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RD Sharma Solutions Class 12 Mathematics Chapter 20 MCQ

RD Sharma Solutions Class 12 Mathematics Chapter 20 MCQ

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 03:18 PM IST

It is quite famous that the solutions of RD Sharma books are an efficient path for any student to follow to ace the subject of mathematics. The Class 12 RD Sharma chapter 20 exercise MCQ solution is one of the examples for it. The solution is constructive in every way out and is also exam-oriented, simple, and essential. The RD Sharma class 12th exercise MCQ can be used for self-practice and evaluation of marks.

This Story also Contains
  1. RD Sharma Class 12 Solutions Chapter20 MCQ Areas Of Bounded Region - Other Exercise
  2. Areas of Bounded Regions Excercise:MCQ
  3. RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter20 MCQ Areas Of Bounded Region - Other Exercise

Areas of Bounded Regions Excercise:MCQ

Areas of Bounded Region exercise multiple choice question 1

Answer:
1(b)
Hint:
Make the graph of 2kx and then apply integration
Given:
y=2kx and x=0,x=2 is 3loge2
Explanation:

Area of the bounded region
=022kxdx
3loge2=022kxdx3loge2=[2kxloge2]023=[2kx]023=(2k)2(2k)03=22k1 [axdx=axlogea][a0=1]
4=22k22=22k2k=2k=1

Areas of Bounded Regions exercise multiple choice question 2

Answer:
(c)
Hint:
Find the bounded region between the parabolas
Given:
y2 = 4x and x2 = 4y
Explanation:

Intersection point is
y2=4x=4(2y)[x2=4y]y2=8yy4=64yy464y=0y(y364)=0y=0,y3=64,y=4x=4
Area between the parabolas
=04(2xx24)dx=1404(8xx2)dx=14[8x3232x33]04=14[163(4)324330+0]=14[163(22)32643]=14[1283643]=163sq.units


Areas of Bounded Regions exercise multiple choice question 3

Answer:
(b)
Hint:
Construct the graph
Given:
y=logex and x - axis, the straight line x = e
Explanation:

Area bounded by the curve, x - axis and x = e
=1elogexdx=[x.logex]1e1ex.1xdx=elogeeloge11edx=e(1)0[x]1e[ Integration by parts ]=e[e1]=ee+1=1sq.units


Areas of Bounded Regions exercise multiple choice question 4

Answer:
(b)
Hint:
Find the region between parabola and line
Given:
y=2x2,x+y=0
Explanation:

To find the point of intersection of
y=2x2 and x+y=0
x=yy=2x2x=2x2x2x2=0x22x+x2=0x(x2)+1(x2)=0(x+1)(x2)=0x=1,2
Area of the region ABCD,
A=12(2x2(x))dx=12(2x2+x)dx=[2xx33+x22]12=[2(2)(2)33+(2)22][2(1)(1)33+(1)22]=[483+2][2+13+12]=683+21312=512=92sq.units


Areas of Bounded Regions exercise multiple choice question 5

Answer:
323(b)
Hint:
Find the region between parabola and y - axis
Given:
x = 4 - y2 and y-axis (x=0)
Explanation:

Area bounded by region
=22(4y2)dy=[4yy33]22=883+883=16163=323

Areas of Bounded Regions exercise multiple choice question 6

Answer:
(a)
Hint:
Find
An>An+1orAn<An+1
Given:
y=(tanx)n,x=0,y=0andx=π4
Explanation:

An=0π4(tanx)ndxAn2=0π4tann2xdx(i)An=0π4tannxdxAn=0π4tann2xtan2dxAn=0π4tann2x(sec2x1)dxAn=0π4tann2xsec2xdx0π4tann2xdxAn+An2=0π4tann2x(sec2x)dx
Let,
t=tanxdt=sec2xdxx=0x=π4t=0t=1An+An2=01tn2dt=[tn1n1]01=1n1

Areas of Bounded Regions exercise multiple choice question 7

Answer:
(c)
Hint:
Integration
Given:
x2+y26x4y+120
yxandx52
Explanation:
x2+y26x4y+120x26x+99+y24y+44+120(x3)2+(y2)210(x3)2+(y2)21(y2)21(x3)2(y2)1(x3)2y2±1(x3)2 As x52,y=21(x3)2
Area,
=252xdx252(21(x3)2)dx=[x22]252[2x(x3)21(x3)2+12sin1(x3)]252=2582[2(12)12(12)×(32)+12sin1(12)sin1(1)]
=981+38+12(π6+π2)=18+38π6=π6+1+38

