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CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 10 - MCQExams.com

Area bounded by the curves satisfying the conditions x225+y2361x5+y6 is given by
  • 15(π2+1) sq.units
  • 154(π21) sq.units
  • 30(π1) sq.unit
  • 152(π2) sq.unit
If A1 and A2 respectively represents the area bounded by the curves f(x,y) 4x2y3x and g (x,y)4x2y|3x| then A1 : A2 equals:
  • 2:1
  • 3:1
  • 1:2
  • 1:3
sinx & cosx meet each other at a number of points and develop symmetrical area. Area of one such region is
  • 42
  • 32
  • 22
  • 2
Area bounded by the curves yx=logx and y2=x2+x equals:
  • 7/12
  • 12/7
  • 7/6
  • 6/7
The area of the plane region bounded by the curves x+2y2=0 and x+3y2=1 is
  • 13
  • 23
  • 43
  • 53
The area enclosed between the curves, x2=y and y2=x is equal to:
  • 13 sq. unit
  • 210(xx2)dx
  • Area enclosed by the region {(x,y):x2yx}
  • Area enclosed by the region {(x,y):x2yx}
The function f(x)=max \{x^{2},(1-x)^{2},2x(1-x) \forall 0\leq x \leq 1\} then area of the region bounded by the curve \mathrm{y}=\mathrm{f}(\mathrm{x}) , \mathrm{x}-axis and \mathrm{x}= 0,\ \mathrm{x} = 1 is equals
  • \dfrac{27}{17}
  • \dfrac{17}{27}
  • \dfrac{18}{17}
  • \dfrac{19}{17}

The ratio in which the area bounded by the curves y^2=12x and x^2=12y is divided by the line x = 3 is

  • 15 : 16
  • 15 : 49
  • 1 : 2
  • None of these
The area bounded by the tangent and normal to the curve y(6-x)=x^{2} at (3,3) and the x-axis is
  • 5
  • 6
  • 15
  • 3
If \left| z- (4 + 4i) \right| \geq  4, then area of the region bounded by the locii of z,\; iz,\; - z and -iz is:
  • 4(4-\pi )
  • 16(4-\pi )
  • 16(\pi -1)
  • 4(\pi -1)
The area of the region bounded by the curve y \displaystyle =\frac{16-x^{2}}{4} and \displaystyle y=sec^{-1}[-sin^{2}x], where [.] stands for the greatest integer function is:
  • (4-\pi )^{3/2}
  • \dfrac{8}{3}(4-\pi )^{3/2}
  • \dfrac{4}{3}(4-\pi )^{3/2}
  • \dfrac{8}{3}(4-\pi )
The parabola y^{2} = 4x and 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 the areas of these parts numbered from top to bottom respectively, then
  • S_{1} : S_{2}\equiv 1 : 1
  • S_{2} : S_{3}\equiv 1 : 2
  • S_{1} : S_{3}\equiv 1 : 1
  • S_{1} : (S_{1} + S_{2})\equiv 1 : 2
The area of the smaller region in which the curve y=\left [ \frac{x^{3}}{100}+\frac{x}{50} \right ], where[.] denotes the greatest integer function, divides the circle \left ( x-2 \right )^{2}+\left ( y+1 \right )^{2}=4, is equal to







  • \frac{2\pi-3\sqrt{3}}{3}sq. units
  • \frac{3\sqrt{3}-\pi}{3}sq. units
  • \frac{4\pi-3\sqrt{3}}{3}sq. units
  • \frac{5\pi-3\sqrt{3}}{3}sq. units
  • \frac{4\pi-3\sqrt{3}}{6}sq. units
Area of the region bounded by the curve y = x^{2} and y = sec^{-1} [sin^{2}x] (where [ . ] denotes the greatest integer function) is
  • \frac{\pi}{3}\sqrt{\pi}
  • \frac{2\pi\sqrt{\pi}}{3}
  • \frac{4\pi\sqrt{\pi}}{3}
  • \frac{6\pi\sqrt{\pi}}{3}
  • \frac{3\pi\sqrt{\pi}}{2}
