The function $f (x) = {sin}^{4} x + {cos}^{4} x$ increases if

0 < x < \frac{π}{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{4} < x < \frac{3 π}{8}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{3 π}{4} < x < \frac{5 π}{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{5 π}{8} < x < \frac{3 π}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{π}{4} < x < \frac{3 π}{8}$
$f (x) = {sin}^{4} x + {cos}^{4} x = ({sin}^{2} x)^{2} + ({cos}^{2} x)^{2}$
$= ({sin}^{2} x + {cos}^{2} x)^{2} - 2 {sin}^{2} x {cos}^{2} x$
$= 1 - \frac{1}{2} {sin}^{2} 2 x$
$\Rightarrow f^{'} (x) = - \frac{1}{2} (2 sin 2 x cos 2 x) (2) = - sin 4 x$
Now for f to be increasing $f^{'} (x) > 0$
$\Rightarrow sin 4 x < 0 \Rightarrow 4 x \in (π, 2 π)$
$\Rightarrow x \in (\frac{π}{4}, \frac{π}{2})$

Suggest Corrections

Similar questions

f (x) = {sin}^{4} x + {cos}^{4} x, x \in (0, π / 4)

f (x) = \frac{\sqrt{2} cos x - 1}{cot x - 1}, x \neq \frac{π}{4}

f (\frac{π}{4})

f (x)

x = \frac{π}{4}

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} {(\frac{6}{5})}^{\frac{t a n 6 x}{t a n 5 x}}; 0 < x < \frac{π}{2} a + 2; x = \frac{π}{2} (1 + | c o t x |)^{\frac{b | t a n x |}{a}}; \frac{π}{2} < x < π \end{matrix}

x = \frac{π}{2}

a

b

f (x) = {sin}^{4} x + {cos}^{4} x \forall x \in [0, π / 2]

f (x) = \frac{{sin}^{4} x + {cos}^{4} x}{x + tan x}

Join BYJU'S Learning Program