Application of Integrals Test

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Application of Integrals Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

Area common to the curve $$y^2 = 16x$$ and $$y = 2x$$, is :

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

B
$$\dfrac{17}{3}$$ sq. units

C
$$\dfrac{19}{3}$$ sq. units

D
$$\dfrac{20}{3}$$ sq. units

Solution

$$y^2=16x,y=2x$$

$$\Rightarrow 4\sqrt{x}=2x$$

$$\sqrt{x}=2\Rightarrow x=4$$

Area$$=\int_{0}^{4}(\sqrt{16x}-2x)dx$$

$$=\int_{0}^{4}(4\sqrt{x}-2x)dx$$

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

$$=\dfrac{8}{3}[8-0]-[16-0]$$

$$=\dfrac{64}{3}-16$$

$$=\dfrac{16}{3}$$sq.units
Question 2
1 / -0

The area bounded by the curve $$y=sin(x-[x]),y=sin1,\,x=1$$ and the x-axis is

A
$$sin1$$

B
$$1-sin1$$

C
$$1+sin1$$

D
$$1-\cos1$$

Solution
Question 3
1 / -0

The area of the region bounded by $$\left| arg\left( z+1 \right) \right| \le \frac { \pi }{ 3 } $$ and $$ \left|z+1 \right| \le \frac { \pi }{ 4 } $$ is given by

A
$$\dfrac{4\pi}{3}$$

B
$$\dfrac{16\pi}{3}$$

C
$$\dfrac{2\pi}{3}$$

D
$$\dfrac{20\pi}{3}$$

Solution
Question 4
1 / -0

The curves $$y = x^{2} - 1, y = 8x - x^{2} - 9$$ at

A
Intersect at right angles at $$(2, 3)$$

B
Touch each other at $$(2, 3)$$

C
Do not intersect at $$(2, 3)$$

D
Intersect at an angle $$\dfrac {\pi}{3}$$

Solution

$$y={ x }^{ 2 }-1=8x-{ x }^{ 2 }-9\\ 2{ x }^{ 2 }-8x+8=0\\ { x }^{ 2 }-4x+4=0\\ { (x-2) }^{ 2 }=0\\ x=2\\ y={ x }^{ 2 }-1={ (2) }^{ 2 }-1=3$$
So, curves intersect at $$(2,3)$$
Now, $${ m }_{ 1 }=\cfrac { dy }{ dx } $$ of $$({ x }^{ 2 }-1)$$ at $$(2,3)$$
$${ m }_{ 1 }=2x=2(2)=4$$
$${ m }_{ 2 }=\cfrac { dy }{ dx } $$ of $$(8x-{ x }^{ 2 }-9)$$ at $$(2,3)$$
$${ m }_{ 2 }=8-2x=8-2(2)=4$$
Since $${ m }_{ 1 }={ m }_{ 2 }$$
we can say that curves touch each other at $$(2,3)$$.
Question 5
1 / -0

The area enclosed between the curves $$y={ ax }^{ 2 }$$ and $$x={ ay }^{ 2 }$$ $$(a>0)$$ is $$1\ sq.unit$$. then $$a=$$

A
$$\dfrac { 1 }{ \sqrt { 3 } } $$

B
$$\dfrac { 2 }{ \sqrt { 3 } } $$

C
$$\dfrac { 4 }{ \sqrt { 3 } } $$

D
$$\sqrt { 3 } $$

Solution

$$\Rightarrow x = a(ax^2)^2$$
$$\Rightarrow x = 0, \dfrac{1}{a} \Rightarrow y = 0, \dfrac{1}{a}$$
$$\therefore$$ Point of intersection are $$(0,0)$$ and $$\left(\dfrac{1}{a} , \dfrac{1}{a} \right)$$.
Thus, required area $$OABCO =$$ Area of curve $$OCBDO -$$ area of curve $$OABDO$$
$$\Rightarrow \displaystyle \int_0^{1/a} \left(\sqrt{\dfrac{x}{a}} - ax^2 \right) dx = 1$$ (given)
$$\Rightarrow \left[\dfrac{1}{\sqrt{a}} . \dfrac{x^{3/2}}{3/2} - \dfrac{ax^3}{3} \right]_0^{1/a} = 1$$
$$\Rightarrow \dfrac{2}{3a^2} - \dfrac{1}{3a^2} = 1$$
$$\Rightarrow a^2 = \dfrac{1}{3} \Rightarrow a = \dfrac{1}{\sqrt{3}} \, \, \, (\because a > 0)$$
Hence, this is the answer.
Question 6
1 / -0

