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CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 12 - MCQExams.com
CBSE
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?
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0%
$$\pi$$
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$$\dfrac{3\pi}{4}$$
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 $$'a'$$ -
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0%
$$\frac{e}{2}$$
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$$e$$
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$$ 1$$
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$$2\left( {e - 1} \right)$$
The area bounded by the curves $$y=xe^{x},y=xe^{-x}$$ and the line $$x=1$$, is
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$$\dfrac {2}{e}$$
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$$1-\dfrac {2}{e}$$
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$$\dfrac {1}{e}$$
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$$1-\dfrac {1}{e}$$
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 -
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0%
$$\dfrac{\pi^2 +4}{\pi^2-2\sqrt{2}}$$
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$$\dfrac{\pi^2-2}{\pi^2+2}$$
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$$\dfrac{\pi^2+6}{\pi^2+3\sqrt{3}}$$
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$$\dfrac{\pi^2+1}{\pi^2-\sqrt{2}}$$
If $$z$$ 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)-
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0%
$$4\pi$$
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$$3\pi$$
0%
$$2\pi$$
0%
$$\pi$$
Area bounded between the curves $$y=\sqrt{4-x^2}$$ and $$y^2=3|x|$$ is/are?
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$$\dfrac{\pi -1}{\sqrt{3}}$$
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$$\dfrac{2\pi -1}{3\sqrt{3}}$$
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$$\dfrac{2\pi -\sqrt{3}}{3}$$
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$$\dfrac{2\pi -\sqrt{3}}{3\sqrt{3}}$$
The area bounded by $$y\frac{{\sin x}}{x},x - $$ axis and the co-ordinate $$x = 0,x = \frac{\pi }{4}$$ is
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0%
$$\frac{\pi }{4}$$
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$$ < \frac{\pi }{4}$$
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$$ > \frac{\pi }{4}$$
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$$ < \int\limits_0^{\pi /4} {\frac{{\tan x}}{x}} $$
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 :
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0%
$$\dfrac{1}{3}$$
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$$\dfrac{1}{2}$$
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$$1$$
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$$2$$
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$${f_5} =$$
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0%
$$4$$
0%
$$5$$
0%
$$6$$
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$$7$$
Area bounded by the curves y= $$[\frac{x^{2}}{64}+2],y=x-1$$ and x=0 above x-axis is where $$[.]$$ denotes greatest $$x$$.
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2 sq. unit
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3 sq. unit
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4 sq. unit
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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
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$$9$$
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$$36$$
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$$18$$
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.
Area bounded by $$y=-x^{2}+6x-5,y=-x^{2}+4x-3$$ and $$y=3x-15$$ for $$x > 1$$, is (in $$sq.\ units$$)
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0%
$$73$$
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$$\dfrac {13}{6}$$
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$$\dfrac {73}{6}$$
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$$13$$
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-
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0%
$$8\pi\ sq.\ units$$
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$$16\pi\ sq.\ units$$
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$$32\pi\ sq.\ units$$
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$$None\ of\ these$$
The area bounded by the curve $$y = f(x)$$ the x-axis & the ordinates x=1 & x=b is $$\left( {b - 1} \right)\,\sin \left( {3b + 4} \right).\,then\,f\left( x \right)is:$$
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$$\left( {x - 1} \right)\cos \left( {3x + 4} \right)$$
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$$\sin (3x + 4)$$
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$$\sin (3x + 4) + 3\left( {x - 1} \right).cos\left( {3x + 4} \right)$$
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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
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$$A_{n+2}+A_{n}=\dfrac{1}{(n+1)}$$
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$$A_{1}=\dfrac{1}{2}\ln { 2 } $$
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$$A_{n}
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$$A_{2}=1=\dfrac{\pi}{4}$$
Area of the region defined by $$||x|+|y||\ge 1$$ and $$x^{2}+y^{2}\le 1$$ is
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$$1$$
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$$2$$
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$$\pi -2$$
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$$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 }$$ is
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$$\log 3$$ sq. units
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$$\log 5$$ sq. units
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$$\log 1$$ sq. unit
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$$\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 } }$$ is:
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$$=\dfrac { \pi }{ { 4 } }$$
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$$<\dfrac { \pi }{ 4 }$$
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$$>\dfrac { \pi }{ 4 }$$
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None of these
Area of the figure bounded by $$x$$ -axis, $$y = \sin ^ { - 1 } x , y = \cos ^ { - 1 } x$$ and the first point intersection from the origin is
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$$2$$ $$\sqrt { 2 }$$
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$$2 \sqrt { 2 } + 1$$
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$$\sqrt { 2 } - 1$$
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$$\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$$ is?
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0%
$$2:1:1$$
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$$1:1:1$$
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$$1:2:1$$
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$$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$$ is
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$$\pi - \dfrac { 8 } { 3 }$$
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$$\pi - \dfrac { 4 \sqrt { 2 } } { 3 }$$
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$$\dfrac { \pi } { 2 } - \dfrac { 2 \sqrt { 2 } } { 3 }$$
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$$\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}$$.
