Question

State True or False.

$-4\le cos2x+3sinx\le \frac{17}{8}$

• A. True
• B. False

Answer 1

The correct option is A True

Let $f\left(x\right)=cos2x+3sinx$
$=1-2si{n}^{2}x+3sinx$
$=1-2[{(sinx-\frac{3}{4})}^2-\frac{9}{16}]$
$=\frac{17}{8}-2{(sinx-\frac{3}{4})}^2$
Now $f\left(x\right)$ is max when, ${(sinx-\frac{3}{4})}^2$ is min i.e., when $sinx-\frac{3}{4}=0$ for $sinx=\frac{3}{4}$
which is possible.
max. $f\left(x\right)=\frac{17}{8}$
Also min-value of ${(sinx-\frac{3}{4})}^2$ is
${(-1-\frac{3}{4})}^2$
i.e.$\frac{49}{16}$
$\therefore min.f\left(x\right)=\left(\frac{17}{8}\right)-\left(\frac{49}{8}\right)=-4$
$-4\le cos2x+3sinx\le \frac{17}{8}$