Definite Integration — Page 6 | JEE Mathematics

Q51

The value of

\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx

is equal to :

A 4

\pi

B 2

\pi

C

\pi

2

D 2

\pi

2

Correct Answer

Option C

Solution

I =

\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx

.....(1) I =

\int\limits_0^{2\pi } {{{\left( {2\pi - x} \right){{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx

......(2) Adding (1) and (2) 2I =

\int\limits_0^{2\pi } {{{2\pi {{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx

$\Rightarrow$ I =

2\pi \int\limits_0^\pi {{{{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx

$\Rightarrow$ I =

2\pi \left[ {\int\limits_0^{\pi /2} {{{{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx + \int\limits_0^{\pi /2} {{{{{\cos }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx} \right]

=

{2\pi \int\limits_0^{\pi /2} 1 dx}

=

2\pi .{\pi \over 2}

=

{\pi ^2}

Q52

If I1 =

\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}} dx

and I2 =

\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx

such that I2 =

\alpha

I1 then

\alpha

equals to :

A

{{5051} \over {5050}}

B

{{5050} \over {5051}}

C

{{5050} \over {5049}}

D

{{5049} \over {5050}}

Correct Answer

Option B

Solution

I2 =

\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx

I2 =

\int\limits_0^1 {\left( {1 - {x^{50}}} \right){{\left( {1 - {x^{50}}} \right)}^{100}}dx}

I2 =

\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} - \int\limits_0^1 {{x^{50}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx}

I2 = I1 -

\int\limits_0^1 {x.{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx}

Now apply IBP I2 = I1 -

\left[ {x\int\limits_0^1 {{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} - \int {{{d\left( x \right)} \over {dx}}.\int {{{d\left( x \right)} \over {dx}}\int\limits_0^1 {{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} } } } \right]

Let (1 – x50) = t $\Rightarrow$ -50x49dx = dt I2 = I1 -

\left[ {x.\left( { - {1 \over {50}}} \right){{{{\left( {1 - {x^{50}}} \right)}^{101}}} \over {101}}} \right]_0^1

-

- \int\limits_0^1 {\left( { - {1 \over {50}}} \right){{{{\left( {1 - {x^{50}}} \right)}^{101}}} \over {101}}dx}

= I1 - 0 -

{ - {1 \over {50}}.{1 \over {101}}{I_2}}

I2 = I1 -

{1 \over {5050}}{I_2}

$\Rightarrow$ I2 +

{1 \over {5050}}{I_2}

= I1 $\Rightarrow$

{{5051} \over {5050}}{I_2}

= I1 $\Rightarrow$ I2 =

{{5050} \over {5051}}

I1 Given I2 = $\alpha$ I1 $\therefore$ $\alpha$ =

{{5050} \over {5051}}

Q53

The integral

\int\limits_1^2 {{e^x}.{x^x}\left( {2 + {{\log }_e}x} \right)} dx

equals :

A e(4e + 1)

B e(2e – 1)

C e(4e – 1)

D 4e2 – 1

Correct Answer

Option C

Solution

\int\limits_1^2 {{e^x}.{x^x}\left( {2 + {{\log }_e}x} \right)} dx

=

\int\limits_1^2 {{e^x}{x^x}\left[ {1 + \left( {1 + {{\log }_e}x} \right)} \right]} dx

=

\int\limits_1^2 {{e^x}\left[ {{x^x} + {x^x}\left( {1 + {{\log }_e}x} \right)} \right]} dx

=

\left[ {{e^x}{x^x}} \right]_1^2

= e2 $\times$ 4 - e $\times$ 1 = 4e2 - e = e(4e - 1) Note :

\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)dx}

= exf(x) + c

Q54

If ƒ(a + b + 1 - x) = ƒ(x), for all x, where a and b are fixed positive real numbers, then

{1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx

is equal to:

A

\int_{a - 1}^{b - 1} {f(x+1)dx}

B

\int_{a + 1}^{b + 1} {f(x + 1)dx}

C

\int_{a - 1}^{b - 1} {f(x)dx}

D

\int_{a + 1}^{b + 1} {f(x)dx}

Correct Answer

Option A

Solution

I =

{1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx

...(1) x $\to$ a + b - x I =

{1 \over {a + b}}\int\limits_a^b {\left( {a + b - x} \right)\left( {f\left( {a + b - x} \right) + f\left( {a + b - x + 1} \right)} \right)dx}