If A_1 is the area bounded by y= \cos x, y = \sin xx=0  and A_2 the area bounded by y = \cos x , y = \sin x , y = 0 in (0,\frac{\pi}{2}) then \displaystyle  \dfrac{A_1}{A_2} equals to :
  • \dfrac{1}{2}
  • \dfrac{1}{\sqrt2}
  • 1
  • None of these
The area bounded by the curves y = sin^{-1} |sin  x| and y = (sin^{-1} | sin  x|)^{2}, where 0\leq x\leq 2\pi , is:
  • \dfrac{1}{3}+\dfrac{\pi ^{2}}{4} sq. units
  • \dfrac{1}{6}+\dfrac{\pi ^{3}}{8} sq. units
  • 2 sq. units
  • \displaystyle \frac{4}{3}+\pi ^{2}\frac{\pi -3}{6}  sq.units
Area lying in the first quadrant and bounded by the circle x^2+y^2=4 and the line x=y\sqrt{3} is:
  • \pi
  • \displaystyle\frac{\pi}{2}
  • \displaystyle\frac{\pi}{3}
  • None of these
Area bounded by the region R\equiv \left \{ (x,y):y^{2}\leq x\leq |y| \right \} is
  • \displaystyle \frac{4}{3}
  • \displaystyle \frac{3}{4}
  • \displaystyle \frac{1}{3}
  • \displaystyle \frac{1}{4}
Area bounded by curves y^2 = x and y = | x | is given by
  • \displaystyle \frac{1}{6}
  • \displaystyle \frac{2}{3}
  • -\displaystyle \frac{1}{6}
  • \displaystyle \frac{1}{3}
If the area bounded by the curve |y|=sin^{-1}|x| and  x=1 is a(\pi+b), then the value a-b is:
  • 1
  • 2
  • 3
  • 4
The area bounded by y=3-|3-x| and \displaystyle y=\frac{6}{|x+1|} is:
  • \dfrac{15}{2} -6 ln 2 sq. units
  • \dfrac{13}{2} -3 ln 2 sq. units
  • \dfrac{13}{2} -6 ln 2 sq. units
  • None of these
The area bounded by y=\sec^ {-1}{x}, y= \text{cosec}^{-1}{x} and the line x-1=0 is:
  • \ln(3+2\sqrt{2})-\displaystyle \frac{\pi}{2}
  • \displaystyle \frac{\pi}{2}+\ln(3+2\sqrt{2})
  • \pi - \ln3
  • \pi + \ln3
The area of the smaller part bounded by the semi-circle \displaystyle y=\sqrt { 4-{ x }^{ 2 } } ,y=x\sqrt { 3 } and x-axis is
  • \displaystyle \frac { \pi  }{ 3 }
  • \displaystyle \frac { 2\pi  }{ 3 }
  • \displaystyle \frac { 4\pi  }{ 3 }
  • none of these
The area enclosed by x^2 + y^2 = 4, y = x^2 + x + 1,  y=\left [ \sin^{2}\displaystyle \frac{x}{4}+\cos\displaystyle \frac{x}{4} \right ] and x-axis (where [.] denotes the greatest integer function) is:
  • \displaystyle \frac{2\pi}{3}+\sqrt{3}-\displaystyle \frac{1}{6}
  • \displaystyle \frac{2\pi}{3}+2\sqrt{3}-\displaystyle \frac{1}{6}
  • 2\sqrt3-\frac{1}{3}
  • \displaystyle \frac{\pi}{3}+\sqrt3
The area bounded by the function f(x)=x^{2}:R^{+}\rightarrow R^{+} and its inverse function is:
  • \dfrac{1}{2}sq.units
  • \dfrac{1}{3}sq.units
  • \dfrac{2}{3}sq.units
  • \dfrac{1}{6}sq.units
Find the area of the region bounded by the curves y= log_{e}x ,  y=\sin ^{4} \pi x , x=0
  • \displaystyle\ \frac{11}{8}sq.units
  • \displaystyle\ \frac{9}{8}sq.units
  • \displaystyle\ \frac{13}{8}sq.units
  • \displaystyle\ \frac{15}{8}sq.units
State the following statement is True or False
The area bounded by the circle x^2 + y^2 = 1, x^2 + y^2 = 4 and the pair of lines \sqrt{3} (x^2 + y^2) = 4xy, is equal to \dfrac{\pi}{2}. The statement is true or false.
