Question

$\mathrm{The}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{function}f\left(x\right)={\cos }^{2}x+{\sin }^{4}x\mathrm{is}$.

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{3}{4}$

Answer 1

The correct option is C $\frac{3}{4}$

Here, the function $f\left(x\right)$ is composed of sine and cosine terms with even powers.
Now, we can write the function as:
$f\left(x\right)={\sin }^{4}x+{\cos }^{2}x\Rightarrow f\left(x\right)={(1-{\cos }^{2}x)}^2+{\cos }^{2}x\Rightarrow f\left(x\right)=1+{\cos }^{4}x-2{\cos }^{2}x+{\cos }^{2}x\Rightarrow f\left(x\right)=1+{\cos }^{4}x-{\cos }^{2}x\Rightarrow f\left(x\right)={({\cos }^{2}x-\frac{1}{2})}^2+\frac{3}{4}\mathrm{We}\mathrm{know}\mathrm{that}{({\cos }^{2}x-\frac{1}{2})}^2\mathrm{takes}\mathrm{only}\mathrm{non}-\mathrm{negative}\mathrm{value}.\mathrm{For}\mathrm{minimum}\mathrm{value}\mathrm{of}f\left(x\right),\mathrm{the}\mathrm{value}\mathrm{of}{\cos }^{2}x-\frac{1}{2}\mathrm{must}\mathrm{be}\mathrm{zero}.{\cos }^{2}x-\frac{1}{2}=0\Rightarrow \cos x=\frac{1}{\sqrt{2}}\mathrm{or}-\frac{1}{\sqrt{2}},\mathrm{So},\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}f\left(x\right)\mathrm{is}\frac{3}{4}.$