MCQ Questions for Class 12 Commerce Maths Application Of Integrals Quiz 13

The area bounded by

$y ^2 = 2x + 1$ and

$x - y - 1 = 0$ is

0%
4/3
0%
8/3
0%
16/3
0%
None of these

Explanation

${ y }^{ 2 }=2x+1$ and

$x-y=1$

${ y }^{ 2 }=2x+1$

$\Rightarrow { y }^{ 2 }=2\left( 1+y \right) +1$

$\Rightarrow { y }^{ 2 }-2y-3=0$

$\Rightarrow { y }^{ 2 }-3y+y-3=0$

$\Rightarrow \left( y-3 \right) \left( y+1 \right) =0$

$\therefore$

$y=3$ or

$-1$

$x=4$ or

$0$

$\therefore$ the curves

${ y }^{ 2 }=2x+1$ &

$x-y=1$ intersect at

$(4,3)$ &

$(0,-1)$ and we have find the area enclosed by the curves

${ y }^{ 2 }=2x+1$ &

$x-y=1$ . From the graph above, it is the shaded region which we have to find

$\therefore$ area enclosed by the graph is

$A=\int _{ -1 }^{ 3 }{ \left\{ \left( y+1 \right) -\dfrac { 1 }{ 2 } \left( { y }^{ 2 }-1 \right) \right\} } dy$

$=\int _{ -1 }^{ 3 }{ \left\{ y+1-\dfrac { { y }^{ 2 } }{ 2 } +\dfrac { 1 }{ 2 } \right\} } dy$

$=\int _{ -1 }^{ 3 }{ \left\{ y-\dfrac { { y }^{ 2 } }{ 2 } +\dfrac { 3 }{ 2 } \right\} dy } ={ \left\{ \dfrac { { y }^{ 2 } }{ 2 } -\dfrac { { y }^{ 3 } }{ 6 } +\dfrac { 3y }{ 2 } \right\} }_{ -1 }^{ 3 }=\dfrac { 16 }{ 3 }$ sq units.

1179522_1300591_ans_8cf3a8028fc84fb6b39a9a7b75125f72.png

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

$12 y = 36 - x ^ { 2 }$ and the tangents drawn to it at the points,where the curve intersects the

$x$ -axis, is :

0%
$18$
0%
$27$
0%
$6$
0%
$12$

The area enclosed by the curves

$y-\sin x+\cos x$ and

$y-\left| \cos x-\sin x \right|$ over the interval

$\{ 0 , \pi / 2 \}$ is given as

$2\sqrt { a } ( \sqrt { b } - c )$ where

$a$ and

$b$ are prime number then the value of

$a,b$ and

$c$ respectively.

0%
$( 2,2,1 )$
0%
$( 2,2,-1 )$
0%
$( 3,2,-1 )$
0%
none of these

The area bounded by the curves

$y = x e ^ { x } , y = x e ^ { - x }$ and the line

$x = 1 ,$ is

0%
$\frac { 2 } { e }$
0%
$1 - \frac { 2 } { e }$
0%
$\frac { 1 } { e }$
0%
$1 - \frac { 1 } { e }$

The area enclosed by the curve

$\left[x+3y\right]=\left[x-2\right]$ where

$x\epsilon\left[3,4\right)$ is (where

$[.]$ denotes greatest integer function.)

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

Explanation

$\text { Given that: } \\$

$\qquad[x+3 y]=[x-2] \text { where } x \in[3,4) \\$

$x \in[3,4)$

$\therefore x-2 \in[1,2) \\$

$\therefore(x-2)=1$

So,

$[x+3 y]=1$

that is for,

$1 \leq x+3 y<2 \quad \& \quad 3 \leq x<4 \\$

$\frac{1-x}{3} \leq y<\frac{2-x}{3} \quad \& \quad 3 \leqslant x<4$

$\therefore \frac{1-x}{3} \leq y$ &

$y<\frac{2-x}{3}$ &

$3 \leqslant x<4$

$\therefore x+3 y \geqslant 1 \quad \& \quad x+3 y<2$

$\& \quad 3 \leqslant x<4$

The plot is as follows:

$\therefore \text { Area }=\text { Area of } A B C D \text { - Area of } \triangle A B P \\$

$\text { - Area of } \triangle Q C D \\$

$1\left(\frac{2}{3}\right)-\frac{1}{2}\left(\frac{1}{3}\right)-\frac{1}{2}\left(\frac{1}{3}\right) \\$

