Question

If $A=si{n}^8\theta +co{s}^{14}\theta ,$ then for all real values of $\theta$

• A. $A\ge 1$
• B. 0 < $A\le 1$
• C. 1 < $2A\le 3$
• D. $Noneofthese$

Answer 1

The correct option is B 0 < $A\le 1$

Given that $A=si{n}^8\theta +co{s}^{14}\theta ,$
we know that $(si{n}^2\theta {)}^4\le si{n}^2\theta and(co{s}^2\theta {)}^7\le co{s}^2\theta$
Adding, we that $si{n}^8\theta +co{s}^{14}\theta \le 1$
or 0 < $si{n}^8\theta +co{s}^{14}\theta \le 1.$ Hence 0 < $A\le 1.$