4} - {1 \over 3}$
2020
Q132
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the region2 : x2 $ \le $ y $ \le $ 3 – 2x}, is :
Show Answer
Correct Answer: D
Explanation:
x
2 $ \le $ y $ \le $ – 2x + 3
$ \Rightarrow $ x
2
= – 2x + 3
$ \Rightarrow $ x
2
+ 2x – 3 = 0
$ \Rightarrow $ (x + 3) (x – 1) = 0
$ \Rightarrow $ x = – 3, x = 1
Area = $\int\limits_{ - 3}^1 {\left( { - 2x + 3 - {x^2}} \right)dx} $
= $\left( { - {x^2} + 3x - {{{x^3}} \over 3}} \right)_{ - 3}^1$
= $\left( { - 1 + 3 - {1 \over 3} + 9 + 9 - 9} \right)$
= 11 - ${{1 \over 3}}$
= ${{{32} \over 3}}$ sq. unit
2020
Q133
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
For a > 0, let the curves C1 : y2 = ax and
C2 : x2 = ay intersect at origin O and a point P.
Let the line x = b (0 < b < a) intersect the chord
OP and the x-axis at points Q and R,
respectively. If the line x = b bisects the area
bounded by the curves, C1 and C2 , and the area of
Show Answer
Correct Answer: A
Explanation:
C
1 : y
2
= ax, C
2 : x
2
= ay (a > 0)
P : intersection point of y
2
= ax and x
2
= ay
$ \Rightarrow $ x
4
= a
2
y
2
$ \Rightarrow $ x
4
= a
2 ax
$ \Rightarrow $ x = a, y = a
$ \therefore $ Point P : (a, a)
Line OP : y = x
$ \Rightarrow $ Point Q = (b, b)
Area $\Delta $OQR = ${1 \over 2}$
$ \Rightarrow $ ${1 \over 2} \times b \times b$ = ${1 \over 2}$
$ \Rightarrow $ b = 1
As line x = b bisect the area between curve
$ \therefore $ ${1 \over 2}\int\limits_0^a {\left( {\sqrt {ax} - {{{x^2}} \over a}} \right)} dx$ = $\int\limits_0^1 {\left( {\sqrt {ax} - {{{x^2}} \over a}} \right)} dx$
$ \Rightarrow $ $\left[ {{1 \over 2}\sqrt a {x^{{3 \over 2}}} \times {2 \over 3} - {1 \over 2}{{{x^3}} \over {3a}}} \right]_0^a$ = $\left[ {\sqrt a {x^{{3 \over 2}}} \times {2 \over 3} - {{{x^3}} \over {3a}}} \right]_0^1$
$ \Rightarrow $ ${{{a^2}} \over 3} - {1 \over 2}{{{a^2}} \over 3}$ - 0 = ${{2\sqrt a } \over 3} - {1 \over {3a}}$ - 0
$ \Rightarrow $ ${{{a^2}} \over 2} + {1 \over a} = 2\sqrt a $
$ \Rightarrow $ ${a^3} + 2 = 4a\sqrt a $
Squareing both sides, we get
$ \Rightarrow $ ${a^6} + 4{a^3} + 4 = 16{a^3}$
$ \Rightarrow $ ${a^6} - 12{a^3} + 4 = $ 0
Hence $a$ satisfy x
6 – 12x
3 + 4 = 0.
2020
Q134
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the region
2 | 4x2 $ \le $ y $ \le $ 8x + 12} is :
Show Answer
Correct Answer: B
Explanation:
For point of intersection
4x
2
= 8x + 12
$ \Rightarrow $ x
2 - 2x - 3 = 0
$ \Rightarrow $ x = –1, 3
Required area = area of the shaded region
= $\int\limits_{ - 1}^3 {\left( {8x + 12 - 4{x^2}} \right)dx} $
= $4\left[ {2.{{{x^2}} \over 2} + 3x - {{{x^3}} \over 3}} \right]_{ - 1}^3$
= (36 + 36 – 36) – (4 – 12 + ${4 \over 3}$)
= ${{128} \over 3}$
2020
Q135
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area of the region, enclosed by the circle x2 + y2 = 2 which is not common to the region bounded by the parabola y2 = x and the straight line y = x, is:
A.
