MCQ Questions for Class 12 Commerce Maths Application Of Integrals Quiz 11

The smaller area enclosed by

$y=f(x)$ , where

$f(x)$ is polynomial of least degree satisfying

$\displaystyle{ \left[ \lim _{ x\rightarrow 0 }{ 1+\frac { f\left( x \right) }{ { x }^{ 3 } } } \right] }^{ \tfrac { 1 }{ x } }=e$ and the circle

$x^2+y^2=2$ above the

$x-$ axis is

0%
$\displaystyle\frac { \pi }{ 2 } +\frac { 3 }{ 5 }$
0%
$\displaystyle\frac { \pi }{ 2 } -\frac { 3 }{ 5 }$
0%
$\dfrac { \pi }{ 2 } -\dfrac { 6 }{ 5 }$
0%
None of these

Explanation

Since

$\displaystyle\lim _{ x\rightarrow 0 }{ { \left[ 1+\frac { f\left( x \right) }{ { x }^{ 3 } } \right] }^{ \tfrac { 1 }{ x } } }$ exists, so

$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { f\left( x \right) }{ { x }^{ 3 } } } =0$

$\therefore f(x)={ a }_{ 4 }{ x }^{ 4 }+{ a }_{ 5 }{ x }^{ 5 }+...+{ a }_{ n }{ x }^{ n },a_n\neq 0,n\ge 4$

Since,

$f(x)$ is of least degree

$\Rightarrow f(x)={ a }_{ 4 }{ x }^{ 4 }$
The graph of

$y=x^4$ and

$x^2+y^2=2$ are shown in the figure

$\therefore$ The required area

$\displaystyle=2\int _{ 0 }^{ 1 }{ \left( \sqrt { 2-{ x }^{ 2 } } -{ x }^{ 4 } \right) } dx=\frac { \pi }{ 2 } +\frac { 3 }{ 5 }$

The area of the region described by

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

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

The area (in square units) of the region bounded by the curves

$x=y^2$ and

$x=3-2y^2$ is

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

Explanation

Required area

$\displaystyle =2\int_0^1(3-2y^2-y^2)dy=6\int_0^1(1-y^2)dy=6\left(y-\dfrac{y^3}{3}\right)_0^1=4$

Since given curves are symmetrical about x-axis

The area (in square units) bounded by the curves

$x\, =\, -2y^2$ and

$x\, =\, 1-3y^2$ is

0%
$\displaystyle \frac{2}{3}$
0%
$1$
0%
$\displaystyle \frac{4}{3}$
0%
$\displaystyle \frac{5}{3}$

Explanation

$x=-2y^2$

$x=1-3y^2$

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

Area between the curves

$= A_1-A_2$

$\displaystyle =\int^1_{-1}(1-3y^2+2y^2)dy$

$=\int^1_{-1}(1-y^2)dy$

$=[y-\dfrac{y^3}{3}]^1_{-1}=\dfrac{4}{3}$

The area of the region bounded by the curves

$x^{2} + y^{2} = 8$ and

$y^{2} = 2x$ is

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

Explanation

Given curves,

$x^{2} + y^{2} = 8$ .....(i)
and

$y^{2} = 2x$ .....(ii)

On solving Eqs. (i) and (ii), we get

$x^{2} + 2x - 8 = 0$

$x^{2} + 4x - 2x - 8 = 0$

$x (x - 4) - 2 (x + 4) = 0$

$(x- 2) (x +4) = 0$

$\therefore x = 2$ and

$y = \pm 2$

$\therefore$ Required area

$= 2[\text {Area of OAP} + \text {Area of PAB}]$

$= 2\left [\displaystyle\int_{0}^{2} \sqrt {2x} dx + \int_{2}^{2\sqrt {2}} \sqrt {8 - x^{2}} dx \right ]$

$= 2\left [\sqrt {2} \left (x^{3/2} \cdot \dfrac {2}{3}\right )^{2}_{0} + \left (\dfrac {x}{2} \sqrt {8 - x^{2}} + \dfrac {8}{2} \sin^{-1} \dfrac {x}{2\sqrt {2}} \right )_{2}^{2\sqrt {2}} \right ]$

$= 2\left [\dfrac {2\sqrt {2}}{3} (2^{3/2}) + 4\times \dfrac {\pi}{2} - 2 - 4\times \dfrac {\pi}{4}\right ]$

$= 2\left [\dfrac {2\sqrt {2}}{3} \cdot 2\sqrt {2} + 2\pi - 2 - \pi \right ]$

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

Area common to the curves

$5x^2 = 0$ and

$2x^2 + 9 = 0$ is equal to

0%
$12 \sqrt 3$
0%
$6 \sqrt3$
0%
$36$
0%
$18$

Explanation

Given curves

$y=5x^2$

$\Rightarrow x^2=\dfrac{y}{5}$ ....(1)
which is a parabola opening upward having vertex at (0,0)

Other curve is

$2x^2+9=y$ .....(2)

$\Rightarrow x^2=\dfrac{y-9}{2}$
which is a parabola having vertex at (0,9).

Now, solving eqn (1) and (2), we get

$5x^2=2x^2+9$

$\Rightarrow 3x^2=9$

$\Rightarrow x=\pm\sqrt{3}$

$\Rightarrow y=15$
Point of intersection of the curves is

$(-\sqrt{3},15)$ and

$(\sqrt{3},15)$

Required area

$=2(ar OAB)$

$=2 \int ^\sqrt 3 _0 (y_1-y_2)dx$

$= 2 \int ^\sqrt 3 _0 \{ (2x^2 + 9) - 5 x^2\} dx$

$2 \int ^\sqrt 3 _ 0 (9-3x^2) dx$

$= 2 \left( 9x - 3 \displaystyle \frac{x^3}{3}\right) ^\sqrt3 _0$

$= 2(9 \sqrt 3 - 3 \sqrt 3 ) = 12 \sqrt 3$ sq. units

The area of the region, bounded by the curves

$y = \sin^{-1} x + x (1 - x)$ and

$y = \sin^{-1} x - x (1 - x)$ in the first quadrant, is

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

Explanation

$\sin^{-1} x$ is defined, if

$-1 \leq x \leq 1$
In first quadrant

$0\leq x\leq 1$ and

$x(1 - x) \geq 0$

$\therefore y = \sin^{-1} x + x (1 - x)$ ...... (i)
Lies above

$y = \sin^{-1} x - x (1 - x)$ ...... (ii)
On solving, we get

$2x (1 - x) = 0$

$\Rightarrow x = 0, 1$

$\therefore$ Required area

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

$=\displaystyle \int_{0}^{1} [\left \{\sin^{-1} x + x(1 - x)\right \} - \left \{\sin^{-1} x - x(1 - x) \right \}]dx$

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

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

If the area bounded by the curves

$y=a{ x }^{ 2 }$ and

$x=a{ y }^{ 2 }$ ,

$(a>0)$ is

$1$ sq.units, then the value of

$a$ is

0%
$\cfrac { 2 }{ 3 }$
0%
$\cfrac { 1 }{ \sqrt 3 }$
0%
$1$
0%
$4$

Explanation

Points of intersection of

$y=ax^2$ and

$x=ay^2$ are

$(0,0)$ and

$\left (\dfrac {1}{a},\dfrac {1}{a}\right)$ .

