1
MCQ (Single Correct Answer)

### JEE Main 2017 (Online) 8th April Morning Slot

The integral $\int_{{\pi \over {12}}}^{{\pi \over 4}} {\,\,{{8\cos 2x} \over {{{\left( {\tan x + \cot x} \right)}^3}}}} \,dx$ equals :
A
${{15} \over {128}}$
B
${{15} \over {64}}$
C
${{13} \over {32}}$
D
${{13} \over {256}}$

## Explanation

tan x  +  cot x

= ${{\sin x} \over {\cos x}}$ + ${{\cos x} \over {\sin x}}$

= ${{{{\sin }^2}x + {{\cos }^2}x} \over {\sin x\,\,\cos x}}$

= ${1 \over {\sin x\,\,\cos x}}$

= ${2 \over {\sin 2x}}$

$\therefore\,\,\,$ $\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8\cos 2x} \over {{{\left( {\tan x + \cot x} \right)}^3}}}} \,\,dx$

= $\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8\cos 2x} \over {{{\left( {{2 \over {\sin 2x}}} \right)}^3}}}} \,\,dx$

= $\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8{{\sin }^3}\,2x.\cos \,\,2x} \over 8}} \,\,dx$

= $\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{\sin }^3}2x.\cos \,2x} \,\,dx$

Let sin 2x = t

$\Rightarrow $$\,\,\, 2 cos2x dx = dt At x = {{\pi \over {12}}}, t = sin {{\pi \over {6}}} = {1 \over 2} At x = {{\pi \over 4}}, t = sin {{\pi \over 2}} = 1. = {1 \over 2} \int\limits_{{1 \over 2}}^1 {{t^3}.dt} = {{1 \over 2}} \left[ {{{{t^4}} \over 4}} \right]_{{1 \over 2}}^1 = {{1 \over 2}} \times {{1 \over 4}} \times [ 14 - ({{1 \over 2}})4] = {{1 \over 8}} \times {{15 \over 16}} = {{15 \over 128}} 2 MCQ (Single Correct Answer) ### JEE Main 2017 (Online) 8th April Morning Slot The area (in sq. units) of the smaller portion enclosed between the curves, x2 + y2 = 4 and y2 = 3x, is : A {1 \over {2\sqrt 3 }} + {\pi \over 3} B {1 \over {\sqrt 3 }} + {{2\pi } \over 3} C {1 \over {2\sqrt 3 }} + {{2\pi } \over 3} D {1 \over {\sqrt 3 }} + {{4\pi } \over 3} 3 MCQ (Single Correct Answer) ### JEE Main 2017 (Online) 9th April Morning Slot If \int\limits_1^2 {{{dx} \over {{{\left( {{x^2} - 2x + 4} \right)}^{{3 \over 2}}}}}} = {k \over {k + 5}}, then k is equal to : A 1 B 2 C 3 D 4 ## Explanation Given, I = \int\limits_1^2 {{{dx} \over {{{\left( {{x^2} - 2x + 4} \right)}^{{3 \over 2}}}}}} = \int\limits_1^2 {{{dx} \over {{{\left[ {{{\left( {x - 1} \right)}^2} + 3} \right]}^{{3 \over 2}}}}}} Let x - 1 = \sqrt 3 tan\theta \Rightarrow$$\,\,\,$ dx = $\sqrt 3$ sec2$\theta$ d$\theta$

When x = 1, then $\theta$ = 0

and when x = 2, $\theta$ = ${\pi \over 6}$

I = $\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left( {3{{\tan }^2}\theta + 3} \right)}^{{3 \over 2}}}}}}$

= $\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left( {3{{\sec }^2}\theta } \right)}^{{3 \over 2}}}}}}$

= $\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {3\sqrt 3 {{\sec }^3}\theta }}}$

= ${1 \over 3}$ $\int\limits_0^{{\pi \over 6}} {{{d\theta } \over {\sec \theta }}}$

= ${1 \over 3}\int\limits_0^{{\pi \over 6}} {\cos \theta \,d\theta }$

= ${1 \over 3}{\left[ {\sin \theta } \right]_\theta }^{{\pi \over 6}}$

= ${1 \over 3} \times {1 \over 2}$

= ${1 \over 6}$

$\therefore\,\,\,$ According to the question,

${k \over {k + 5}}$ = ${1 \over 6}$

$\Rightarrow $$\,\,\, 6k = k + 5 \Rightarrow$$\,\,\,$ k = 1
4
MCQ (Single Correct Answer)

### JEE Main 2017 (Online) 9th April Morning Slot

If    $\mathop {\lim }\limits_{n \to \infty } \,\,{{{1^a} + {2^a} + ...... + {n^a}} \over {{{(n + 1)}^{a - 1}}\left[ {\left( {na + 1} \right) + \left( {na + 2} \right) + ..... + \left( {na + n} \right)} \right]}} = {1 \over {60}}$

for some positive real number a, then a is equal to :
A
7
B
8
C
${{15} \over 2}$
D
${{17} \over 2}$

## Explanation

$\mathop {\lim }\limits_{n \to \infty } {{{1 \over {\left( {a + 1} \right)}}\,.\,{n^{a + 1}} + {a_1}{n^a} + {a_2}{n^{a - 1}} + .\,.\,.\,.} \over {{{\left( {n + 1} \right)}^{a - 1}}.\,{n^2}\left( {a + {{1 + {1 \over n}} \over 2}} \right)}} = {1 \over {60}}$

$\Rightarrow$   $\mathop {\lim }\limits_{n \to \infty } {{{{\left( {{1 \over n}} \right)}^2} + {{\left( {{2 \over n}} \right)}^a} + ....... + {{\left( {{n \over n}} \right)}^a}} \over {{{\left( {n + 1} \right)}^{a - 1}}\left[ {{a^2}a + {{n\left( {n + 1} \right)} \over 2}} \right]}} = {1 \over {60}}$

$= {{\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)}^a}} } \over {{{\left( {1 + {1 \over n}} \right)}^{a - 1}}\left[ {a + {1 \over 2}\left( {1 + {1 \over n}} \right)} \right]}} = {1 \over {60}}$

$= {{\int_0^1 {{x^a}dx} } \over {\left( {a + {1 \over 2}} \right)}} = {1 \over {60}} = {{{1 \over {a + 1}}} \over {a + {1 \over 2}}} = {1 \over {60}}$

$\Rightarrow$  ${{{1 \over {a + 1}}} \over {\left( {a + {1 \over 2}} \right)}} = {1 \over {60}} \Rightarrow \left( {a + 1} \right)\left( {2a + 1} \right) = 120$

$\Rightarrow$  2a2 + 3a $-$ 119 $=$ 0

$\Rightarrow$   2a2 + 17a $-$ 14a $-$ 119 $=$ 0

$\Rightarrow$   (a $-$ 7) (2a + 17) $=$ 0

$\Rightarrow$   a $=$ 7, $-$ ${{17} \over 2}$

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