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
$${{13} \over {256}}$$

Answer


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, $$-$$ $${{17} \over 2}$$

Questions Asked from Definite Integrals and Applications of Integrals

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