Question

Range of ${\sin ^2}\theta + {\cos ^4}\theta = $
A) $1 \leqslant A \leqslant 2$
B) $\dfrac{3}{4} \leqslant A \leqslant 1$
C) $\dfrac{{13}}{6} \leqslant A \leqslant 1$
D) None of these

Answer 1

Hint:
It is known that range of $ - 1 \leqslant \cos \theta \leqslant 1$. 1st we will convert $\sin \theta $ into $\cos \theta $ by using ${\sin ^2}\theta + {\cos ^2}\theta = 1$then ${\sin ^2}\theta + {\cos ^4}\theta = 1 - {\cos ^2}\theta + {\cos ^4}\theta $. After this we will convert this square form by complete square method. after that using range of $\cos \theta $ we will proceed further and convert it into ${\sin ^2}\theta + {\cos ^4}\theta $

Complete step by step solution:
 ${\sin ^2}\theta + {\cos ^4}\theta = $
We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$, substituting this in equation we get
$ \Rightarrow 1 - {\cos ^2}\theta + {\cos ^4}\theta $
We can also write it as
$ \Rightarrow 1 - 2 \times \dfrac{1}{2} \times {\cos ^2}\theta + {\cos ^4}\theta + \dfrac{1}{4} - \dfrac{1}{4}$
$ \Rightarrow {\left( {{{\cos }^2}\theta - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}$ $\left[ {\because {{\left( {a - b} \right)}^2} = {a^2} + {b^2} - 2ab} \right]$
It is known that
$ - 1 \leqslant \cos \theta \leqslant 1$
On squaring we get
$0 \leqslant {\cos ^2}\theta \leqslant 1$
Subtracting $\dfrac{1}{2}$ we get
$ - \dfrac{1}{2} \leqslant {\cos ^2}\theta - \dfrac{1}{2} \leqslant \dfrac{1}{2}$
Again, on squaring we get
$0 \leqslant {\left( {{{\cos }^2}\theta - \dfrac{1}{2}} \right)^2} \leqslant \dfrac{1}{4}$
Then adding $\dfrac{3}{4}$ we get
$\dfrac{3}{4} \leqslant {\left( {{{\cos }^2}\theta - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \leqslant 1$
There for range of ${\sin ^2}\theta + {\cos ^4}\theta = $ $\dfrac{3}{4} \leqslant A \leqslant 1$

Hence, option B is the correct option.

Note:
Point to remember are, in these types of questions we will convert all trigonometric ratios in a single form. Range of $ - 1 \leqslant \cos \theta \leqslant 1$, $ - 1 \leqslant \sin \theta \leqslant 1$. While converting division should be avoided as far as possible it will reduce the possible value.