Hence,

$\displaystyle \int _0 ^{\tfrac1a} \left (\sqrt {\dfrac {x}{a}}-ax^2\right)dx=1$

$\Rightarrow \displaystyle\cfrac{2x^{\tfrac32}}{3\sqrt a} \big|_0^{\tfrac1a} - \cfrac{ax^3}3\bigr|_0^{\tfrac1a} =1$

$\Rightarrow \cfrac{2}{3a^2} - \cfrac1{3a^2} =1$

$\Rightarrow \cfrac{1}{3a^2} =1$

$\Rightarrow a=\dfrac {1}{\sqrt3}$ ....As

$(a>0)$

Area bounded by curve

$y=x^2$ and

$y=2-x^2$ is ?

0%
$\dfrac{8}{3}$ sq units
0%
$\dfrac{3}{8}$ sq units
0%
$\dfrac{3}{2}$ sq units
0%
None of these

Explanation

$y=x^2$ and

$y=2-x^2$

now both he curve intersect each other ,

from both the equation

$x^2=2-x^2$

$x=+1,-1$

now area bounded by both curve is

$A =\displaystyle \int_{-1}^{1}(2-x^2)-\displaystyle \int_{-1}{1}(x^2)$

because both are even function hence

$A =2[\displaystyle \int_{0}^{1}(2-x^2)-\displaystyle \int_{0}{1}(x^2)$ ]

$A=2[(2x-\dfrac{x^3}{3})-(\dfrac{x^3}{3})]$

now on putting upper and lower value of limit ,we will get

$A=\dfrac{8}{3}$

If the line

$x = \alpha$ divides the area of region

$R=\left\{ \left( x,y \right) \in { R }^{ 2 }:{ x }^{ 3 }\le y\le x,0\le x\le 1 \right\}$ into two equal parts, then

0%
$2{ \alpha }^{ 4 }-4{ \alpha }^{ 2 }+1=0$
0%
${ \alpha }^{ 4 }+4{ \alpha }^{ 2 }-1=0$
0%
$0 < \alpha \le \dfrac { 1 }{ 2 }$
0%
$\dfrac { 1 }{ 2 } < \alpha < 1$

Explanation

Area

$(x=0\to x=\alpha) =$ Area of

$\triangle OAB -$ Area of

$\triangle OCB$

$\displaystyle= \dfrac12\alpha\cdot\alpha - \int_0^\alpha x^2 dx$

$=\cfrac{\alpha^2}2 - \cfrac{\alpha^2}4$

Area

$(0=\alpha \to x=1) =$ Area of

$\Box BAEF -$ Area of

$\Box BCEF$

$\cfrac12(\alpha+1)(1-\alpha) -\displaystyle \int_\alpha^1x^3\ dx$

$\cfrac{1-\alpha^2}{2} - \left(\cfrac14-\cfrac{\alpha^2}4\right)$

According to the question:

$\cfrac{\alpha^2}2-\cfrac{\alpha^4}4 = \cfrac12-\cfrac{\alpha^2}2 - \cfrac14 + \cfrac{\alpha^4}4$

$\Rightarrow 2\alpha^4-4\alpha^2+1=0$ ... Option A

$f(a) = 2\alpha^2 - 4\alpha^2 + 1 = 0$

$\alpha = 0, f(a) = 1$

$\alpha = 1, f(\alpha) = -1$

$\alpha =\cfrac12, f(a) = \cfrac18 > 0$

Therefore, Root lies in

$\alpha \in \left(\cfrac12,1\right)$ .. Option D.

The area bounded by the parabolas

$y^2 = 4a(x + a)$ and

$y^2 = - 4a (x - a)$ is

0%
$\dfrac{16}{3} a^2$ sq units
0%
$\dfrac{8}{3}$ sq units
0%
$\dfrac{4}{3} a^2$ sq units
0%
None of these

Explanation

$y^{2}=4a\left ( x+a \right )$

$y=\pm \sqrt{4ax+4a^{2}}$

As the area bounded by this curve is in -ve

$x$ quadrant . Therfore

$y=-\sqrt{-4ax+4a^{2}}$

Area bounded by this curve will be

$\int_{-a}^{0}\sqrt{4ax+4a^{2}dx}$

Area bounded by this curve will be

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

Similarly , The area bounded by

$y^{2}=-4a\left ( x-a \right )$ curve will be

$\int_{0}^{a}\sqrt{-4ax+4a^{2}dx}$

Area =

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

thus , total bounded area will be

$\dfrac {16 }{3} a^{2}$

The area of the region bounded by the curves

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

$x = 2$ is

0%
$\dfrac {3}{\log 2} - \dfrac {4}{3}$
0%
$\dfrac {3}{\log 2} - \dfrac {4}{9}$
0%
$\dfrac {3}{2} - \dfrac {\log 2}{9}$
0%
None of these

Explanation

Given curves are

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

$x = 2$

$\therefore$ Required area

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

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

$= \left [\dfrac {2^{x}}{\log 2} - x^{2} + \dfrac {x^{3}}{3}\right ]_{0}^{2}$

$= \dfrac {4}{\log 2} - 4 + \dfrac {8}{3} - \dfrac {1}{\log 2}$

$= \dfrac {3}{\log 2} - \dfrac {4}{3}$ .

The area of the portion of the circle

${ x }^{ 2 }+{ y }^{ 2 }=64$ which is exterior to the parabola

${ y }^{ 2 }=12x$ , is

0%
$\left( 8\pi -\sqrt { 3 } \right)$ sq units
0%
$\dfrac { 16 }{ 3 } \left( 8-\sqrt { 3 } \right)$ sq units
0%
$\dfrac { 16 }{ 3 } \left( 8\pi -\sqrt { 3 } \right)$ sq units
0%
None of the above

Explanation

Required shaded area

$=$

$=Area\, of \, circle -2\left[ \displaystyle\int _{ 0 }^{ 4 }{ 2\sqrt { 3 } \sqrt { x } dx } +\displaystyle\int _{ 4 }^{ 8 }{ \sqrt { 64-{ x }^{ 2 } } dx } \right]$

$=64\pi -2\left[ { \left( 2\sqrt { 3 } { x }^{ { 3 }/{ 2 } }\times \dfrac { 2 }{ 3 } \right) }_{ 0 }^{ 4 }+{ \left( \dfrac { x }{ 2 } \sqrt { 64-{ x }^{ 2 } } +\dfrac { 64 }{ 2 } \sin ^{ -1 }{ \dfrac { x }{ 8 } } \right) }_{ 4 }^{ 8 } \right]$

$=64\pi -\dfrac { 64 }{ \sqrt { 3 } } -32\pi +16\sqrt { 3 } +\dfrac { 32\pi }{ 3 }$

$=\dfrac { 128\pi }{ 3 } -\dfrac { 16\sqrt { 3 } }{ 3 }$

$=\dfrac { 16 }{ 3 } \left( 8\pi -\sqrt { 3 } \right)$ sq units.