$=\frac{2}{3}-\frac{1}{3}$

$\text { doired Area } =\frac{1}{3} \text { sq. units }$

Area bounded by the parabola

${ x }^{ 2 }=36y$ and its latus rectum is

0%
116
0%
616
0%
216
0%
126

If the area between the curves y=kx2 and x=ky2 is 1
then k is

0%
1/\sqrt { 2 }
0%
1/\sqrt { 3 }
0%
1/\sqrt { 4 }
0%
1

Area of region bounded by

$[x]^2=[y]^2$ if

$x\in [1,5]$ where [.] represents the greatest integer function is-

0%
$10$ sq.units
0%
$8$ sq.units
0%
$6$ sq.units
0%
$5$ sq.units

$\left[ \begin{array} { c c c } { 4 a ^ { 2 } } & { 4 a } & { 1 } \\ { 4 b ^ { 2 } } & { 4 b } & { 1 } \\ { 4 c ^ { 2 } } & { 4 c } & { 1 } \end{array} \right] \left[ \begin{array} { c } { f ( - 1 ) } \\ { f ( 1 ) } \\ { f ( 2 ) } \end{array} \right] = \left[ \begin{array} { c } { 3 a ^ { 2 } + 3 a } \\ { 3 b ^ { 2 } + 3 b } \\ { 3 c ^ { 2 } + 3 c } \end{array} \right]$

$f ( x )$ is a quadratic function and its maximum value occurs at a point

$V. A$ is a point of intersection of

$y = f ( x )$ with

$x$ -axis and point

$B$ is such that chord

$AB$ subtends a right angled at

$V .$ Find The area enclosed by

$f ( x )$ and chord

$A B .$

0%
$\dfrac { 125 } { 3 }$ sq unit
0%
$\dfrac { 115 } { 3 }$ sq unit
0%
$\dfrac { 120 } { 3 }$ sq unit
0%
$\dfrac { 130 } { 3 }$ sq unit

The area of the closed figure bounded by

$y = x , y = - x \&$ the tangent to the curve

$y = \sqrt { x ^ { 2 } - 5 }$ at the point

$( 3,2 )$ is:

0%
$5$
0%
$\frac { 15 } { 2 }$
0%
$10$
0%
$\frac { 35 } { 2 }$

The area of the figure bounded by the curves

$\displaystyle y = lnx$ and

$y$ =

$(lnx)^2$ is

0%
e + 1
0%
e-1
0%
3- e
0%
1

The area between the parabola

${ y }^{ 2 }=4x$ , normal at one end of latusreetum and X-axis sin sq. units is

0%
$\frac { 1 }{ 3 }$
0%
$\frac { 2 }{ 3 }$
0%
$\frac { 10 }{ 3 }$
0%
$\frac { 4 }{ 3 }$

${ A }_{ n }$ is the area bounded by y=x and y=

${ x }^{ n },n\epsilon N$ , then

${ A }_{}.{ A }_{ 3 }...{ A }_{ n }=$

0%
$\dfrac { 1 }{ n\left( n+1 \right) }$
0%
$\frac { 1 }{ { 2 }^{ n }n\left( n+1 \right) }$
0%
$\frac { 1 }{ { 2 }^{ n-1 }n\left( n+1 \right) }$
0%
$\frac { 1 }{ { 2 }^{ n-2 }n\left( n+1 \right) }$

The area of the region

$x+y\le 6$ ,

$x^2+y^2\le 6y$ and

$y^2\le 8x$ is

0%
$\cfrac{1}{72}(27\pi - 5)$ sq.units
0%
$\cfrac{1}{12}(27\pi +2)$ sq.units
0%
$\cfrac{1}{12}(27\pi -2)$ sq.units
0%
none of these

The area (in sq units) of the region

$\left\{ {\left( {x,y} \right):x \geqslant 0,x + y \leqslant 3,{x^2} \leqslant 4y\,\,and\,\,y \leqslant 1 + \sqrt x } \right\}$ is

0%
59 / 12
0%
3 / 2
0%
7 / 3
0%
5 / 2

The area of the region

$A = \{ ( x , y ) : 0 \leq y \leq x | + 1 \text { and } - 1 \leq x \leq 1 \}$ in sq. units, is:

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

Area of the region enclosed between the curves

$x={ y }^{ 2 }-1$ and

$x=|y|\sqrt { 1-{ y }^{ 2 } }$ is

0%
$1$ sq. units
0%
$\dfrac{4}{3}$ sq. units
0%
$\dfrac{2}{3}$ sq. units
0%
$2$ sq. units