The area bounded by the curves $$y=f(x)$$, the x-axis and the ordinates $$x=1$$ and $$x=\beta $$ is $$(\beta -1)\sin(3\beta +4)$$. Then $$f(x)$$ is

A
$$(x-1)\cos(3x+4)$$

B
$$\sin(3x+4)$$

C
$$\sin(3x+4)+3(x-1)\cos(3x+4)$$

D
$$\sin(3x+4)+x$$

Solution
Question 7
1 / -0

Two vertices of a rectangle are on the positive x-axis. The other two vertices lie on the lines $$y=4x$$ and $$y=-5x+6$$. Then the maximum area of the rectangle is?

A
$$\dfrac{2}{3}$$

B
$$\dfrac{2}{4}$$

C
$$\dfrac{1}{3}$$

D
$$\dfrac{4}{3}$$

Solution

Let the two vertex lying be $$(a,0)$$ and $$(b,0)$$ where $$a<b$$
Since $$y=4x$$ is containing coordinate corresponding to $$(a,0)$$ ; we get
Point to be $$(a,4a)$$
and the other with $$y=-5x+6$$ , we get point as $$(b,-5b+6)$$
Now as $$b-a=-5b+6-4a$$ (Length)
$$a+2d=2$$
Area of Rectangle$$=$$ Length $$\times$$ breadth $$=(b-a)(4a)$$
Area$$=(b-a)(4a)=(\cfrac { 2-a }{ 2 } -a)(4a)=(2-3a)(2a)$$
Area $$=4a-6{ a }^{ 2 }$$
For Maxima/Minima,
$$\cfrac { d(Area) }{ da } =0\quad \Rightarrow 4-12a=0\\ a=\cfrac { 1 }{ 3 } $$
Area$$=(4a-{ 6a }^{ 2 })=\cfrac { 4 }{ 3 } -\cfrac { 6 }{ 9 } =\cfrac { 2 }{ 3 } $$ sq. unit
Question 8
1 / -0

The area of the region bounded by the curve $${a^4}{y^2} = \left( {2a - x} \right){x^5}$$ is to that curve whose radius is $$a$$, is given by the ration.

A
$$5:4$$

B
$$5 : 8$$

C
$$2 : 3$$

D
$$3 : 2$$

Solution

$$ {{a}^{4}}{{y}^{2}}=\left( 2a-x \right){{x}^{5}} $$
$$ cut\text{ off x-axis when y=0} $$
$$ \left( 2a-x \right){{x}^{5}}=0 $$
$$ A_{1}^{{}}=\int\limits_{0}^{2a}{\dfrac{\sqrt{\left( 2a-x \right)}{{x}^{5/2}}}{{{a}^{2}}}}dx $$
$$ put\,x=\,2a{{\sin }^{2}}\theta $$
$$ dx=4a\sin \theta \cos \theta d\theta $$
$$ \therefore A_{1}^{{}}=\int\limits_{0}^{\pi /2}{\dfrac{\sqrt{2a}\cos \theta {{\left( 2a \right)}^{5/2}}{{\sin }^{5}}\theta \times 4a\sin \theta \cos \theta }{{{a}^{2}}}}d\theta $$
$$ \,\,\,\,\,\,\,\,\,=32{{a}^{2}}\int\limits_{0}^{\pi /2}{{{\sin }^{6}}\theta {{\cos }^{2}}\theta d\theta } $$
$$ \,\,\,\,\,\,\,=32{{a}^{2}}\dfrac{\left( 5.3.1 \right)\left( 1 \right)}{8.6.4.2}\dfrac{\pi }{2}\,\,\,\,\left[ by\,Walli'sformula \right] $$
$$ \,\,\,\,\,\,=\dfrac{5\pi {{a}^{2}}}{8} $$
$$ Area\,of\,circle\,A_{2}^{{}}=\pi {{a}^{2}} $$
$$ \dfrac{A_{1}^{{}}}{A_{2}^{{}}}=\dfrac{5}{8} $$
$$ A_{1}^{{}}:A_{2}^{{}}=5:8 $$
Question 9
1 / -0

The area enclosed between the curves $$y=a{ x }^{ 2 }$$ and $$x=a{ y }^{ 2 }$$ $$\\ (a>0)$$ is $$1sq.unit$$. then $$a=$$

