Integrals Test

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Integrals Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$\int_{a}^{b}(x-a)^{3}(b-x)^{4} d x$$ is

A
$$\dfrac{(b-a)^{4}}{6^{4}}$$

B
$$\dfrac{(b-a)^{8}}{280}$$

C
$$\dfrac{(b-a)^{7}}{7^{3}}$$

D
None of these

Solution

Let $$I=\int_{a}^{b}(x-a)^{3}(b-x)^{4} d x$$
Put $$x=a \cos ^{2} \theta+b \sin ^{2} \theta, \Rightarrow d x=2(b-a) \sin \theta \cos \theta d \theta$$
then $$\int_{a}^{b}(x-a)^{3}(b-x)^{4} d x$$
$$=2(b-a) \int_{0}^{\pi / 2}\left(a \cos ^{2} \theta+b \sin ^{2} \theta-a\right)^{3}\left(b-a \cos ^{2} \theta\right)\\$$
$$=2(b-a)^{8} \int_{0}^{\pi / 2} \sin ^{7} \theta \cos ^{9} \theta d \theta\\$$
$$=2(b-a)^{8} \int_{0}^{\pi / 2} \sin ^{7} \theta\left(1-\sin ^{2} \theta\right)^{4} \cos \theta d \theta\\$$
$$=2(b-a)^{8} \int_{0}^{1} x^{7}\left(1-x^{2}\right)^{4} d x\\$$
$$=2(b-a)^{8} \int_{0}^{1} x^{7}\left(1-x^{2}\right)^{4} d x\\$$
$$=2(b-a)^{8} \int_{0}^{1} x^{7}\left(1-4 x^{2}+6 x^{4}-4 x^{6}+x^{8}\right) d x\\$$
$$=2(b-a)^{8}\left[\dfrac{1}{8}-\dfrac{4}{10}+\dfrac{6}{12}-\dfrac{4}{14}+\dfrac{1}{16}\right]=\dfrac{(b-a)^{8}}{280}$$
Question 2
1 / -0

Given $$I_{m}=\displaystyle \int_{1}^{e}(\log x)^{m} d x .$$ If $$\dfrac{I_{m}}{K}+\dfrac{I_{m-2}}{L}=e,$$ then the values of $$K$$ and $$L$$ are

A
$$\dfrac{1}{1-m}, \dfrac{1}{m}$$

B
$$(1-m), \dfrac{1}{m}$$

C
$$\dfrac{1}{1-m}, \dfrac{m(m-2)}{m-1}$$

D
$$\dfrac{m}{m-1}, m-2$$

Solution

$$I_{m}= \displaystyle \int_{1}^{e}(\log x)^{m} d x \\$$
$$\therefore I_{m}=\left[x(\log x)^{m}\right]_{1}^{e}-\int_{1}^{e} x \dfrac{m(\log x)^{m-1}}{x} d x \text { (integrating by parts) }$$

$$\Rightarrow \dot{I}_{m}=e-m \displaystyle \int_{1}^{e}(\log x)^{m-1} d x=e-m I_{m-1}\\$$

Replacing $$m$$ by $$m-1\\$$
$$I_{m-1}=e-(m-1) I_{m-2}\\$$
From equations $$(1)$$ and $$(2),$$ we have
$$I_{m}=e-m\left[e-(m-1) I_{m-2}\right]\\$$
$$\Rightarrow I_{m}-m(m-1) I_{m-1}=e(1-m) \\$$
$$\Rightarrow \dfrac{I_{m}}{1-m}+m I_{m-2}=e \\$$
$$\Rightarrow K=1-m \text { and } L=\dfrac{1}{m}$$
Question 3
1 / -0

