1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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 $$\cos {}^2x = t = {1 \over 3}$$

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

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

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

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

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

$$ = {2 \over 9} - 1$$

$$ = - {7 \over 9}$$

Questions Asked from Trigonometric Functions & Equations

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