CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 14 - MCQExams.com

The area bounded by the curves $$  | x| +| y | > 1 $$ and $$x^2 + y^2 < 1 $$ is 
  • $$2$$ sq units
  • $$ \pi $$ sq units
  • $$ ( \pi - 2) $$ sq units
  • $$ ( \pi +2) sq units $$
The area of the region enclosed by the curve $$ \mid y \mid =-(1- \mid x \mid)^2 +5 $$, is
  • $$ \dfrac{8}{3} (7 + 5 \sqrt{5}) $$ sq units
  • $$ \dfrac{2}{3} (7 + 5 \sqrt{5}) $$ sq units
  • $$ \dfrac{2}{3} (5 \sqrt{5}-7) $$ sq units
  • None of these
The area bounded by the curve $$y= \dfrac{3}{\mid x \mid}$$ and $$y+ \mid 2-x \mid =2$$ is
  • $$ \dfrac{4-log27}{3} $$
  • $$ 2-log3 $$
  • $$ 2+log3 $$
  • None of these
The area of the figure bounded by $$y=sinx, \ y=cosx$$ is the first quadrant is
  • $$ 2(\sqrt{2-1}) $$
  • $$ \sqrt{3}+1 $$
  • $$ 2(\sqrt{3}-1) $$
  • None of these
The area bounded by the curves $$y=x^2 +2$$ and $$y=2 \mid x \mid -cosx +x$$ is 
  • $$ \dfrac{2}{3} $$
  • $$ \dfrac{8}{3} $$
  • $$ \dfrac{4}{3} $$
  • $$ \dfrac{1}{3} $$
The areas of the figure into which the curve $$y^2=6x$$ divides the circle $$x^2 + y^2 = 16$$ are in the ratio
  • $$ \dfrac{2}{3} $$
  • $$ \dfrac{4 \pi - \sqrt{3}}{8 \pi + \sqrt{3}} $$
  • $$ \dfrac{4 \pi + \sqrt{3}}{8 \pi - \sqrt{3}} $$
  • None of these
The area bounded by the curve $$f(x) = \mid \mid tanx + cotx \mid - \mid tanx - cotx \mid \mid$$ between the lines $$x=0, \ x= \dfrac{\pi}{2}$$ and the $$X-$$axis is
  • $$ log4 $$
  • $$ log \sqrt{2] $$
  • $$ 2log2$$
  • $$ \sqrt{2} log2$$
The area bounded by the curve $$y^2 =4x$$ and the circle $$x^2 + y^2 -2x -3=0$$ is 
  • $$ 2 \pi + \dfrac{8}{3} $$
  • $$ 4 \pi + \dfrac{8}{3} $$
  • $$ \pi + \dfrac{8}{3} $$
  • $$ \pi - \dfrac{8}{3} $$
The area of the region defined by $$1 \leq \mid x-2 \mid + \mid y+ 1 \mid \leq 2$$ is
  • $$2$$
  • $$4$$
  • $$6$$
  • None of these
The area of the region defined by $$ \mid \mid x \mid - \mid y \mid \mid \leq 1 $$  and $$ x^2  + y^2 \leq 1 $$ in the $$xy$$ plane is
  • $$\pi-2$$
  • $$2 \pi$$
  • $$3 \pi$$
  • $$1$$
If the length of latus rectum of ellipse
$$ E_1 : 4(x+y-1)^2 + 2(x-y+3)^2 =8 $$
and $$ E_2: \dfrac{x^2}{p} + \dfrac{y^2}{p^2}=1, \ (0< p<1)$$ are equal, then area of ellipse $$ E_2, $$ is
  • $$ \dfrac{\pi}{2} $$
  • $$ \dfrac{\pi}{\sqrt{2}} $$
  • $$ \dfrac{\pi}{2 \sqrt{2}} $$
  • $$ \dfrac{\pi}{4} $$
If $$f(x)=x-1$$ and $$g(x) = \mid f( \mid x \mid ) - 2 \mid$$, then the area bounded by $$y=g(x)$$ and the curve $$x^2 -4y+8=0$$ is equal to
  • $$ \dfrac{4}{3} (4 \sqrt{2} - 5)$$
  • $$ \dfrac{4}{3} (4 \sqrt{2} - 3)$$