  • True
  • False
Find the area bounded by the curves \displaystyle\ y=\sqrt{1-x^{2}} and \displaystyle\ y=x^{3}-x . Also find the ratio in which the y-axis divide this area
  • \displaystyle\ \frac{\pi }{2} , \displaystyle\ \frac{\pi -1}{\pi +1}
  • \displaystyle\ \frac{\pi }{4} , \displaystyle\ \frac{\pi -1}{\pi +1}
  • \displaystyle\ \frac{\pi }{2} , \displaystyle\ \frac{\pi +1}{\pi -1}
  • None of these
Area bounded by the curve y = \sin^{-1} (\sin x), y = x^2 -\pi x is
  • \dfrac{\pi ^2}{4}+\pi ^3
  • \dfrac{\pi ^2}{4}+\dfrac{\pi ^3}{2}
  • \dfrac{\pi ^3}{6}
  • \dfrac{\pi ^2}{4}+\dfrac{\pi ^3}{6}
Find the area enclosed the curves : y=ex\log { x } and \displaystyle y=\frac { \log { x }  }{ ex } where \log { e } =1
  • \displaystyle\frac { { e }^{ 2 }-5 }{ 4e }
  • \displaystyle\frac { { e }^{ 2 }+5 }{ 4e }
  • \displaystyle\frac { { e }^{ 2 }-3 }{ 2e }
  • \displaystyle\frac { { e }^{ 2 }+3 }{ 2e }
Sketch the region bounded by the curves  \displaystyle y=x^{2} &  \displaystyle\ y= 2/(1+x^{2}). Find the area:
  • \displaystyle\ \pi -\frac{2}{3}
  • \displaystyle\ \pi -\frac{1}{3}
  • \displaystyle\ \pi -\frac{5}{3}
  • \displaystyle\ \pi -\frac{7}{3}
The ratio in which the area bounded by the curves y^{2}=4x and x^{2}=4y is divided by the line x=1 is
  • 64 : 49
  • 15 : 34
  • 15 : 49
  • none of these
Find the equation of the line passing through the origin and dividing the curvilinear triangle with vertex at the origin, bounded by the curves y=2x-x^2, y=0 and x=1 into two parts of equal area.
  • y=2x/3
  • y=x/3
  • y=2x/5
  • y=5x/3
For the curve \displaystyle f(x)=\frac{1}{1+x^2}, let two points on it be A(\alpha ,f(\alpha )), B \left ( -\dfrac{1}{\alpha }, f\left (- \dfrac{1}{\alpha } \right )\right) (\alpha > 0). Find the minimum area bounded by the line segments OA, OB and f(x), where 'O' is the origin.
  • \displaystyle \frac{(\pi -1)}{2}
  • \dfrac{\pi}{2}
  • \dfrac{(\pi-2)}{2}
  • Maximum area is always infinite
Find the area of the region enclosed between the two circles \displaystyle\ x^{2}+y^{2}=1 & (x-1)^{2}+y^{2}=1
  • \displaystyle\ \frac{\pi }{6}-\frac{\sqrt{3}}{2} sq.units
  • \displaystyle\ \frac{\pi }{3}-\frac{\sqrt{3}}{2} sq.units
  • \displaystyle\ \frac{\pi }{6}-\frac{\sqrt{3}}{4} sq.units
  • \displaystyle\ \frac{\pi }{3}-\frac{\sqrt{3}}{4} sq.units
The area bounded by the region by the curves \displaystyle \begin{vmatrix}x\end{vmatrix} =1-y^{2} and \displaystyle \begin{vmatrix}x\end{vmatrix}+\begin{vmatrix}y\end{vmatrix}=1 is
  • \displaystyle \frac{1}{3}
  • 1
  • \displaystyle \frac{1}{2}
  • \displaystyle \frac{2}{3}
Compute the area of the curvilinear triangle bounded by the y-axis & the curve, \displaystyle\ y=\tan x &  \displaystyle\ y=(2/3) \cos x
  • \displaystyle\ \frac{1}{3}+ ln\left [ \frac{\sqrt{3}}{2} \right ]sq.units
  • \displaystyle\ \frac{1}{3}- ln\left [ \frac{\sqrt{3}}{2} \right ]sq.units
  • \displaystyle\ \frac{2}{3}+ ln\left [ \frac{\sqrt{3}}{2} \right ]sq.units
