CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 6 - MCQExams.com

From a piece of cardboard, in the shape of a trapezium ABCD, and AB$$||$$CD and $$\angle$$BCD$$=90^o$$, quarter circle is removed. Given AB$$=$$BC$$=3.5$$cm and DE$$=2$$cm. Calculate the area of the remaining piece of the cardboard.(Take $$\pi$$ to be $$\dfrac{22}{7}$$)
1141303_5801d059cb2c4b69a1726ef30e9a6058.jpg
  • $$9.625cm^2$$
  • $$6.125cm^2$$
  • $$2.625cm^2$$
  • None of these
The area between the curves y=$$x^{2}$$ and $$y=\frac{2}{1+x^{2}}$$ is-
  • $$\pi -\frac{1}{3}$$
  • $$\pi -2$$
  • $$\pi -\frac{2}{3}$$
  • $$\pi +\frac{2}{3}$$
The area (in sq. units) of the region $$\left\{ {\left( {x,y} \right):{y^2} \ge 2x\,and\,{x^2} + {y^2} \le 4x,x \ge 0} \right\}$$ is
  • $$\pi - \frac{4}{3}$$
  • $$\pi - \frac{8}{3}$$
  • $$\pi - \frac{{4\sqrt 2 }}{3}$$
  • $$\frac{\pi }{2} - \frac{{2\sqrt 2 }}{3}$$
The area of the region $$[(x, y) : x^{2} + y^{2} \leq 1 \leq x + y|$$ is
  • $$\dfrac {\pi}{5}$$
  • $$\dfrac {\pi}{4}$$
  • $$\dfrac {\pi^{2}}{3}$$
  • $$\dfrac {\pi}{4} - \dfrac {1}{2}$$
The area of the region bounded by the x-axis and the curves
$$y = \tan x\left( { - \frac{\pi }{3} \le x \le \frac{\pi }{3}} \right),and\,y = \cot x\left( {\frac{\pi }{6} \le x \le \frac{{3\pi }}{2}} \right)$$ is
  • $$log 2$$
  • $$2log2$$
  • $$log\sqrt 2 $$
  • $$log\left( {\frac{3}{2}} \right)$$
The area of the region bounded by the curves $$y=ex\log x$$ and $$y=\dfrac{\log x}{ex}$$ is
  • $$\dfrac{e^{2}-5}{4e}$$
  • $$\dfrac{e^{2}+1}{2e}$$
  • $$\dfrac{e^{2}}{2}$$
  • $$None\ of\ these$$
Area bounded  by the curve $${x^2} = 4y$$ and the straight line $$x=4y-2$$ is 
  • $$\frac{8}{9}$$ sq. unit
  • $$\frac{9}{8}$$ sq. unit
  • $$\frac{4}{3}$$ sq. unit
  • None of these
If the area enclosed between $$y=m{x}^{2}$$ and $$x=n{y}^{2}$$ is $$\cfrac{1}{3}$$ sq. units, then $$m,n$$ can be roots of (where $$m,n$$ are non zero real numbers)
  • $$2{x}^{2}+5x+1=0$$
  • $$2{x}^{2}+3x-2=0$$
  • $$2{x}^{2}+3x+2=0$$
  • $$2{x}^{2}+5x-1=0$$
In the given figure, a square $$OABC$$ has been inscribed in the quadrant $$OPBQ$$. If $$OA=20cm$$ then the area of the shaded region is $$\left[take\pi=3.14\right]$$
1168351_05db577288cb4de3b938a97f0b42ea5a.png
  • $$214cm^{2}$$
  • $$228cm^{2}$$
  • $$222cm^{2}$$
  • $$242cm^{2}$$
The area bounded by $$y=x^2,x=y^2$$ is
  • $$1 $$
  • $$\dfrac 16$$
  • $$\dfrac 34$$
  • $$None\ of\ these$$
The area of the plane region bounded by the curve $$x + 2 y ^ { 2 } = 0$$ and $$x + 3 y ^ { 2 } = 1$$ is equal to:
