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1

### JEE Main 2021 (Online) 27th August Morning Shift

If $${U_n} = \left( {1 + {1 \over {{n^2}}}} \right)\left( {1 + {{{2^2}} \over {{n^2}}}} \right)^2.....\left( {1 + {{{n^2}} \over {{n^2}}}} \right)^n$$, then $$\mathop {\lim }\limits_{n \to \infty } {({U_n})^{{{ - 4} \over {{n^2}}}}}$$ is equal to :
A
$${{{e^2}} \over {16}}$$
B
$${4 \over e}$$
C
$${{16} \over {{e^2}}}$$
D
$${4 \over {{e^2}}}$$

## Explanation

$${U_n} = \prod\limits_{r = 1}^n {{{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}}$$

$$L = \mathop {\lim }\limits_{n \to \infty } {({U_n})^{ - 4/{n^2}}}$$

$$\log L = \mathop {\lim }\limits_{n \to \infty } {{ - 4} \over {{n^2}}}\sum\limits_{r = 1}^n {\log {{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}}$$

$$\Rightarrow \log L = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n { - {{4r} \over n}.{1 \over n}\log \left( {1 + {{{r^2}} \over {{n^2}}}} \right)}$$

$$\Rightarrow \log L = - 4\int\limits_0^1 {x\log (1 + {x^2})\,dx}$$

put 1 + x2 = t

Now, 2xdx = dt

$$= - 2\int\limits_1^2 {\log (t)dt = - 2[t\log t - t]_1^2}$$

$$\Rightarrow \log L = - 2(2\log 2 - 1)$$

$$\therefore$$ $$L = {e^{ - 2(2\log 2 - 1)}}$$

$$= {e^{ - 2\left( {\log \left( {{4 \over e}} \right)} \right)}}$$

$$= {e^{\log {{\left( {{4 \over e}} \right)}^2}}}$$

$$= {\left( {{e \over 4}} \right)^2} = {{{e^2}} \over {16}}$$
2

### JEE Main 2021 (Online) 26th August Evening Shift

The value of $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{1 + {{\sin }^2}x} \over {1 + {\pi ^{\sin x}}}}} \right)} \,dx$$ is
A
$${\pi \over 2}$$
B
$${{5\pi } \over 4}$$
C
$${{3\pi } \over 4}$$
D
$${{3\pi } \over 2}$$

## Explanation

$$I = \int\limits_0^{{\pi \over 2}} {{{(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}} + {{{\pi ^{\sin x}}(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}}} dx$$

$$I = \int_0^{{\pi \over 2}} {(1 + {{\sin }^2}x)\,dx}$$

$$I = {\pi \over 2} + {\pi \over 2}.{1 \over 2} = {{3\pi } \over 4}$$
3

### JEE Main 2021 (Online) 26th August Evening Shift

If the value of the integral
$$\int\limits_0^5 {{{x + [x]} \over {{e^{x - [x]}}}}dx = \alpha {e^{ - 1}} + \beta }$$, where $$\alpha$$, $$\beta$$ $$\in$$ R, 5$$\alpha$$ + 6$$\beta$$ = 0, and [x] denotes the greatest integer less than or equal to x; then the value of ($$\alpha$$ + $$\beta$$)2 is equal to :
A
100
B
25
C
16
D
36

## Explanation

$$I = \int\limits_0^5 {{{x + [x]} \over {{e^{x - [x]}}}}dx = \alpha {e^{ - 1}} + \beta }$$

$$I = \int\limits_0^1 {{x \over {{e^x}}}dx + \int\limits_1^2 {{{x + 1} \over {{e^{x - 1}}}}dx + \int\limits_2^3 {{{x + 2} \over {{e^{x - 2}}}}dx + \int\limits_3^4 {{{x + 3} \over {{e^{x - 3}}}}dx + \int\limits_4^5 {{{x + 4} \over {{e^{x - 4}}}}dx} } } } }$$

