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Area Under Curves

JEE Mathematics · 107 questions · Page 7 of 11 · Click an option or "Show Solution" to reveal answer

Q61
The area of the smaller region enclosed by the curves y2=8x+4y^{2}=8 x+4 and x2+y2+43x−4=0x^{2}+y^{2}+4 \sqrt{3} x-4=0 is equal to
A 13(2−123+8π)\dfrac{1}{3}(2-12 \sqrt{3}+8 \pi)
B 13(2−123+6π)\dfrac{1}{3}(2-12 \sqrt{3}+6 \pi)
C 13(4−123+8π)\dfrac{1}{3}(4-12 \sqrt{3}+8 \pi)
D 13(4−123+6π)\dfrac{1}{3}(4-12 \sqrt{3}+6 \pi)
Correct Answer
Option C
Solution
cos⁡θ=234=32⇒θ=30∘\cos \theta = {{2\sqrt 3 } \over 4} = {{\sqrt 3 } \over 2} \Rightarrow \theta = 30^\circ

Area of the required region

=23(4×12)+42×π6−12×4×23= {2 \over 3}\left( {4 \times {1 \over 2}} \right) + {4^2} \times {\pi \over 6} - {1 \over 2} \times 4 \times 2\sqrt 3
=43+8π3−43=13{4−123+8π}= {4 \over 3} + {{8\pi } \over 3} - 4\sqrt 3 = {1 \over 3}\left\{ {4 - 12\sqrt 3 + 8\pi } \right\}
Q62
The area of the region enclosed by y≤4x2,x2≤9yy \leq 4 x^{2}, x^{2} \leq 9 y and y≤4y \leq 4, is equal to :
A 403\dfrac{40}{3}
B 563\dfrac{56}{3}
C 1123\dfrac{112}{3}
D 803\dfrac{80}{3}
Correct Answer
Option D
Solution
y≤4x2, x2≤9y, y≤4y \le 4{x^2},\,{x^2} \le 9y,\,y \le 4

So, required area

A=2∫04(3y−12y)dyA = 2\int_0^4 {\left( {3\sqrt y - {1 \over 2}\sqrt y } \right)dy}
=2 . 52        [23y32]04= 2\,.\,{5 \over 2}\,\,\,\,\,\,\,\,\left[ {{2 \over 3}{y^{{3 \over 2}}}} \right]_0^4
=103          [432−0]=803= {{10} \over 3}\,\,\,\,\,\,\,\,\,\,\left[ {{4^{{3 \over 2}}} - 0} \right] = {{80} \over 3}
Q63
The area enclosed by the curves y=log⁡e(x+e2),x=log⁡e(2y)y=\log _{e}\left(x+\mathrm{e}^{2}\right), x=\log _{e}\left(\dfrac{2}{y}\right) and x=log⁡e2x=\log _{\mathrm{e}} 2, above the line y=1y=1 is:
A 2+e−log⁡e22+\mathrm{e}-\log _{\mathrm{e}} 2
B 1+e−log⁡e21+e-\log _{e} 2
C e−log⁡e2e-\log _{e} 2
D 1+log⁡e21+\log _{e} 2
Correct Answer
Option B
Solution

According to NTA, the required region A2A_2 which is shaded in crossed lines and comes out to be

A2=∫12(ln⁡2y−ey+e2)dy=1+e−ln⁡2A_2=\int_1^2\left(\ln \frac{2}{y}-e^y+e^2\right) d y=1+e-\ln 2

But according to us the required region A1A_1 comes out to be shaded in parallel lines, which can be obtained as

A1=∫0ln⁡2(ln⁡(x+e2)−2e−x)dx={(x+e2)ln⁡(x+e2)−x+2e−x}∣0ln⁡2=(ln⁡2+e2)ln⁡(ln⁡2+e2)−ln⁡2+1−2e2−2=(ln⁡2+e2)ln⁡(ln⁡2+e2)−ln⁡2−2e2−1\begin{aligned} A_1 &=\int_0^{\ln 2}\left(\ln \left(x+e^2\right)-2 e^{-x}\right) d x \\\\ &=\left.\left\{\left(x+e^2\right) \ln \left(x+e^2\right)-x+2 e^{-x}\right\}\right|_0 ^{\ln 2} \\\\ &=\left(\ln 2+e^2\right) \ln \left(\ln 2+e^2\right)-\ln 2+1 \\\\ & \quad-2 e^2-2 \\\\ &=\left(\ln 2+e^2\right) \ln \left(\ln 2+e^2\right)-\ln 2-2 e^2-1 \end{aligned}

Not given in any option. The region asked in the question is bounded by three curves

y=ln⁡(x+e2)x=ln⁡(2y)x=ln⁡2\begin{aligned} &y=\ln \left(x+e^2\right) \\\\ &x=\ln \left(\frac{2}{y}\right) \\\\ &x=\ln 2 \end{aligned}

There is only one region which satisfies above requirement and which also lies above line y=1y=1 Line y=1y=1 may or may not be the boundary of the region.