If $$\int_{0}^{f(x)} t^{2} d t=x \cos \pi x,$$ then $$f^{\prime}({9})$$ is

A
$$-\dfrac{1}{9}$$

B
$$-\dfrac{1}{3}$$

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

D
non-existent

Solution

Given $$\int_{0}^{f(x)} t^{2} d t=x \cos \pi x\\$$
Differentiating with respect to $$x$$.
$$\left.\Rightarrow \dfrac{t^{3}}{3}\right|_{0} ^{f(x)}=x \cos \pi x\\$$
$$\Rightarrow[f(x)]^{3}=3 x \cos \pi x..........(1)\\$$
$$\Rightarrow[f(9)]^{3}=-27\\$$
$$\Rightarrow f(9)=-3\\$$
Also, differentiating equation (1) w.r.t. $$x$$, we get
$$[f(x)]^{2} f^{\prime}(x)=\cos \pi x-x \pi \sin \pi x\\$$
$$\Rightarrow[f(9)]^{2} f^{\prime}(9)=-1\\$$
$$\Rightarrow f^{\prime}(9)=-\dfrac{1}{(f(9))^{2}}=-\dfrac{1}{9}$$
Question 4
1 / -0

The value of the definite integral $$\int_{2}^{4}(x(3-x)(4+x)(6-x)$$ $$(10-x)+\sin x) d x$$ equals

A
$$\cos 2+\cos 4$$

B
$$\cos 2-\cos 4$$

C
$$\sin 2+\sin 4$$

D
$$\sin 2-\sin 4$$

Solution

Let $$\left.I=\int_{2}^{4}(x(3-x)(4+x)(6-x)(10-x)+\sin x) d x\right) $$ .........(i)
$$=\int_{2}^{4}((6-x)(3-(6-x))(4+(6-x))(6-(6-x)) $$ .......... $$\because \int_{a}^b f(x) \ dx= \int _{a}^{b} f(a+b-x) \ dx$$

$$=\int_{2}^{4}((6-x)(x-3)(10-x) x(4+x)+\sin (6-x)) d x$$ .......... (ii)

Adding equations (1) and $$(2),$$ we get
$$2 I=\int_{2}^{4}(\sin x+\sin (6-x)) d x \\$$
$$\quad=(-\cos x+\cos (6-x))_{2}^{4} \\$$
$$\quad=-\cos 4+\cos 2+\cos 2-\cos 4 \\$$
$$\quad=2(\cos 2-\cos 4) \\$$
$$\Rightarrow \quad I=\cos 2-\cos 4$$
Question 5
1 / -0

Given, $$f(x) = \begin{vmatrix} 0 & {x^2 - \sin x} & {\cos x - 2} \\ {\sin x -x^2} & 0 & {1 - 2x} \\ {2 - \cos x} & {2x - 1} & 0 \end{vmatrix}$$, then $$\displaystyle \int f(x) dx$$ is equal to

A
$$\frac{x^3}{3} - x^2 \sin x + \sin 2x + C$$

B
$$\frac{x^3}{3} - x^2 \sin x + \cos 2x + C$$

C
$$\frac{x^3}{3} - x^2 \cos x + \cos 2x + C$$

D
None of the above

Solution

We have, $$f(x) = \begin{vmatrix} 0 & {x^2 - \sin x} & {\cos x - 2} \\ {\sin x -x^2} & 0 & {1 - 2x} \\ {2 - \cos x} & {2x - 1} & 0 \end{vmatrix}$$
$$\Rightarrow$$       $$f(x) = \begin{vmatrix} 0 & {\sin x - x^2} & {2 - \cos x} \\ {x^2 - \sin} & 0 & {2x - 1} \\ {\cos x - 2} & {1 - 2x} & 0 \end{vmatrix} $$
[Interchanging rows and columns]
$$\Rightarrow$$       $$f(x) = (-1)^3 \begin{vmatrix} 0 & {x^2 - \sin x} & {\cos x - 2} \\ {\sin x -x^2} & 0 & {1 - 2x} \\ {2 - \cos x} & {2x - 1} & 0 \end{vmatrix}$$
[Taking (-1) common from each column]
$$\Rightarrow$$       $$f(x) = - f(x)$$
$$\Rightarrow$$       $$f(x) = 0$$
$$\Rightarrow$$ $$ \displaystyle \int f(x) dx = 0$$
Question 6
1 / -0

Let $$f(x)=\displaystyle \int_{-1}^{x}e^{t^2}dt$$ and $$h(x)=f(1+g(x))$$ where $$g(x)$$ is defined for all $$x, g'(x)$$ exists for all $$x,$$ and $$g(x) < 0$$ for $$x>0.$$ If $$h'(1)=e$$ and $$g'(1)=1,$$ then the possible values which $$g(1)$$ can take