$R=((x,y):|x|\le |y|\quad and\quad { x }^{ 2 }+{ y }^{ 2 }\le 1)is$

0%
$\dfrac { 3\pi }{ 8 } s.q.units$
0%
$\dfrac { \pi }{ 8 } s.q.units$
0%
$\dfrac { 5\pi }{ 8 } s.q.units$
0%
$\dfrac { \pi }{ 2 } s.q.units$

The area of the region:

$A = \left\{ {\left( {x,y} \right)} \right\}:0\underline < y\underline < \left| x \right| + 1\,\,{\text{and}}\,\,\left. { - 1\underline < x\underline < 1} \right\}$ in sq. units, is

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

Area enclosed by

$|x-1|+|y+1|=1$ .

0%
2
0%
4
0%
1
0%
8

Area bounded by Curve

${ y }^{ 2 }=4x,y$ axis and line y=3 is :

0%
7/4 Sq.unit
0%
9/4 Sq. unit
0%
5/4 Sq. unit
0%
11/4 Sq. unit

The area of the figure bounded by the curves

$y=|x-1|$ and

$y=3-|x|$ is-

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

Explanation

${\textbf{Step - 1: Drawing a graph}}$

${\text{y = |x - 1|}}$

$\Rightarrow {\text{ y = 1 - x if x < 1 and x - 1 if x}} \geqslant {\text{1}}$

${\text{y = 3 - |x|}}$

$\Rightarrow {\text{ y = 3 + x if x < 0 and 3 - x if x}} \geqslant 0$

${\textbf{Step - 2: Calculating area under the curve}}$

$\text{Required area =A=}$

$\int\limits_{{\text{ - 1}}}^{\text{0}} {\left( {\left( {{\text{3 + x}}} \right){\text{ - }}\left( {{\text{1 - x}}} \right)} \right){\text{ dx}}} {\text{ + }}\int\limits_{\text{0}}^{\text{1}} {\left( {{\text{(3 - x) - (1 - x)}}} \right){\text{ dx + }}\int\limits_{\text{1}}^{\text{2}} {\left( {{\text{(3 - x) - (x - 1)}}} \right)} } {\text{ dx}}$

$\Rightarrow {\text{ A = }}\int\limits_{{\text{ - 1}}}^{\text{0}} {{\text{(3 + x - 1 + x) dx}}} {\text{ + }}\int\limits_{\text{0}}^{\text{1}} {{\text{(3 - x - 1 + x) dx + }}\int\limits_{\text{1}}^{\text{2}} {{\text{(3 - x - x + 1) dx}}} }$

$\Rightarrow {\text{ A = }}\int\limits_{{\text{ - 1}}}^{\text{0}} {{\text{(2 + 2x) dx}}} {\text{ + }}\int\limits_{\text{0}}^{\text{1}} {{\text{2 dx + }}\int\limits_{\text{1}}^{\text{2}} {{\text{(4 - 2x) dx}}} }$

$\Rightarrow {\text{ A = }}\left[ {{\text{2x + }}\dfrac{{{\text{2}}{{\text{x}}^{\text{2}}}}}{{\text{2}}}} \right]_{{\text{ - 1}}}^{\text{0}}{\text{ + }}\left[ {{\text{2x}}} \right]_{\text{0}}^{\text{1}}{\text{ + }}\left[ {{\text{4x - }}\dfrac{{{\text{2}}{{\text{x}}^{\text{2}}}}}{{\text{2}}}} \right]_{\text{1}}^{\text{2}}$

$\Rightarrow {\text{ A = }}\left[ {{\text{2x + }}{{\text{x}}^{\text{2}}}} \right]_{{\text{ - 1}}}^{\text{0}}{\text{ + }}\left[ {{\text{2x}}} \right]_{\text{0}}^{\text{1}}{\text{ + }}\left[ {{\text{4x - }}{{\text{x}}^{\text{2}}}} \right]_{\text{1}}^{\text{2}}$

