1

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

### 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 $= \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {\cos \,\,dx}$

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

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

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

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

### JEE Main 2018 (Offline)

The value of $\int\limits_{ - \pi /2}^{\pi /2} {{{{{\sin }^2}x} \over {1 + {2^x}}}} dx$ is
A
${\pi \over 4}$
B
${\pi \over 8}$
C
${\pi \over 2}$
D
${4\pi }$

## Explanation

As we know,

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

Let $I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}x} \over {1 + {2^x}}}} \,dx$

$\Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}\left( { - {\pi \over 2} + {\pi \over 2} - x} \right)} \over {1 + {2^{ - {\pi \over 2} + {\pi \over 2} - x}}}}}$

$\Rightarrow\,\,\,\, I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\sin }^2}x} \over {1 + {2^{ - x}}}}} dx$

$\Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \,\,dx$

$\therefore\,\,\,$ $2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{{{\sin }^2}x} \over {1 + {2^x}}} + {{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \right)} \,\,dx$

$\Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x\left( {{{1 + {2^x}} \over {1 + {2^x}}}} \right)} \,\,dx$

$\Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx$

$\Rightarrow \,\,\,\,2I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,$ [ as sin2 x is an even function ]

$\Rightarrow \,\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,\,$

[ as $\,\,\,\,$ $\int\limits_0^a {f\left( a \right)\,\,dx = \int\limits_0^a {f\left( {a - x} \right)\,dx]} }$

So, $\,\,\,$ $I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}\left( {{b \over 2} - x} \right)} \,\,dx\,\,\,\,$

$\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,\,dx\,\,\,\,$

$\therefore\,\,\,\,$ $\,\,\,\,2I = \int\limits_0^{{\pi \over 2}} {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \,\,dx\,\,\,\,$

$\Rightarrow \,\,\,\,2I = \int\limits_0^{{\pi \over 2}} \, \,\,dx\,\,\,\,$

$\Rightarrow \,\,\,\,2I = {\left[ x \right]_0}^{{\pi \over 2}}$

$\Rightarrow \,\,\,\,2I = {\pi \over 2}$

$\Rightarrow \,\,\,\,I = {\pi \over 4}$
4

### JEE Main 2018 (Online) 15th April Morning Slot

The area (in sq. units) of the region

{x $\in$ R : x $\ge$ 0, y $\ge$ 0, y $\ge$ x $-$ 2  and y $\le$ $\sqrt x$}, is :
A
${{13} \over 3}$
B
${{8} \over 3}$
C
${{10} \over 3}$
D
${{5} \over 3}$

## Explanation

y = $\sqrt x$

y = x $-$ 2

$\therefore\,\,\,$ $\sqrt x$ = x $-$ 2

$\Rightarrow $$\,\,\, x = x2 - 4x + 4 x2 - 5x + 4 = 0 x2 - 4x - x + 4 = 0 \Rightarrow$$\,\,\,$ x(x $-$ 4) $-$ (x $-$ 4) = 0

$\Rightarrow$$\,\,\,$ (x $-$ 4) (x $-$ 1) = 0

$\therefore\,\,\,$ x = 4, 1

and y = 2, $-$ 1

$\therefore\,\,\,$ Their point of intersection (4, 2) and (1, $-$ 1)

Required area is shown in the shaded figure.

$\therefore\,\,\,$ Required area

= $\int\limits_0^2 {\sqrt x \,dx + \int\limits_2^4 {\left( {\sqrt x - x + 2} \right)\,dx} }$

= $\int\limits_0^4 {\sqrt x \,dx + \int\limits_2^4 {\left( {2 - x} \right)\,dx} }$

= $\left[ {{2 \over 3}{x^{{3 \over 2}}}} \right]_0^4 + \left[ {2x - {{{x^2}} \over 2}} \right]_2^4$

= ${2 \over 3}\left( 8 \right)$ + 2$\left( {4 - 2} \right)$ $-$ ${1 \over 2}$ (16 $-$ 4)

= ${{16} \over 3}$ + 4 $-$ 6

= ${{10} \over 3}$