MCQ Questions for Class 12 Commerce Maths Application Of Integrals Quiz 12

The area bounded by circles

$x^2+y^2=r^2$ ,

$r=1, 2$ and rays given by

$2x^2-3xy-2y^2=0$ (

$y > 0$ ) is?

0%
$\pi$
0%
$\dfrac{3\pi}{4}$
0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{4}$

Consider two curves

${C_1}\,:\,y = \frac{1}{x}\,and\,{C_2}:\,y = \,\ell nx$ on the

$xy$ plane. Let

${D_1}$ denotes the region surrounded by

${C_1},{C_2}$ and the lines

$x=1$ and

${D_2}$ denotes the region the region surrounded by

${C_1},{D_2}$ and the line

$x=a$ . If a

${D_1}={D_2}$ then the value of

$'a'$ -

0%
$\frac{e}{2}$
0%
$e$
0%
$1$
0%
$2\left( {e - 1} \right)$

The area bounded by the curves

$y=xe^{x},y=xe^{-x}$ and the line

$x=1$ , is

0%
$\dfrac {2}{e}$
0%
$1-\dfrac {2}{e}$
0%
$\dfrac {1}{e}$
0%
$1-\dfrac {1}{e}$

The ratio of the areas of two regions of the curve

$C_1 : 4x^2 + \pi^2y^2 = 4\pi^2$ divided by the curve

$C_2 : y = -sgn \left(x - \dfrac{\pi}{2}\right) \cos x$ (where sgn(x) denotes signum function) is -

0%
$\dfrac{\pi^2 +4}{\pi^2-2\sqrt{2}}$
0%
$\dfrac{\pi^2-2}{\pi^2+2}$
0%
$\dfrac{\pi^2+6}{\pi^2+3\sqrt{3}}$
0%
$\dfrac{\pi^2+1}{\pi^2-\sqrt{2}}$

$z$ is not purely real then area bounded by curves

$lm\left(z+\dfrac{1}{z}\right) = 0$ and

$|z-1| = 2$ is (in square units)-

0%
$4\pi$
0%
$3\pi$
0%
$2\pi$
0%
$\pi$

Area bounded between the curves

$y=\sqrt{4-x^2}$ and

$y^2=3|x|$ is/are?

0%
$\dfrac{\pi -1}{\sqrt{3}}$
0%
$\dfrac{2\pi -1}{3\sqrt{3}}$
0%
$\dfrac{2\pi -\sqrt{3}}{3}$
0%
$\dfrac{2\pi -\sqrt{3}}{3\sqrt{3}}$

The area bounded by

$y\frac{{\sin x}}{x},x -$ axis and the co-ordinate

$x = 0,x = \frac{\pi }{4}$ is

0%
$\frac{\pi }{4}$
0%
$< \frac{\pi }{4}$
0%
$> \frac{\pi }{4}$
0%
$< \int\limits_0^{\pi /4} {\frac{{\tan x}}{x}}$

The graphs of

$f(x)=x^{2}$ and

$g(x)=cx^{3}(c>0)$ intersect at the points

$(0, 0)$ and

$(\dfrac{1}{c}, \dfrac{1}{c^{2}})$ . If the region which lies between these graphs and over the interval

$[0, \dfrac{1}{c}]$ has the area equal to

$(\dfrac{2}{3})sq.\ units$ , then the value of

$c$ is :

0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$
0%
$1$
0%
$2$

Let function

$f_n$ be the number of way in which a positive integer n can be written as an ordered sum of several positive integers. For example, for

$n=3$ ,

${f_3} = 3,since, 3 = 3, 3 = 2 + 1$ and

$3 = 1+1+1$ . Then

${f_5} =$

0%
$4$
0%
$5$
0%
$6$
0%
$7$

Area bounded by the curves y=

$[\frac{x^{2}}{64}+2],y=x-1$ and x=0 above x-axis is where

$[.]$ denotes greatest

$x$ .

0%
2 sq. unit
0%
3 sq. unit
0%
4 sq. unit
0%
none of these

The area (in square units) bounded by the curve

$y=\sqrt{x},2y=x+3=0$ , x-axis, and lying in the first quadrant is

0%
$9$
0%
$36$
0%
$18$
0%
none

Explanation

Finding intersection of

$y=\sqrt{x}$ and

$2y=x+3$ ,

$2\sqrt{x}=x+3$

$\Rightarrow 4x=x^2+9+6x$

$\Rightarrow x^2+2x+9$

$D=4-36 < 0$

$\Rightarrow$ No intersection

Area bounded by these curves is not bounded, hence it years the infinite.