A
$$0$$

B
$$-1$$

C
$$-2$$

D
$$-4$$

Solution

Given, $$f(x)=\displaystyle \int_{-1}^{x}e^{t^2}dt; h(x)=f(1+g(x)); g(x)<0$$ for $$x>0$$

Now, $$h(x)=\displaystyle \int_{-1}^{1+g(x)}e^{t^2}dt\quad (given)$$
Differentiating, we get $$h'(x)=e^{(1+g(x))^2}.g'(x)$$
Now, $$h'(1)=e$$
$$\therefore e^{(1+g(1))^2}.g'(1)=e$$
$$\Rightarrow (1+g(1))^2=1$$
$$\Rightarrow 1+g(1)=\pm1$$
$$\Rightarrow g(1)=0\,(not\, possible)$$
or $$g(1)=-2$$
Question 7
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Let $$f : R \rightarrow R$$ be a function as $$f(x) = (x - 1)(x + 2)(x - 3)(x - 6) - 100$$. If $$g(x)$$ is a polynomial of degree $$\leq 3$$ such that $$\displaystyle \int \frac{g(x)}{f(x)} dx$$ does not contain any logarithm function and $$g(-2) = 10$$. Then

$$\displaystyle \int \frac{g(x)}{f(x)} dx$$, equals

A
$$\tan^{-1} \left ( \frac{x - 2}{2}
\right ) + c$$

B
$$\tan^{-1} \left ( \frac{x - 1}{1}
\right ) + c$$

C
$$\tan^{-1} (x) + c$$

D
None of these

Solution

Here, $$f(x) = (x - 1)(x + 2)(x - 3)(x - 6) - 100$$
         $$ = (x^2 + 4x + 3) (x^2 - 4x - 12) - 100$$
          $$ = (x^2 - 4x)^2 - 9(x^2 - 4x) - 136$$
          $$ = (x^2 - 4x + 8)(x^2 - 4x + 17)$$
$$\displaystyle \int \frac{g(x)}{f(x)} = \frac{g(x)}{(x^2 - 4x - 17)(x^2 - 4x + 8)}$$
   $$ = \frac{Ax + B}{x^2 - 4x - 17} + \frac{Cx + D}{x^2 - 4x + 8}$$
Clearly, $$A, B$$ and $$C$$ must be zero.
$$\therefore \frac{g(x)}{(x^2 - 4x - 17)(x^2 - 4x + 8)} = \frac{D}{x^2 - 4x + 8}$$
$$\therefore$$ $$g(x) = D (x^2 - 4x - 17)$$
$$g(-2) = D (4 + 8 - 17) = -10$$ [given]
$$\Rightarrow$$ $$\frac{g(x)}{f(x)} = \frac{2 (x^2 - 4x - 17)}{(x^2 - 4x - 17)(x^2 - 4x + 8)} = \frac{2}{x^2 - 4x + 8}$$
$$\therefore$$ $$\displaystyle \int \frac{g(x)}{f(x)} dx = \displaystyle \int \frac{2}{x^2 - 4x + 8}dx = 2 \displaystyle \int \frac{dx}{(x - 2)^2 + (2)^2}$$
      $$= 2. \frac{1}{2} \tan^{-1} \left ( \frac{x - 2}{2} \right ) + C = \tan^{-1} \left ( \frac{x - 2}{2} \right ) + C$$
Question 8
1 / -0

If $$\displaystyle \int_{-2}^{-1} (ax^2-5)dx $$ and $$5+\displaystyle \int_{1}^{2} (bx+c)dx=0, $$ then

A
$$ax^2-bx+c=0 $$ has atleast one root in (1,2)

B
$$ax^2-bx+c=0 $$ has atleast one root in (-2,-1)

C
$$ax^2+bx+c=0 $$ has atleast one root in (-2,-1)

D
None of the above

Solution
Question 9
1 / -0

The value of the definite integral
$$\displaystyle \int_{0}^{\infty} \dfrac{dx}{(1+x^a)(1+x^2)}(a>0)$$ is

