MCQ Questions for Class 12 Commerce Maths Application Of Integrals Quiz 7

${\int}_{0}^{1}\left(4x^{3}=f(x)\right)f(x)dx=\dfrac{4}{7}$ , then the area of region bounded by

$y=f(x),x-$ axis and the line

$x=$ and

$x=2$ is

0%
$\dfrac{11}{2}$
0%
$\dfrac{13}{2}$
0%
$\dfrac{15}{2}$
0%
$\dfrac{17}{2}$

The are boundede by the curve

$y=x^{2},y=-x$ and

$y^{2}=4x-3$ is

$k$ , them the value of

$9k$ is

0%
$2$
0%
$3$
0%
$0$
0%
$4$

If area bounded by to curves

$y^2 = 4ax$ and y=mx is

$\dfrac{a^2}{3}$ , then the value of m is

0%
$2$
0%
$-1$
0%
$\dfrac{1}{2}$
0%
none of these

Explanation

Given,

$y^2=4ax,y=mx$ curves intersect at

$\left ( 4\dfrac{a}{m^2},\dfrac{a}{m} \right )$

Area enclosed is,

$A=\displaystyle \int_{0}^{4\frac{a}{m^2}}(\sqrt{4ax}-mx)dx$

$\therefore \displaystyle \int_{0}^{4\frac{a}{m^2}}(\sqrt{4ax}-mx)dx=\dfrac{a^2}{3}$ .... given

$\left[2\sqrt a \ \dfrac{x^{\tfrac 32 }}{\tfrac 32 }- m \dfrac {x^2} 2 \right]^{4\frac{a}{m^2}}_0=\dfrac{a^2}{3}$

$\dfrac {4\sqrt a} 3 {({4\frac{a}{m^2}})^{\tfrac 3 2}}-\dfrac m 2 {(4\frac{a}{m^2})}^2-0-0=\dfrac{a^2}{3}$

$\dfrac{32 a ^2} {3m^3}-\dfrac{8 a ^2} {m^3}=\dfrac{a^2}{3}$

$\dfrac{8}{3}\dfrac{a^2}{m^3}=\dfrac{a^2}{3}$

$m^3=8$

$\therefore m=2$

1490097_1252277_ans_4d83f8396de6430c9fd8d17b40c99fc3.png

The area bounded by curves

$3 x^2 + 5 y= 32$ and

$y = |x-2|$ is

0%
25
0%
33/2
0%
17/2
0%
33

The area of the plane region bounded by the curves

$x+{ 2y }^{ 2 }=0$ and

$x+{ 3y }^{ 2 }=1$ is equal to

0%
$\cfrac { 5 }{ 3 } sq.unit$
0%
$\cfrac { 1 }{ 3 } sq.unit$
0%
$\cfrac { 2 }{ 3 } sq.unit$
0%
$\cfrac { 4 }{ 3 } sq.unit$

Explanation

The given curves are

$x+2y^2=0$ &

$x+3y^2=1$

$1$ st curve

$\Rightarrow x^2+2y^2=0$

$\Rightarrow x^2=-2y^2$

$\therefore y^2=-x/2$ (parabola)

$2nd$ curve

$\Rightarrow x+3y^2=1$

$\Rightarrow 3y^2=1-x$

$y^2=(1-x)/3$ (parabola)

So, area between parabolas is required

Solving

$-\dfrac{x}{2}=\dfrac{(1-x)}{3}$

$\Rightarrow -3x=2-2x$

$\Rightarrow -3x+2x=2$

$\Rightarrow -x=2$

$=x=-2$

$\therefore y=1, -1$

$(-2, -1)$ &

$(-2, 1)$ are points of intersection.

Required area

$=2|\displaystyle\int^1_0(-2y^2-1+3y^2)dy|$

$=2|\displaystyle\int ^1_0(y^2-1)dy|=2\left[y-\dfrac{y^3}{3}\right]^1_0$

$=2|\left(\dfrac{1}{3}-1\right)|=2\times 2/3=4/3$ square units.