  • $$-\dfrac{4}{3}$$
  • $$\dfrac{4}{3}$$
  • $$\dfrac{2}{3}$$
  • None of these
The maximum area of the triangle whose sides $$a, b \, and \, c$$ satisfy $$0 \le a \le 1, \, 1 \le b \le 2$$ and $$2 \le c \le 3$$
  • $$1$$
  • $$\frac {1}{2}$$
  • $$2$$
  • $$\frac {3}{2}$$
The area enclosed between the curve $${y^2} = 4x$$ and line $$y = x$$is
  • $$\frac {8}{3}$$
  • $$\frac {4}{3}$$
  • $$\frac {2}{3}$$
  • $$\frac {1}{2}$$
If the area anclosed by $$f\left( x \right) = \sin x + \cos x,y = a$$ between two consecutive points of extremum is minimum, then the value of a is
  • $$0$$
  • $$-1$$
  • $$1$$
  • $$2$$
The area of the region bounded by the curve $$y = 2x - {x^2}$$ and the line $$y = x$$ is 
  • $$\dfrac{1}{2}$$
  • $$\dfrac{1}{3}$$
  • $$\dfrac{1}{4}$$
  • $$\dfrac{1}{6}$$
Area bounded by the curve $$y = \sin ^ { - 1 } x , y - a x i s$$ and $$y = \cos ^ { - 1 } x$$ is equal to
  • $$( 2 + \sqrt { 2 } )$$ sq. unit
  • $$( 2 - \sqrt { 2 } )$$ sq. unit
  • $$( 1 + \sqrt { 2 } )$$ sq. unit
  • $$( \sqrt { 2 } - 1 )$$ sq. unit
The area bounded by $$|x|=1-y^{2}$$ and $$|x|+|y|=1$$ is
  • $$\dfrac{1}{3}$$
  • $$\dfrac{1}{2}$$
  • $$\dfrac{2}{3}$$
  • $$1$$
Find the area enclosed between $$y=1+\left| x+1 \right| ,y=4$$.
  • $$9$$ sq. units
  • $$25$$ sq. units
  • $$18$$ sq. units
  • $$21$$ sq. units
Area bounded by the curves $$y=x\sin x\ and x-axis$$ between $$x=0\ and x=2\pi$$ is 
  • $$2\pi$$
  • $$3\pi$$
  • $$4\pi$$
  • None of these
The area of the figure bounded by the parabola $$(y-2)^2=x-1$$, the tangent to it at the point with the ordinate $$x=3$$, and the x-axis is
  • $$7$$ sq. units
  • $$6$$ sq. units
  • $$9$$ sq. units
  • None of these
The area bounded by the curve $$y=\cos x$$ and $$y=\sin x$$ between the ordinates $$x=0$$ and $$x=\dfrac {3\pi}{2}$$ is
  • $$4\sqrt {2}+2$$
  • $$4\sqrt {2}-1$$
  • $$4\sqrt {2}+1$$
  • $$4\sqrt {2}-2$$
Let $$y=g(x)$$ be the inverse of a bijective mapping $$f:R\rightarrow R\quad f(x)=3{ x }^{ 3 }+2x$$. The area bounded by graph of $$g(x)$$, the axis and the ordinate at $$x=5$$ is
  • $$\displaystyle\frac { 5 }{ 4 } $$
  • $$\displaystyle\frac { 7 }{ 4 } $$
  • $$\displaystyle\frac { 9 }{ 4 } $$
  • $$\displaystyle\frac { 13 }{ 4 } $$
The area bounded by the curves $$|x|+|y|\ge1$$ and $$x^2+y^2\le1$$ is 
  • 2 sq unit
  • $$\pi$$ sq unit
  • $$(\pi-2)$$ sq unit
  • None of these
The area of the figure bounded by the parabola $$x = - 2 y ^ { 2 } \text { and } x = 1 - 3 y ^ { 2 } $$ is ?