I =

{1 \over {a + b}}\int\limits_a^b {\left( {a + b - x} \right)\left( {f\left( {x + 1} \right) + f\left( x \right)} \right)dx}

.....(2) [As ƒ(x) = ƒ(a + b + 1 - x) $\Rightarrow$ ƒ(x + 1) = ƒ(a + b - x)] Adding (1) and (2) we get 2I =

{{a + b} \over {a + b}}\int\limits_a^b {\left( {f\left( {x + 1} \right) + f\left( x \right)} \right)dx}

$\Rightarrow$ I =

{1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( {x + 1} \right)dx}

$\Rightarrow$ I =

{1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( {a + b - x + 1} \right)dx}

$\Rightarrow$ I =

{1 \over 2}\int\limits_a^b {f\left( x \right)dx} + {1 \over 2}\int\limits_a^b {f\left( x \right)dx}

$\Rightarrow$ I =

\int\limits_a^b {f\left( x \right)dx}

Let x = t + 1 $\therefore$ I =

\int\limits_{a - 1}^{b - 1} {f\left( {t + 1} \right)dt}

=

\int\limits_{a - 1}^{b - 1} {f\left( {x + 1} \right)dx}

Q55

The value of

\alpha

for which

4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5

, is:

A

{\log _e}2

B

{\log _e}\sqrt 2

C

{\log _e}\left( {{4 \over 3}} \right)

D

{\log _e}\left( {{3 \over 2}} \right)

Correct Answer

Option A

Solution

4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5

$\Rightarrow$

4\alpha \int\limits_{ - 1}^0 {{e^{\alpha x}}dx} + 4\alpha \int\limits_0^2 {{e^{ - \alpha x}}dx}

= 5 $\Rightarrow$

4\alpha \left[ {{{{e^{\alpha x}}} \over \alpha }} \right]_{ - 1}^0 + 4\alpha \left[ {{{{e^{ - \alpha x}}} \over { - \alpha }}} \right]_0^2

= 5 $\Rightarrow$

4{e^{ - 2\alpha }} + 4{e^{ - \alpha }} - 3

= 0 Let e- $\alpha$ = t $\therefore$ 4t2 + 4t - 3 = 0 $\Rightarrow$ t =

{1 \over 2}

$\therefore$ e- $\alpha$ =

{1 \over 2}

$\Rightarrow$ $\alpha$ =

{\log _e}2

Q56

If

\theta

1 and

\theta

2 be respectively the smallest and the largest values of

\theta

in (0, 2

\pi

) - {

\pi

} which satisfy the equation, 2cot2

\theta

-

{5 \over {\sin \theta }}

+ 4 = 0, then

\int\limits_{{\theta _1}}^{{\theta _2}} {{{\cos }^2}3\theta d\theta }

is equal to :

A

{\pi \over 9}

B

{{2\pi } \over 3}

C

{{\pi } \over 3}

D

{\pi \over 3} + {1 \over 6}

Correct Answer

Option C

Solution

2cot2 $\theta$ -

{5 \over {\sin \theta }}

+ 4 = 0 $\Rightarrow$

2{{{{\cos }^2}\theta } \over {{{\sin }^2}\theta }} - {5 \over {\sin \theta }} + 4

= 0 $\Rightarrow$ 2sin2 $\theta$ – 5sin $\theta$ + 2 = 0 $\Rightarrow$ (2sin $\theta$ – 1)(sin $\theta$ – 2) = 0 $\therefore$ sin $\theta$ =

{1 \over 2}

[ sin $\theta$ = 2 not possible] $\therefore$ $\theta$ 1 =

{\pi \over 6}

and $\theta$ 2 =

{{5\pi } \over 6}

as $\theta$ 1

<

$\theta$ 2 $\therefore$ I =

\int\limits_{{\pi \over 6}}^{{{5\pi } \over 6}} {{{\cos }^2}3\theta d\theta }

=

\int\limits_{{\pi \over 6}}^{{{5\pi } \over 6}} {{{1 + \cos 6\theta } \over 2}d\theta }