1177766_1261095_ans_ff0a3acea7eb4efe95d87816d34a25f5.jpg

The area of the figure formed by

$|x| + |y| = 2$ is (in sq. units)

0%
$2$
0%
$4$
0%
$6$
0%
$8$

The area bounded by the curve

$y=\ln (x)$ and the lines

$y=\ln (3),y=0$ and

$x=0$ is equal

0%
$3$
0%
$3\ln (3)-2$
0%
$3\ln (3)+2$
0%
$2$

The area (in sq.units) of the region described by

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

0%
$\dfrac{15}{64}$
0%
$\dfrac{9}{32}$
0%
$\dfrac{7}{32}$
0%
$\dfrac{5}{64}$

The area bounded by the curves

$y = \log _ { e } x$ and

$y = \left( \log _ { e } x \right) ^ { 2 }$ is

0%
$3 - e$
0%
$e-3$
0%
$\frac { 1 } { 2 } ( 3 - e )$
0%
$\frac { 1 } { 2 } ( e - 3 )$

Explanation

$\begin{array}{l} y=\left( { { { \log }_{ e } }x } \right) \, \, \, \, \, \, \, ----\left( 1 \right) \\ y={ \left( { { { \log }_{ e } }x } \right) ^{ 2 } }\, \, \, \, \, -----\left( 2 \right) \\ For\, the\, po{ { int } }\, of\, { { int } }er\sec tion \\ { \left( { { { \log }_{ e } }x } \right) ^{ 2 } }\, \, =\left( { { { \log }_{ e } }x } \right) \\ { \left( { { { \log }_{ e } }x } \right) ^{ 2 } }\, \, -\left( { { { \log }_{ e } }x } \right) =0 \\ \left( { { { \log }_{ e } }x } \right) \left[ { \left( { { { \log }_{ e } }x } \right) \, -1 } \right] \, \\ \left( { { { \log }_{ e } }x } \right) =0\, \, and\, \left( { { { \log }_{ e } }x } \right) =1 \\ x=1\, \, and\, \, x=e \\ area\, bounded \\ =\left| { \int _{ 1 }^{ e }{ \left\{ { { { \left( { { { \log }_{ e } }x } \right) }^{ 2 } }\, -\left( { { { \log }_{ e } }x } \right) } \right\} dx } } \right| \, \, \\ consider \\ I=\int { \left\{ { { { \left( { { { \log }_{ e } }x } \right) }^{ 2 } }\, -\left( { { { \log }_{ e } }x } \right) } \right\} dx } \\ { \log _{ e } }x=t \\ x={ e^{ t } } \\ dx={ e^{ t.dt } } \\ Now, \\ I=\int { { e^{ t } }\left\{ { { t^{ 2 } }-t } \right\} dt } \\ ={ e^{ t } }\left\{ { { t^{ 2 } }-t } \right\} -\left\{ { \left( { 2t-1 } \right) { e^{ t } }-2{ e^{ t } } } \right\} \, \, \, \, \left[ { by\, { { int } }ergration\, by\, parts } \right] \\ ={ e^{ t } }\left[ { { t^{ 2 } }-t-2t+1+2 } \right] \\ ={ e^{ t } }\left[ { { t^{ 2 } }-3t+3 } \right] \\ =x\left[ { \log { { \left( x \right) }^{ 2 } } -3\log \left( x \right) +3 } \right] \\ Now, \\ \left| { \int _{ 1 }^{ e }{ \left\{ { { { \left( { { { \log }_{ e } }x } \right) }^{ 2 } }\, -\left( { { { \log }_{ e } }x } \right) } \right\} dx } } \right| \, \, \\ =\left| { \left[ { x\left\{ { loh{ { \left( x \right) }^{ 2 } } } \right\} -3\log \left( x \right) +3 } \right] _{ 1 }^{ e } } \right| \\ =\left| { e\left( { 1-3+3 } \right) -1\left( { 0-0+3 } \right) } \right| \\ =\left| { e-3 } \right| \\ =3-e \end{array}$

Hence, the option

$B$ is the correct answer.

The area common to the parabola

$y=2{ x }^{ 2 }\quad$ and

$\quad y={ x }^{ 2 }+4$

0%
$\dfrac { 2 }{ 3 } sq.\quad units\quad$
0%
$\dfrac { 3 }{ 2 } sq.\quad units\quad$
0%
$\dfrac { 32 }{ 3 } sq.\quad units\quad$
0%
none of these.