  • \displaystyle\ \frac{1}{3}+ ln\left [ \frac{\sqrt{1}}{2} \right ]sq.units
The area lying in the first quadrant inside the circle { x }^{ 2 }+{ y }^{ 2 }=12 and bounded by the parabolas { y }^{ 2 }=4x,{ x }^{ 2 }=4y is:
  • \displaystyle 2\left( \frac { \sqrt { 2 }  }{ 3 } +\frac { 3 }{ 2 } \sin ^{ -1 }{ \frac { 1 }{ 3 }  }  \right)
  • \displaystyle 4\left( \frac { \sqrt { 2 }  }{ 3 } +\frac { 3 }{ 2 } \sin ^{ -1 }{ \frac { 1 }{ 3 }  }  \right)
  • \displaystyle \left( \frac { \sqrt { 2 }  }{ 3 } +\frac { 3 }{ 2 } \sin ^{ -1 }{ \frac { 1 }{ 3 }  }  \right)
  • none of these
A polynomial function f(x) satisfies the condition f(x+1) =f(x)+2x+1. Find f(x) if f(0)=1. Find also the equations of the pair of tangents from the origin on the curve y=f(x) and compute the area enclosed by the curve and the pair of tangents.
  • f(x)=x^2+1;, y=\pm 2x; , \displaystyle A=\frac{2}{3} sq.units
  • f(x)=x^2-1;, y=\pm 2x; , \displaystyle A=\frac{2}{3} sq.units
  • f(x)=x^2+1;, y=\pm 2x; , \displaystyle A=\frac{3}{2} sq.units
  • f(x)=x^2-1;, y=\pm 2x; , \displaystyle A=\frac{3}{2} sq.units
If f(x) be an increasing function defined on [a, b] then
max {f(t) such that a\leq t\leq x, a\leq x\leq b}=f(x)  & min {f(t), a\leq t\leq x, a\leq x\leq b}=f(a) and if f(x) be decreasing function defined on [a, b] then
max {f(t), a\leq t\leq x, a\leq x\leq b}=f(a),
min {f(t), a\leq t\leq x, a\leq x\leq b}=f(x).
On the basis of above information answer the following questions.
Let \displaystyle f\left ( x \right )=min \left \{ 1, 1-\cos x, 2\sin x \right \} then \displaystyle \int_{0}^{\pi}f\left ( x \right )dx is
  • \displaystyle \frac{\pi }{3}+1-\sqrt{3}
  • \displaystyle \frac{2\pi }{3}-1+\sqrt{3}
  • \displaystyle \frac{5\pi }{6}+1-\sqrt{3}
  • \displaystyle \frac{\pi }{6}+1-\sqrt{3}
The area included between the curve { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 } and \displaystyle \sqrt { \left| x \right|  } +\sqrt { \left| y \right|  } =\sqrt { a } \left( a>0 \right) is:
  • \displaystyle \left( \pi +\frac { 2 }{ 3 }  \right) { a }^{ 2 }
  • \displaystyle \left( \pi -\frac { 2 }{ 3 }  \right) { a }^{ 2 }
  • \displaystyle \frac { 2 }{ 3 } { a }^{ 2 }
  • \displaystyle \frac { 2\pi }{ 3 } { a }^{ 2 }
Let y=f(x) be the given curve and x=a, x=b be two ordinates then area bounded by the curve y=f(x), the axis of x between the ordinates x=a & x=b, is given by definite integral
\int_{a}^{b}ydx or \int_{a}^{b}f\left ( x \right )dx and the area bounded by the curve x=f(y), the axis of y & two abscissae y=c & y=d is given by \int_{c}^{d}xdy or \int_{c}^{d}f\left ( x \right )dy. Again if we consider two curves y=f(x), y=g(x) where f\left ( x \right )\geq g\left ( x \right ) in the interval [a, b] where x=a & x=b are the points of intersection of these two curves Shown by the graph given
Then area bounded by these two curves is given by
\int_{a}^{b}\left [ f\left ( x \right )-g\left ( x \right ) \right ]dx
On the basis of above information answer the following questions.

