Question

The minimum and maximum values of
${\sin }^{4}x+{\cos }^{4}x$ are

• A. $\frac{1}{2},\frac{3}{2}$
• B. $\frac{1}{2},1$
• C. $1,\frac{3}{2}$
• D. 1,2

Answer 1

The correct option is B $\frac{1}{2},1$

Let $f\left(x\right)={\cos }^{4}x+{\sin }^{4}x$
$=({\cos }^{2}x{)}^2+({\sin }^{2}x{)}^2$
$=({\cos }^{2}x+{\sin }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x.................[a^2+b^2=(a+b{)}^2-2ab]$
$=1-\frac{(\sin 2x{)}^2}{2}$
for max value of $f\left(x\right),\sin 2x=0$
$f\left(x\right)=1-0=1$
for min. value of $f\left(x\right),\sin x=1or-1$
$f\left(x\right)=1-\frac{1}{2}=\frac{1}{2}$