$\Rightarrow {\text{ A = [0 - ( - 2 + 1)] + [2 - 0] + [(8 - 4) - (4 - 1)]}}$

$\Rightarrow {\text{ A = 4 sq}}{\text{. units}}$

$\textbf{Hence option C is correct}$

The area of the region bounded by the curve

$x=\sin ^{-1}y$ , the x-axis and the line |x|=1 is

0%
$2-2\cos 1$
0%
$1-\cos 1$
0%
$1-2\cos 1$
0%
none of these

The area enclosed within the curve is

$\left| x \right| + \left| y \right| = 1$ is

0%
$\sqrt 2$
0%
$1$
0%
$\sqrt 3$
0%
$2$

The area enclosed between the curve

${x^2} + {y^2} = 16$ and the coordinates axes in the first quadrant is

0%
$\pi \,\,sq.\,\,units$
0%
$2\pi \,\,sq.\,\,units$
0%
$3\pi \,\,sq.\,\,units$
0%
$4\pi \,\,sq.\,\,units$

The area bounded by the curve

$y=\begin{cases} x^{ 2 };x<0 \\ x;x\ge 0 \end{cases}$ and the line

$y=4$ is

0%
$\cfrac{20}{41}$
0%
$\cfrac{20}{147}$
0%
$\cfrac{40}{3}$
0%
$\cfrac{20}{21}$

The area of the region bounded by the curve

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

0%
$\frac { 9 }{ 4 }$
0%
9
0%
$\frac { 9 }{ 2 }$
0%
$\frac { 9 }{ 7 }$

y= f(x) is a function which satisfies -
(i)f(0)=0 (ii)f''(x)= f'(x) and (iii) f'(0)=1
then the area bounded by the graph of y = f(x), the lines x=0,x-1=0 and y+1=0 is -

0%
e
0%
e-2
0%
e-1
0%
e+1

Area of the region bounded by curves y=x log x and

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

0%
1/12
0%
7/12
0%
5/12
0%
none of these

Area of the region bounded by the curve

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

$x = 1$ is

0%
$\frac {5} {4}$
0%
$\frac {3} {4}$
0%
$\frac {4} {5}$
0%
$\frac {4} {3}$

Area enclosed by the curve

$y = f ( x )$ defined parametrical as

$x = \frac { 1 - t ^ { 2 } } { 1 + t ^ { 2 } } , y = \frac { 2 t } { 1 + t ^ { 2 } }$

0%
$\pi$ sq. units
0%
$\frac { \pi } { 2 }$ sq. units
0%
$\frac { 3 \pi } { 4 }$ sq. units
0%
$\frac { 3 \pi } { 2 }$ sq. units

The area (in sq units) of the region bounded by the curve

$y=\sqrt { x }$ and the lines

$y=0,y=x-2$ , is

0%
$\frac { 10 }{ 3 }$
0%
$\frac { 8 }{ 3 }$
0%
$\frac { 4 }{ 3 }$
0%
$\frac { 16 }{ 3 }$

The area bounded by the curve y=

${ x }^{ 3 },$ x-axis and two ordinates x=1 to x=2 equal to

0%
15/2 sq.unit
0%
15/4 sq.unit
0%
17/2 sq.unit
0%
17/4 sq.unit

Area enclosed by the graph of the function

$y=l{ n }^{ 2 }x-1$ lying in the

${ 4 }^{ th }$ quadrant is

0%
$\frac { 2 }{ e }$
0%
$\frac { 4 }{ e }$
0%
$2(e+\frac { 1 }{ e } )$
0%
$4(e-\frac { 1 }{ e } )$

Area bounded by curves

$x=\sqrt{y-1}$
and y=x+1 is -

0%
$\frac{1}{3} s q \cdot u n i t$
0%
$\frac{8}{3} s q \cdot u n i t$
0%
$\frac{1}{6} s q \cdot u n i t$
0%
$5sq.unit$

Explanation

$\text { given: } \mathrm{eq}^{n} \text { of curves are: } \\$

$x=\sqrt{y-1} \rightarrow y=x^{2}+1 \\$

$y=x+1$

$\text { So, the area enclosed :- } \\$

$A =\int_{0}^{1}(x+1) d x-\int_{0}^{1}\left(x^{2}+1\right) d x \\$

$=\left[\frac{x^{2}}{2}+x\right]_{0}^{1}-\left(\frac{x^{3}}{3}+x\right)^{1} \\$

$=\left(\left(\frac{1}{2}+1\right)-0\right)-\left(\left(\frac{1}{3}+1\right)-0\right) \\$