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

B
$$\dfrac{\pi}{2}$$

C
$${\pi}$$

D
Some function of a

Solution
Question 10
1 / -0

Directions For Questions

[passage-header]undefined[/passage-header]If $$I_n = \displaystyle \int_{-\pi}^{\pi} \dfrac{sinnx}{(1+\pi^x)sinx}dx, $$ $$n=0,1,2,..., $$ then[passage-footer]undefined[/passage-footer]

...view full instructions

the value of $$I_{n+2}-I_n$$ is equal to

A
$$n\pi$$

B
$$\pi$$

C
$$-\pi$$

D
$$0$$

Solution

Given $$ I_n = \displaystyle \int_{-\pi}^{\pi} \dfrac{sinnx}{(1+\pi^x)sinx}dx \qquad ...(i)$$
Using $$\displaystyle \int_{a}^{b}f(x)dx=\displaystyle \int_{a}^{b}f(b+a-x)dx,$$ we get
$$I_n=\displaystyle \int_{-\pi}^{\pi} \dfrac{\pi^x sinnx}{(1+\pi^x) sinx}dx\qquad..(ii)$$

On adding Eqs.(i) and (ii), we have
$$2I_n=\displaystyle \int_{-\pi}^{\pi} \dfrac{sinnx}{sinx}dx=2\displaystyle \int_{0}^{\pi} \dfrac{sinnx}{sinx}dx [\because f(x)=\dfrac{sinnx}{sinx}$$ is an even function]
$$\Rightarrow I_n=\displaystyle \int_{0}^{\pi}\dfrac{sinnx}{sinx}dx$$

Now, $$I_{n+2}-I_n=\displaystyle \int_{0}^{\pi} \dfrac{sin(n+2)x-sinnx}{sinx}dx$$
$$=\displaystyle \int_{0}^{\pi} \dfrac{2cos(n+1)x.sinx}{sinx}dx$$
$$=2\displaystyle \int_{0}^{\pi}cos(n+1)dx=2\left[\dfrac{sin(n+1)x}{(n+1)}\right]_{0}^{\pi}=0\qquad...(iii)$$
$$\therefore I_{n+2}-I_n=0$$

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Re-Attempt Weekly Quiz Competition

Integrals Test - 44

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Integrals Test - 44

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SHARING IS CARING

Question 1

$$\dfrac{(b-a)^{4}}{6^{4}}$$

$$\dfrac{(b-a)^{8}}{280}$$

$$\dfrac{(b-a)^{7}}{7^{3}}$$

None of these

Solution

Question 2

$$\dfrac{1}{1-m}, \dfrac{1}{m}$$

$$(1-m), \dfrac{1}{m}$$

$$\dfrac{1}{1-m}, \dfrac{m(m-2)}{m-1}$$

$$\dfrac{m}{m-1}, m-2$$

Solution

Question 3

$$-\dfrac{1}{9}$$

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

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

non-existent

Solution

Question 4

$$\cos 2+\cos 4$$

$$\cos 2-\cos 4$$

$$\sin 2+\sin 4$$

$$\sin 2-\sin 4$$

Solution

Question 5

$$\frac{x^3}{3} - x^2 \sin x + \sin 2x + C$$

$$\frac{x^3}{3} - x^2 \sin x + \cos 2x + C$$

$$\frac{x^3}{3} - x^2 \cos x + \cos 2x + C$$

None of the above

Solution

Question 6

$$0$$

$$-1$$

$$-2$$

$$-4$$

Solution

Question 7

$$\tan^{-1} \left ( \frac{x - 2}{2} \right ) + c$$

$$\tan^{-1} \left ( \frac{x - 1}{1} \right ) + c$$

$$\tan^{-1} (x) + c$$

None of these

Solution

Question 8

$$ax^2-bx+c=0 $$ has atleast one root in (1,2)

$$ax^2-bx+c=0 $$ has atleast one root in (-2,-1)

$$ax^2+bx+c=0 $$ has atleast one root in (-2,-1)

None of the above

Solution

Question 9

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

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

$${\pi}$$

Some function of a

Solution

Question 10

$$n\pi$$

$$\pi$$

$$-\pi$$

$$0$$

Solution

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$$\tan^{-1} \left ( \frac{x - 2}{2}
\right ) + c$$

$$\tan^{-1} \left ( \frac{x - 1}{1}
\right ) + c$$