  • $$ \dfrac{8}{3} (4 \sqrt{2} - 3)$$
  • $$ \dfrac{8}{3} (4 \sqrt{2} - 5)$$
If the area bounded by the curve $$ y=x^2 +1, y=x $$ and the pair of lines $$ x^2 + y^2 + 2xy - 4x -4y +3 =0 $$ is $$K$$ units, then the area of the region bounded by the curve $$y=x^2 +1, \ y= \sqrt{x-1}$$ and the pair of lines $$ (x+y-1)(x+y-3)=0 ,$$ is
  • $$K$$
  • $$2K$$
  • $$ \dfrac{K}{2} $$
  • None of these
Area bounded by the ellipse $$ \dfrac{x^2}{4} + \dfrac{y^2}{9} =1$$ is equal to
  • $$6 \pi$$ sq units
  • $$3 \pi$$ sq units
  • $$12 \pi$$ sq units
  • $$2 \pi$$ sq units
Area bounded by $$y=f^{-1}(x)$$ and tangent and normal drawn to it at the points with abscissae $$\pi$$ and $$2 \pi$$, where $$f(x)=sin x- x$$ is
  • $$\dfrac{ \pi^2}{2} -1$$
  • $$\dfrac{ \pi^2}{2} -2$$
  • $$\dfrac{ \pi^2}{2} -4$$
  • $$\dfrac{ \pi^2}{2}$$
A point $$P$$ lying inside the curve $$y= \sqrt{2ax-x^2}$$ is moving such that its shortest distance from the curve at any position is greater than its distance from $$X-$$axis. The point $$P$$ enclose a region whose area is equal to
  • $$ \frac{\pi a^2}{2} $$
  • $$ \frac{ a^2}{3} $$
  • $$ \frac{2 a^2}{3} $$
  • $$ \bigg( \frac{3 \pi -4}{6} \bigg) a^2 $$
The area bounded by $$y=2- \mid 2-x \mid$$ and $$y= \dfrac{3}{\mid x \mid}$$ is
  • $$ \dfrac{4-3 ln3}{2} $$
  • $$\dfrac{19}{8} - 3 ln 2$$
  • $$ \dfrac{3}{2} + ln 3$$
  • $$ \dfrac{1}{2} + ln 3$$
The area of the region bounded between the curves $$y=e \mid \mid x \mid \ ln \mid x \mid \mid, , \ x^2 + y^2 - 2( \mid x \mid + \mid y \mid) + 1 \geq 0 $$ and $$X-$$axis where $$ \mid x \mid \leq 1,$$ if $$ \alpha$$ is the $$x-$$coordinate of the point of intersection of curves in 1st quadrant, is
  • $$ 4 \bigg[ \int^\alpha_0 \ e \ x \ ln x \ dx + \int^1_\alpha (1- \sqrt{1-(x-1)^2)} \ dx \bigg] $$
  • $$ 4 \bigg[ \int^\alpha_0 \ e \ x \ ln x \ dx + \int^\alpha_1 (1- \sqrt{1-(x-1)^2)} \ dx \bigg] $$
  • $$ 4 \bigg[ - \int^\alpha_0 \ e \ x \ ln x \ dx + \int^1_\alpha (1- \sqrt{1-(x-1)^2)} \ dx \bigg] $$
  • $$ 2 \bigg[ \int^\alpha_0 \ e \ x \ ln x \ dx + \int^\alpha_1 (1- \sqrt{1-(x-1)^2)} \ dx \bigg] $$
Area of the region enclosed between the curves $$x=y^2 -1$$ and $$x= \mid y \mid \sqrt{1-y^2}$$ is
  • $$1$$
  • $$\dfrac{4}{3}$$
  • $$\dfrac{2}{3}$$
  • $$2$$
Area of the loop described as $$x=\dfrac{t}{3}(6-t), \ y=\dfrac{t^2}{8}(6-t)$$ is
  • $$ \dfrac{27}{5} $$
  • $$ \dfrac{24}{5} $$
  • $$ \dfrac{27}{6} $$
  • $$ \dfrac{21}{5} $$
Then, the absolute area enclosed by $$y=f(x)$$ and $$y=g(x)$$ is given by