Area common to the curves

$y^{2} = ax$ and

$x^{2} + y^{2} = 4ax$ is equal to

0%
$(9\sqrt {3} + 4\pi) \dfrac {a^{2}}{3}$
0%
$(9\sqrt {3} + 4\pi)a^{2}$
0%
$(9\sqrt {3} - 4\pi) \dfrac {a^{2}}{3}$
0%
None of these

Explanation

Area common to the curves

${ y }^{ 2 }=a_{ x }\quad { x }^{ 2 }+{ y }^{ 2 }=4ax$ is equal to :

${ x }^{ 2 }+{ y }^{ 2 }=4ax\\ { y }^{ 2 }=a_{ x }\\ x=0,3a\\ y=0,\pm \sqrt { 3 } a$

Shaded region

$\int _{ 0 }^{ 3 }{ \sqrt { ax } dx } +\int _{ 3a }^{ 4a }{ \sqrt { 4ax-{ x }^{ 2 } } dx } \\ =\left[ \cfrac { \sqrt { a } { x }^{ \cfrac { 3 }{ 2 } } }{ \cfrac { 3 }{ 2 } } \right] _{ 0 }^{ 3a }+\int _{ 3a }^{ 4a }{ \sqrt { (2a)^{ 2 }-(x-2a)^{ 2 } } dx } \\ =\cfrac { 2.\sqrt { a } .3\sqrt { 3 } .a\sqrt { a } }{ 3 } +\left[ \cfrac { x-2a }{ 2 } \sqrt { 4a_{ x }-{ x }^{ 2 } } +\cfrac { 4{ a }^{ 2 } }{ 2 } \sin ^{ -1 }{ \cfrac { x-2a }{ 2a } } \right] _{ 3a }^{ 4a }\\ =2\sqrt { 3 } { a }^{ 3 }+\left[ 0+2{ a }^{ 2 }\left( \cfrac { \pi }{ 2 } \right) -\cfrac { \sqrt { 3 } { a }^{ 2 } }{ 2 } -2{ a }^{ 2 }\left( \cfrac { \pi }{ b } \right) \right] \\ =2\sqrt { 3 } { a }^{ 3 }-\cfrac { \sqrt { 3 } { a }^{ 2 } }{ 2 } +\pi { a }^{ 2 }-\cfrac { \pi }{ 3 } { a }^{ 2 }\\ =\cfrac { 3\sqrt { 3 } { a }^{ 2 } }{ 2 } +\cfrac { 2\pi { a }^{ 2 } }{ 3 } =\left[ 9\sqrt { 3 } +4\pi \right] \cfrac { { a }^{ 2 } }{ 6 }$

Required area is :

$=2(shaded\quad area)$

$=2\left( \left[ 9\sqrt { 3 } +4\pi \right] \cfrac { { a }^{ 2 } }{ 6 } \right) \\ =\left[ 9\sqrt { 3 } +4\pi \right] \cfrac { { a }^{ 2 } }{ 6 } sq.units$

The area bounded by

$x^2+y^2-2x=0$ &

$y=\sin\displaystyle\frac{\pi x}{2}$ in the upper half of the circle is?

0%
$\displaystyle\frac{\pi}{2}-\frac{4}{\pi}$
0%
$\displaystyle\frac{\pi}{4}-\frac{2}{\pi}$
0%
$\displaystyle \pi -\frac{8}{\pi}$
0%
None

Explanation

The area bonded by

${ x }^{ 2 }+{ y }^{ 2 }-2x=0\quad y=\sin { \cfrac { \pi x }{ 2 } }$

In the upper half of the circle is

Required Area area = shaded area

$=\cfrac { 1 }{ 2 } \pi { r }^{ 2 }-\int _{ 0 }^{ 2 }{ ydx } \\ =\cfrac { 1 }{ 2 } \pi { (1) }^{ 2 }-\int _{ 0 }^{ 2 }{ \cfrac { \sin { \pi x } }{ 2 } dx } \\ =\cfrac { \pi }{ 2 } +\left[ \cfrac { \cos { \cfrac { \pi x }{ 2 } } }{ \cfrac { \pi }{ 2 } } \right] _{ 0 }^{ 2 }=\cfrac { \pi }{ 2 } +\cfrac { 2 }{ \pi } \left[ \cos { \cfrac { \pi x }{ 2 } } \right] _{ 0 }^{ 2 }\\ =\cfrac { \pi }{ 2 } +\cfrac { 2 }{ \pi } \left[ \cos { \pi } -\cos { 0 } \right] =\cfrac { \pi }{ 2 } +\cfrac { 2 }{ \pi } -2\\ =\cfrac { \pi }{ 2 } +\cfrac { 4 }{ \pi } sq.\quad units$

Hence the correct answer is

$\cfrac { \pi }{ 2 } +\cfrac { 4 }{ \pi } sq.\quad units$

On the real line R, we define two functions f and g as follows:

$f(x) = min [x - [x], 1 - x + [x]]$ ,

$g(x) = max [x - [x], 1 - x + [x]]$ ,
where [x] denotes the largest integer not exceeding x.

The positive integer n for which

$\displaystyle \int_{0}^{n}{(g(x) - f(x) ) dx = 100}$ is?

0%
$100$
0%
$193$
0%
$200$
0%
$202$

Explanation

Given,

$f(x) = min [x - [x], 1 - x + [x]]=min[f,1-f]$

and

$g(x) = max [x - [x], 1 - x + [x]]=max[f,1-f]$

where

$f$ is the fractional part of the function.

and

$f$ is always

$0\le f <1$ .

$\therefore$ if

$0<f<0.5$ , then

$f(x) = f$

$g(x) =1-f$

and If

$0.5<f<1$ , then

$f(x) = 1-f$

$g(x) =f$

$\displaystyle \int_0^n(g(x)-f(x))dx=$

$\displaystyle n\int_0^1(g(x)-f(x))dx$

$=\displaystyle n\int_0^{0.5}(1-2x)dx+$

$\displaystyle n\int_{0.5}^1(2x-1)dx=n(0.5-0.25)+n(0.75-0.5)$

$0.5 n=100\implies n=200$

The parabola

$y^2=4x+1$ divides the disc

$x^2+y^2\leq 1$ into two regions with areas

$A_1$ and

$A_2$ . Then

$|A_1-A_2|$ equals.

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

Explanation

$\Rightarrow$ Area of semi-circle

$=\pi/2$

suppose the area of the parabola between

$-1/4$ to

$0=P$ .

Thus,

$A_1=\pi/2-P$ and

$A_2=\pi/2+P$

$\Rightarrow A_2-A_1=2P=\displaystyle2\times 2\int_{-1/4}^0(\sqrt{4x+1})dx$

$\Rightarrow A_2-A_1=\dfrac{4\times2(4x+1)^{3/2}}{3\times 4}=2/3(1-0)=2/3$

The area bounded by the curves

$y = \sin x, y = \cos x$ and x-axis from

$x = 0$ to

$x = \pi /2$ is

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

The area bounded by min (|x|, |y|) = 2 and max (|x|, |y|) = 4 is

0%
8 sq unit
0%
16 sq unit
0%
24 sq unit
0%
32 sq unit

Explanation

$\to$ represents

$max (|x|, |y|) = 4$

$\to$ represents

$min (|x|, |y|)=2$

To solve this question we first need to understand the meaning of

$min(|x|,|y|)=2$ and

$max(|x|,|y|)=4$

$min(|x|,|y|)=2$ means that either

$|x|=2$ and

$|y|\geq 2$ or

$|y|=2$ and

$|x|\geq 2$

Similarly,

$max(|x|,|y|)=4$ means that either

$|x|=4$ and

$|y|\leq 4$ or

$|y|=4$ and

$|x|\leq 4$

As can be seen from the image of the graph plotted for this, we can see 4 squares each of size

$2\times2$

So, total area=

$4\times(2\times2)=16$

Hence, correct answer is option

$B$

The area of the region

$\left\lfloor x \right\rfloor +\left\lfloor y \right\rfloor =1,-1\le x\le 1$ and

$xy\le 1/2$

0%
rational
0%
$\cfrac { 1 }{ 2 } \left( \cfrac { 3 }{ 2 } +\ln { 2 } \right)$
0%
$\cfrac { 1 }{ 2 } \left( \cfrac { 5 }{ 2 } +\ln { 2 } \right)$
0%
Irrational