$=\left(\frac{3}{2}-\frac{4}{3}\right) \text { sq. Urits } \\$

$\text { Ans. } A =\frac{1}{6} sq . \text { units }$

The area of the region bounded by the curves

$y = x^2$ and

$y = |x|$ is

0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{5}{6}$
0%
$\dfrac{5}{3}$

Area bounded by the curves

$y = \sin x ,$ tangent drawn to it at

$x = 0$ and the line

$x = \frac { \pi } { 2 }$ is equal to

0%
$\frac { \pi ^ { 2 } - 4 } { 2 }$ sq.units
0%
$\frac { \pi ^ { 2 } - 4 } { 4 }$ sq.units
0%
$\frac { \pi ^ { 2 } - 2 } { 4 }$ sq.units
0%
$\frac { \pi ^ { 2 } - 2 } { 2 }$ sq.units

The area enclosed by the line y = x + 1, X- axis and the lines x = -3 and x = 3 is

0%
5
0%
7
0%
10
0%
9

The area of region bounded by

$x = 0,2 x - 3 y = - 6$ and

$2 x + 3 y = 18$ is

0%
$2$ square unit
0%
$4$ square unit
0%
$6$ square unit
0%
$9$ square unit
0%
None of these

The area of the region
A=

$\left [ \left ( x,y \right ) :0\leq y\leq x\left | x \right |+1and -1\leq x\leq 1\right ]$ . in sq. units, is:

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

The area bounded by the parabolas

$y = {\left( {x + 1} \right)^2}\;and\;y = {\left( {x - 1} \right)^2}$ and the line

$y = \frac{1}{4}$ is

0%
${\text{4sq}}{\text{.}}\;{\text{units}}$
0%
$\frac{1}{6}\;{\text{sq}}{\text{.}}\;{\text{units}}$
0%
$\frac{4}{3}\;{\text{sq}}{\text{.}}\;{\text{units}}$
0%
$\frac{1}{3}\;{\text{sq}}{\text{.}}\;{\text{units}}$

The area enclosed by parabola

${ y }^{ 2 }=64x$ and its latus-rectum is

$\lambda$ then

$3\lambda$ then

$3\lambda$ =

0%
2048
0%
2408
0%
2804
0%
2084

The area enclosed between the curves

$y = a x ^ { 2 }$ and

$x = a y ^ { 2 } ( a > 0 )$ is

$1$ sq. unit, then the value of

$a$ is

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

the volume of a solid obtained by revolving about y-axis enclosed between the ellipse

${x}^{2}+9{y}^{2}=9$ and the straight line

$x+3y=3$ in the first quadrant is

0%
$3\pi$
0%
$4\pi$
0%
$6\pi$
0%
$9\pi$

The area of the region lying between the line

$x-y+2=0$ and the curve

$x=\sqrt{y}$ is

0%
$9$
0%
$\dfrac {9}{2}$
0%
$\dfrac {10}{3}$
0%
none

Explanation

Given

$x-y+2$ and

$x=\sqrt{y}$

$x^{2}=y$

$\therefore$ Required area

$A=\int_{0}^{2}[x-x^{2}+2]dx$

This represents the point of intersection of curves

so, by equating the equations we get,

$x+2=x^{2}$

$\Rightarrow x^{2}-x-2=0$

$\Rightarrow x^{2}-2x+x-2=0$

$\Rightarrow x(x-2)+1(x-2)=0$

$\Rightarrow (x-2)(x+1)=0$

$\Rightarrow x=-1,2$

Since x> 0 x=1 is not valid

hence a=2

$A=\int_{0}^{2}[x+2-x^{2}]dx$

$A=\int_{0}^{2} xdx +\int_{0}^{2}2dx-\int_{0}^{2}x^{2}dx$

$A=[\frac{x^{2}}{2}+2x-\frac{x^{3}}{3}]_{0}^{2}$

$A=2+4-\frac{8}{3}=\frac{6+12-8}{3}$

$A=\frac{10}{3}$

1213804_1505685_ans_d6a745079ace42c49e8c0c34f901a480.png

Area bounded by curve

$y = (x - 1)(x - 2)(x - 3)$ and x-axis between lines

$x = 0, x = 3$

0%
$5/2$
0%
$11/4$
0%
$1$
0%
$3$

As shown in the figure of an ellipse

$\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ . The area of shaded region is .......
.