Areas of Bounded Regions exercise multiple choice question 8

Answer:
(a)
Hint:
Integration
Given:
y=loge(x+e),x=loge(1y),xaxis
Explanation:
xaxis(y=0)0=loge(x+e)x+e=e0x+e=1x=1ex=loge10x=
Area,
=1e0[log(x+e)]dx+0[ex]dx=[(x+e)log(x+e)]1e0[(x+e)x+e]1e0limk[ex]0k=e[11][1]+limk[ex]0k=1+(10)=2

Areas of Bounded Regions exercise multiple choice question 9

Answer:
(c)
Hint:
Integration
Given:
(y-2)2 = x-1, the tangent to it at the point with ordinate 3 and x-axis.
Explanation:
(y-2)2 = x-1
Tangent,
2(y2)dydx=1dydx=12(y2)
At y = 3
dydx=12
Therefore, the equation of tangent at (2,3)
y3=12(x2)x2y+4=0x=2y4
The area of bounded region
=03[(y2)2+1(2y4)]dy=03[y26y+9]dy=[y336y22+3y]03=333=9


Areas of Bounded Regions exercise multiple choice question 10

Answer:
(a)
Hint:
Integration
Given:
y=sinx,x=0andx=π,xaxis
Explanation:
0πsinxdx=[cosx]0π=cosπ+cos0=1+1=2


Areas of Bounded Regions exercise multiple choice question 11

Answer:
(b)
Hint:
Integration
Given:
y2=4ax,x2=4ay
Explanation:
(x24a)2=4axx416a2=4axx464a3x=0x(x364a3)=0x=0x=4a
Area,
=04a[2axx24a]dx=[2ax3232x312a]04a=43a(4a)32(4a)312a0=43a2a3216a23=32a2316a23=16a23


Areas of Bounded Regions exercise multiple choice question 12

Answer:
(b)
Hint:
Integration
Given:
y = x4 - 2x3 + x2 + 3 , x - axis and minima of y
Explanation:
Finding the minima of y
y=x42x3+x2+3dydx=4x36x2+2xd2ydx2=12x212x+2
Now,
dydx=04x36x2+2x=02x(2x23x+1)=0x=02x23x+1=0(2x1)(x1)=0x=0,1,12
At x = 0
d2ydx2=2>0
At x = 1
d2ydx2=1212+2=2>0
At
x=12d2ydx2=1<0

So,f(x) has minimum value at

x=12

The minimum value is y = 1

1=x42x3+x2+3x42x3+x2+2=0x=1

01(x42x3+x2+3)dx=[x552(x4x)+x33+3x]01=1512+13+3=0=9130


Areas of Bounded Regions exercise multiple choice question 13

Answer:
(b)
Hint:
Integration
Given:
y2=4ax, latus rectum =a, x-axis (y=0)
Explanation:

y = 0
x = 0
=0a2axdx=2a0axdx Area =2a[x3232]0a=43a[a]32=4a23 sq.units 


Areas of Bounded Regions exercise multiple choice question 14

Answer:
(c)
Hint:
Integration
Given:
[(x,y):x2+y21x+y]
Explanation:

x2+y2=1y2=1x2y=1x2,0<x<1x+y=1y=1x
Area
=01(1x2)(1x)dx=[x21x2+12sin1x2x+x22]01=12sin1(12)1+12=12sin1(12)12=12(π61)


Areas of Bounded Regions exercise 5 multiple choice question 15

Answer:
(c)
Hint:
Integration
Given:
y=2x2,y=x2+4
Explanation:
2x2=x2+4x2=4x=±2 Area =22(2x2+42x2)dx=22(x2+4)dx
=[x33+4x]22=83+883+8=16163=48163=323 sq.units 


Areas of Bounded Regions exercise multiple choice question 16

Answer:
(d)
Hint:
Integration
Given:
y=x2+1,x+y=3
Explanation:
Intersection points are
x+y=3x+x2+1=3x2+x2=0x2+2xx2=0x(x+2)1(x+2)=0x=1,x=2
Area
=21[(3x)(x2+1)]dx=21[2xx2]dx=[2xx22x33]21=21213[442+83]=3213+683=92