1131049_1208099_ans_06641c605f994a78808de686e25bf7f9.jpg

Area bounded by

$y=-x^{2}+6x-5,y=-x^{2}+4x-3$ and

$y=3x-15$ for

$x > 1$ , is (in

$sq.\ units$ )

0%
$73$
0%
$\dfrac {13}{6}$
0%
$\dfrac {73}{6}$
0%
$13$

In a system of three curves

$C_{1}, C_{2}$ and

$C_{3}, C_{1}$ is a circle whose equation is

$x^{2}+y^{2}=4$ .

$C_{2}$ is the locus of orthogonal tangents drawn on

$C_{1}. C_{3}$ is the intersection of perpendicular tangents drawn on

$C_{2}$ . Area enclosed between the curve

$C_{2}$ and

$C_{3}$ is-

0%
$8\pi\ sq.\ units$
0%
$16\pi\ sq.\ units$
0%
$32\pi\ sq.\ units$
0%
$None\ of\ these$

The area bounded by the curve

$y = f(x)$ the x-axis & the ordinates x=1 & x=b is

$\left( {b - 1} \right)\,\sin \left( {3b + 4} \right).\,then\,f\left( x \right)is:$

0%
$\left( {x - 1} \right)\cos \left( {3x + 4} \right)$
0%
$\sin (3x + 4)$
0%
$\sin (3x + 4) + 3\left( {x - 1} \right).cos\left( {3x + 4} \right)$
0%
none

Let

$A_{n}$ be the area bounded by the curve

$y=(\tan x)^{n}$ and the lines

$x=0,y=0$ and

$4x-\pi=0$ , where

0%
$A_{n+2}+A_{n}=\dfrac{1}{(n+1)}$
0%
$A_{1}=\dfrac{1}{2}\ln { 2 }$
0%
$$A_{n}
0%
$A_{2}=1=\dfrac{\pi}{4}$

Area of the region defined by

$||x|+|y||\ge 1$ and

$x^{2}+y^{2}\le 1$ is

0%
$1$
0%
$2$
0%
$\pi -2$
0%
$2\pi -1$

The area of a region bounded by

$X$ -axis and the curves defined by

$y = \tan x$

$0 \leq x \leq \frac { \pi } { 4 }$ and

$y = \cot x , \frac { \pi } { 4 } \leq x \leq \frac { \pi } { 2 }$ is

0%
$\log 3$ sq. units
0%
$\log 5$ sq. units
0%
$\log 1$ sq. unit
0%
$\log 2$ sq. unit

The area bounded by the curve

$y = \dfrac { \sin { x } }{ { x } } , x-$ axis and the ordinates

$x=0,x=\dfrac { \pi }{ { 4 } }$ is:

0%
$=\dfrac { \pi }{ { 4 } }$
0%
$<\dfrac { \pi }{ 4 }$
0%
$>\dfrac { \pi }{ 4 }$
0%
None of these

Area of the figure bounded by

$x$ -axis,

$y = \sin ^ { - 1 } x , y = \cos ^ { - 1 } x$ and the first point intersection from the origin is

0%
$2$ $\sqrt { 2 }$
0%
$2 \sqrt { 2 } + 1$
0%
$\sqrt { 2 } - 1$
0%
$\sqrt { 2 } + 1$

The parabolas

$y^2=4x, x^2=4y$ divide the square region bounded by the lines

$x=4$ ,

$y=4$ and the coordinate axes. If

$S_1, S_2, S_3$ are respectively the area of these parts numbered from top to bottom then

$S_1 : S_2 : S_3$ is?

0%
$2:1:1$
0%
$1:1:1$
0%
$1:2:1$
0%
$1:2:3$

The area (in sq.units) of the region

$\left\{ ( x , y ) :{ y} ^ { 2 } \ge 2 x\right.$ and

${x} ^ { 2 } +{ y} ^ { 2 } \le 4 x , x \ge 0 , y \ge 0$ is

0%
$\pi - \dfrac { 8 } { 3 }$
0%
$\pi - \dfrac { 4 \sqrt { 2 } } { 3 }$
0%
$\dfrac { \pi } { 2 } - \dfrac { 2 \sqrt { 2 } } { 3 }$
0%
$\pi - \dfrac { 4 } { 3 }$