${1 \over 6}\left( {24\pi - 1} \right)$
B.
${1 \over 3}\left( {12\pi - 1} \right)$
C.
${1 \over 3}\left( {6\pi - 1} \right)$
D.
${1 \over 6}\left( {12\pi - 1} \right)$
Show Answer
Correct Answer: D
Explanation:
Required area = Area of circle – $\int\limits_0^1 {\left( {\sqrt x - x} \right)dx} $
= $\pi $r
2 - $\left( {{{2{x^{3/2}}} \over 3} - {{{x^2}} \over 2}} \right)_0^1$
= $\pi {\left( {\sqrt 2 } \right)^2}$ - ${1 \over 6}$
= ${1 \over 6}\left( {12\pi - 1} \right)$
2019
Q136
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the area (in sq. units) bounded by the parabola y2
= 4$\lambda $x and the line y = $\lambda $x, $\lambda $ > 0, is ${1 \over 9}$
, then $\lambda $ is equal to :
Show Answer
Correct Answer: D
Explanation:
y
2 = 4$\lambda $x and y = $\lambda x$
If $\lambda $ > 0 then
Hence $\int\limits_0^{4/\lambda } {(2\sqrt \lambda \sqrt x } - \lambda x)dx = {1 \over 9}$
$ \Rightarrow $ ${\left( {{{2\sqrt \lambda {x^{3/2}}} \over {3/2}} - {{\lambda {x^2}} \over 2}} \right)^{{4 \over \lambda }}} = {1 \over 9}$
$ \Rightarrow $ ${4 \over 3}\sqrt \lambda {8 \over {{\lambda ^{3/2}}}} - \lambda {8 \over {{\lambda ^2}}} = {1 \over 9}$
$ \Rightarrow $ ${{32} \over {3\lambda }} - {8 \over \lambda } = {1 \over 9}$
$ \Rightarrow $ $ {8 \over 3\lambda } = {1 \over 9}$
$ \Rightarrow $ $\lambda $ = 24
2019
Q137
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the area (in sq. units) of the region {(x, y) : y2
$ \le $ 4x, x + y $ \le $ 1, x $ \ge $ 0, y $ \ge $ 0} is a $\sqrt 2 $ + b, then a – b is equal
to :
Show Answer
Correct Answer: C
Explanation:
Let P be the point common to x + y = 1 and y
2 = 4x
Then y
2 = 4(1-y) $ \Rightarrow $ y
2 - 4y - 4 = 0
$ \Rightarrow y = {{ - 4 \pm \sqrt {16 + 16} } \over 2}$
$ \Rightarrow y = - 2 + 2\sqrt 2 $
Then P is (3 - $ 2\sqrt 2$, -2 + $ 2\sqrt 2$).
Hence shaded area = Area of region (OPN) + area of ($\Delta OPQ$)
$ \Rightarrow {\int\limits_0^{3 - 2\sqrt 2 } {2\sqrt {xdx} + {1 \over 2}\left[ {1 - (3 - 2\sqrt 2 )} \right]} ^2}$
$ \Rightarrow {2 \over 3}2\left( {\sqrt 2 - 1} \right)\left( {3 - 2\sqrt 2 } \right) + {1 \over 2}{\left[ {2(\sqrt 2 - 1)} \right]^2}$
$ \Rightarrow {4 \over 3}\left\{ { - 7 + 5\sqrt 2 } \right\} + 2\left( {3 - 2\sqrt 2 } \right)$
$ \Rightarrow \left( {{{20} \over 3} - 4} \right)\sqrt 2 + 6 - {{28} \over 3}$
$ \Rightarrow {8 \over 3}\sqrt 2 - {{10} \over 3}$
Then a = ${8 \over 3}$ and b = ${-10 \over 3}$, so a - b = 6
2019
Q138
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq.units) of the region bounded by the curves y = 2x
and y = |x + 1|, in the first quadrant is :
C.
${3 \over 2} - {1 \over {\log _e^2}}$
D.