Explanation

NOTE:

$xy\leq1/2$ is equation of region below the parabola

$y\leq1/2x$ and above the co-ordinate axis (shown by blue)

$\bf{CASE-1:}$

$x\in[-1,0)$

$\left \lfloor{x}\right \rfloor=-1$ , so we want

$\left \lfloor{y}\right \rfloor=2$

$\rightarrow$ we want

$y\in[2,3)$

$\rightarrow$ This region is depited by full red square in image

$\rightarrow Area_1=1*1 = 1$

$\bf{CASE-2}$

$x\in[0,1)$

$\left \lfloor{x}\right \rfloor=0$ , so we want

$\left \lfloor{y}\right \rfloor=1$

$\rightarrow$ we want

$y\in[1,2)$

$\rightarrow$ so this case depicts a square (with lower-left vertex at

$(0,1)$ and upper-right vertex at

$(1,2)$ ) on co-ordinate axis, but note that

$xy\leq1/2$ cuts of this square as shown in image

We can find the shaded red region by doing,

$Area_2= \int_{1}^{2} \dfrac{1}{2y}dy$ ,treating x as a function of y, (

$x=1/2y$ )

$\dfrac{1}{2}$

$\int_{1}^{2}\dfrac{1}{y}dy=\dfrac{1}{2} (ln(2)-ln(1))=\dfrac{ln(2)}{2}$

Total area=

$Area_1+Area_2=1+ln(2)/2=\dfrac{1}{2}(2+ln(2)) \rightarrow IRRATIONAL$

Note: ln(n) -> IRRATIONAL for all integers n

$\geq$ 2

Area bounded by the curves

$\displaystyle y = \left[ \frac{x^2}{64} + 2 \right]$ ([.] denotes the greatest integer function)

$y = x - 1$ and

$x = 0$ above the x-axis is

0%
2
0%
3
0%
4
0%
none of these

The area bounded by the curves

$y = \dfrac {1}{4} |4 - x^{2}|$ and

$y = 7 -|x|$ is

0%
$18$
0%
$32$
0%
$36$
0%
$64$

Explanation

The graphs can be drawn as shown in the figure.

We need to find the area of the shaded portion. Since the area is symmetric about the y-axis, we will find the area on the right-hand side of the y-axis and later, multiply it by 2.

First, we find the area bounded by the straight line and the x-axis. Then, we subtract the area bounded by the parabola and the x-axis from our previous result.

Now, point A is the intersection point of both the graphs for

$x>0$ . So, we have

$\frac{x^2}{4} - 1 = 7-x$

$\implies x^2-4 = 28 - 4x$

$\implies x^2 +4x -32 = 0 \implies (x+8)(x-4) = 0$

$\implies x = 4$ (since

$x>0$ )

$\implies y = 7-x= 7-4 = 3$

So,

$A \equiv (4,3)$

Area bounded by the straight line and the x-axis

$=\int _{ 0 }^{ 4 } (7-x)dx= \left( 7x - \frac { x^{ 2 } }{ 2 } \right) _{ 0 }^{ 4 }=28-\frac { 4^{ 2 } }{ 2 } =20$

Area bounded by the parabola

$=\int _{ 0 }^{ 2 }{ \left( 1-\frac { x^{ 2 } }{ 4 } \right) } dx+\int _{ 2 }^{ 4 }{ \left( \frac { x^{ 2 } }{ 4 } -1 \right) dx } ={ \left( x-\frac { x^{ 3 } }{ 12 } \right) }_{ 0 }^{ 2 }+{ \left( \frac { x^{ 3 } }{ 12 } -x \right) }_{ 2 }^4=4$

So, the shaded area on the right hand side becomes

$(20-4)=16$

$\therefore$ The total area

$=2 \times 16 = 32$

Consider two curves

$C_1 : (y - \sqrt 3)^2 = 4 ( x - \sqrt2)$ and

$C_2 : x^2 + y^2 = ( 6 + 2 \sqrt2 ) x + 2 \sqrt{3y} - 6 ( 1 + \sqrt2)$ then

0%
$C_1 and C-2$ touch each other only at one point.
0%
$C_1 and C-2$ touch each other exactly at two points.
0%
$C_1 and C-2$ intersect (but do not touch) at exactly two points.
0%
$C_1 and C-2$ neither intersect nor touch each other.

What is the area of the region bounded by the parabola

${ y }^{ 2 }=6(x-1)$ and

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

0%
$\cfrac { \sqrt { 6 } }{ 3 }$
0%
$\cfrac { 2\sqrt { 6 } }{ 3 }$
0%
$\cfrac { 4\sqrt { 6 } }{ 3 }$
0%
$\cfrac { 5\sqrt { 6 } }{ 3 }$

Explanation

Solving

$y^2=6(x-1)$ and

$y^2=3x$ we get

$6(x-1)=3x$

$\Rightarrow x=2$

Hence

$y=\pm \sqrt{6}$

$y^2=6(x-1)\Rightarrow x=1+\dfrac{y^2}{6}$ and

$y^2=3x \Rightarrow x=\dfrac{y^2}{3}$

Area

$=\int_{-\sqrt{6}}^{\sqrt{6}}\left(1+\dfrac{y^2}{6}-\dfrac{y^2}{3}\right)dy$

$=2\int_{0}^{\sqrt{6}}\left(1-\dfrac{y^2}{6}\right)dy$

$=2\left[y-\dfrac{y^3}{18}\right]_{0}^{\sqrt{6}}$

$=2 \times \dfrac{2\sqrt{6}}{3}=\dfrac{4\sqrt{6}}{3}$

Area

$=\dfrac{4\sqrt{6}}{3}$

The area bounded by the curves

$x= a \cos^3t, y= a \sin^3 t$ is

0%
$\dfrac{3\pi a^2}{8}$
0%
$\dfrac{3\pi a^2}{16}$
0%
$\dfrac{3\pi a^2}{32}$
0%
None of the above

Explanation

$x=a\cos^{3}t$ ,

$y=a\sin^{2}t$

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

$A=\int_{0}^{2\pi}{x}dy$

$A=a^{2}\int_{0}^{2\pi}{\cos^{3}t\times3\sin^{2}t\times\cos t}dt$

$=3a^{2}\int_{0}^{2\pi}{\cos^{2}t\times\sin^{2}t}dt$

$\rightarrow(1)$

similarily

$A=\int_{0}^{2\pi}{y}dx$

$=3a^{2}\int_{0}^{2\pi}{\sin^{2}t\times\cos^{2}t}dt$

$\rightarrow(2)$

Adding (1) and (2)

$$2A=3a^{2}\int_{0}^{2\pi}{\cos^{2}t\times\sin^{2}t}dt

$A=\dfrac{3a^{2}}{8}\int_{0}^{2\pi}{\sin^{2}2}dt$

$\rightarrow(3)$

Similarily

$A=\dfrac{3a^{2}}{8}\int_{0}^{2\pi}{\cos^{2}t}dt$

$\rightarrow(4)$

By putting

$t=t+\dfrac{\pi}{4}$ in (3)

Adding (3) and (4)

$2A=\dfrac{3a^{2}}{8}\int_{0}^{2\pi}{dt}$

$A=\dfrac{3a^{2}}{8}\pi$

996393_1028220_ans_b75b683b148d4d4790c7831f300d4719.JPG

The area of the region lying between the line x-y+2=0 and the curve x=

$\sqrt y$ .