Areas of Bounded Regions exercise multiple choice question 17

Answer:
(b)
Hint:
Integration
Given:
 The ratio of y=cosx and y=cos2x from x=0 to x=π3
Explanation:
Area of cosx=A1=0π3cosxdx=[sinx]0π3=sinπ2sin0=32
Area of cos2x=A2=0π3cos2xdx=sin2π3sin0=12sin(ππ3)0=sinπ3=322=34
A1:A2=(32)(34)=2:1


Areas of Bounded Regions exercise multiple choice question 18

Answer:
(d)
Hint:
Integration
Given:
y=cosx,xaxis,x=0,x=2π
Explanation:

0π2cosx+π23π2(cosx)dx+3π22π(cosx)dx=[sinx]0π2+[sinx]π23π2+[sinx]3π22π=sinπ2sin0[sin3π2sinπ2]+sin2πsin3π2=1+1+1+1=4


Areas of Bounded Regions exercise multiple choice question 19

Answer:
(a)
Hint:
Integration
Given:
y2=x,2y=x
Explanation:
Intersection point
y2=xy2=2yy22y=0y(y2)=0y=0,y=2x=4
Area
=04(xx2)dx=[23x32]04[x24]04=23×84=43 sq.units 


Areas of Bounded Regions exercise multiple choice question 20

Answer:
(c)
Hint:
Integration
Given:
y=4xx2 and xaxis(y=0)
Explanation:
y=00=4xx2x24x=0x(x4)=0x=0,x=4
Area
=04(4xx2)dx=[4x22x33]04=2(4)2(4)330+0=32643=96643=323 sq.units 

Areas of Bounded Regions exercise multiple choice question 21

Answer:
(b)
Hint:
Integration
Given:
y2(2ax)=x3 and x=2a above xaxis
Explanation:
x=2ay2(2ax)=x3y2=x32axy=x32ax
Let
x=2asin2θdx=4asinθcosθ
Area
=02ax32axdx=0π2(2a)3(sin2θ)32a2asin2θ(4asinθcosθdθ)[Whenx=0,θ=0andx=2a,θ=π2]=0π28a3sin6θ2acos2θ(4asinθcosθdθ)[1sin2θ=cos2θ]
=8a20π2sin6θ(sinθdθ)=8a20π2sin4θdθ[sin4xdx=3x814sin2x+132sin4x+c]=8a2[3θ814sin2θ+132sin4θ]0π2=8a2[3(π2)814sinπ+132sin2π]0=8a2×3(π2)8=32πa2


Areas of Bounded Regions exercise multiple choice question 22

Answer:
(b)
Hint:
Integration
Given:
x2=4y,x=2 and xaxis(y=0)
Explanation:
y=0x2=4(0)x=0 Area =02x24dx=[x312]02=812=23

Areas of Bounded Regions exercise multiple choice question 23

Answer:
(c)
Hint:
Integration
Given:
 Area =(b1)sin(3b+4)x=1,x=b
Explanation:
 Area =1bf(x)dx(b1)sin(3b+4)=1bf(x)dx
Differentiate both sides with respect to b,
f(b)=3(b1)cos(3b+4)+sin(3b+4) Put b=xf(x)=3(x1)cos(3x+4)+sin(3x+4)


Areas of Bounded Regions exercise multiple choice question 24

Answer:
(a)
Hint:
Integration
Given:
y2=8x,x2=8y
Explanation:
y2=8xx=y28x2=8y(y28)2=8y(y464)=8yy4512y=0y(y3512)=0y=0,y3=512y3=83y=8
 Area =08(8xx28)dx=8x08xdx1808x2dx=8[x3232]0818[x33]08=238[8320]124[830]=2×823823=643 sq.units 

Areas of Bounded Regions exercise multiple choice question 25

Answer:
(c)
Hint:
Integration
Given:
y2=8x,xaxis(y=0) and latus rectum y2=4×2x, latus rectum =2Area =2028xdx=4202xdx=42[x3232]02=42×23(2320)=823×22=323

