Question

If $A={\sin }^{2}x+{\cos }^{4}x$ then for all real $x$-

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

Answer 1

Answer: (C)

$-1\le A\le 1$

$A={\sin }^{2}x+{\cos }^{4}x={\cos }^{4}x+{\cos }^{2}x-1$

$A=({\cos }^{2}x+\frac{1}{2})-5/4$

Now, $-1\le \cos x\le 1⟹0\le {\cos }^{2}x\le 1$

$\frac{-4}{4}\le ({\cos }^{2}x+\frac{1}{2}{)}^2-\frac{5}{4}\le 4/4$

So, $-1\le A\le 1$