A
$$\dfrac { 1 }{ \sqrt { 3 } } $$

B
$$\dfrac { 2 }{ \sqrt { 3 } } $$

C
$$\dfrac { 4 }{ \sqrt { 3 } } $$

D
$$\sqrt { 3 } $$

Solution

$$y=a(ay^2)^2\\y=a^3y^4\\y(y^3a^3-1)=0$$
$$y=0,1/a$$
$$x=ay^2\\x=a/a^2$$
$$x=1/a$$
$$\\ A=[\int _{ 0 }^{ 1/a }{ \sqrt { \cfrac { x }{ a } } -a{ x }^{ 2 } } dx\\ A={ [\cfrac { 2{ x }^{ 3/2 } }{ 3\sqrt { a } } -\cfrac { a{ x }^{ 3 } }{ 3 } ] }_{ 0 }^{ 1/a }\\ A=\cfrac { 2{ (1/a) }^{ 3/2 } }{ 3\sqrt { a } } -\cfrac { a{ (1/a) }^{ 3 } }{ 3 } \\ A=\cfrac { 2 }{ 3{ a }^{ 2 } } -\cfrac { 1 }{ 3{ a }^{ 2 } } \\ A=\cfrac { 1 }{ 3{ a }^{ 2 } } $$
given $$A=1\\\cfrac{1}{3a^2}=1\\3a^2=1\\a^2=1/3\\a=\sqrt{1/3}$$
Question 10
1 / -0

If area bounded by $$f(x)=x^{\frac{1}{3}}(x-1)$$ $$x-$$axis is A then find the value of $$28A$$.

A
$$5$$

B
$$6$$

C
$$7$$

D
$$9$$

Solution

We have,
$$f(x)=y=x^{1/3}(x-1)$$

Since, $$x-axis$$ means $$y=0$$

Therefore,
$$f(x)=y=x^{1/3}(x-1)=0$$
$$x=0,1$$

Area bounded by this function is $$A$$.
$$A=\left|\displaystyle \int_{0}^{1}x^{1/3}(x-1)dx\right|$$

$$A=\left|\displaystyle \int_{0}^{1}(x^{4/3}-x^{1/3})dx\right|$$

$$A=\left|\displaystyle \left(\dfrac{x^{7/3}}{7/3}-\dfrac{x^{4/3}}{4/3}\right)_{0}^{1}\right|$$

$$A=\left|\displaystyle \left(\dfrac{3}{7}(1)-\dfrac{3}{4}(1)\right)\right|$$

$$A=\left|\displaystyle \left(\dfrac{12-21}{28}\right)\right|$$

$$A=\left|\displaystyle \left(\dfrac{-9}{28}\right)\right|$$

$$A=\dfrac{9}{28}$$

Therefore,
$$28A=28\times \dfrac{9}{28}=9$$

Hence, this is the answer.

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Application of Integrals Test - 33

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Application of Integrals Test - 33

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SHARING IS CARING

Question 1

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

$$\dfrac{17}{3}$$ sq. units

$$\dfrac{19}{3}$$ sq. units

$$\dfrac{20}{3}$$ sq. units

Solution

Question 2

$$sin1$$

$$1-sin1$$

$$1+sin1$$

$$1-\cos1$$

Solution

Question 3

$$\dfrac{4\pi}{3}$$

$$\dfrac{16\pi}{3}$$

$$\dfrac{2\pi}{3}$$

$$\dfrac{20\pi}{3}$$

Solution

Question 4

Intersect at right angles at $$(2, 3)$$

Touch each other at $$(2, 3)$$

Do not intersect at $$(2, 3)$$

Intersect at an angle $$\dfrac {\pi}{3}$$

Solution

Question 5

$$\dfrac { 1 }{ \sqrt { 3 } } $$

$$\dfrac { 2 }{ \sqrt { 3 } } $$

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

$$\sqrt { 3 } $$

Solution

Question 6

$$(x-1)\cos(3x+4)$$

$$\sin(3x+4)$$

$$\sin(3x+4)+3(x-1)\cos(3x+4)$$

$$\sin(3x+4)+x$$

Solution

Question 7

$$\dfrac{2}{3}$$

$$\dfrac{2}{4}$$

$$\dfrac{1}{3}$$

$$\dfrac{4}{3}$$

Solution

Question 8

$$5:4$$

$$5 : 8$$

$$2 : 3$$

$$3 : 2$$

Solution

Question 9

$$\dfrac { 1 }{ \sqrt { 3 } } $$

$$\dfrac { 2 }{ \sqrt { 3 } } $$

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

$$\sqrt { 3 } $$

Solution

Question 10

$$5$$

$$6$$

$$7$$

$$9$$

Solution

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