Q64
The area of the region {(x,y):∣x−1∣≤y≤5−x2}\left\{(x, y):|x-1| \leq y \leq \sqrt{5-x^{2}}\right\} is equal to :
A 52sin⁡−1(35)−12\dfrac{5}{2} \sin ^{-1}\left(\dfrac{3}{5}\right)-\dfrac{1}{2}
B 5π4−32\dfrac{5 \pi}{4}-\dfrac{3}{2}
C 3π4+32\dfrac{3 \pi}{4}+\dfrac{3}{2}
D 5π4−12\dfrac{5 \pi}{4}-\dfrac{1}{2}
Correct Answer
Option D
Solution
A=∫−11(5−x2−(1−x))dx+∫12(5−x2−(x−1))dxA = \int\limits_{ - 1}^1 {\left( {\sqrt {5 - {x^2}} - (1 - x)} \right)dx + \int\limits_1^2 {\left( {\sqrt {5 - {x^2}} - (x - 1)} \right)dx} }
A=2(x25−x2+52sin⁡−1x5)−2x∣01+x25−x2+52sin⁡−1x5−x22+x∣12\left. {A = 2\left( {{x \over 2}\sqrt {5 - {x^2}} + {5 \over 2}{{\sin }^{ - 1}}{x \over {\sqrt 5 }}} \right) - 2x} \right|_0^1 + \left. {{x \over 2}\sqrt {5 - {x^2}} + {5 \over 2}{{\sin }^{ - 1}}{x \over {\sqrt 5 }} - {{{x^2}} \over 2} + x} \right|_1^2
=(5π4−12)= \left( {{{5\pi } \over 4} - {1 \over 2}} \right)

sq. units

Q65
The area of the region given by {(x,y):xy≤8,1≤y≤x2}\{ (x,y):xy \le 8,1 \le y \le {x^2}\} is :
A 16log⁡e2−14316{\log _e}2 - {{14} \over 3}
B 8log⁡e2−1338{\log _e}2 - {{13} \over 3}
C 16log⁡e2+7316{\log _e}2 + {7 \over 3}
D 8log⁡e2+768{\log _e}2 + {7 \over 6}
Correct Answer
Option A
Solution

 Required area =∫12(x2−1)dx+∫28(8x−1)dx=(x33−x)∣12+(8ln⁡x−x)∣28=[(83−2)−(13−1)]+[8ln⁡8−8−(8ln⁡2−2)]=43+8ln⁡4−6=8ln⁡4−143=16ln⁡2−143\begin{aligned} & \text{ Required area }=\int_1^2\left(x^2-1\right) d x+\int_2^8\left(\dfrac{8}{x}-1\right) d x \\\\ & =\left.\left(\dfrac{x^3}{3}-x\right)\right|_1 ^2+\left.(8 \ln x-x)\right|_2 ^8 \\\\ & =\left[\left(\dfrac{8}{3}-2\right)-\left(\dfrac{1}{3}-1\right)\right]+[8 \ln 8-8-(8 \ln 2-2)] \\\\ & =\dfrac{4}{3}+8 \ln 4-6 \\\\ & =8 \ln 4-\dfrac{14}{3} \\\\ & =16 \ln 2-\dfrac{14}{3}\end{aligned}

Q66
Let qq be the maximum integral value of pp in [0,10][0,10] for which the roots of the equation x2−px+54p=0x^2-p x+\dfrac{5}{4} p=0 are rational. Then the area of the region {(x,y):0≤y≤(x−q)2,0≤x≤q}\left\{(x, y): 0 \leq y \leq(x-q)^2, 0 \leq x \leq q\right\} is :
A 1253\dfrac{125}{3}
B 243
C 164
D 25
Correct Answer
Option B
Solution

Given equation :

4x2−4px+5p=04{x^2} - 4px + 5p = 0

for rational roots, D must be perfect square

D=16p2−4×4×5p=16p(p−5)D = 16{p^2} - 4 \times 4 \times 5p = 16p(p - 5)

