Question

If $A={\sin }^8\theta +{\cos }^{14}\theta$, then for all real values of $\theta$

• A. $A\ge 1$
• B. $0<A\le 1$
• C. $1<2A\le 3$
• D. None of these

Answer 1

The correct option is B

$0<A\le 1$

Step1. Construction :

Given, $A={\sin }^8\theta +{\cos }^{14}\theta$

We know that, $0\le {\sin }^2\theta \le 1$

$$\begin{gathered} \therefore {\left({\sin }^2\theta \right)}^4\le {\sin }^2\theta \\ \Rightarrow {\sin }^8\theta \le {\sin }^2\theta \end{gathered}$$

And also $0\le {\cos }^2\theta \le 1$

$$\begin{gathered} \therefore {\left({\cos }^2\theta \right)}^7\le {\cos }^2\theta \\ \Rightarrow {\cos }^{14}\theta \le {\cos }^2\theta \end{gathered}$$

Step2. Find the condition on$\mathit{A}$ for all real values of $\theta$:

Now, ${\sin }^8\theta +{\cos }^{14}\theta \le {\sin }^2\theta +{\cos }^2\theta$

$\Rightarrow A\le 1$

When $A>0$ consider $A=0$

Thus, $\sin \theta =\cos \theta =0$ which is not possible.

Therefore, $0<A\le 1$

Hence the correct option is B.