  • $$1 / 6$$
  • $$2/3$$
  • $$3/2$$
  • $$4/3$$
The area bounded by parabola $$y^{2}=x$$, straight line $$y=4$$ and  $$y-axis$$ is-
  • $$\dfrac{16}{3}$$
  • $$7\sqrt{2}$$
  • $$\dfrac{32}{3}$$
  • $$\dfrac{64}{3}$$
The angle between the curves $$y=sin x$$ and $$ y=cos x$$ is : 
  • $$tan^{-1}(2\sqrt{2})$$
  • $$tan^{-1}(3\sqrt{2})$$
  • $$tan^{-1}(3\sqrt{3})$$
  • $$tan^{-1}(5\sqrt{2})$$
The area bounded by $$\dfrac { | x | } { a } + \dfrac { | y | } { b } = 1$$ where $$a > 0$$ and $$b > 0$$ is
  • $$\dfrac { 1 } { 2 } a b$$
  • $$ab$$
  • $$2ab$$
  • $$4ab$$
Area bounded by $$y=x^2$$and line $$y=x$$
  • 2/3
  • 1/6
  • 1/3
  • 1/4
The area bounded by the curve $$y=(x+1)^2,y=(x-1)^2$$ and the line $$y=0$$ is
  • $$\dfrac{1}{6}$$
  • $$\dfrac{2}{3}$$
  • $$\dfrac{1}{4}$$
  • $$\dfrac{1}{3}$$
The area (in $$sq\ unit$$) of the region $$\left\{(x,y):y^{2}\ge 2x\ and\ x^{2}+y^{2} \le 4x,x \ge 0,y\ \ge 0\right\}$$ is:-
  • $$\dfrac {\pi}{2}+\dfrac {2\sqrt {2}}{3}$$
  • $$\pi-\dfrac {4}{3}$$
  • $$\pi-\dfrac {8}{3}$$
  • $$\pi-\dfrac {4\sqrt {2}}{3}$$
Area common to the cutve $$y=\sqrt {9-x^{2}}$$ and $$x^{2}-y^{2}=6x$$ is:
  • $$\dfrac {\pi +\sqrt {3}}{4}$$
  • $$\dfrac {\pi -\sqrt {3}}{4}$$
  • $$3\left(\pi +\dfrac {\sqrt {3}}{4}\right)$$
  • $$None\ of\ these$$
If the point $$(\lambda, \lambda +1)$$ lies inside the region bounded by the curve $$x=\sqrt{25-y^2}$$ and y-axis, then $$\lambda$$ belongs to the interval.
  • $$(-1, 3)$$
  • $$(-4, 3)$$
  • $$(-\infty, -4)\cup (3, \infty)$$
  • None of these
The area enclosed by the curves $$y=x^{2},y=x^{3},x=0$$ and $$x=p$$, where $$p > 1$$, is $$\dfrac{1}{6}$$. then $$p$$ equals 
  • $$8/3$$
  • $$16/3$$
  • $$4/3$$
  • $$2$$
Area of the region bounded by the curve $$y=e^x$$ and lines $$x=0$$ and $$y=e$$ is?
  • $$e-1$$
  • $$1$$
  • $$2$$
  • $$e^2-1$$
The area (in sq units) of the region $$\{ (x,y):{ y }^{ 2 }\ge 2x$$ and $${ x }^{ 2 }+{ y }^{ 2 }\le 4x ,\chi \ge 0, Y\ge 0\}$$ 
  • $$\pi -\frac { 4 }{ 3 } $$
  • $$\pi -\frac { 8 }{ 3 } $$
  • $$\pi -\frac { 4\sqrt { 2 } }{ 3 } $$
  • $$-\frac { 2\sqrt { 2 } }{ 3 } $$
The area bounded by the curve $$xy^{2}=1$$ and the lines $$x=1$$, $$x=2$$ is
  • $$4\left( {\sqrt 2 - 1} \right)$$
  • $$4\left( {\sqrt 2 + 1} \right)$$
  • $$2\left( {\sqrt 2 - 1} \right)$$
  • $$2\left( {\sqrt 2 + 1} \right)$$
The area in square units bounded by the curves $$y = x ^ { 3 } , y = x ^ { 2 }$$ and the ordinates $$x = 1 , x = 2$$ is
  • $$\frac { 17 } { 12 }$$
  • $$\frac { 12 } { 13 }$$
  • $$\frac { 2 } { 7 }$$
  • $$\frac { 7 } { 2 }$$
The area bounded by the curves $$y=\sqrt{-x}$$ and $$x=-\sqrt{-y}$$, where $$x,y\le0$$, is equal to
  • $$\dfrac{2}{3}sq. unit$$
  • $$\dfrac{1}{3}sq. unit$$
  • $$\dfrac{1}{2}sq. unit$$
  • $$Cannot\ be\ determined$$
Let $$T$$ be the triangle with vertices $$\left (0,0\right), \left (0,{c}^{2}\right)\ and \left (c,{c}^{2}\right)$$ and let $$R$$ be the region between $$y=cx$$ and $$y={x}^{2}\ where c>0$$ then 
  • Area $$\left( R \right) =\dfrac { { c }^{ 3 } }{ 6 }$$