$\log _e^2 + {3 \over 2}$
Show Answer
Correct Answer: C
Explanation:
Required area = $\int\limits_0^1 {((x + 1) - {2^x})dx} $
$ \Rightarrow \left( {{{{x^2}} \over 2} + x - {{{2^x}} \over {{{\log }_e}2}}} \right)_0^1$
$ \Rightarrow {3 \over 2} - {1 \over {{{\log }_e}2}}$
2019
Q139
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the region
Show Answer
Correct Answer: B
Explanation:
y
2 = 2x ...........(1)
and x = y + 4 .............(2)
Solving (1) and (2)
(x - 4)
2 = 2x
$ \Rightarrow $ x
2 - 10x + 16 = 0
$ \Rightarrow $ x = 8, 2 and y = 4, -2
Integrating in y direction from A to B
Required area (A) = $\int\limits_{ - 2}^4 {\left( {y + 4 - {{{y^2}} \over 2}} \right)dy} $
= $\left[ {{{{y^2}} \over 2} + 4y - {{{y^3}} \over 6}} \right]_{ - 2}^4$
= ${\left( {{{16} \over 2} + 16 - {{64} \over 6}} \right)}$ - ${\left( {{4 \over 2} - 8 + {8 \over 6}} \right)}$
= 30 - 12 sq. unit
= 18 sq. unit
2019
Q140
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the region
2 $ \le $ y $ \le $ x + 2} is
Show Answer
Correct Answer: D
Explanation:
Parabola : x
2 = y
Straight line : y = x + 2
$ \therefore $ x
2 = x + 2
$ \Rightarrow $ x
2 - x - 2 = 0
x = -1, 2
$ \therefore $ y = 1, 4
Required area = $\int\limits_{ - 1}^2 {\left[ {\left( {x + 2} \right) - {x^2}} \right]} dx$
= $\left[ {{{{x^2}} \over 2} + 2x - {{{x^3}} \over 3}} \right]_{ - 1}^2$
= $\left( {2 + 4 - {8 \over 3}} \right) - \left( {{1 \over 2} - 2 + {1 \over 3}} \right)$
= ${{9 \over 2}}$
2019
Q141
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let S($\alpha $) = {(x, y) : y2
$ \le $ x, 0 $ \le $ x $ \le $ $\alpha $} and A($\alpha $)
is area of the region S($\alpha $). If for a $\lambda $, 0 < $\lambda $ < 4,
A($\lambda $) : A(4) = 2 : 5, then $\lambda $ equals
A.
$2{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}$
B.
$2{\left( {{2 \over {5}}} \right)^{{1 \over 3}}}$
C.
$4{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}$
D.
$4{\left( {{2 \over {5}}} \right)^{{1 \over 3}}}$
Show Answer
Correct Answer: C
Explanation:
A($\lambda $) = $2\int\limits_0^\lambda {\sqrt x } dx$
= $2\left[ {{{{x^{{3 \over 2}}}} \over {{3 \over 2}}}} \right]_0^\lambda $
= ${4 \over 3}{\lambda ^{{3 \over 2}}}$
$ \therefore $ A(4) = ${4 \over 3}{\left( 4 \right)^{{3 \over 2}}}$
Given, ${{A\left( \lambda \right)} \over {A\left( 4 \right)}} = {2 \over 5}$
$ \Rightarrow $ ${{{4 \over 3}{{\left( \lambda \right)}^{{3 \over 2}}}} \over {{4 \over 3}{{\left( 4 \right)}^{{3 \over 2}}}}} = {2 \over 5}$
$ \Rightarrow $ ${\lambda ^{{3 \over 2}}} = {2 \over 5} \times 8$
$ \Rightarrow $ $\lambda $ = ${\left( {{{16} \over 5}} \right)^{{2 \over 3}}}$
$ \Rightarrow $ $\lambda $ = ${\left( {{{256} \over {25}}} \right)^{{1 \over 3}}}$
$ \Rightarrow $ $\lambda $ = ${\left( {{{{4^3} \times 4} \over {25}}} \right)^{{1 \over 3}}}$
= $4{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}$
2019
Q142
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the region2 + 3x} is :
Show Answer
Correct Answer: A
Explanation:
When y = 4 then,
x
2 + 3x = 4
$ \Rightarrow $ x
2 + 3x - 4 = 0
$ \Rightarrow $ x
2 + 4x - x - 4 = 0
$ \Rightarrow $ x(x + 4) - (x + 4) = 0
$ \Rightarrow $ (x + 4)(x - 1) = 0
$ \Rightarrow $ x = 1, - 4
As 0 $ \le $ x $ \le $ 3,
so possible value of x = 1.