Explanation

Find the P.O.I of the parabolas equate the equations

$y = 2x^2$ and

$y = x^2 + y$ we get

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

$x^2 = 4$

$x = \pm 2$

$y = 8$

$\therefore$ POI are

$A(-2, 8)$ &

$C(2, 8)$

$\therefore$ the required are ABCD

$A = \displaystyle \int_{-2}^2 (y_1 - y_2) dx$ (where

$y_1 = x^2 + 4 \, \& \, y_2 = 2x^2$ )

$= \displaystyle \int_{-2}^2 (x^2 + 4 - 2x^2) dx = \int_{-2}^2 (4 - x^2) dx = \left(4x - \dfrac{x^3}{3} \right)^2_{-2}$

$\left[4(2) - \dfrac{(2)^3}{3} \right] - \left[4(-2) - \dfrac{(-2)^3}{3} \right] = \left[8 - \dfrac{8}{3} \right] - \left[8 + \dfrac{8}{3} \right]$

$= \dfrac{32}{3}$ sq. units

The area bounded by a the curves y=x(1-/nX) and positive X-axis between

$X={ e }^{ -1 }$ amd X=e is:-

0%
$\left( \frac { { e }^{ 2 }-{ 4e }^{ -2 } }{ 5 } \right)$
0%
$\left( \frac { { e }^{ 2 }-{ 5e }^{ -2 } }{ 4 } \right)$
0%
$\left( \frac { { 4e }^{ 2 }-{ e }^{ -2 } }{ 5 } \right)$
0%
$\left( \frac { { 5e }^{ 2 }-{ e }^{ -2 } }{ 4 } \right)$

The area bounded by the curves

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

$y=x$ is (in

$sq.units$ ) is

0%
$28$
0%
$32$
0%
$4$
0%
$8$

The area bounded by the curve

$y={ x }^{ 2 }$ , X=axis and the ordinates z=1, z=3 is ____________.

0%
$\dfrac { 26 }{ 3 } sq.units$
0%
$\dfrac { 28 }{ 3 } sq.unit$
0%
$\dfrac { 1 }{ 3 } sq.units$
0%
$9\quad sq.units$

The area bounded by the curves

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

$y=x$ is (in sq.units) is

0%
$28$
0%
$32$
0%
$4$
0%
$8$

The area bounded by the curve y = log x, X-axis and the ordinates x =1, x =2 is

0%
log 4 sq. units
0%
log 2 sq units
0%
(log 4 - 1) sq.units
0%
(log 4 + 1)sq. units

The area enclosed by the curves

$xy^{2}=a^{2}(a-x)$ and

$(a-x)y^{2}=a^{2}x$ is

0%
$(\pi-2)a^{2}\ sq.units$
0%
$(4-\pi)a^{2}\ sq.units$
0%
$(\pi a^{2}/3\ sq.units$
0%
$\dfrac{\pi+a^{2}}{4}\ sq.units$

The area bounded by the curved

${ y }^{ 2 }=16x$ and the line x=4 is ___________________________.

0%
$\frac { 128 }{ 3 } sq-units$
0%
$\frac { 64 }{ 3 } squnits$
0%
$\frac { 32 }{ 3 } sq-units$
0%
$\frac { 16 }{ 3 } sq-units$

