Question

If $A={\sin }^2\theta +{\cos }^4\theta$, then for all real value of $\theta$

• A. $1\le A\le 2$
• B. $\frac{3}{4}\le A\le 1$
• C. $\frac{13}{16}\le A\le 1$
• D. $\frac{3}{4}\le A\le \frac{13}{16}$

Answer 1

Answer: (B)

$\frac{3}{4}\le A\le 1$

$A=si{n}^2\theta +co{s}^4\theta$
$=si{n}^2\theta +co{s}^2\theta .co{s}^2\theta$
$=si{n}^2\theta +co{s}^2\theta (1-si{n}^2\theta )$
$=si{n}^2\theta +co{s}^2\theta -si{n}^2\theta co{s}^2\theta$
$=1-\frac{1}{4}(2sin\theta cos\theta {)}^2$
$=1-\frac{1}{4}si{n}^{2}2\theta$
$Min.\left(si{n}^{2}2\theta \right)=0$
$Max.\left(si{n}^{2}2\theta \right)=1$
$Min.\left(A\right)=1-\frac{1}{4}.Max\left(si{n}^{2}2\theta \right)$
$=1-\frac{1}{4}.1=\frac{3}{4}$
$Max.\left(A\right)=1-\frac{1}{4}.Min.\left(si{n}^{2}2\theta \right)$
$=1-\frac{1}{4}.0=1$
$\therefore \frac{3}{4}\le A\le 1$
Hence, option 'B' is correct.