Explanation

Consider

$y^2=2x$ and

$x^2+y^2=4x$ , on solving,

we get

$x^2+2x=4x$

$x^2=2x\Rightarrow x=0, 2$

The integral lies between

$0$ and

$2$

$y^2=2x$

$y^2+x^2=4x$

$\Rightarrow y=\sqrt{2x}$

$\Rightarrow y=\sqrt{4x-x^2}$

Area

$=\displaystyle\int^2_0\sqrt{4x-x^2}-\sqrt{2x}dx$

$=\displaystyle\int^2_0\sqrt{2^2-(2-x)^2}-\displaystyle\int^2_0\sqrt{2x}dx$

$=\left[\dfrac{x-2}{2}\sqrt{4x-x^2}-\dfrac{4}{2}\sin^{-1}\left(\dfrac{x-2}{2}\right)\right]^2_0-\left[\dfrac{\sqrt{2}}{3/2}x^{3/2}\right]^2_0$

$=\left|2\sin^{-1}(-1)\right|-\dfrac{2\sqrt{2}}{3}\cdot 2\sqrt{2}$

$=\pi -\dfrac{8}{3}$ .

1213937_1245590_ans_c310c9417689413393c2f17b77e159f0.jpg

Find the area of shaded portion

0%
$90\ cm^{2}$
0%
$80\ cm^{2}$
0%
$110\ cm^{2}$
0%
$70\ cm^{2}$

Explanation

Area of rectangle

$=18\times 10$

$=180sq\;cm$

$\Rightarrow$ Triangle area

$DEF=\dfrac{1}{2}\times 10\times 6$

$=30\;sq\;cm$

$\Rightarrow$ Triangle area

$BCE=\dfrac{1}{2}\times 10\times 8$

$=40\;sq\;cm$

$\Rightarrow$ Area shaded region

$=180-30-40$

$=110\;sq\;cm$

Hence, the answer is

$110\;sq\;cm.$

1187904_1222859_ans_daecbc17ba6847da97d203adc8ce073a.png

The area bounded by the curves

$\sqrt{x}+\sqrt{y}=1$ and

${x}+{y}=1$ is ?

0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{5}{6}$
0%
$\dfrac{1}{4}$

Explanation

$\textbf{Step -1: Finding the point of intersection.}$

$\text{Find the points of intersection of the curves}$

$\implies \sqrt x=1-\sqrt y$

$\implies x=(1-\sqrt y)^2=1+y-2\sqrt y$

$\implies 1+y-2\sqrt y +y=1$

$\implies 2y=2\sqrt y$

$\implies y=0,1$

$\implies \text{Point of intersections are }(1,0)\text{ and }(0,1)$

$\textbf{Step -2: Calculating the area .}$

$\sqrt y=1-\sqrt x$

$\implies y=1+x-2\sqrt x$

$\implies f_1(x)=x-2\sqrt x+1$

$\implies f_2(x)=1-x$

$\implies \text{Area}=\int_0^1(f_1(x)-f_2(x))dx$

$\implies \int_0^1(2x-2\sqrt x)dx$

$\implies x^2-2\dfrac{x^{\frac32}}{\frac32}|_0^1$

$\implies |1-\dfrac43|=|-\dfrac13|=\dfrac13$

$\textbf{Hence, The required area is }\mathbf{\dfrac13}\textbf{ and correct answer is option A .}$

Area bounded by the curves

$y=\cos^{-1}(\sin x)$ and

$y=\sin^{-1}(\sin x)$ in the interval

$[0, \pi]$ is

0%
$\dfrac{\pi^{2}}{16}$
0%
$\dfrac{\pi^{2}}{32}$
0%
$\dfrac{\pi^{2}}{4}$
0%
$\dfrac{\pi^{2}}{8}$

Area bounded between asymptomes of curves

$f(x)$ and

$f^{-1}(x)$ is

0%
$4$
0%
$9$
0%
$16$
0%
$25$

The area of the region bounded by the X-axis and the curves defined by

$y=tanx\left( \dfrac { -\pi }{ 3 } \le x\le \dfrac { \pi }{ 3 } \right) and\quad y=cotx\left( \dfrac { \pi }{ 6 } \le x\le \dfrac { 3\pi }{ 2 } \right)$

0%
$log\dfrac { 3 }{ 2 }$
0%
$log\sqrt { \dfrac { 3 }{ 2 } }$
0%
$2log\dfrac { 3 }{ 2 }$
0%
$log\left( \dfrac { 3 }{ \sqrt { 2 } } \right)$