Areas of Bounded Regions exercise multiple choice question 26

Answer:
(d)
Hint:
Integration
Given:
y=x3,xaxis,x=2,x=1
Explanation:

Area OAB
=01x3dx=[x44]01=14
Area OCD
=20x3dx=[x4]20=4
As area is always positive
=4
Total area
=4+14=174


Areas of Bounded Regions exercise multiple choice question 27

Answer:
(c)
Hint:
Integration
Given:
y=x|x|,x=1,x=1
Explanation:

y=x|x|={x2x>0x2x<0
Area of bounded region,
=11x|x|dx
Area AOB,
=10x2dx=[x33]10=13
Area is always positive,
=(13)=13
Area OCD,
=01x2dx=[x33]01=13
Total area
=13+13=23


Areas of Bounded Regions exercise multiple choice question 28

Answer:
(b)
Hint:
Integration
Given:
y=cosx,y=sinx,0xπ2
Explanation:

Intersection point,
cosx=sinxtanx=1x=π4,0xπ2
Area AOB
=0π4(cosxsinx)dx=[sinx+cosx]0π4=sinπ4sin0+cosπ4cos0=12+0+121=1+22=1+2=21


Areas of Bounded Regions exercise multiple choice question 29

Answer:
(c)
Hint:
Integration
Given:
x2+y2=16,y2=6x
Explanation:

Area OABC
=2 Area OAB=[ Area OAD+ Area ABD](i)
Intersection points are
x2+y2=16x2+6x=16[y2=6x]x2+6x16=0x2+8x2x16=0x(x+8)2(x+8)=0x2=0x+8=0x=2,x=8
Not possible in first and third quadrant
x=2
From (i)
Area OABC,
=2[026xdx+2416x2dx]=2[6x3232]02+2[x216x2+162sin1x4]24=436[(2)321]+2[4216(4)2+8sin144][221622+8sin124]
=463(22)+2[0+2×π2128sin112]=1633+2[4π238π6]=1633+8π438π3=433+16π3=43(4π+3) sq.units 
Area of circle
=πr2=π(4)2=16π
Exterior area
=16π43(4π+3)=16π163π43=43(8π3)sq.units

Areas of Bounded Regions exercise multiple choice question 30

Answer:
(b)
Hint:
Integration
Given:
x2 + y2 = 4 and x + y = 2
Explanation:

x2 + y2 = 4
r = 2
Area OAB,
=024x2(2x)dx=024x22+xdx=024x2dx+02(x2)dx=[x24x2+42sin1x2]02+[x222x]02=2244+42sin1220+2sin10+4240+0=2sin11+24=2×π22=π2

Areas of Bounded Regions exercise multiple choice question 31

Answer:
(b)
Hint:
Integration
Given:
y2 = 4x and y = 2x
Explanation:

y=2xy2=4a(2x)2=4x4x2=4x4x24x=04x(x1)=0x=0,x=1
Area OAB
=01(232x)dx=[2x32322x22]01=[43x32x2]01=43(1)32(1)20+0=431=13


Areas of Bounded Regions exercise multiple choice question 32

Answer:
(a)
Hint:
Integration
Given:
x2 + y2 = 4 and x = 0, x = 2
Explanation:

x2+y2=4y=4x2
Area OAB,
=024x2dx=[x24x2+42sin1x2]02=2244+2sin12202sin10=2sin11=2(π2)=π

Areas of Bounded Regions exercise multiple choice question 33

Answer:
(b)
Hint:
Integration
Given:
y2 =4x, y-axis (x = 0) and y=3
Explanation:

x=0y2=4(0)=0y=0 and y=39=4xx=94
Area CAB
=094(32x)dx=[3x2x3232]094=[3x43x32]094=3(94)43(94)320+0=27443((32)2)32=27443×278=27492=94

Areas of Bounded Regions exercise multiple choice question 34

Answer:
(d)
Hint:
Area of the circle = 4 X (Area of circle in first quadrant)
Given:
x2 + y2 = 2
Explanation:

x2+y2=2y2=2x2[r2=2r=2]y=2x2
Area of the circle = 4 X (Area of OAB)
=4022x2dx=4[x22x2+22sin1x2]02=4[222(2)2+1sin1220220sin10]=4[sin11]=4(π2)=2π sq.units 