$ \therefore $ y = x
2 + 3x parabola cut the line y = 4 at x = 1.
Required area
= $\int\limits_0^1 {\left( {{x^2} + 3x} \right)} dx$ + 2 $ \times $ 4
= $\left[ {{{{x^3}} \over 3} + 3\left( {{{{x^2}} \over 2}} \right)} \right]_0^1$ + 8
= ${\left( {{1 \over 3} + {3 \over 2}} \right)}$ + 8
= ${{11} \over 6} + 8$
= ${{59} \over 6}$
2019
Q143
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the region bounded by the parabola, y = x2 + 2 and the lines, y = x + 1, x = 0 and x = 3, is
Show Answer
Correct Answer: B
Explanation:
Required area
$ = \int\limits_0^3 {\left( {{x^2} + 2} \right)dx - {1 \over 2}.5.3 = 9 + 6 - {{15} \over 2}} {}$
$ = {{15} \over 2}$
2019
Q144
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) in the first quadrant bounded by the parabola, y = x2 + 1, the tangent to it at the point (2, 5) and the coordinate axes is :
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Correct Answer: D
Explanation:
Area $ = \int\limits_0^2 {\left( {{x^2} + 1} \right)dx - {1 \over 2}} \left( {{5 \over 4}} \right)\left( 5 \right) = {{37} \over {24}}$
2019
Q145
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 is :
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Correct Answer: D
Explanation:
x = 4y $-$ 2 & x
2 = 4y
$ \Rightarrow $ x
2 = x + 2 $ \Rightarrow $ x
2 $-$ x $-$ 2 = 0
x = 2, $-$ 1
So, $\int\limits_{ - 1}^2 {\left( {{{x + 2} \over 4} - {{{x^2}} \over 4}} \right)\,dx = {9 \over 8}} $
2019
Q146
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the area enclosed between the curves y = kx2 and x = ky2 , (k > 0), is 1 square unit. Then k is -
B.
${{\sqrt 3 } \over 2}$
C.
${2 \over {\sqrt 3 }}$
D.
${1 \over {\sqrt 3 }}$
Show Answer
Correct Answer: D
Explanation:
Area bounded by
2 = 4ax & x2 = 4by, a, b $ \ne $ 0
2 $ = {1 \over 3}$
2019
Q147
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area of the region
Show Answer
Correct Answer: B
Explanation:
Required area
$ = \int\limits_{ - 1}^1 {\left( {x\left| x \right| + 1} \right)} dx$
$ = 0 + \left( x \right)_{ - 1}^1 = 2$
2019
Q148
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
The area (in sq. units) bounded by the parabolae y = x2 – 1, the tangent at the point (2, 3) to it and the y-axis is :
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Correct Answer: C
Explanation:
Equation of tangent at (2, 3) on the parabola y = x
2 $-$ 1 is
${{y + 3} \over 2} = 2x - 1$
$ \Rightarrow $ y + 3 = 4x $-$ 2
$ \Rightarrow $ y = 4x $-$ 5
When x = 0 then for the tangent y = $-$ 5
$ \therefore $ Tangent cuts x y axis at (0, $-$ 5) point.
$ \therefore $ Area of the bounded region is
= $\int\limits_{ - 5}^3 {{{y + 5} \over 4}} \,\,\,dy - \int\limits_{ - 1}^3 {\sqrt {y + 1} } \,\,\,dy$
= ${1 \over 4}\left[ {{{{y^2}} \over 2} + 5y} \right]_{ - 5}^3 - \left[ {{2 \over 3} \times {{\left( {y + 1} \right)}^{{3 \over 2}}}} \right]_{ - 1}^3$
${1 \over 4}\left[ {\left( {{9 \over 2} + 15} \right) - \left( {{{25} \over 2} - 25} \right)} \right] - {2 \over 3}{\left( 4 \right)^{{3 \over 2}}}$
= ${1 \over 4}\left[ {{{93} \over 2} + {{25} \over 2}} \right] - {2 \over 3} \times 8$
= ${1 \over 4} \times {{64} \over 2} - {{16} \over 3}$
= $8 - {{16} \over 3}$
= ${8 \over 3}$
2018
Q149
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
If the area of the region bounded by the curves, $y = {x^2},y = {1 \over x}$ and the lines y = 0 and x= t (t >1) is 1 sq. unit, then t is equal to :
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Correct Answer: D
Explanation:
Point of intersection of y = x
2 and y = ${1 \over x}$.