  • $$ \overset{n}{\underset{r=0}{\Sigma}} \ \int^{x_{r+1}}_{x_r} \ (-1)^r. h(x) dx$$
  • $$ \overset{n}{\underset{r=0}{\Sigma}} \ \int^{x_{r+1}}_{x_r} \ (-1)^{r+1}. h(x) dx$$
  • $$2 \overset{n}{\underset{r=0}{\Sigma}} \ \int^{x_{r+1}}_{x_r} \ (-1)^r. h(x) dx$$
  • $$\dfrac{1}{2} \ \overset{n}{\underset{r=0}{\Sigma}} \ \int^{x_{r+1}}_{x_r} \ (-1)^{r+1}. h(x) dx$$
The area enclosed by the asteroid $$ \bigg( \dfrac{x}{a} \bigg)^{2/3} + \bigg( \dfrac{y}{a} \bigg)^{2/3} =1$$ is
  • $$ \dfrac{3}{4} a^2 \pi $$
  • $$ \dfrac{3}{18} \pi a^2 $$
  • $$ \dfrac{3}{8} \pi a^2 $$
  • $$ \dfrac{3}{4} a \pi $$
The area enclosed by the curves $$x = a \sin^3t$$ and $$y = a \cos^3 t$$ is equal to
  • $$\displaystyle 12a^2 \int_0^{\pi/2} \cos^4 t \sin^2t \ dt$$
  • $$\displaystyle 12a \int_0^{\pi/2} \cos^2 t \sin^4t \ dt$$
  • $$\displaystyle 2 \int_{-a}^{a} (a^{2/3} - x^{2/3})^{3/2}dx$$
  • $$\displaystyle 4 \int_{0}^{a} (a^{2/3} - x^{2/3})^{3/2}dx$$
The area enclosed by the ellipse $$\dfrac {x^{2}}{a^{2}} + \dfrac {y^{2}}{b^{2}} = 1$$ is equal to
  • $$\pi^{2} ab$$
  • $$\pi ab$$
  • $$\pi a^{2}b$$
  • $$\pi ab^{2}$$
The value of the parameter a such that the area bounded by $$y=a^2x^2+ax+1$$, coordinate axes and the line $$x=1$$ attains its least value is equal to
  • $$-\frac{1}{4}$$ sq. units
  • $$-\frac{1}{2}$$ sq. units
  • $$-\frac{3}{4}$$ sq. units
  • $$-1$$ sq. units
Let the functions $$f:R\rightarrow R$$ and $$𝑔:R\rightarrow R$$ be defined by $$f(x)=e^{ x-1 }-e^{ -|x-1| }$$ and $$g(x)=\dfrac{1}{2}(e^{x-1}+e^{1-x})$$. Then the area of the region in the first quadrant bounded by the curves $$𝑦=𝑓(𝑥), 𝑦=𝑔(𝑥) $$ and $$x=0$$ is 
  • $$(2-\sqrt{3})+\dfrac{1}{2}(e-e^{-1})$$
  • $$(2+\sqrt{3})+\dfrac{1}{2}(e-e^{-1})$$
  • $$(2-\sqrt{3})+\dfrac{1}{2}(e+e^{-1})$$
  • $$(2+\sqrt{3})+\dfrac{1}{2}(e+e^{-1})$$
Consider two regions
$$R_1$$ : Point $$P$$ is nearer to $$(i, 0)$$ than to $$x = -1$$.
$$R_2$$ : Point $$P$$ is nearer to $$(0, 0)$$ than to $$(8, 0)$$.
Statement 1 : Area of the region common to $$R_1$$ and $$R_2$$ is $$\dfrac{128}{3}$$ sq. units.

Statement 2 : Area bounded by $$x = 4 \sqrt{y}$$ and $$y = 4 $$ is $$\dfrac{32}{3}$$ sq. units.
  • If both Statement are correct and Statement 2 is the correct explanation of Statement 1
  • If both Statement are correct and Statement 2 is not the correct explanation of Statement 1
  • If Statement 1 is correct and Statement 2 is incorrect
  • If Statement 1 is incorrect and Statement 2 is correct
0:0:1


Answered Not Answered Not Visited Correct : 0 Incorrect : 0

Practice Class 12 Commerce Maths Quiz Questions and Answers