1

### JEE Main 2016 (Online) 9th April Morning Slot

If  m and M are the minimum and the maximum values of

4 + ${1 \over 2}$ sin2 2x $-$ 2cos4 x, x $\in$ R, then M $-$ m is equal to :
A
${{15} \over 4}$
B
${{9} \over 4}$
C
${{7} \over 4}$
D
${{1} \over 4}$

## Explanation

Given,

4 + ${1 \over 2}$ sin2 2x $-$ 2cos4 x

= 4 + ${1 \over 2}$ (2sinx cosx)2 $-$ 2cos4x

= 4 + ${1 \over 2}$ $\times$ 4 sin2x cos2x $-$ 2cos4 x

= 4 + 2 (1 $-$ cos2x) cos2x $-$ 2cos4 x

= 4 + 2 cos2x $-$ 4cos4x

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

= $-$ 4 $\left\{ {\cos } \right.$4x $-$ 2 . ${1 \over 4}$ . cos2x + $\left. {{1 \over {16}} - {1 \over {16}} - 1} \right\}$

= $-$ 4 $\left\{ {{{\left( {{{\cos }^2}x - {1 \over 4}} \right)}^2} - {{17} \over {16}}} \right\}$

We know,

O $\le$ cos2x $\le$ 1

$\Rightarrow$   $- {1 \over 4}$ $\le$cos2x $- {1 \over 4}$ $\le$ ${3 \over 4}$

$\Rightarrow$   O $\le$ ${\left( {{{\cos }^2}x - {1 \over 4}} \right)^2}$ $\le$ ${9 \over {16}}$

$\Rightarrow$   $-$ ${17 \over {16}}$ $\le$ ${\left( {{{\cos }^2}x - {1 \over 4}} \right)^2}$ $-$ ${{17} \over {16}}$ $\le$ ${9 \over {16}}$ $-$ ${{17} \over {16}}$

$\Rightarrow$    $-$ ${{17} \over {16}}$ $\le$ ${\left( {{{\cos }^2}x - {1 \over 4}} \right)^2}$ $-$ ${{17} \over {16}}$ $\le$ $-$ ${{1} \over {2}}$

$\Rightarrow$   ${{17} \over {4}}$ $\ge$ $-$ 4 $\left\{ {{{\left( {{{\cos }^2}x - {1 \over 4}} \right)}^2} - {{17} \over {16}}} \right\} \ge 2$

$\therefore$   Maximum value, M = ${{17} \over 4}$

Minimum value, m = ${{1} \over 2}$

$\therefore$   M $-$ m = ${{17} \over 4}$ $-$ ${{1} \over 2}$ = ${{15} \over 4}$
2

### JEE Main 2016 (Online) 10th April Morning Slot

Let f(x) = sin4x + cos4 x. Then f is an increasing function in the interval :
A
$\left[ {0,{\pi \over 4}} \right]$
B
$\left[ {{\pi \over 4},{\pi \over 2}} \right]$
C
$\left[ {{\pi \over 2},{{5\pi } \over 8}} \right]$
D
$\left[ {{{5\pi } \over 8},{{3\pi } \over 4}} \right]$

## Explanation

f(x) = sin4x + cos4x

$\therefore$   f'(x) = 4sin3x cosx + 4cos3x ($-$ sinx)

= 4sinx cosx (sin2x $-$ cos2x)

= $-$ 2sin2x cos2x

= $-$ sin4x

As, f(x) is increasing function when f'(x) > 0

$\Rightarrow$    $-$ sin4x > 0

$\Rightarrow$   sin4x < 0

$\therefore$   $\pi$ < 4x < 2$\pi$

${\pi \over 4} < x < {\pi \over 2}$

$\therefore$   x $\in$ $\left( {{\pi \over 4},{\pi \over 2}} \right)$
3

### JEE Main 2017 (Offline)

Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If $\angle$BPC = $\beta$, then tan$\beta$ is equal to:
A
${1 \over 4}$
B
${2 \over 9}$
C
${4 \over 9}$
D
${6 \over 7}$

## Explanation

Let the height of tower $AB = x$ and $LCPA = \propto$

From the diagram you can see,

$\tan \left( { \propto + \beta } \right) = {x \over {2x}} = {1 \over 2}$

we know,

$\tan \left( { \propto + \beta } \right) = {{\tan \propto + \tan \beta } \over {1 - \tan \propto \tan \beta }}$

$\therefore\,\,\,$ ${{\tan \propto + \tan \beta } \over {1 - \tan \propto \tan \beta }} = {1 \over 2}....\left( 1 \right)$

From the diagram,

$\tan \widehat \propto = {{x/2} \over {2x}} = {1 \over 4}......\left( 2 \right)$

Putting value of $\tan \propto$ in eq$(1)$,

${{{1 \over 4} + \tan \beta } \over {1 - {1 \over 4}\tan \beta }} = {1 \over 2}$

$\Rightarrow 1 - {1 \over 4}\tan \beta$ $= {1 \over 2} + 2\tan \beta$

$\Rightarrow {{9\tan \beta } \over 4} = {1 \over 2}$

$\Rightarrow \tan \beta = {2 \over 9}$
4

### JEE Main 2017 (Offline)

If $5\left( {{{\tan }^2}x - {{\cos }^2}x} \right) = 2\cos 2x + 9$,

then the value of $\cos 4x$ is
A
${1 \over 3}$
B
${2 \over 9}$
C
$- {7 \over 9}$
D
$- {3 \over 5}$

## Explanation

Given that,

$5\left( {{{\tan }^2}x - {{\cos }^2}x} \right) = 2\cos 2x + 9$

$\Rightarrow 5\left( {{{{{\sin }^2}x} \over {{{\cos }^2}x}} - {{\cos }^2}x} \right) = 2\left( {2{{\cos }^2}x - 1} \right) + 9$

Let ${\cos ^2}x = t,$ then we have

$5\left( {{{1 - t} \over t} - t} \right) = 2\left( {2t - 1} \right) + 9$

$\Rightarrow 5\left( {{{1 - t - {t^2}} \over t}} \right) = 4t - 2 + 9$

$\Rightarrow 5 - 5t - 5{t^2} = 4{t^2} + 7t$

$\Rightarrow 9{t^2} + 12t - 5 = 0$

$\Rightarrow 9{t^2} + 15t - 3t - 5 = 0$

$\Rightarrow 3t\left( {3t + 5} \right) - 1\left( {3t + 5} \right) = 0$

$\Rightarrow \left( {3t + 5} \right)\left( {3t - 1} \right) = 0$

$\therefore$ $t = {1 \over 3}$ and $t = - {5 \over 3}$

If $t = - {5 \over 3}$ then ${\cos ^2}x$ is negative

So , $t$ can not be $- {5 \over 3}$.

So, correct value of $t = {1 \over 3}$ then $\cos {}^2x = t = {1 \over 3}$

$\therefore\,\,\,$ $\cos 4x$

$= 2{\cos ^2}2x - 1$

$= 2{\left[ {2{{\cos }^2}x - 1} \right]^2} - 1$

$= 2.{\left[ {2.{1 \over 3} - 1} \right]^2} - 1$

$= 2.{\left( { - {1 \over 3}} \right)^2} - 1$

$= {2 \over 9} - 1$

$= - {7 \over 9}$