put y = ${1 \over x}$ in y = x
2 , then we get,
${1 \over x} = {x^2}$
$ \Rightarrow $ $\,\,\,$ x
3 $-$ 1 = 0
$ \Rightarrow $ $\,\,\,$ x = 1
$\therefore\,\,\,$ y = 1
$\therefore\,\,\,$ point B = (1, 1)
Area of region ABCDA
= $\int\limits_0^1 {{x^2}} $ dx + $\int\limits_1^t {{1 \over x}} $ dx
$=$ $\left[ {{{{x^3}} \over 3}} \right]_0^1$ + $\left[ {\ell n\,x} \right]_1^t$
= ${1 \over 3}$ + $\ell n\,t$ $-$ $\ell n$ 1
= ${1 \over 3}$ + $\ell n\,t$ [ as $\ell n$ 1 = 0]
given this Area = 1 sq unit.
$\therefore\,\,\,$ ${1 \over 3}$ + $\ell n\,t$ = 1
$ \Rightarrow $ $\ell n\,t$ = ${2 \over 3}$
$ \Rightarrow $ t = e${^{{2 \over 3}}}$
2018
Q150
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826
14 Mar 2026
Let g(x) = cosx2 , f(x) = $\sqrt x $ and $\alpha ,\beta \left( {\alpha < \beta } \right)$ be the roots of the quadratic equation 18x2 - 9$\pi $x + ${\pi ^2}$ = 0. Then the area (in sq. units) bounded by the curve
A.
${1 \over 2}\left( {\sqrt 2 - 1} \right)$
B.
${1 \over 2}\left( {\sqrt 3 - 1} \right)$
C.
${1 \over 2}\left( {\sqrt 3 + 1} \right)$
D.
${1 \over 2}\left( {\sqrt 3 - \sqrt 2 } \right)$
Show Answer
Correct Answer: B
Explanation:
Given quadratic equation,
$18{x^2} - 9\pi x + {\pi ^2} = 0$
$ \Rightarrow \,\,\,18{x^2} - 6\pi x - 3\pi x + {\pi ^2} = 0$
$ \Rightarrow \,\,\,\,6x\left( {3x - \pi } \right) - \pi \left( {3x - \pi } \right) = 0$
$ \Rightarrow \,\,\,\,\left( {3x - \pi } \right)\left( {6x - \pi } \right) = 0$
$\therefore$ $\,\,\,\,x = {\pi \over 3},{\pi \over 6}$
as $\,\,\,\, \propto < B$
$\therefore$ $\,\,\,\, \propto = {\pi \over 6}$ $\,\,\,\,$ and $\,\,\,\,$ $\beta = {\pi \over 3}$
Given, $g\left( x \right) = \cos {x^2}$ and $ + \left( x \right) = \sqrt x $
$y = \left( {gof} \right)x$
$ = \,\,\,\,\,g\left( {f\left( x \right)} \right)$
$ = \,\,\,\,\cos \left( {f{{\left( x \right)}^2}} \right)$
$ = \,\,\,\,\cos {\left( {\sqrt x} \right)^2}$
$ = \,\,\,\,\cos x$
So, the required area in the curve is
Area $ = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {\cos \,\,dx} $
$ = \left[ {\sin x} \right]_{{\pi \over 6}}^{{\pi \over 3}}$
$ = \sin {\pi \over 3} - \sin {\pi \over 6}$
$ = {{\sqrt 3} \over 2} - {1 \over 2}$
$ = {{\sqrt 3 - 1} \over 2}$
For point of intersection
$A = \int\limits_{{1 \over 2}}^1 {({x^2} + 1)dx + } $$\int\limits_1^2 {} $${(x + 1)dx}$
C1 : y2
= ax, C2 : x2
= ay (a > 0)
y2 = 4$\lambda $x and y = $\lambda x$
Let P be the point common to x + y = 1 and y2 = 4x