=

{1 \over 2}\left[ {\theta + {{\sin 6\theta } \over 2}} \right]_{{\pi \over 6}}^{{{5\pi } \over 6}}

=

{{\pi \over 3}}

Q57

If

I = \int\limits_1^2 {{{dx} \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}}

, then :

A

{1 \over 16} < {I^2} < {1 \over 9}

B

{1 \over 8} < {I^2} < {1 \over 4}

C

{1 \over 9} < {I^2} < {1 \over 8}

D

{1 \over 6} < {I^2} < {1 \over 2}

Correct Answer

Option C

Solution

Let f(x) =

{1 \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}

f'(x) =

- {1 \over 2}{{ - 6{x^2} - 18x + 12} \over {2{{\left( {2{x^3} - 9{x^2} + 12x + 4} \right)}^{3/2}}}}

=

{{ - 6\left( {x - 1} \right)\left( {x - 2} \right)} \over {2{{\left( {2{x^3} - 9{x^2} + 12x + 4} \right)}^{3/2}}}}

fmax = f(1) fmin = f(2) f(1) =

{1 \over {\sqrt {2 - 9 + 12 + 4} }}

=

{1 \over {\sqrt 9 }}

=

{1 \over 3}

f(2) =

{1 \over {\sqrt {16 - 36 + 24 + 4} }}

=

{1 \over {\sqrt 8 }}

$\therefore$

{1 \over 3} < {I} <

{1 \over {\sqrt 8 }}

So

{1 \over 9} < {I^2} < {1 \over 8}

Q58

If for all real triplets (a, b, c), ƒ(x) = a + bx + cx2; then

\int\limits_0^1 {f(x)dx}

is equal to :

A

{1 \over 6}\left\{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right\}

B

2\left\{ 3{f(1) + 2f\left( {{1 \over 2}} \right)} \right\}

C

{1 \over 3}\left\{ {f(0) + f\left( {{1 \over 2}} \right)} \right\}

D

{1 \over 2}\left\{ {f(1) + 3f\left( {{1 \over 2}} \right)} \right\}

Correct Answer

Option A

Solution

ƒ(x) = a + bx + cx2

\int\limits_0^1 {f\left( x \right)dx}

=

\left[ {ax + {{b{x^2}} \over 2} + {{c{x^3}} \over 3}} \right]_0^1

=

{a + {b \over 2} + {c \over 3}}

=

{1 \over 6}\left[ {6a + 3b + c} \right]

f(1) = a + b + c f(0) = a

f\left( {{1 \over 2}} \right) = a + {b \over 2} + {c \over 4}

By checking each option you have to find the solution.

{1 \over 6}\left\{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right\}

=

{1 \over 6}\left[ {a + a + b + c + 4\left( {a + {b \over 2} + {c \over 4}} \right)} \right]

=

{1 \over 6}\left[ {6a + 3b + c} \right]

$\therefore$ Option (A) is correct option.

Q59

Let a function ƒ : [0, 5]

\to

R be continuous, ƒ(1) = 3 and F be defined as :

F(x) = \int\limits_1^x {{t^2}g(t)dt}

, where

g(t) = \int\limits_1^t {f(u)du}

Then for the function F, the point x = 1 is :

A a point of inflection.

B a point of local maxima.

C a point of local minima.

D not a critical point.

Correct Answer

Option C

Solution

F(x) = \int\limits_1^x {{t^2}g(t)dt}

$\Rightarrow$ F'(x) = x2g(x) = x2

\int\limits_1^t {f(u)du}

$\therefore$ F'(1) = (1)(0) = 0 Now, F''(x) = 2xg(x) + x2g'(x) F''(1) = 2g(1) + g'(1) = 0 + g'(1) = 3 [ As g'(t) = f(t); g'(1) = f'(1) = 3 ] So, at x = 1, F'(1) = 0 and F"(1) = 3 > 0 $\therefore$ For the function f(x), x = 1 is a point of local minima.

Q60

\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}

is equal to

A

- {1 \over 5}

B

- {1 \over 10}

C 0

D

{1 \over 10}

Correct Answer

Option C

Solution

\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}

This is in

{0 \over 0}

form. So apply newton leibniz rule

\mathop {\lim }\limits_{x \to 0} {{x.\sin \left( {10x} \right) - 0} \over 1}

= 0