Application of Integrals Test

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Application of Integrals Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

The area of the region bounded by the x-axis and the curves
$$y = \tan x\left( { - \frac{\pi }{3} \le x \le \frac{\pi }{3}} \right),and\,y = \cot x\left( {\frac{\pi }{6} \le x \le \frac{{3\pi }}{2}} \right)$$ is

A
$$log 2$$

B
$$2log2$$

C
$$log\sqrt 2 $$

D
$$log\left( {\frac{3}{2}} \right)$$

Solution
Question 2
1 / -0

The area of the region bounded by the curves $$y=ex\log x$$ and $$y=\dfrac{\log x}{ex}$$ is

A
$$\dfrac{e^{2}-5}{4e}$$

B
$$\dfrac{e^{2}+1}{2e}$$

C
$$\dfrac{e^{2}}{2}$$

D
$$None\ of\ these$$

Solution
Question 3
1 / -0

Area bounded by the curve $${x^2} = 4y$$ and the straight line $$x=4y-2$$ is

A
$$\frac{8}{9}$$ sq. unit

B
$$\frac{9}{8}$$ sq. unit

C
$$\frac{4}{3}$$ sq. unit

D
None of these

Solution
Question 4
1 / -0

If the area enclosed between $$y=m{x}^{2}$$ and $$x=n{y}^{2}$$ is $$\cfrac{1}{3}$$ sq. units, then $$m,n$$ can be roots of (where $$m,n$$ are non zero real numbers)

A
$$2{x}^{2}+5x+1=0$$

B
$$2{x}^{2}+3x-2=0$$

C
$$2{x}^{2}+3x+2=0$$

D
$$2{x}^{2}+5x-1=0$$

Solution

$$y=mx^{2}$$
$$y^{2}=m^{2}x^{4}$$
$$x=nm^{2}x^{1/3}$$
$$x = \left( \dfrac{1}{nm^{2}} \right)^{1/3}$$
$$y=m \dfrac{1}{(nm^{2})^{2/3}}$$
$$y= m\dfrac{1}{n^{2/3} m^{4/3}}$$
$$y=\dfrac{1}{(n^{2}m)^{1/3}}$$
$$(1/nm^{2})^{1/3}$$
Area$$= \int_{0} \sqrt{\dfrac{x}{n}}- mx^{2}$$
$$=\left[ \right]^{(1/nm^{2})^{1/3}}$$
$$=\dfrac{2}{3\sqrt{n}} \left( \dfrac{1}{nm^{2}} \right)^{\dfrac{1}{3}. \dfrac{3}{2}}- \dfrac{m}{3} \left( \dfrac{1}{nm^{2}} \right)^{3 \times \dfrac{1}{3}}$$
$$=\dfrac{2}{3\sqrt{n}} \dfrac{1}{\sqrt{n}m}- \dfrac{m}{3} \dfrac{1}{nm^{2}}$$
$$=\dfrac{2}{3nm}- \dfrac{1}{3nm}$$
$$\dfrac{1}{3}= \dfrac{1}{3nm}$$
$$nm=1$$
Product of roots $$=1$$
Question 5
1 / -0

In the given figure, a square $$OABC$$ has been inscribed in the quadrant $$OPBQ$$. If $$OA=20cm$$ then the area of the shaded region is $$\left[take\pi=3.14\right]$$

A
$$214cm^{2}$$

B
$$228cm^{2}$$

C
$$222cm^{2}$$

D
$$242cm^{2}$$

Solution

Area of shaded region $$=Area\ of \left[OPBQ\right]-Area\ of \left[OABC\right]$$

$$=\dfrac{1}{4}\pi {r}^{2}-{a}^{2}$$

$$\left[ \because OA=a=20cm\\ r=OB=OA\sqrt { 2 } =20\sqrt { 2 } cm \right] $$

$$=\dfrac{1}{4}\pi {\left(20\sqrt{2}\right)}^{2}-{20}^{2}$$

$$=200\pi-400=228\ {cm}^{2}$$
Question 6
1 / -0

The area bounded by $$y=x^2,x=y^2$$ is

A
$$1 $$

B
$$\dfrac 16$$

C
$$\dfrac 34$$

D
$$None\ of\ these$$

Solution

$$y=x^2\\y^2=x\implies y=\sqrt x$$

The curves intersect at $$(0,0) $$ and $$(1,1)$$

Area between the curves is given by

$$\displaystyle \int _0^1 \sqrt x-x^2dx\\\left.\dfrac 23 x^{3/2}+\dfrac{x^3}{3} \right|_0^1\\\dfrac 23+\dfrac13=1$$
Question 7
1 / -0

The area of the plane region bounded by the curve $$x + 2 y ^ { 2 } = 0$$ and $$x + 3 y ^ { 2 } = 1$$ is equal to:

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

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

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

D
None of these

Solution

The given curves are $$x+2y^2 = 0$$ and $$x+3y^2 = 1$$.

The first curve is $$x+2y^2 = 0$$ which implies $$2y^2 = -x$$ and hence
$$y^2 = -x/2$$, which is a parabola.

Second curve $$x+3y^2 = 1 $$which is again a parabola as $$y^2 = -1/3 (x-1)$$.

Hence we are required to find the area between the two parabolas.
On solving the equations of the two parabolas, we get the points of intersection as $$(-2, 1) $$and $$(-2, -1)$$.

Hence we set$$ x = -2$$ and $$y = 1, -1$$.

Hence the required area is$$\int(-2y^2 -1 + 3y^2)dy$$ , where the integral runs from 0 to 1.

= $$\int(y^2 -1) dy$$ , where the integral runs from 0 to 1.

= $$(1/3 -1) $$

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

Hence the area of the region bounded by the curves is $$\dfrac{4}{3}$$.
Question 8
1 / -0

The maximum area of the triangle whose sides $$a, b \, and \, c$$ satisfy $$0 \le a \le 1, \, 1 \le b \le 2$$ and $$2 \le c \le 3$$

A
$$1$$

B
$$\frac {1}{2}$$

C
$$2$$

D
$$\frac {3}{2}$$

Solution
Question 9
1 / -0

If the area anclosed by $$f\left( x \right) = \sin x + \cos x,y = a$$ between two consecutive points of extremum is minimum, then the value of a is

A
$$0$$

B
$$-1$$

C
$$1$$

D
$$2$$
Question 10
1 / -0

The area enclosed between the curve $${y^2} = 4x$$ and line $$y = x$$is

A
$$\frac {8}{3}$$

B
$$\frac {4}{3}$$

C
$$\frac {2}{3}$$

D
$$\frac {1}{2}$$

Solution