The area of the region bounded by the X-axis and the curves defined by

$y=\tan { x } \left( \dfrac { -\pi }{ 3 } \le x\le \dfrac { \pi }{ 3 } \right)$ and

$y=\cot { x } \left( \dfrac { \pi }{ 6 } \le x\le \dfrac { 3\pi }{ 2 } \right)$

0%
$\log { \dfrac { 3 }{ 2 } }$
0%
$\log { \sqrt { \dfrac { 3 }{ 2 } } }$
0%
$2\log { \dfrac { 3 }{ 2 } } c$
0%
$\log { \left( \dfrac { 3 }{ \sqrt { 2 } } \right) }$

Area bounded by

$y|y|-x|x|=1,\ y|y|+x|x|=1$ and

$y=|x|$ is

0%
$\dfrac{\pi}{2}$
0%
${\pi}$
0%
$\dfrac{\pi}{4}$
0%
$None\ of\ these$

The area bounded by the curves is

$\sqrt{\left|x\right|}+\sqrt{\left|y\right|}=\sqrt{a}$ and

$x^{2}+y^{2}=a^{2}$ (where

$a>0$ ) is

0%
$\left(\pi-\dfrac{2}{3}\right)a^{2}\ sq\ units$
0%
$\left(\pi+\dfrac{2}{3}\right)a^{2}\ sq\ units$
0%
$\left(\pi+\dfrac{2}{3}\right)a^{3}\ sq\ units$
0%
$\left(\pi-\dfrac{2}{3}\right)a^{3}\ sq\ units$

The area enclosed between the curves

$y=\left|x^{3}\right|$ and

$x=y^{3}$ is

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{8}$
0%
$\dfrac{1}{16}$

Area bounded by the loop of the curve

$x( x+ {y ^2})= {x}^{3}- {y}^{2}$ equals

0%
$\cfrac { \pi} {2}$
0%
$1- \cfrac { \pi} {4}$
0%
$2- \cfrac { \pi} {2}$
0%
$\pi$

If the slope of a tangent to the curve

$y=f(x)$ is

$4x+3$ . The curve passes through the point

$(1, 5)$ then area bounded by the curve, and the line

$x=1$ in first quadrant is?

0%
$\dfrac{11}{6}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{13}{6}$
0%
$\dfrac{6}{13}$

What is the area of a plane figure bounded by the points of the lines max

$(x,y)=1$ and

$x^{2}+y^{2}=1$ ?

0%
$4-\pi \ sq.\ units$
0%
$\dfrac{\pi}{3} \ sq.\ units$
0%
$1-\dfrac{\pi}{4}\ sq.\ units$
0%
$4+ \pi \ sq.\ units$

Let

$f\left( x \right) =maximum\quad \left\{ { x }^{ 2 },\left( 1-x \right) ^{ 2 },2x\left( 1-x \right) \right\}$ where

$x\epsilon \left[ 0,1 \right]$ . The area of the region bounded by the curve

$v$ and the lines

$y=0,x=0,x=1$

0%
$\frac { 17 }{ 27 }$
0%
$\frac { 27 }{ 17 }$
0%
$\frac { 17 }{ 9 }$
0%
None of these

The area of the region bounded by the limits x = 0,

$x\quad =\quad \frac { \pi }{ 2 }$ and f(x)=sinx, g(x) = cos x is:-

0%
$2(\sqrt { 2 } +1)$
0%
$\sqrt { 3 } -1$
0%
$2(\sqrt { 3 } -1)$
0%
$2(\sqrt { 2 } -1)$

The area bounded by the curve

${ x }^{ 2/3 }+{ y }^{ 2/3 }={ a }^{ 2/3 },+vex-axis\& +vey-axis\quad is$ :-

0%
$\dfrac { { \pi a }^{ 2 } }{ 32 }$
0%
$\dfrac { 3{ \pi a }^{ 2 } }{ 32 }$
0%
$\dfrac { 5{ \pi a }^{ 2 } }{ 32 }$
0%
$\dfrac { 3{ \pi a }^{ 2 } }{ 16 }$

The area bounded by the curve

$y^2=4x$ with the line x=1,x=9 is

0%
$\cfrac {436}{15}$
0%
$\cfrac {208}{3}$
0%
$\cfrac {236}{5}$
0%
$\cfrac {340}{13}$