$A_{m}$ represents the area bounded by the curve

$y=\ln x^{m}$ ., the

$x-$ axis and the lines

$x=1$ and

$x=2$ , then

$A_{m}+m\ A_{m-1}$ is

0%
$m$
0%
$m^{2}$
0%
$m^{2}/2$
0%
$m^{2}-1$

The area of the region bounded by the curve

$y=x^{2}-3x$ with

$y \le 0$ is

0%
$3$
0%
$\dfrac {9}{2}$
0%
$\dfrac {5}{2}$
0%
$none\ of\ these$

Explanation

REF.Image

when x=0

$\Rightarrow$ y0

when y=0

$\Rightarrow$ x(x-3)=0

x=0 and x=3

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

$\frac{dy}{dx}=2x-3\Rightarrow x=1.5$

min value at x=1.5

$y=\frac{3}{2}(\frac{3}{2}-3)$

$=\frac{3}{2}(\frac{-3}{2})=\frac{-9}{4}=-2.25$

Area of shaded region =

$\int y.dx$

$=\int_{0}^{3}(x^{2}-3x)dx$

$=(\frac{x^{3}}{3}-\frac{3x^{2}}{2})^{3}_{0}$

$=\frac{3^{3}}{3}-3\times \frac{3^{2}}{2}=9-\frac{9\times 3}{2}=-4.5$

considering magnitude = 4.5

1222715_1296827_ans_864f232226984da1b1397b83dd274396.jpg

If a curve

$y = a \sqrt { x } +$ bx passes through the point

$( 1,2 )$ and the area bounded by the curve, line

$x = 4$ and

$x$ axis is

$8$ square units, then

0%
$a = 3 , b = - 1$
0%
$a = 3 , b = 1$
0%
$a = - 3 , b = 1$
0%
$a = - 3 , b = - 1$

The area bounded by the circle

$x^{2}+y^{2}=8$ , the parabola

$x^{2}=2y$ and the line

$y=x$ in first quadrant is

$\dfrac{2}{3}+k\pi$ , where

$k=$

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

The area enclosed between the curve

$y=\log_{e}\left(x+e\right)$ and the coordinate axes is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$A=\int_{1-e}^{0} ln(x+e)dx$

$=\int_{1-e}^{0} x^{0} ln (x+e)dx$

$(x ln(x+e))-\int_{1-e}^{0}\frac{x}{x+e}dx$

$-\int_{1-e}^{0}\frac{x+e-e}{x+e}dx$

$=\int_{1-e}^{0}(\frac{e}{x+e})dx$

$=[e ln(x+e)-x]_{1-e}^{0}=1$

1204240_1296377_ans_331eb95141d54c64b9d4aaa84e0594ed.jpg

The area of the region formed by

${ x }^{ 2 }+{ y }^{ 2 }-6x-4y+12\le 0,y\le x\quad and\quad x\quad \le \quad \dfrac { 5 }{ 2 } is$

0%
$\frac { \pi }{ 6 } -\frac { \sqrt { 3}+1 }{ 8 }$
0%
$\frac { \pi }{ 6 } +\frac { \sqrt { 3}+1 }{ 8 }$
0%
$\frac { \pi }{ 6 } -\frac { \sqrt { 3}-1 }{ 8 }$
0%
none of these

Area bounded by

$y=2x^{2}$ and

$y=\dfrac{4}{(1+x^{2})}$ will be (in sq units)

0%
$(2\pi+4/3)$
0%
$(2\pi-4/3)$
0%
$4/3-2\tan^{-1}2+\pi/2$
0%
$4/3-8\tan^{-1}2+2\pi$

Explanation

$\begin{array}{l} Po{ { int } }\, of\, intersection\, : \\ 2{ x^{ 2 } }=\dfrac { 4 }{ { 1+{ x^{ 2 } } } } \\ \Rightarrow { x^{ 4 } }+{ x^{ 2 } }-2=0 \\ \Rightarrow { x^{ 4 } }+2{ x^{ 2 } }-{ x^{ 2 } }-2=0 \\ \Rightarrow \left( { { x^{ 2 } }-1 } \right) \left( { { x^{ 2 } }+2 } \right) =0 \\ Then \\ Area\, bounded\, by\, y=2{ x^{ 2 } }\, and\, y=\dfrac { 4 }{ { 1+{ x^{ 2 } } } } \\ 2\int _{ 0 }^{ 1 }{ \left( { \dfrac { 4 }{ { 1+{ x^{ 2 } } } } -2{ x^{ 2 } } } \right) }dx=2\left[ { 4{ { \tan }^{ -1 } }x-\dfrac { { 2{ x^{ 3 } } } }{ 3 } } \right] _{ 0 }^{ 1 } \\ =2\left[ { 4\times \dfrac { \pi }{ 4 } -\dfrac { 2 }{ 3 } } \right] \\ =2\left[ { \pi -\dfrac { 2 }{ 3 } } \right] =2\pi -\dfrac { 4 }{ 3 } \\ Hence,\, option\, \, \, B\, \, is\, the\, correct\, answer. \end{array}$
1216600_1308671_ans_cc248d4277894ce1ba61262826dac1b3.PNG

