You visited us 1 times! Enjoying our articles? Unlock Full Access!

Range of Trigonometric Expressions

A = sin 2 x ...

Question

$A = {sin}^{2} x + {cos}^{4} x, then for all real x :$

1 \leq A \leq 2

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

\frac{3}{4} \leq A \leq \frac{13}{16}

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

\frac{3}{4} \leq A \leq 1

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

\frac{13}{16} \leq A \leq 1

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

Open in App

Solution

The correct option is C $\frac{3}{4} \leq A \leq 1$
$s i n^{2} x + c o s^{4} x + c o s^{4} x$
= $s i n^{2} x + (1 - s i n^{2} x)^{2}$
= $s i n^{4} x - s i n^{2} x + 1$
$= (s i n^{2} x - \frac{1}{2})^{2} + \frac{3}{4}$
we know that
$0 \leq s i n^{2} x \leq 1$
$\Rightarrow 0 \leq (s i n x - \frac{1}{2})^{2} \leq \frac{1}{4}$
$\Rightarrow \frac{3}{4} \leq (s i n x - \frac{1}{2})^{2} + \frac{3}{4} \leq 1$
$\Rightarrow \frac{3}{4} \leq s i n^{2} x + c o s^{4} x \leq 1$

Suggest Corrections

Similar questions

{\{{(\frac{1}{3})}^{2}\}}^{4}

{(\frac{1}{3})}^{6}

{(\frac{1}{3})}^{8}

{(\frac{1}{3})}^{24}

{(\frac{1}{3})}^{16}

{\{{(\frac{1}{3})}^{2}\}}^{4} = ?

{(\frac{1}{3})}^{6}

{(\frac{1}{3})}^{8}

{(\frac{1}{3})}^{16}

{(\frac{1}{3})}^{24}

\frac{1095}{1168}

\frac{17}{26}

\frac{25}{26}

\frac{13}{16}

\frac{15}{16}

If a : b = 2 : 5, then \frac{3 a + 2 b}{4 a + b} = (a) \frac{16}{13} (b) \frac{13}{16} (c) \frac{25}{22} (d) \frac{20}{21}

\frac{13}{16}

\frac{5}{8}

\frac{1}{5}

\frac{3}{4}

Join BYJU'S Learning Program