0%
9
0%
9/2
0%
10/3
0%
none of these

$Let\quad f(x)=2-\left| x-1 \right| and\quad g(x)={ \left( x-1 \right) }^{ 2 },\quad then\quad$

0%
area bounded by $f(x)$ and $g(x)$ is $\cfrac { 7 }{ 6 }$
0%
area bounded by $f(x)$ and $g(x)$ is $\cfrac { 7 }{ 3 }$
0%
area bounded by $f(x)$ $g(x)$ and $x-$ axis is $\cfrac { 5 }{ 3 }$
0%
area bounded by $f(x)$ $g(x)$ and $x-$ axis is $\cfrac { 5 }{ 6 }$

Explanation

Area of parabola w.r.t x axis

$=g(X)$

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

$\Rightarrow$

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

Area of

$f(x)$

$=2\times1+\dfrac{1}{2}\times2\times1$

$\Rightarrow$

$2+1=3$

Area bounded by

$f(x)$ and

$g(x)$

$=3-\dfrac{2}{3}=\dfrac{7}{3}$

993376_1018452_ans_d307f54179e74ea58afbbdc96d888503.png

In the square ABCD, the "shaded" region is the intersection of two circular regions centered at B and D respectively. If AB= 10, then what is the area of the shaded region?

0%
$25(\pi-2)$
0%
$50(\pi-2)$
0%
$25\pi$
0%
$50\pi$
0%
$40\pi (5-\sqrt{2})$

The area bounded by the curves

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

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

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

Area common to the circle

$x^{2}+y^{2}=64$ and the parabola

$y^{2}=4x$ is

0%
$\dfrac{16}{3}(4\pi + \sqrt{3})$
0%
$\dfrac{16}{3}(8\pi + \sqrt{3})$
0%
$\dfrac{16}{3}(4\pi - \sqrt{3})$
0%
$none\ of\ these$

$f\left(x\right)=$ max

$\left\{\sin{x},\cos{x},\dfrac{1}{2}\right\}$ , then the area of the region bounded by the curves

$y=f\left(x\right),x-$ axis

$y-$ axis and

$x=2\pi$ is

0%
$\left(\dfrac{5\pi}{12}+3\right)$ .sq.unit
0%
$\left(\dfrac{5\pi}{12}+\sqrt{2}\right)$ .sq.unit
0%
$\left(\dfrac{5\pi}{12}+\sqrt{3}\right)$ .sq.unit
0%
$\left(\dfrac{5\pi}{12}+\sqrt{2}+\sqrt{3}\right)$ .sq.unit

Explanation

$\because f\left(x\right)=$ max

$\left\{\sin{x},\cos{x},\dfrac{1}{2}\right\}$
Interval value of

$f\left(x\right)$
For

$0\le x<\dfrac{\pi}{4}, \cos{x}$
For

$\dfrac{\pi}{4}\le x<\dfrac{5\pi}{6}, \sin{x}$
For

$\dfrac{5\pi}{6}\le x<\dfrac{5\pi}{3}, \dfrac{1}{2}$
For

$\dfrac{5\pi}{3}\le x<2\pi, \cos{x}$
Hence, required area

$=\int_{0}^{\frac{\pi}{4}}{\cos{x}dx}+\int_{\frac{\pi}{4}}^{\frac{5\pi}{6}}{\sin{x}dx}+\int_{\frac{5\pi}{6}}^{\frac{5\pi}{3}}{\dfrac{1}{2}dx}+\int_{\frac{5\pi}{3}}^{2\pi}{\cos{x}dx}$

$=\left[\sin{x}\right]_{0}^{\frac{\pi}{4}}-\left[\cos{x}\right]_{\frac{\pi}{4}}^{\frac{5\pi}{6}}+\dfrac{1}{2}\left[x\right]_{\frac{5\pi}{6}}^{\frac{5\pi}{3}}+\left[\sin{x}\right]_{\frac{5\pi}{3}}^{2\pi}$

$=\left(\dfrac{1}{\sqrt{2}}-0\right)-\left(-\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\right)+\dfrac{1}{2}\left(\dfrac{5\pi}{3}-\dfrac{5\pi}{6}\right)+\left(0+\dfrac{\sqrt{3}}{2}\right)$
On simplification, we get

$=\left(\dfrac{5\pi}{12}+\sqrt{2}+\sqrt{3}\right)$ .sq.unit.
959581_1040931_ans_125967c8eb8641e6b4b39b45876a5b09.png

The parabola

$y=\dfrac{x^2}{2}$ divides the circle

$x^2+y^2=8$ into two parts. Find the area of both parts.

0%
$6\pi+\dfrac{4}{3}$ , $2\pi-\dfrac{4}{3}$
0%
$6\pi-\dfrac{4}{3}$ , $2\pi+\dfrac{4}{3}$
0%
$6\pi+\dfrac{2}{3}$ , $2\pi-\dfrac{2}{3}$
0%
$6\pi-\dfrac{2}{3}$ , $2\pi+\dfrac{2}{3}$

Explanation

Let us first find the x-values of the points of intersection using the given equations of parabola

$y=\dfrac {x^2}{2}$ and circle

$x^2+y^2=8$ as follows:

$x^{ 2 }+y^{ 2 }=8\\ \Rightarrow x^{ 2 }+\left( \dfrac { x^{ 2 } }{ 2 } \right) ^{ 2 }=8\quad \quad \quad \quad \quad \left( \because \quad y=\dfrac { x^{ 2 } }{ 2 } \right) \\ \Rightarrow x^{ 2 }+\frac { x^{ 4 } }{ 4 } =8\\ \Rightarrow 4x^{ 2 }+x^{ 4 }=32$

$\Rightarrow x^{ 4 }+4x^{ 2 }-32=0\\ \Rightarrow x^{ 4 }+8x^{ 2 }-4x^{ 2 }-32=0\\ \Rightarrow x^{ 2 }(x^{ 2 }+8)-4(x^{ 2 }+8)=0\\ \Rightarrow (x^{ 2 }-4)(x^{ 2 }+8)=0\\ \Rightarrow (x^{ 2 }-4)=0,\quad (x^{ 2 }+8)=0\\ \Rightarrow x^{ 2 }=4,\quad x^{ 2 }=-8\\ \Rightarrow x=\pm \sqrt { 4 } \\ \Rightarrow x=\pm 2$

Let

$A_1$ be the area of the region inside the circle and above the parabola and

$A_2$ be the area of the region inside the circle and below the parabola. Then we have,

${ A }_{ 1 }=\int _{ -2 }^{ 2 }{ \left( \sqrt { 8-{ x }^{ 2 } } -\dfrac { 1 }{ 2 } { x }^{ 2 } \right) } dx\\ =2\int _{ 0 }^{ 2 }{ \left( \sqrt { 8-{ x }^{ 2 } } -\dfrac { 1 }{ 2 } { x }^{ 2 } \right) } dx\\ =2\left[ \dfrac { 1 }{ 2 } \times 8\sin ^{ -1 }{ \left( \dfrac { 2 }{ \sqrt { 8 } } \right) } +\dfrac { 1 }{ 2 } \times 2\sqrt { 8-{ 2 }^{ 2 } } -\dfrac { 1 }{ 2 } { \left[ \dfrac { 1 }{ 3 } { x }^{ 3 } \right] }_{ 0 }^{ 2 } \right] \\ =8\sin ^{ -1 }{ \left( \dfrac { 1 }{ \sqrt { 2 } } \right) } +2\sqrt { 4 } -\dfrac { 8 }{ 3 }$

$=8\times \dfrac { \pi }{ 4 } +4-\dfrac { 8 }{ 3 } \\ =2\pi +\dfrac { 4 }{ 3 }$

We know that the area of the circle is

$\pi r^2$ , therefore the area of the circle

$x^2+y^2=8$ with radius

$r=\sqrt {8}$ is:

$\pi \left( \sqrt { 8 } \right) ^{ 2 }=8\pi$

Thus, we have

$A_{ 2 }=8\pi -\left( 2\pi +\dfrac { 4 }{ 3 } \right) =6\pi -\dfrac { 4 }{ 3 }$

Hence, the area of parabola and circle is

$2\pi +\dfrac { 4 }{ 3 }$ and

$6\pi -\dfrac { 4 }{ 3 }$ respectively.