1216600_1308671_ans_cc248d4277894ce1ba61262826dac1b3.PNG

$Letf(x)={ sin }^{ -1 }(sin\quad x)+{ cos }^{ -1 }(\quad cos\quad x),\quad g(x)=mx\quad and\quad h(x)=x\quad$ are three functions. Now g(x) is divided area between f(x),x=

$\pi$ and y=0 into two equal parts.
The area bounded by the curve y=f(x), x=

$\pi$ and y=0 is:

0%
$\dfrac { { \pi }^{ 2 } }{ 4 } sq.\quad units$
0%
${ \pi }^{ 2 }sq.units$
0%
$\dfrac { { \pi }^{ 2 } }{ 8 } sq.\quad units$
0%
$2{ \pi }^{ 2 }sq.units$

Explanation

Given,

$f(x)=\sin^{-1}\sin x+\cos^{-1}\cos x=x+x=2x$

$x=\pi$

$y=0$

The above 3 forms a right angled triagle, with center as one vertex.

$Area=\dfrac{1}{2} \times \pi times 2\pi$

$=\pi ^2sq.units$

The area of the region bounded by the curves

$1-y^{2}= \left | x \right | and \left | x \right |+\left | y \right |= 1$ is

0%
$\frac{1}{3}sq. unit$
0%
$\frac{2}{3}sq. unit$
0%
$\frac{4}{3}sq. unit$
0%
$1 sq.unit$

Find the area of the region enclosed by the curves

$y=x\ \log x$ and

$y=2x-2x^{2}$ .

0%
$1/12$
0%
$1/4$
0%
$2/12$
0%
$7/12$

Area of the region bounded by

$x^{2}+y^{2}-6y\leq 0$ and

$3y\leq x^{2}$ is

0%
$\frac{9\pi }{2}-12$
0%
$\frac{9\pi }{4}-6$
0%
$9\pi$ -24
0%
$\frac{9\pi }{2}+6$

The area enclosed by the curves y = cosx - sin x and y = [socx - sin x] and between x = 0 and

$x =\dfrac{\pi}{2}$ is

0%
$2(\sqrt{2} + 1)$ sq. units
0%
$2(\sqrt{2} - 1)$ sq. units
0%
$(\sqrt{2} - 1)$ sq. units
0%
$(\sqrt{2} + 1)$ sq. units

The area bounded by the parabola

${{\text{y}}^{\text{2}}}{\text{ = 4}}\;{\text{ax}}\;{\text{and}}\;{{\text{x}}^{\text{2}}}\;{\text{ = }}\;{\text{4ay}}\;$ is

0%
$\dfrac{{8{a^2}}}{3}$
0%
$\dfrac{{16{a^2}}}{3}$
0%
$\dfrac{{32{a^2}}}{3}$
0%
$\dfrac{{64{a^2}}}{3}$

The area (in sq. units) bounded by the parabola

$y = x^{2} - 1$ , the tangent at the point

$(2, 3)$ to it and the y-axis is

0%
$\dfrac {14}{3}$
0%
$\dfrac {56}{3}$
0%
$\dfrac {8}{3}$
0%
$\dfrac {32}{3}$

Explanation

Here,

$y = x^{2} - 1$

So, Equation of tangent at

$(2, 3)$ on

$\Rightarrow$

$y =x^{2} - 1$ , is

$y = (4x - 5) ......(i)$

$\therefore$ Required shaded area

$\displaystyle = area (\triangle ABC) - \int_{-1}^{3} \sqrt {y + 1} dy$

$= \dfrac {1}{2} \cdot (8)\cdot (2) - \dfrac {2}{3} \left ((y + 1)^{3/2}\right )_{-1}^{3}$

$= 8 - \dfrac {16}{3} = \dfrac {8}{3}$ (square units).

