MCQ
If $A = {\sin ^2}x + {\cos ^4}x$,then for all real $x :$
- A$1 \le A \le 2$
- B$\frac{{13}}{{16}} \le A \le 1$
- ✓$\frac{3}{4} \le A \le 1$
- D$\frac{3}{4} \le A \le \frac{{13}}{{16}}$
✓
Answer
Correct option: C.
$\frac{3}{4} \le A \le 1$
c
$A=\sin ^{2} x+\cos ^{2} x$
$A=\sin ^{2} x+\cos ^{2} x$
We have $\cos ^{4} x \leq \cos ^{2} x$
$\sin ^{2} x=\sin ^{2} x$
Adding $\sin ^{2} x+\cos ^{4} x \leq \sin ^{2} x+\cos ^{2} x$
$\therefore A \leq 1$
Again $A=t+(1-t)^{2}=t^{2}-t+1, t \geq 0$,
where minimum is $3 / 4$
Thus $3 / 4 \leq A \leq 1$.
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