The area enclosed by the curve

$y=\sqrt{(4-x^2)}, y\geq \sqrt{2}\sin\left(\dfrac{x\pi}{2\sqrt{2}}\right)$ and x-axis is divided by y-axis in the ratio.

0%
$\dfrac{\pi^2-8}{\pi^2+8}$
0%
$\dfrac{\pi^2-4}{\pi^2+4}$
0%
$\dfrac{\pi -4}{\pi +4}$
0%
$\dfrac{2\pi^2}{\pi^2+2\pi -8}$

Explanation

$y=\sqrt{4-x^{2}} \ldots$ (given)

$\cdots(i)$

$\Rightarrow y^{2}-4 x^{2}$

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

$y \geq \sqrt{2} \sin \left(\dfrac{\pi x}{2 \sqrt{2}}\right)$ (given) ... (ii)

To find the period,

$=\frac{2 \not \pi}{\not \pi} \times 2 \sqrt{2}$

$=4 \sqrt{2} \Rightarrow$ period.

$=\dfrac{4 \sqrt{2}}{2}=2 \sqrt{2}$

$\quad=2 \times 1.732$

$\quad[3.4]$

To find the intersection point using (i) and (ii)

$x=\sqrt{2}$

Area of circle

$=\pi r^{2}=\pi \times y=4 \pi$

Area of circle

$=\pi=A_{1}$ .. (iii)

from (-2,0) to (0,0)

$A_{2}=\int_{0}^{\sqrt{2}} \sqrt{4-x^{2}}-\sqrt{2} \sin \left(\dfrac{\pi}{2 \sqrt{2}} x\right) d x$

$\Rightarrow \int_{0}^{\sqrt{2}} \sqrt{4-x^{2}} d x-\sqrt{2} \int_{0}^{\sqrt{2}} \sin \left[\dfrac{\pi x}{2 \sqrt{2}}\right] d x$

$\left[\begin{array}{ll}\therefore & \sqrt{a^{2}-x^{2}} & d x=\dfrac{x}{2} \sqrt{a^{2}-x^{2}}+\dfrac{a^{2}}{2} \sin ^{-1} \dfrac{x}{a}\end{array}\right]$

$=\left.\left[\dfrac{x}{2} \sqrt{4-x^{2}}+\dfrac{4}{2} \sin ^{-1} \dfrac{x}{2}\right]_{0}^{\sqrt{2}}-\dfrac{\sqrt{2}}{\left(\dfrac{\pi}{2 \sqrt{2}}\right)}[-\operatorname{cos}\left(\dfrac{\pi x}{2 \sqrt{2}}\right)\right]_{0}^{\sqrt{2}}$

$=\left[\dfrac{2}{2} \pi \sqrt{2}+\dfrac{4}{2} \sin ^{-1} \left[\dfrac{1}{\sqrt{2}}\right]\right]+\dfrac{2 \times 2}{\pi}\left|\cos \left(\dfrac{\pi x}{2 \sqrt{2}}\right)\right|_{0}$

$=\left[1+\dfrac{4}{2} \sin ^{-1}\left[\dfrac{1}{\sqrt{2}}\right]\right]+\dfrac{4}{\pi}\left[\cos \left(\dfrac{\pi}{2\sqrt 2} \times \sqrt2 )-\operatorname{cos} 0\right]\right.$

$\Rightarrow[1+\pi / 2-4 / \pi]$

$\Rightarrow \dfrac{2 \pi+\pi^{2}-8}{2 \pi}$ .. (iv)

Using equation (iii) and (iv).

$\dfrac{A_{1}}{A_{2}}=\dfrac{\pi \times 2 \pi}{2 \pi+\pi^{2}-8}=\dfrac{2 \pi^{2}}{2 \pi+\pi^{2}-8}$ Answer (D)

1990070_1048889_ans_16b40053f05a445a87f916aed591a971.png

$k=2$ then

$f\left(x\right)$ attains point of inflection at

0%
$0$
0%
$\sqrt{2}$
0%
$-\sqrt{2}$
0%
None of these

The area of the region enclosed between by the

${x^2} + {y^2} = 16$ and the parabola

${y^2} = 6x$ .

0%
$\dfrac{2}{3} (\sqrt 3 + 4\pi)$ sq. units
0%
$\dfrac{4}{3} (\sqrt 3 + 4\pi)$ sq. units
0%
$\dfrac{2}{3} (\sqrt 3 + 8\pi)$ sq. units
0%
$\dfrac{4}{3} (\sqrt 3 + 8\pi)$ sq. units

Explanation

Point of intersection of the parabola and the circle is obtained by solving the equations:

${x}^{2}+{y}^{2}=16$ and

${y}^{2}=6x$

$\Rightarrow\,{x}^{2}+6x-16=0$

$\Rightarrow\,{x}^{2}+8x-2x-16=0$

$\Rightarrow\,x\left(x+8\right)-2\left(x+8\right)=0$

$\Rightarrow\,\left(x-2\right)\left(x+8\right)=0$

$\Rightarrow\,x=2,\,x=-8$

$\therefore\,x=2$ is the only possible solution(from the fig.)

$\therefore\,$ when

$x=2,\,y=\pm\sqrt{6\times 2}=\pm\,2\sqrt{3}$

$\therefore\,B\left(2,2\sqrt{3}\right)$ and

${B}^{\prime}\left(2,-2\sqrt{3}\right)$ are the points of intersection of parabola and the circle.

Required area

$=area\,of\,OBA{B}^{\prime}O=2\,area\,of \,OBAO$

$=2\left[area\,of\,OBDO+area\,of\,DBAD\right]$

$=2\left[\displaystyle\int_{0}^{2}{\sqrt{6x}dx}+\displaystyle\int_{2}^{4}{\sqrt{16-{x}^{2}}dx}\right]$

$=2\left[\sqrt{6}\left[\dfrac{{x}^{\frac{3}{2}}}{\dfrac{3}{2}}\right]_{0}^{2}+\left[\dfrac{1}{2}x\sqrt{16-{x}^{2}}+\dfrac{1}{2}\times 16{\sin}^{-1}{\dfrac{x}{4}}\right]_{2}^{4}\right]$

$=2\left[\left[\sqrt{6}\times\dfrac{2}{3}\times{2}^{\frac{3}{2}}-0\right]+\left[\dfrac{1}{2}\times\,4\sqrt{16-16}+\dfrac{1}{2}\times\,16{\sin}^{-1}{\dfrac{4}{4}}-\dfrac{1}{2}\times\,2\sqrt{16-4}-\dfrac{1}{2}\times\,16{\sin}^{-1}{\dfrac{1}{2}}\right]\right]$

