Application of Integrals Test

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Application of Integrals Test - 59

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Question 1
1 / -0

Area common to the curves $$y^{2} = ax$$ and $$x^{2} + y^{2} = 4ax$$ is equal to

A
$$(9\sqrt {3} + 4\pi) \dfrac {a^{2}}{3}$$

B
$$(9\sqrt {3} + 4\pi)a^{2}$$

C
$$(9\sqrt {3} - 4\pi) \dfrac {a^{2}}{3}$$

D
None of these

Solution

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

A
$$\displaystyle\frac{\pi}{2}-\frac{4}{\pi}$$

B
$$\displaystyle\frac{\pi}{4}-\frac{2}{\pi}$$

C
$$\displaystyle \pi -\frac{8}{\pi}$$

D
None

Solution

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 $$\cfrac { \pi }{ 2 } +\cfrac { 4 }{ \pi } sq.\quad units$$
Question 3
1 / -0

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?

A
$$100$$

B
$$193$$

C
$$200$$

D
$$202$$

Solution

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
$$f(x) = 1-f$$
$$g(x) =f$$

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

$$=\displaystyle n\int_0^{0.5}(1-2x)dx+$$$$\displaystyle n\int_{0.5}^1(2x-1)dx=n(0.5-0.25)+n(0.75-0.5)$$

$$ 0.5 n=100\implies n=200$$
Question 4
1 / -0

The area in sq.units bounded by the hyperbola $$xy={ c }^{ 2 }$$, the x-axis and the ordinates at $$x=a$$ and $$x=b$$ ($$0< a< b)$$ is

A
$${ c }^{ 2 }\log _{ e }{ \left( \cfrac { a }{ b } \right) } $$

B
$${ c }^{ 2 }\log _{ e }{ \left( \cfrac { b }{ a } \right) } $$

C
$${ c }^{ 2 }\log _{ e }{ ab } $$

D
$$c{ \left( \log _{ e }{ a(a+b) } \right) }^{ 2 }$$

Solution
Question 5
1 / -0

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.

A
$$\displaystyle\frac{1}{3}$$

B
$$\displaystyle\frac{2}{3}$$

C
$$\displaystyle\frac{\pi}{4}$$

D
$$\displaystyle\frac{\pi}{3}$$

Solution

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

$$\Rightarrow A_2-A_1=\dfrac{4\times2(4x+1)^{3/2}}{3\times 4}=2/3(1-0)=2/3$$
Question 6
1 / -0

The area bounded by the curves $$y = \sin x, y = \cos x$$ and x-axis from $$x = 0$$ to $$x = \pi /2$$ is

A
$$2 + \sqrt {2}$$

B
$$\sqrt {2}$$

C
$$2$$

D
$$2 - \sqrt {2}$$

Solution
Question 7
1 / -0

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

A
8 sq unit

B
16 sq unit

C
24 sq unit

D
32 sq unit

Solution

$$\to$$ 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 $$B$$
Question 8
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The area bounded by the curves $$y = \dfrac {1}{4} |4 - x^{2}|$$ and $$y = 7 -|x|$$ is

A
$$18$$

B
$$32$$

C
$$36$$

D
$$64$$

Solution

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 $$=2 \times 16 = 32$$
Question 9
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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

A
$$C_1 and C-2$$ touch each other only at one point.

B
$$C_1 and C-2$$ touch each other exactly at two points.

C
$$C_1 and C-2$$ intersect (but do not touch) at exactly two points.

D
$$C_1 and C-2$$ neither intersect nor touch each other.
Question 10
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The area of the region $$\left\lfloor x \right\rfloor +\left\lfloor y \right\rfloor =1,-1\le x\le 1$$ and $$xy\le 1/2$$

A
rational

B
$$\cfrac { 1 }{ 2 } \left( \cfrac { 3 }{ 2 } +\ln { 2 } \right) $$

C
$$\cfrac { 1 }{ 2 } \left( \cfrac { 5 }{ 2 } +\ln { 2 } \right) $$

D
Irrational

Solution

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