1141380_1329190_ans_613acca5ef984b2b9eb3b109292f525d.jpg

The area ehclosed by the curves y = f(x) and y =g(x), where f9x) = max

${x , x^2}$ and g(x) = min

${x, x^2}$ opver the interval [0,1] is

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

The region in the

$xy$ - plane is bounded by curve

$y=\sqrt {(25-x^2)}$ and the line

$y=0$ . If the point

$(a,a+1)$ lies in the interior of the region, then

0%
$a \in \left( { - 4,3} \right)$
0%
$a \in (- \infty, -1) \cup (3, \infty)$
0%
$a \in (-1,3)$
0%
None of these

The area of the region bounded by the parabolas

$y^2= and x^2 = y, is$

0%
$\dfrac{1}{3}$ q.units
0%
$\dfrac{8}{3}$ q.units
0%
$\dfrac{16}{3}$ q.units
0%
$\dfrac{4}{3}$ q.units

If the area of the region bounded by the curves,

$y=x^{2},y=\frac{1}{x}$ and the lines y=0 and x=t (t > 1) is 1 sq. unit, then t is equal to:

0%
${\dfrac{10}{3}}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{7}{3}$
0%
$\dfrac{11}{3}$

The area (in sq. units) in the first quadrant bounded by the parabola,

$y = x^2 + 1$ , the tangent to it at the point

$(2, 5)$ and the coordinate axes is:-

0%
$\dfrac{14}{3}$
0%
$\dfrac{187}{24}$
0%
$\dfrac{37}{24}$
0%
$\dfrac{8}{3}$

The slope of the tangent to the curve y =f(x) at a point (x, Y) is 2x + 1 and the curve passes through (1, 2) The area of the region bounded by the curve, the x-axis and the line x= 1 is -

0%
5/3 units
0%
5/6 units
0%
6/5 units
0%
6 units

The area (in sq. units) of the region

$\{ { x },{ y }):{ y }^{ { 2 } }\geq 2{ x }$ and

$x ^ { 2 } + y ^ { 2 } \leq 4 x , x \geq 0 , y \geq 0 \}$ is :

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

The area of the region

$A=[(x,y):0\le y\le x|x|+1$ and

$-1\le x\le 1]$ in sq . units is :

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

Explanation

The graph is as follows:

$\displaystyle \int_{-1}^0(-x^2+1)dx+\displaystyle \int_0^1(x^2+1)dx=2$
1142868_1329318_ans_daece77cacf84b69a4359f1ef1320929.png

The area (in sq. units) of the region bounded by the parabola,

$y = x ^ { 2 } + 2$ and the lines,

$y = x + 1 , x = 0$ and

$x = 3 ,$ is :

0%
$\dfrac { 15 } { 4 }$
0%
$\dfrac { 15 } { 2 }$
0%
$\dfrac { 21 } { 2 }$
0%
$\dfrac { 17 } { 4 }$

Explanation

Req. area

$= \int _ { 0 } ^ { 3 } \left( x ^ { 2 } + 2 \right) d x - \dfrac { 1 } { 2 } \cdot 5.3 = 9 + 6 - \dfrac { 15 } { 2 } = \dfrac { 15 } { 2 }$
1142967_1331308_ans_91c5b5693d4c41579e9c89d43101928f.png

If the area enclosed between the curves

$y=kx^2$ and

$x=ky^2$ ,

$(k > 0)$ , is

$1$ square unit. Then

$k$ is?

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

Explanation

Area bounded by

$y^2=4ax$ &

$x^2=4by$ , a, b

$\neq 0$ is

$\left|\dfrac{16ab}{3}\right|$
by using formula:

$4a=\dfrac{1}{k}=4b, k > 0$
Area

$=\left|\dfrac{16\cdot \dfrac{1}{4k}\cdot \dfrac{1}{4k}}{3}\right|=1$

$\Rightarrow k^2=\dfrac{1}{3}$

$\Rightarrow k=\dfrac{1}{\sqrt{3}}$ .

The area of the region bounded by

$y=\left | x-1 \right | and \,\,y=1$ is

0%
1
0%
2
0%
1/2
0%
None of these

The area of the region

$\left\{ ( x , y ) : x ^ { 2 } + y ^ { 2 } \leq 1 \leq x + y \right\}$ is

0%
$\dfrac { \pi ^ { 2 } } { 5 } \text { unit } ^ { 2 }$
0%
$\dfrac { \pi ^ { 2 } } { 2 } \text { unit } ^ { 2 }$
0%
$\dfrac { \pi ^ { 2 } } { 3 } \text { unit } ^ { 2 }$
0%
$\left( \dfrac { \pi } { 4 } - \dfrac { 1 } { 2 } \right)\text { unit } ^ { 2 }$

