1

### 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
${1 \over {\sqrt 3 }} + {{4\pi } \over 3}$
2

### 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
3

### 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, $-$ ${{17} \over 2}$
4

### JEE Main 2018 (Offline)

Let g(x) = cosx2, f(x) = $\sqrt x$ and $\alpha ,\beta \left( {\alpha < \beta } \right)$ be the roots of the quadratic equation 18x2 - 9$\pi$x + ${\pi ^2}$ = 0. Then the area (in sq. units) bounded by the curve
y = (gof)(x) and the lines $x = \alpha$, $x = \beta$ and y = 0 is
A
${1 \over 2}\left( {\sqrt 2 - 1} \right)$
B
${1 \over 2}\left( {\sqrt 3 - 1} \right)$
C
${1 \over 2}\left( {\sqrt 3 + 1} \right)$
D
${1 \over 2}\left( {\sqrt 3 - \sqrt 2 } \right)$

## Explanation

$18{x^2} - 9\pi x + {\pi ^2} = 0$

$\Rightarrow \,\,\,18{x^2} - 6\pi x - 3\pi x + {\pi ^2} = 0$

$\Rightarrow \,\,\,\,6x\left( {3x - \pi } \right) - \pi \left( {3x - \pi } \right) = 0$

$\Rightarrow \,\,\,\,\left( {3x - \pi } \right)\left( {6x - \pi } \right) = 0$

$\therefore$ $\,\,\,\,x = {\pi \over 3},{\pi \over 6}$

as $\,\,\,\, \propto < B$

$\therefore$ $\,\,\,\, \propto = {\pi \over 6}$ $\,\,\,\,$ and $\,\,\,\,$ $\beta = {\pi \over 3}$

Given, $g\left( x \right) = \cos {x^2}$ and $+ \left( x \right) = \sqrt x$

$y = \left( {gof} \right)x$

$= \,\,\,\,\,g\left( {f\left( x \right)} \right)$

$= \,\,\,\,\cos \left( {f{{\left( x \right)}^2}} \right)$

$= \,\,\,\,\cos {\left( {\sqrt x} \right)^2}$

$= \,\,\,\,\cos x$

So, the required area in the curve is Area $= \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {\cos \,\,dx}$

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

$= \sin {\pi \over 3} - \sin {\pi \over 6}$

$= {{\sqrt 3} \over 2} - {1 \over 2}$

$= {{\sqrt 3 - 1} \over 2}$