$=2\left[\dfrac{8\sqrt{3}}{3}+\dfrac{8\pi}{2}-2\sqrt{3}-\dfrac{8\pi}{6}\right]$

$=2\left[\dfrac{8\sqrt{3}-6\sqrt{3}}{3}+8\left(\dfrac{\pi}{2}-\dfrac{\pi}{6}\right)\right]$

$=2\left[\dfrac{2\sqrt{3}}{3}+8\times\dfrac{2\pi}{6}\right]$

$=\dfrac{4\sqrt{3}}{3}+\dfrac{16\pi}{3}$

1495741_1051360_ans_c9349aa70c524f45a5352689e333bc59.png

Area bounded by

$|x-1| \le 2$ and

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

0%
$6 \sqrt{2}+\dfrac{1}{2}$ In $|3+2\sqrt{2}|$
0%
$6 \sqrt{2}+\dfrac{1}{2}$ In $|3-2\sqrt{2}|$
0%
$6 \sqrt{2}-$ In $|3+2\sqrt{2}|$
0%
$none\ of\ these$

Explanation

Given :

$|x-1|\leq 2$ and

$x^2-y^2=1$

We have to find area bounded by both curve.

As we know

$|x|\leq a$ means

$-a \leq x \leq a$

$|x-1|\leq 2$

$-2\leq x-1\leq 2$

$-2+1\leq x\leq 2+1$

$\boxed{-1 \leq x \leq 3}$

Area of curve

$|x-1|\leq 2$

Area

$=2 \int_{3}^{1} \sqrt {x^2-1}dx$

$\because \int \sqrt {x^2-a^2}dx=\dfrac{x}{2}=\dfrac{a^2}{2}ln(x+\sqrt{x^2-a^2})$

$=2\left[\dfrac{x}{2}\sqrt{x^2-1}-\dfrac{1}{2}in (x+\sqrt{x^2-1})\right]^3_1$

$2\left[\dfrac{3}{2}\sqrt8-\dfrac{1}{2}in (3+\sqrt 8)\right]-2\left[\dfrac{1}{2}\sqrt 0-\dfrac{1}{2}in (1+0)\right]$

$6\sqrt 2 in (3+2\sqrt 2)-0$

$A=6\sqrt 2-in (3+2\sqrt 2)$

2110210_1040258_ans_17ad4f6e7bb34edcb1a8ae80ff2182af.png

Find area curved by three circles

0%
$(5\pi-3\sqrt{3})units^{2}$
0%
$(5\pi+4\sqrt{3})units^{2}$
0%
$(5\pi+3\sqrt{3})units^{2}$
0%
$(5\pi+3\sqrt{2})units^{2}$

The whole area of the curves

$x=a\cos^3t, y=b\sin^3t$ is given by?

0%
$\dfrac{3}{8}\pi ab$
0%
$\dfrac{5}{8}\pi ab$
0%
$\dfrac{1}{8}\pi ab$
0%
None of these

Explanation

$x=a{ \cos }^{ 3 }\theta$

$y=b{ \sin }^{ 3 }\theta$

To get the area,

$A=\int _{ 0 }^{ 2x }{ x } dy$

$=\int _{ 0 }^{ 2\pi }{ a } { \cos }^{ 3 }\theta 3b{ \sin }^{ 2 }\theta \cos\theta d\theta$

$=3ab\int _{ 0 }^{ 2\pi }{ { \sin }^{ 2 }\theta { \cos }^{ 4 }\theta } d\theta \quad \longrightarrow \left( 1 \right)$

Substituting

$\theta \rightarrow \theta +\pi /2$ in

$1$ ,

$A=3ab\int _{ 0 }^{ 2\pi }{ { \cos }^{ 2 }\theta { \sin }^{ 4 }\theta } d\theta \quad \longrightarrow \left( 2 \right)$

Adding

$(1)$ and

$(2)$

$2A=3ab\int _{ 0 }^{ 2\pi }{ { \cos }^{ 2 }\theta } { \sin }^{ 2 }\theta d\theta \quad \longrightarrow \left( 3 \right)$

$\left\{ { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta =1 \right\}$

Multiplying

$(3)$ by

$(4)$

$8A=3ab\int _{ 0 }^{ 2\pi }{ { \sin }^{ 2 }2\theta } d\theta \quad \longrightarrow \left( 4 \right)$

Substituting

$\theta \rightarrow \theta +\pi /4$ in

$(4)$

$8A=3ab\int _{ 0 }^{ 2\pi }{ { \cos }^{ 2 }2\theta } d\theta \quad \longrightarrow \left( 5 \right)$

Adding

$(4)$ and

$(5)$

$16A=3ab\int _{ 0 }^{ 2\pi }{ 1 } d\theta$

$=6ab\pi$

$A=\dfrac { 3\pi ab }{ 8 }$

The triangle formed by the tangent to the parabola

$y^2=4x$ at the point whose abscissa lies in the interval

$\left[a^2, 4a^2\right]$ , the ordinate and the x-axis, has the greatest area equal to?

0%
$12a^3$
0%
$8a^3$
0%
$16a^3$
0%
$20a^3$

Consider the two curves

${ C }_{ 1 } :{ y }^{ 2 }=4x$

${ C }_{ 2 } : { x }^{ 2 }+ { y }^{ 2 } - 6x + 1 = 0$

Then, the area of region between these curves?

0%
$\dfrac{20}{3}-2\pi$
0%
$\dfrac{10}{3}-2\pi$
0%
$\dfrac{20}{3}-\pi$
0%
$\dfrac{10}{3}-\pi$

Area bounded by the curves

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

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

0%
$e-2$
0%
$3-e$
0%
$e$
0%
$e-1$

The area enclosed between the curve

$y=x^3$ and

$y=\sqrt{x}$ is

0%
$\dfrac{5}{3}$
0%
$\dfrac{5}{4}$
0%
$\dfrac{5}{12}$
0%
None of these

Explanation

Given two curves are

$y=x^3$

and

$y=\sqrt x$

Both curves intersect at point

$(1,1)$

Now equating the curves we get,

$x^3=\sqrt x;$ when

$x=1$

$\therefore$ The curve represented by

$y=\sqrt x$ is upward

to the curve represented by

$y=x^3$

$\therefore$ Area enclosed by the curves= Area of the shaded region

$\Rightarrow$ Area enclosed by the curves

$=\int_{1}^{0}(\sqrt x-x^3)dx$

$=\\int_{1}^{0}x\dfrac{1}{2}dx-\int_{2}^{0}x^3dx$

$=\left[\dfrac{x^{3/2}}{\dfrac{3}{2}}\right]^1_0-\left[\dfrac{x^4}{4}\right]^1_0$

$=\left[\dfrac{(1)^{3/2}}{\dfrac{3}{2}}-0\right]-\left[\dfrac{(1)^4}{4}-0$\right]$

$=\dfrac{1}{\dfrac{3}{2}}-\dfrac{1}{4}$

$$=\dfrac{2}{3}-\dfrac{1}{4}$$

$=\dfrac{8-3}{12}$

$\therefore$ Area enclosed by the curves

$=\dfrac{5}{12}$ sq. units

2115402_1103689_ans_7d9ae18871124180bcd435fe6d49d39e.png

The area between the curves y=tan x, cot x and axis in the interval

$\left[0,\pi \right/2$ ]is ?

