JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 20)
Let f(x) = sin4x + cos4 x. Then f is an increasing function in the interval :
$$] 0, \frac{\pi}{4}[$$
$$] \frac{\pi}{4}, \frac{\pi}{2}[$$
$$] \frac{\pi}{2}, \frac{5 \pi}{8}[$$
$$] \frac{5 \pi}{8}, \frac{3 \pi}{4}[$$
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 $$ \in $$ $$\left( {{\pi \over 4},{\pi \over 2}} \right)$$
$$ \therefore $$ 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 $$ \in $$ $$\left( {{\pi \over 4},{\pi \over 2}} \right)$$
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