So, max. Integral value of

p=9p = 9

for making D is perfect square ∴\therefore

q=9q = 9

Area of shared region

=∫09(x−9)2dx= \int\limits_0^9 {{{(x - 9)}^2}dx}
=(x−9)33∣09=933=243= \left. {{{{{(x - 9)}^3}} \over 3}} \right|_0^9 = {{{9^3}} \over 3} = 243

sq. units

Q67
The area of the region A={(x,y):∣cos⁡x−sin⁡x∣≤y≤sin⁡x,0≤x≤π2}A = \left\{ {(x,y):\left| {\cos x - \sin x} \right| \le y \le \sin x,0 \le x \le {\pi \over 2}} \right\} is
A 5+22−4.5\sqrt 5 + 2\sqrt 2 - 4.5
B 1−32+451 - {3 \over {\sqrt 2 }} + {4 \over {\sqrt 5 }}
C 5−22+1\sqrt 5 - 2\sqrt 2 + 1
D 35−32+1{3 \over {\sqrt 5 }} - {3 \over {\sqrt 2 }} + 1
Correct Answer
Option C
Solution
∣cos⁡x−sin⁡x∣≤y≤sin⁡x|\cos x-\sin x| \leq y \leq \sin x

Intersection point of cos⁡x−sin⁡x=sin⁡x\cos x-\sin x=\sin x

⇒tan⁡x=12\Rightarrow \tan x=\frac{1}{2}

Let ψ=tan⁡−112\psi=\tan ^{-1} \dfrac{1}{2} So, tan⁡ψ=12,sin⁡ψ=15,cos⁡ψ=25\tan \psi=\dfrac{1}{2}, \sin \psi=\dfrac{1}{\sqrt{5}}, \cos \psi=\dfrac{2}{\sqrt{5}} Area

=∫tan⁡−112π4(sin⁡x−(cos⁡x−sin⁡x))dx+∫π4π2(sin⁡x−(sin⁡x−cos⁡x))dx= \int\limits_{{{\tan }^{ - 1}}{1 \over 2}}^{{\pi \over 4}} {(\sin x - (\cos x - \sin x))dx + \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {(\sin x - (\sin x - \cos x))dx} }
=∫tan⁡−112π/4(2sin⁡x−cos⁡x)dx+∫π/4π/2cos⁡xdx=\int\limits_{{{\tan }^{ - 1}}{1 \over 2}}^{\pi / 4}(2 \sin x-\cos x) d x+\int\limits_{\pi / 4}^{\pi / 2} \cos x d x
=[−2cos⁡x−sin⁡x]tan⁡−112π/4+[sin⁡x]π/4π/2=[-2 \cos x-\sin x]_{{{\tan }^{ - 1}}{1 \over 2}}^{\pi / 4}+[\sin x]_{\pi / 4}^{\pi / 2}
=−2−12+2cos⁡(tan⁡−112)+sin⁡(tan⁡−112)+(1−12)=-\sqrt{2}-\frac{1}{\sqrt{2}}+2 \cos ({{{\tan }^{ - 1}}{1 \over 2}}) +\sin ({{{\tan }^{ - 1}}{1 \over 2}})+\left(1-\frac{1}{\sqrt{2}}\right)

=−2−12+2(25)+(15)+1−12=-\sqrt{2}-\dfrac{1}{\sqrt{2}}+2\left(\dfrac{2}{\sqrt{5}}\right)+\left(\dfrac{1}{\sqrt{5}}\right)+1-\dfrac{1}{\sqrt{2}}

=5−22+1= \sqrt 5 - 2\sqrt 2 + 1
Q68
Let Δ\Delta be the area of the region {(x,y)∈R2:x2+y2≤21,y2≤4x,x≥1}\left\{ {(x,y) \in {R^2}:{x^2} + {y^2} \le 21,{y^2} \le 4x,x \ge 1} \right\}. Then 12(Δ−21sin⁡−127){1 \over 2}\left( {\Delta - 21{{\sin }^{ - 1}}{2 \over {\sqrt 7 }}} \right) is equal to
A 23−132\sqrt 3 - {1 \over 3}
B 23−232\sqrt 3 - {2 \over 3}
C 3−43\sqrt 3 - {4 \over 3}
D 3−23\sqrt 3 - {2 \over 3}
Correct Answer
Option C
Solution

Required area =2∫132xdx+∫321(21−x2)dx=2 \int\limits_{1}^{3} 2 \sqrt{x} d x+\int_{3}^{\sqrt{21}} \sqrt{\left(21-x^{2}\right)} d x =2([2(x3/23/2)∣13]+[x221−x2+212sin⁡−1(x21)]321)=2\left(\left[\left.2\left(\dfrac{x^{3 / 2}}{3 / 2}\right)\right|_{1} ^{3}\right]+\left[\dfrac{x}{2} \sqrt{21-x^{2}}+\dfrac{21}{2} \sin ^{-1}\left(\dfrac{x}{\sqrt{21}}\right)\right]_{3}^{\sqrt{21}}\right) =23+21π2−83−21sin⁡−137=Δ=2 \sqrt{3}+\dfrac{21 \pi}{2}-\dfrac{8}{3}-21 \sin ^{-1} \sqrt{\dfrac{3}{7}}=\Delta ∴12(Δ−21sin⁡−1(27))=3−43\therefore \dfrac{1}{2}\left(\Delta-21 \sin ^{-1}\left(\dfrac{2}{\sqrt{7}}\right)\right)=\sqrt{3}-\dfrac{4}{3}