Let $$I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5}$$

Here, $${I_2} = \int\limits_1^2 {{{x + 1} \over {{e^{x - 1}}}}dx}$$ Put $$x = t + 1 \Rightarrow dx = dt$$

$$= \int\limits_0^1 {{{t + 2} \over {{e^t}}}dt = \int\limits_0^1 {{t \over {{e^t}}}dt + \int\limits_0^1 {{2 \over {{e^t}}}dt} } }$$

$${I_2} = {I_1} + 2\int\limits_0^1 {{e^{ - t}}dt = {I_1} + 2(1 - {e^{ - 1}})}$$

Similarly,

$${I_3} = {I_1} + 4(1 - {e^{ - 1}})$$

$${I_4} = {I_1} + 6(1 - {e^{ - 1}})$$

$${I_5} = {I_1} + 8(1 - {e^{ - 1}})$$

$$I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5} = 5{I_1} + (2 + 4 + 6 + 8)(1 - {e^{ - 1}})$$

$$= 5{I_1} + 20(1 - {e^{ - 1}})$$

$${I_1} = \int_0^1 {x{e^{ - 1}}dx = - [{e^{ - x}}(x + 1)_0^1 = 1 - 2{e^{ - 1}}}$$

$$\therefore$$ $$5{I_1} + 20(1 - {e^{ - 1}}) = 5(1 - 2{e^{ - 1}}) + 20(1 - {e^{ - 1}}) = 25 - 30{e^{ - 1}}$$

$$\therefore$$ $$\alpha$$ = $$-$$30, $$\beta$$ = 25

Also it satisfy $$5\alpha + 6\beta = 0$$

Now, $${(\alpha + \beta )^2} = {( - 30 + 25)^2} = {( - 5)^2} = 25$$

4

### JEE Main 2021 (Online) 26th August Morning Shift

The value of $$\int\limits_{{{ - 1} \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {{{\left( {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right)}^{{1 \over 2}}}dx}$$ is :
A
loge 4
B
loge 16
C
2loge 16
D
4loge (3 + 2$${\sqrt 2 }$$)

## Explanation

$$\sqrt {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2}$$

$$= \sqrt {{{\left( {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right)}^2}}$$

$$= \left| {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right|$$

$$= \left| {{{{{(x + 1)}^2} - {{(x - 1)}^2}} \over {{x^2} - 1}}} \right|$$

$$= \left| {{{{x^2} + 2x + 1 - {x^2} + 2x - 1} \over {{x^2} - 1}}} \right|$$

$$= \left| {{{4x} \over {{x^2} - 1}}} \right|$$

$$\therefore$$ $$I = \int\limits_{ - {1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {{{\left[ {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right]}^{{1 \over 2}}}dx}$$

$$I = \int\limits_{{1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx}$$

Let $$f(x) = \left| {{{4x} \over {{x^2} - 1}}} \right|$$

$$\therefore$$ $$f( - x) = \left| {{{4( - x)} \over {{{( - x)}^2} - 1}}} \right| = \left| {{{ - 4x} \over {{x^2} - 1}}} \right| = \left| {{{4x} \over {{x^2} - 1}}} \right|$$

$$\Rightarrow f(x) = f( - x)$$

$$\therefore$$ f(x) is a even function.

$$\therefore$$ $$I = 2\int_0^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx}$$

Using property, If f(x) is an even function then,

$$\int_{ - a}^a {f(x) = 2\int_0^a {f(x)dx} }$$

x > 0 when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$

$$\Rightarrow$$ x2 > 0 when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$

$$\Rightarrow$$ x2 $$-$$ 1 < 0 when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$

$$\Rightarrow {1 \over {{x^2} - 1}} < 0$$ when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$

$$\Rightarrow {{4x} \over {{x^2} - 1}} < 0$$ when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$

$$\therefore$$ $$\left| {{{4x} \over {{x^2} - 1}}} \right| = - {{4x} \over {{x^2} - 1}}$$ when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$

$$\therefore$$ $$I = 2\int_0^{{1 \over {\sqrt 2 }}} { - \left( {{{4x} \over {{x^2} - 1}}} \right)dx}$$

$$= - 4\int_0^{{1 \over {\sqrt 2 }}} {{{2x} \over {{x^2} - 1}}dx}$$

$$= - 4\left[ {{{\log }_e}\left| {{x^2} - 1} \right|} \right]_0^{{1 \over {\sqrt 2 }}}$$

$$= - 4\left[ {\log \left| {{1 \over 2} - 1} \right| - \log \left| { - 1} \right|} \right]$$

$$= - 4{\log _e}\left| { - {1 \over 2}} \right|$$

$$= - 4{\log _e}{1 \over 2}$$

$$= - 4\log _e^{{2^{ - 1}}}$$

$$= 4\log _e^2$$

$$= \log _e^{{2^4}}$$

$$= \log _e^{16}$$

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