The area of the region bounded by the parabola y =

$x^2$ 3x with y 0 is

0%
$3$
0%
$\dfrac{3}{2}$
0%
$\dfrac{9}{2}$
0%
${9}{}$

The area of the quadrilateral formed by the tangents at the endpoints of the latus recta to the ellipse,

$\dfrac{x^{2}}{9}+\dfrac{y^{2}}{5}=1$ is

0%
$\dfrac{27}{4}$
0%
$18$
0%
$\dfrac{27}{2}$
0%
$27$

Explanation

Given equation of ellipse is

$\dfrac{{x}^{2}}{9}+\dfrac{{y}^{2}}{5}=1$ ......

$(1)$

$\therefore {a}^{2}=9,{b}^{2}=5$

$\Rightarrow a=3,b=\sqrt{5}$

Now,

$e=\sqrt{1-\dfrac{{b}^{2}}{{a}^{2}}}=\sqrt{1-\dfrac{5}{9}}=\dfrac{2}{3}$

Foci

$=\left(\pm ae,0\right)=\left(\pm 2,0\right)$

and

$\dfrac{{b}^{2}}{a}=\dfrac{5}{3}$

$\therefore$ Extremities of one of latus rectum are

$\left(2,\dfrac{5}{3}\right)$ and

$\left(2,\dfrac{-5}{3}\right)$

$\therefore$ Equation of tangent at

$\left(2,\dfrac{5}{3}\right)$ is

$\dfrac{x\left(2\right)}{9}+\dfrac{y\left(\dfrac{5}{3}\right)}{5}=1$

$2x+3y=9$ .......

$(2)$

Eqn

$(2)$ intersects

$X$ and

$Y$ axes at

$\left(\dfrac{9}{2},0\right)$ and

$\left(0,3\right)$ respectively.

$\therefore$ Area of quadrilateral

$4\times$ Area of

$\triangle{POQ}$

$=4\times\dfrac{1}{2}\times\dfrac{9}{2}\times 3=27$ sq.units.

The area bounded by curve

$y=x^{2}-1$ and tangents to it at

$(2,3)$ and

$y-$ axis is

0%
$8/3$
0%
$2/3$
0%
$4/3$
0%
$1/3$

Explanation

$x-axis:(-1,0)$

Area

$=\int_{-1}^{0}(x^2-1)dx$

$=\left [ \dfrac{x^3}{3}-x \right ]_{-1}^0$

$=\left [ \dfrac{-1}{3}-(-1) \right ]-[0]$

$=-\dfrac{1}{3}+1$

$=\dfrac{2}{3}$ sq. units

The area bounded by the curves

$x+2|y|=1$ and

$x=0$ is?

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

Explanation

$x+2\left| y \right| =1$ and

$x=0$

$y=+ve$

$x+2y=1$

$y=-ve$

$x-2y=1$

Area of triangle

$=\cfrac{1}{2}\times base\times height\\=\cfrac{1}{2}\times 1\times 1\\=\cfrac{1}{2}\text{ }unit$

Area included between

${ y }=\dfrac { { x }^{ { 2 } } }{ 4{ a } }$ and

$y = \dfrac { 8 a ^ { 3 } } { x ^ { 2 } + 4 a ^ { 2 } }$ is

0%
$\dfrac { a ^ { 2 } } { 3 } ( 6 \pi - 4 )$
0%
$\dfrac { a ^ { 2 } } { 3 } ( 4 \pi + 3 )$
0%
$\dfrac { a ^ { 2 } } { 3 } ( 8 \pi + 3 )$
0%
None of these

The area of the figure formed by

$a|x|+b|y|+c=0$ , is

0%
$\dfrac{c^{2}}{|ab|}$
0%
$\dfrac{2c^{2}}{|ab|}$
0%
$\dfrac{c^{2}}{2|ab|}$
0%
$None\ of\ these$

The area of the region bounded by the curves

$y = \sin x$ and

$y = \cos x ,$ and lying between the lines

$x = \dfrac { \pi } { 4 }$ and

$x = \dfrac { 5 \pi } { 4 } ,$ is

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

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 7 - 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 7 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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