Q69
Let [x][x] denote the greatest integer ≤x\le x. Consider the function f(x)=max⁡{x2,1+[x]}f(x) = \max \left\{ {{x^2},1 + [x]} \right\}. Then the value of the integral ∫02f(x)dx\int\limits_0^2 {f(x)dx} is
A 5+423{{5 + 4\sqrt 2 } \over 3}
B 4+523{{4 + 5\sqrt 2 } \over 3}
C 8+423{{8 + 4\sqrt 2 } \over 3}
D 1+523{{1 + 5\sqrt 2 } \over 3}
Correct Answer
Option A
Solution

Combining both the graph we get, Both graph cuts each other at

y=2y=2

and

y=1y=1
∴x2=2\therefore x^2=2
⇒x=2\Rightarrow x=\sqrt2

Two graphs cuts each other at

x=1x=1

and

x=2x=\sqrt2

In 0 to 1,

f(x)=max⁡(x2,1+[x])=1f(x) = \max ({x^2},1 + [x]) = 1

In 1 to

2\sqrt2

,

f(x)=max⁡(x2,1+[x])=2f(x) = \max ({x^2},1 + [x]) = 2

In

2\sqrt2

to 2,

f(x)=max⁡(x2,1+[x])=x2f(x) = \max ({x^2},1 + [x]) = {x^2}

∴\therefore

∫02f(x)dx\int\limits_0^2 {f(x)dx}
=∫011 dx+∫122 dx+∫22x2 dx= \int\limits_0^1 {1\,dx + \int\limits_1^{\sqrt 2 } {2\,dx + \int\limits_{\sqrt 2 }^2 {{x^2}\,dx} } }
=[x]01+2[x]12+[x33]22= \left[ x \right]_0^1 + 2\left[ x \right]_1^{\sqrt 2 } + \left[ {{{{x^3}} \over 3}} \right]_{\sqrt 2 }^2
=1+2(2−1)+83−223= 1 + 2(\sqrt 2 - 1) + {8 \over 3} - {{2\sqrt 2 } \over 3}
=83−1+22−223= {8 \over 3} - 1 + 2\sqrt 2 - {{2\sqrt 2 } \over 3}
=53+62−223= {5 \over 3} + {{6\sqrt 2 - 2\sqrt 2 } \over 3}
=53+423= {5 \over 3} + {{4\sqrt 2 } \over 3}
=5+423= {{5 + 4\sqrt 2 } \over 3}
Q70
Let A={(x,y)∈R2:y≥0,2x≤y≤4−(x−1)2}A=\left\{(x, y) \in \mathbb{R}^{2}: y \geq 0,2 x \leq y \leq \sqrt{4-(x-1)^{2}}\right\} and B={(x,y)∈R×R:0≤y≤min⁡{2x,4−(x−1)2}}.  B=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: 0 \leq y \leq \min \left\{2 x, \sqrt{4-(x-1)^{2}}\right\}\right\} \text{. } . Then the ratio of the area of A to the area of B is
A ππ+1\dfrac{\pi}{\pi+1}
B π−1π+1\dfrac{\pi-1}{\pi+1}
C ππ−1\dfrac{\pi}{\pi-1}
D π+1π−1\dfrac{\pi+1}{\pi-1}
Correct Answer
Option B
Solution

y2+(x−1)2=4y^{2}+(x-1)^{2}=4 shaded portion == circular (OABC)(\mathrm{OABC})

−Ar⁡(ΔOAB)=π(4)4−12(2)(1)A=(π−1)\begin{aligned} & -\operatorname{Ar}(\Delta \mathrm{OAB}) \\\\ & =\frac{\pi(4)}{4}-\frac{1}{2}(2)(1) \\\\ & \mathrm{A}=(\pi-1) \end{aligned}

Area B=Ar⁡(△AOB)+B=\operatorname{Ar}(\triangle \mathrm{AOB})+ Area of arc of circle (ABC)(\mathrm{ABC})

=12(1)(2)+π(2)24=π+1=\frac{1}{2}(1)(2)+\frac{\pi(2)^{2}}{4}=\pi+1
AB=π−1π+1\frac{\mathrm{A}}{\mathrm{B}}=\frac{\pi-1}{\pi+1}
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