Question

If $p={\sin }^{2}x+{\cos }^{4}x$, then

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

Answer 1

The correct option is A

$\frac{3}{4}\le p\le 1$

Find the range of $p$

The range of $\sin x$ and $\cos x$ is $\left[-1,1\right]$.

Therefore, the range of ${\sin }^{2}x$ and ${\cos }^{2}x$ is $\left[0,1\right]$.

Given, $p={\sin }^{2}x+{\cos }^{4}x$
$$\begin{gathered} ={\sin }^{2}x+{(1-{\sin }^{2}x)}^2 \\ ={\sin }^{4}x+{\sin }^{2}x–2{\sin }^{2}x+1 \\ ={\sin }^{4}x–{\sin }^{2}x+1 \\ ={\left({\sin }^{2}x-\frac{1}{2}\right)}^{2^{}}+\frac{3}{4} \\ \Rightarrow p\ge 3/4 \end{gathered}$$

or it can also be written like
$$\begin{gathered} p={\sin }^{2}x+{\cos }^{4}x \\ ={\sin }^{2}x+{\cos }^{2}x(1-{\sin }^{2}x) \\ =({\sin }^{2}x+{\cos }^{2}x)–{\sin }^{2}x{\cos }^{2}x \\ =1–{\sin }^{2}x{\cos }^{2}x \\ \Rightarrow p\le 1 \end{gathered}$$
Therefore, $\frac{3}{4}\le p\le 1$

Hence $\frac{3}{4}\le p\le 1$ and therefore option (A) is correct.