0%
$12\pi$
0%
$3(\pi -2)$
0%
$4(\pi -2)$
0%
$12(\pi -2)$

For

$a > 0$ , let the curves

$C_1 : y^2 = ax$ and

$C_2 : x^2 = ay$ intersect at origin

$O$ and a point

$P$ . Let the line

$x = b (0 < b < a)$ intersect the chord

$OP$ and the x-axis at points

$Q$ and

$R$ , rspectively. If the line

$x=b$ bisects the area bounded by the curves,

$C_1$ and

$C_2$ , and the area of

$\Delta OQR = \dfrac{1}{2}$ , then '

$a$ ' satisfies the equation:

0%
$x^6-12x^3 + 4 = 0$
0%
$x^6 + 6x^33 - 4 = 0$
0%
$x^6 - 12x^3 + 4 = 0$
0%
$x^6 - 6x^3 + 4 = 0$

Explanation

area

$\Delta OQR = \dfrac{1}{2}.....given$

For coordinates of

$p$

Curves :

$x^2 = ay, y^2 = ax$

$\Rightarrow y^2 = \left(\dfrac{x^2}{a} \right)^2 = ax$

$\dfrac{x^4}{a^2} = ax$

$x^4 = a^3x$

$x(x^3 - a^3) = 0$

$\Rightarrow x = 0, x=a....(i)$

$\Rightarrow y^2 = ax$

$\therefore y = 0, a....(from \ i)$

$\therefore$ equation of chord

$OP$

$O(0,0)\equiv(x_1,y_1) \ and \ P(a,a)\equiv(x_2,y_2)$

$y - y_1 = \dfrac{y_2-y_1}{x_2-x_1} (x-x_1)$

$y - 0 = \dfrac{a-0}{a-0} (x-o)$

$y = x$

$\therefore$ At point

$Q$

$x = b$ and

$y = b$

$\Rightarrow Q \equiv (b,b)$

$\therefore$ ar

$\Delta OQR = \dfrac{1}{2} \times b \times b = \dfrac{b^2}{2} = \dfrac{1}{2}$

$\Rightarrow b^2 = 1 \Rightarrow b=1$

Now, area bounded by the curves

$= \displaystyle \int_0^a \left(\sqrt{ax} - \dfrac{x^2}{a}\right)dx = \int_0^a a^{1/2} x^{1/2}dx - \int_0^a \dfrac{x^2}{a} dx$

$=a^{1/2}\left[\dfrac{x^{3/2}}{3/2}\right]_0^a-\dfrac{1}{a}\left[\dfrac{x^3}{3}\right]_0^a$

$=\dfrac{2}{3}a^{1/2}. a^{3/2} - \dfrac{1}{a} \dfrac{a^3}{3} = \dfrac{2}{3} a^2 - \dfrac{1}{3}a^2$

$= \dfrac{1}{3}a^2$

$\therefore \displaystyle \int_0^b \left(\sqrt{ax} - \dfrac{x^2}{a}\right)dx = \dfrac{1}{2} \times \dfrac{1}{3}a^2 = \dfrac{a^2}{6}$

$\dfrac{2}{3} a^{1/2} [x^{3/2}]_0^b - \dfrac{1}{3a} [x^3]_0^b = \dfrac{a^2}{6}$

$\Rightarrow \dfrac{2}{3}a^{1/2} b^{3/2} - \dfrac{1}{3a}b^3 = \dfrac{a^2}{6}$

Now after putting value of

$b$ .

we get,

$\dfrac{2}{3}a^{1/2} - \dfrac{1}{3a} = \dfrac{a^2}{6}$

$2 a^{3/2} -1 = \dfrac{3a^3}{6} = \dfrac{a^3}{2}$

$4a^{3/2} - 2 =a^3 \Rightarrow 4a^{3/2} = a^3 + 2$

Now squaring on both the side

$\Rightarrow a^6 + 4a^3 + 4 = 16a^3$

$\Rightarrow a^6 - 12a^3 + 4 = 0$

$\therefore$ a will satisfy

$x^6 - 12x^3 + 4 = 0$

$\therefore$

$\boxed{x^6 - 12x^3 + 4 = 0}....Answer$

Hence option

$'C'$ is the answer.

1477256_1697509_ans_2194c041b61a434db5fb2f0cc1847c9e.png

The area bounded by the curve

$y =x^4 -2x^3 + x^2 + 3$ the axis of abscissas and two oridnates corrsponding to the point of minimum of function y(x) is

0%
10/3
0%
27/10
0%
21/10
0%
None of these

The area of the region bounded by

$y^2=2x +1$ and

$x-y-1=0$ is

0%
$\dfrac{2}{3}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{8}{3}$
0%
$\dfrac{16}{3}$

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 13 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 13 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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