0%
log $2$
0%
log $3$
0%
log $5$
0%
none of these

Explanation

$y=tanx,y=cotx$ are symmetrical about

$x=\frac { \pi }{ 4 }$ ,where they intersect

$\frac { \pi }{ 4 } Req Area=2|∫tanxdx|$

$\frac { \pi }{ 4 } |2∫cotxdx|=2\times \left| \left( lnsec\frac { \pi }{ 4 } \right) -lnsec0 \right|$

$\frac { \pi }{ 22 } \times |ln\sqrt { 2 } -0|=2\times \frac { 1 }{ 2 } \times ln2=ln2$

The area bounded by the curves

$y = \sin \left( {x - \left[ x \right]} \right),\,y = \sin 1$ and the x-axis is

0%
$\sin 1$
0%
$1 - \sin 1$
0%
$1 + \sin 1$
0%
None of these

The area (in square units) bounded by the curves

$y = {\cos ^{ - 1}}\left| {\cos \,x} \right|$ and

$y = {\left( {{{\cos }^{ - 1}}\left| {\cos \,x} \right|} \right)^2},x \in \left[ {0,\pi } \right]$ is

0%
$\frac{4}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} - 1} \right)$
0%
$\frac{4}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} + 1} \right)$
0%
$\frac{2}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} - 1} \right)$
0%
$\frac{2}{3} + \frac{{{\pi ^2}}}{4}\left( {\frac{\pi }{3} + 1} \right)$

The area bounded by the curves

$y = sin^{-1} |sin \, x|$ and

$y = (sin^{-1} | sin \, x|)^2 , \, 0 \le x \le 2 \pi$ is

0%
$\left(\dfrac{\pi^3}{3} + \dfrac{4}{3} \right)$ sq. unit
0%
$\left(\dfrac{\pi^3}{6} - \dfrac{\pi^2}{2} + \dfrac{4}{3} \right)$ sq. unit
0%
$\left(\dfrac{\pi^2}{2} - \dfrac{4}{3} \right)$ sq. unit
0%
$\left(\dfrac{\pi^2}{6} - \dfrac{\pi}{4} + \dfrac{4}{3} \right)$ sq. unit

Explanation

$y=\sin ^{-1}|\sin n |$ and

$y=(\sin ^{-1}|\sin n|^2)^2$

Required area

$=4\int_{1}^{0}[x-(\sin^{-1}(\sin n))^2]dn+4\int_{\dfrac{\pi}{1}}^{1}[(\sin^{-1}(\sin n))^2-n]dn$

$=4\left(\dfrac{1}{2}\right)-4\int_{1}^{0}(\sin ^{-1}(\sin x))^2dn+4\int_{\dfrac{\pi}{2}}^{1}(\sin ^{-1}(\sin n))^2dn-\dfrac{4}{2}\left(\dfrac{\pi^2}{4}-1\right)^1$

$=4-\dfrac{\pi}{2}-4\int_{1}^{0}(\sin ^{-1}(\sin n))^2dn +4\int_{\dfrac{\pi}{2}}^{0}(\sin ^{-1}(\sin n))^2dn$

$=4-\dfrac{\pi^2}{2}-4\int_{1}^{0}n^2dn+4\int_{\dfrac{\pi}{2}}^{1}n^2dn$

$=4-\dfrac{\pi^2}{2}-\dfrac{4}{3}+\dfrac{4}{3}\left[\dfrac{\pi^3}{8}-1^1\right]$

$=4-\dfrac{\pi^2}{2}-\dfrac{8}{3}+\dfrac{\pi^3}{6}$

$=\boxed{\left(\dfrac{4}{3}-\dfrac{\pi^2}{2}+\dfrac{\pi^3}{6}\right)sq\, unit}$

2115724_1110488_ans_565d69175341405892b7831843e851b7.png

The area bounded by the curves

${y^2} = 4x$ and

${x^2} = 4y$ is :

0%
$\frac{{32}}{3}$
0%
$\frac{{16}}{3}$
0%
$\frac{8}{3}$
0%
$0$

Explanation

$x^2=4y$

$y^2=4x$

$x=2\sqrt{y}$

$=4\times 2\sqrt{y}$

$y^2=8\sqrt{y}$

$\Rightarrow y^4-8y=0$

then

$y=0, 4$

$x^2=4y$

we get

$x=4, 0$

So, the points of intersection are

$(0, 0)$ &

$(4, 4)$

Area

$=\displaystyle\int^4_0\displaystyle\int^{x^2/4}_{2\sqrt{x}}dydx=\displaystyle\int^4_0[y]^{x^2/4}_{2\sqrt{x}}dx$

$=\displaystyle\int^4_0\left[\dfrac{x^2}{4}-2\sqrt{x}\right]dx$

$=\left[\dfrac{x^3}{12}-\dfrac{4}{3}\cdot x^{3/2}\right]^4_0$

$=\left[\dfrac{4^3}{12}-\dfrac{4}{3}(4)^{3/2}\right]$

$=\dfrac{16}{3}$ sq. units.

The area bounded by

$y=2-\left| 2-x \right|$ and

$y=\frac { 3 }{ \left| x \right| }$ is :

0%
$\frac { 4+3\ell n3 }{ 2 }$
0%
$\frac { 4-3\ell n3 }{ 2 }$
0%
$\frac { 3 }{ 2 } +\ell n3$
0%
$\frac { 1 }{ 2 } +\ell n3$

Explanation

$y=2-\left|2-x\right|$

$y=\dfrac{3}{\left|x\right|}$

$1$ can be rewriiten as

$y=x if x<+2$

On solving these two equation, we get

$x=\sqrt{3}$ and

$x=3$

$Area=\displaystyle \int_{\sqrt{3}}^{3}{\left({y}_{1}-{y}_{2}\right)}dx$

$=\displaystyle \int_{\sqrt{3}}^{3}{\left(2-\left|2-x\right|\right)}-\dfrac{3}{\left|x\right|}dx$

$=\displaystyle \int_{\sqrt{3}}^{2}{x-\dfrac{3}{\left|x\right|}+\int_{2}^{3}{4-x-\dfrac{3}{\left|x\right|}}}dx$

${\left[\dfrac{{x}^{2}}{2}-3\log_{e}{x}\right]}_{\sqrt{3}}^{2}+{\left[4x-\dfrac{{x}^{2}}{2}-3\log_{e}{x}\right]}_{2}^{3}$

$=\dfrac{1}{2}+3\log_{e}{\dfrac{\sqrt{3}}{2}}+12-\dfrac{9}{2}-3\log_{e}{3}-8+2+3\log_{e}{2}$

$=\dfrac{1}{2}+3\ln{\dfrac{\sqrt{3}}{2}}+\dfrac{3}{2}3\ln{\dfrac{2}{3}}$

$2+3\ln{\dfrac{\sqrt{3}}{2}}=2-3\ln{\sqrt{3}}=\dfrac{4-3\ln{3}}{2}$

Area of the region defined by

$1\ \le |x|+|y|$ and

$x^{2}-2x+1 \le 1-y^{2}$ is

$k \pi$ then

$k=.....sq$ units

0%
$\dfrac {3}{4}$
0%
$\dfrac {7}{6}$
0%
$\dfrac {128}{5}$
0%
$\dfrac {10}{3}$

The maximum area bounded by the curves

${y^2} = 4ax,\,\,\,\,y = ax\,\,a$ and

$y = \frac{x}{a}\,\,,1 \le a \le 2$ is

0%
$44 sq. units$
0%
$74 sq. units$
0%
$84 sq.units$
0%
$114 sq. units$

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 11 - 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 11 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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