Areas of Bounded Regions exercise multiple choice question 35

Answer:
(b)
Hint:
Area in all quadrants are equal. So, we do 4 X Area of one quadrant
Given:
x2a2+y2b2=1
Explanation:

Area of the ellipse
=4× Area of OAB =4×0abaa2x2dx(i)x2a2+y2b2=1y2b2=1x2a2y2b2=a2x2a2y=baa2x2
From (i)
=4ba0aa2x2dx=4ba[x2a2x2+a22sin1xa]0a=4ba[a2a2a2+a22sin1aa02a20a22sin10a]=4ba[0+a22sin1100]=4ba[a22(π2)]=πab


Areas of Bounded Regions exercise multiple choice question 36

Answer:
(b)
Hint:
Find intersection point then area
Given:
y = x2 and y = 16
Explanation:

y=1616=x2x=±4
Area of the region
=4416x2dx=[16xx23]44=16(4+4)[(4)33(4)33]=128[2×4×4×43]=1281283=2563

Areas of Bounded Regions exercise multiple choice question 37

Answer:
(b)
Hint:
a2x2dx=x2a2x2+a22sin1(xa)+c
Given:
In first quadrant, x- axis (y = 0), y = x, x2 + y2 = 32
Explanation:

Intersection point,
y=xx2+x2=322x2=32x2=16x=±4
x = -4 is not possible as we have to find in first quadrant and in first quadrant x > 0
x=4y2+x2=32x2=32x=42
Now, y = x
Therefore, y = 4
Area of ABCD = Area of ABD + Area of BCD
=04xdx+44232x2dx=[x22]04+[x232x2+12(32)sin1x42]442=8+42232(42)2+12(32)sin142424232(4)216sin1442=8+0+16sin112321616sin112
=8+16(π2)2(4)16(π4)=8+8π84π=4π sq.units 


Areas of Bounded Regions exercise multiple choice question 38

Answer:
(c)
Hint:
Put the limit y = -1,1
Given:
x = 2y + 3, y = 1, y = -1
Explanation:
Area of the bounded region
=11(2y+3)dy=[2y22+3y]11=[y2+3y]11=1+3(1)23(1)=41+3=6 sq. units 


Areas of Bounded Regions exercise multiple choice question 39

Answer:
(d)
Hint:
Find the intersection point
Given:
x2=4yx=4y2(i)
Explanation:
Intersection points are
x2=4y(4y2)2=4y16y2+416y=4y16y220y+4=04y25y+1=04y24yy+1=04y(y1)1(y1)=0(y1)(4y1)=0y=1,y=14y=1x=2y=14x=4(14)2=1[From (i)]
Area
=12[(x+24)x24]dx=[x28+2x4x212]12=14[(2+483)(122+13)]=14[103(76)]=14(92)=98 sq.units 


Areas of Bounded Regions exercise multiple choice question 40

Answer:
(a)
Hint:
a2x2dx=x2a2x2+a22sin1(xa)+c
Given:
y=16x2,xaxis(y=0)
Explanation:
y=00=16x216x2=0x2=16x=±4
Area bounded by the curve
=4416x2dx=[x216x2+422sin1x4]44=[421642+422sin144][4216(4)2+422sin1(44)][sin11=π2]=0+8sin110+8sin11=8×2×π2=8π sq.units 

Areas of Bounded Regions exercise multiple choice question 41

Answer:
(a)
Hint:
Find the graph between y2 = x and 2y = x
Given:
y2 = x and 2y = x
Explanation:

Intersection point,
y=x2y2=x(x2)2=xx24=xx24x=0x(x4)=0x=0,x=4
Area of the bounded region ABC
=04(xx2)dx=[x3232x24]04=23(4)32(4)24=43 sq.units 


The RD Sharma class 12 solution of Area of bounded region exercise MCQ is used by thousands of students and teachers for the practical knowledge of maths. The RD Sharma class 12th exercise MCQ consists of a total of 41 questions that covers up all the essential concepts of the chapter that is mentioned below:

  • Algorithm to find Area using vertical and horizontal stripes

  • Method to find the Area between two curves

  • The Area between two curves using vertical and horizontal stripes

  • Area of the region

  • Area of the region bounded by the curve

  • Area of the region bounded by a parabola

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