Find the area of shaded portion
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$$90\ cm^{2}$$
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$$80\ cm^{2}$$
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$$110\ cm^{2}$$
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$$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.$$
The area bounded by the curves $$\sqrt{x}+\sqrt{y}=1$$ and $${x}+{y}=1$$ is ?
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0%
$$\dfrac{1}{3}$$
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$$\dfrac{1}{6}$$
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$$\dfrac{1}{2}$$
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$$\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]$$ is
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0%
$$\dfrac{\pi^{2}}{16}$$
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$$\dfrac{\pi^{2}}{32}$$
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$$\dfrac{\pi^{2}}{4}$$
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$$\dfrac{\pi^{2}}{8}$$
Area bounded between asymptomes of curves $$f(x)$$ and $$f^{-1}(x)$$ is
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0%
$$4$$
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$$9$$
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$$16$$
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$$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) $$
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0%
$$log\dfrac { 3 }{ 2 } $$
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$$log\sqrt { \dfrac { 3 }{ 2 } } $$
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$$2log\dfrac { 3 }{ 2 } $$
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$$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) $$
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0%
$$\log { \dfrac { 3 }{ 2 } } $$
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$$\log { \sqrt { \dfrac { 3 }{ 2 } } } $$
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$$2\log { \dfrac { 3 }{ 2 } } c$$
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$$\log { \left( \dfrac { 3 }{ \sqrt { 2 } } \right) } $$
Area bounded by $$y|y|-x|x|=1,\ y|y|+x|x|=1$$ and $$y=|x|$$ is
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0%
$$\dfrac{\pi}{2}$$
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$${\pi}$$
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$$\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$$) is
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$$\left(\pi-\dfrac{2}{3}\right)a^{2}\ sq\ units$$
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$$\left(\pi+\dfrac{2}{3}\right)a^{2}\ sq\ units$$
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$$\left(\pi+\dfrac{2}{3}\right)a^{3}\ sq\ units$$
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$$\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}$$ is
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0%
$$\dfrac{1}{2}$$
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$$\dfrac{1}{4}$$
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$$\dfrac{1}{8}$$
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$$\dfrac{1}{16}$$
Area bounded by the loop of the curve $$ x( x+ {y ^2})= {x}^{3}- {y}^{2} $$ equals
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$$ \cfrac { \pi} {2} $$
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$$1- \cfrac { \pi} {4} $$
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$$2- \cfrac { \pi} {2} $$
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$$ \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$$ in first quadrant is?
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0%
$$\dfrac{11}{6}$$
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$$\dfrac{1}{6}$$
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$$\dfrac{13}{6}$$
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$$\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$$?
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$$4-\pi \ sq.\ units$$
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$$\dfrac{\pi}{3} \ sq.\ units$$
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$$1-\dfrac{\pi}{4}\ sq.\ units$$
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$$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$$
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$$\frac { 17 }{ 27 } $$
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$$\frac { 27 }{ 17 } $$
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$$\frac { 17 }{ 9 } $$
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None of these
The area of the region bounded by the limits x = 0,$$x\quad =\quad \frac { \pi }{ 2 } $$ and f(x)=sinx, g(x) = cos x is:-
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$$2(\sqrt { 2 } +1)$$
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$$\sqrt { 3 } -1$$
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$$2(\sqrt { 3 } -1)$$
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$$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$$ :-
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0%
$$\dfrac { { \pi a }^{ 2 } }{ 32 } $$
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$$\dfrac { 3{ \pi a }^{ 2 } }{ 32 } $$
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$$\dfrac { 5{ \pi a }^{ 2 } }{ 32 } $$
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$$\dfrac { 3{ \pi a }^{ 2 } }{ 16 } $$
The area bounded by the curve $$y^2=4x$$ with the line x=1,x=9 is
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0%
$$\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$$ is
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0%
$$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$$ is
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0%
$$\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\} $$ is:
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0%
$$\frac { { \pi }^{ 2 } }{ 5 } sq.$$ unit
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$$\frac { { \pi }^{ 2 } }{ 2 } sq.$$ unit
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$$\frac { { \pi }^{ 2 } }{ 4 } sq.$$ unit
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$$\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\}$$ is:
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0%
$$\dfrac { 59 }{ 12 }$$
0%
$$\dfrac { 3 }{ 2 }$$
0%
$$\dfrac { 7 }{ 3 }$$
0%
$$\dfrac { 5 }{ 2 }$$
The area enclosed by $$|x - 1| + |y - 3| = 1$$ is equal to
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0%
4 sq. units
0%
6 sq. units
0%
1 sq. units
0%
2 sq. units
Area of the contained between the parabola $$x^2=4y$$ and the curve $$y=\dfrac{8}{x^2+4}$$ is $$2\pi-K$$ then K=
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0%
$$\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$$ is equal to :
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0%
$$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\},$$ is:
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0%
$$\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 -$$
axis is
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0%
$$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$$.
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$$2-\dfrac{6}{7}=\dfrac{8}{7}$$
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$$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 }$$ is
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0%
$$\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}$$ is
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0%
$$\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$$ is
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0%
$$4\sqrt{3} - 2 \,log(2 + \sqrt{3})$$
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$$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})$$
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