  • Area of $$R=\dfrac {e^{x}}3{3}$$
  • $$\lim_{c \rightarrow 0}\dfrac {Area(T)}{Area(R)}=3$$
  • $$\lim_{c \rightarrow 0}\dfrac {Area(T)}{Area(R)}=\dfrac {3}{2}$$
Let $$f(x)=minimum (x+1,\sqrt{1-x}) $$ for all $$x \le 1$$. Then the area bounded by $$y=f(x)$$ and the x-axis is
  • $$\dfrac{7}{3}$$ sq. units
  • $$\dfrac{1}{6}$$ sq. units
  • $$\dfrac{11}{6}$$ sq. units
  • $$\dfrac{7}{6}$$ sq. units
Tangents are drawn from a point $$P$$ to a parabola $$y^{2}=4ax$$. The area enclosed by the tangents and the corresponding chord of contact is $$4a^{2}$$. Then point $$P$$ satisfies
  • $$y^{2}=4ax$$
  • $$y^{2}=2a(x+a)$$
  • $$y^{2}=4a(x-a)$$
  • $$y^{2}=4a(x+a)$$
The area of the figure bounded by the curves $$y ^ { 2 } = 2 x + 1$$ and $$x - y - 1 = 0$$ is 
  • $$\dfrac {2}{3}$$
  • $$\dfrac {4}{3}$$
  • $$\dfrac {8}{3}$$
  • $$\dfrac {16}{3}$$
Area (in $$sq.$$ unit) of region bounded by $$y=2\cos x,\ y=3\tan x$$ and $$y-$$axis is
  • $$1+3ln \left(\dfrac {2}{\sqrt {3}}\right)$$
  • $$1+\dfrac {3}{2}ln3-3ln2$$
  • $$1+\dfrac {3}{2}ln3-ln2$$
  • $$ln \left(\dfrac {3}{2}\right)$$
Let $$f\left( x \right)$$ be a non-negative continuous function such that the area bounded by the curve $$y= f\left( x \right) $$, x-axis and the ordinates $$x=\cfrac { \pi  }{ 4 } $$, $$x=\beta >\cfrac { \pi  }{ 4 } $$ is $$\left( \beta \sin { \beta  } +\cfrac { \pi  }{ 4 } \cos { \beta  } +\sqrt { 2 } \beta -\cfrac { \pi  }{ 2 }  \right) $$. Then $$f\left( \cfrac { \pi  }{ 2 }  \right) $$ is
  • $$\left( 1-\dfrac { \pi }{ 4 } -\sqrt { 2 } \right)$$
  • $$\left( 1-\dfrac { \pi }{ 4 } +\sqrt { 2 } \right)$$
  • $$\left( \cfrac { \pi }{ 4 } +\sqrt { 2 } -1 \right)$$
  • $$\left( \cfrac { \pi }{ 4 } -\sqrt { 2 } +1 \right)$$
The area of (in sq. units ) of the region described by $$ A={(x,y): x^2+y^2 \leq 1\ and\ y^2 \leq 1-x }$$
  • $$\dfrac{\pi}{2}+\dfrac{4}{3}$$
  • $$\dfrac{\pi}{2}-\dfrac{4}{3}$$
  • $$\dfrac{\pi}{2}-\dfrac{2}{3}$$
  • $$\dfrac{\pi}{2}+\dfrac{2}{3}$$
The area of the region bounded by the curves $$y=|x-1|$$ and $$y=3-|x|$$ is?
  • $$2$$ sq. units
  • $$3$$ sq. units
  • $$4$$ sq. units
  • $$6$$ sq. units
The area bounded by $$y=|x-1|$$, $$y=0$$ and $$|x|=2$$ is?
  • $$4$$
  • $$5$$
  • $$3$$
  • $$10$$
Area of the  region containing all points $$(x, y)$$ satisfying $$0\le y\le \sqrt{4-x^{2}}, y \le x^{2}+x+1$$ and $$y=\left[\sin^{2}\dfrac{\pi}{4}+\cos\dfrac{x}{4}\right]$$ is equal to ( where [.] denotes the greatest integer function ). 
  • $$\dfrac{4\pi-1}{6}+\sqrt{3}$$
  • $$\dfrac{4\pi+1}{6}+\sqrt{3}$$
  • $$\dfrac{2\pi-1}{6}+\sqrt{3}$$
  • $$\dfrac{2\pi+1}{6}+\sqrt{3}$$
The area bounded by $$x^2=4ay$$ and $$y=2a$$ is?
  • $$\dfrac{16\sqrt{2}a^2}{3}$$
  • $$\dfrac{16a^2}{3}$$
  • $$\dfrac{8a^2}{3}$$
  • $$\dfrac{8\sqrt{2}a^2}{3}$$
The area enclosed between the curves $${y^2} = x$$ and $$y = |x|\;is$$ - 
  • $$\dfrac {2}{3}$$
  • $$1$$
  • $$\dfrac {1}{6}$$
  • $$\dfrac {1}{3}$$
0:0:1


Answered Not Answered Not Visited Correct : 0 Incorrect : 0

Practice Class 12 Commerce Maths Quiz Questions and Answers