The area bounded by the curve

$y= x+\sin x$ and its inverse function between the ordinates

$x= 0$ and

$x= 2\pi$ is

0%
$8 \pi$ sq unit
0%
$4 \pi$ sq unit
0%
$8$ sq unit
0%
None of these

The area of the region enclosed by

$y={ x }^{ 3 }-{ 2x }^{ 2 }+2$ and

$y=3x+2$ is

0%
$\frac { 71 }{ 6 }$
0%
14
0%
$\frac { 39 }{ 3 }$
0%
$\frac { 71 }{ 3 }$

The area of region

$\{ (x,y):{ x }^{ 2 }+{ y }^{ 2 }\le 1\le x+y\}$ is:

0%
$\frac { { \pi }^{ 2 } }{ 5 } sq.$ unit
0%
$\frac { { \pi }^{ 2 } }{ 2 } sq.$ unit
0%
$\frac { { \pi }^{ 2 } }{ 4 } sq.$ unit
0%
$\left( \frac { \pi }{ 4 } -\frac { 1 }{ 2 } \right) sq.$ unit

The area

$(in sq. Units)$ of the region

$\left\{ \left( x,y \right) :x\ge 0,x+y\le 3,{ x }^{ 2 }\le 4y and y\le 1+\sqrt { x } \right\}$ is:

0%
$\dfrac { 59 }{ 12 }$
0%
$\dfrac { 3 }{ 2 }$
0%
$\dfrac { 7 }{ 3 }$
0%
$\dfrac { 5 }{ 2 }$

The area enclosed by

$|x - 1| + |y - 3| = 1$ is equal to

0%
4 sq. units
0%
6 sq. units
0%
1 sq. units
0%
2 sq. units

Area of the contained between the parabola

$x^2=4y$ and the curve

$y=\dfrac{8}{x^2+4}$ is

$2\pi-K$ then K=

0%
$\dfrac{2}{3}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{8}{3}$
0%
$\dfrac{1}{3}$

If the area of the region bounded by the curves,

$y=x^{2}, y=\dfrac{1}{x}$ and the lines

$y=0$ and

$x=t(t>1)$ is

$1$ sq. units, then

$t$ is equal to :

0%
$e^{\dfrac{3}{2}}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{3}{2}$
0%
$e^{\dfrac{2}{3}}$

The area (in sq. units)of the region

$\left\{ {x \in R:x \ge 0,y \ge 0,y \ge x - 2\,and\,y \le \sqrt x } \right\},$ is:

0%
$\frac{{13}}{3}$
0%
$\frac{{8}}{3}$
0%
$\frac{{10}}{3}$
0%
$\frac{{5}}{3}$

The area between the curve

$y = 4 + 3 x - x ^ { 2 }$ and

$x -$ axis is

0%
$125/ 6$
0%
$125/ 3$
0%
$125/$
0%
None

Find area of region represented by

$3x+4y > 12, 4x+3y > 12$ and

$x+y < 4$ .

0%
$2-\dfrac{6}{7}=\dfrac{8}{7}$
0%
$2+\dfrac{6}{7}=\dfrac{8}{7}$
0%
$2+\dfrac{6}{7}=\dfrac{7}{8}$
0%
$\dfrac{6}{7}=\dfrac{7}{8}$

The area of the region bounded by the curves

$y = ex \log x$ and

$y = \dfrac { \log x } { ex }$ is

0%
$\dfrac { e } { 4 } - \dfrac { 5 } { 4 e }$
0%
$\dfrac { e } { 4 } + \dfrac { 5 } { 4 e }$
0%
$\dfrac { e } { 3 } - \dfrac { 5 } { 4 e }$
0%
$5e$

The area (in square units) of the region described by

$A={(x,y):y\ge x^{2}-5x+4,x+y\ge 1,y\le 0}$ is

0%
$\dfrac {19}{6}$
0%
$\dfrac {17}{6}$
0%
$\dfrac {7}{2}$
0%
$\dfrac {13}{6}$

The area bounded by the hyperbola

$x^2 - y^2 = 4$ between the lines

$x = 2$ and

$x = 4$ is

0%
$4\sqrt{3} - 2 \,log(2 + \sqrt{3})$
0%
$8\sqrt{3} - 4 \,log(2 - \sqrt{3})$
0%
$8\sqrt{3} - 4 \,log(2 + \sqrt{3})$
0%
$4\sqrt{3} - 2 \,log(2 - \sqrt{3})$

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 12 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 12 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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