The area bounded by parabolas y=x^{2}+2x+1 & y=x^{2}-2x+1 and the line \displaystyle y=\frac{1}{4} is equal to

161838_6c80fc7958864f1f961bdcd5221bb036.png
  • \displaystyle \frac{2}{3} square unit
  • \displaystyle \frac{1}{3} square unit
  • \displaystyle \frac{3}{2} square unit
  • \displaystyle \frac{1}{2} square unit
Let \displaystyle f\left ( x \right )=\min \left \{ x+1,\sqrt{1-x} \right \} then area bounded by \displaystyle y=f\left ( x \right ) and x-axis is:
  • \displaystyle \frac{7}{6}
  • \displaystyle \frac{5}{6}
  • \displaystyle \frac{1}{6}
  • \displaystyle \frac{11}{6}
If f(x) be an increasing function defined on [a, b] then
max {f(t) such that a\leq t\leq x, a\leq x\leq b}=f(x)  & min {f(t), a\leq t\leq x, a\leq x\leq b}=f(a) and if f(x) be decreasing function defined on [a, b] then
max {f(t), a\leq t\leq x, a\leq x\leq b}=f(a),
min {f(t), a\leq t\leq x, a\leq x\leq b}=f(x).
On the basis of above information answer the following questions.
\int_{0}^{\pi /2}min\left \{ \sin x, \cos x \right \}dx equals
  • 2\left ( \sqrt{3}-1 \right )
  • \sqrt{2}\left ( \sqrt{2}-1 \right )
  • \left ( \sqrt{3}-1 \right )
  • \sqrt{2}\left ( \sqrt{2}+1 \right )
If f(x) be an increasing function defined on [a, b] then
max {f(t) such that a\leq t\leq x, a\leq x\leq b}=f(x)  & min {f(t), a\leq t\leq x, a\leq x\leq b}=f(a) and if f(x) be decreasing function defined on [a, b] then
max {f(t), a\leq t\leq x, a\leq x\leq b}=f(a),
min {f(t), a\leq t\leq x, a\leq x\leq b}=f(x).
On the basis of above information answer the following questions.
\int_{0}^{\pi }max\left \{ \sin x, \cos x \right \}dx is equal to
  • \sqrt{2}-1
  • \displaystyle 1-\frac{1}{\sqrt{2}}
  • \displaystyle 1+\frac{1}{\sqrt{2}}
  • None of these
The ratio of the area's bounded by the curves \displaystyle y^{2}=12x and \displaystyle x^{2}=12y is divided by the line \displaystyle x=3 is
  • 15 : 49
  • 9 : 15
  • 7 : 15
  • 7 : 5
The function \displaystyle f\left ( x \right )=\max \left \{ x^{2},\left ( 1-x \right )^{2},2x\left ( 1-x \right ) \forall 0\leq x\leq 1\right \} then area of the region bounded by the curve \displaystyle y=f\left ( x \right ) , x-axis and \displaystyle x=0,x=1 is equals,
  • \displaystyle \frac{27}{17}
  • \displaystyle \frac{9}{17}
  • \displaystyle \frac{18}{17}
  • None of these
The area bounded by \displaystyle x=a\cos ^{3}\theta,y=a\sin ^{3}\theta is:
  • \displaystyle \frac{3\pi a^{2}}{16}
  • \displaystyle \frac{3\pi a^{2}}{8}
  • \displaystyle \frac{3\pi a^{2}}{32}
  • \displaystyle 3\pi a^{2}
The area bounded by \displaystyle y=\frac{3x^{2}}{4} and the line \displaystyle 3x-2y+12=0 is:
  • 9
  • 18
  • 27
  • None of these
Find the area bounded by the curves \displaystyle x = y^{2} and \displaystyle x = 3-2y^{2}
  • 2 sq. units
  • 4 sq. units
  • 6 sq. units
  • 8 sq. units
0:0:2


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