Question

If ${\sin }^x\left(a\right)+{\cos }^x\left(a\right)\ge 1$, $0<a<\frac{\pi }{2}$, then

• A. $x\in [2,\infty )$
• B. $x\in \left(-\infty ,2\right)$
• C. $x\in [-1,1)$
• D. None of these

Answer 1

The correct option is B

$x\in \left(-\infty ,2\right)$

Explanation for the correct option

Step 1: Simplification for the solution

As per the given inequality, we have

${\sin }^x\left(a\right)+{\cos }^x\left(a\right)\ge 1$

We know that ${\sin }^2\left(a\right)+{\cos }^2\left(a\right)=1$,

$$\begin{gathered} \therefore {\sin }^x\left(a\right)+{\cos }^x\left(a\right)\ge {\sin }^2\left(a\right)+{\cos }^2\left(a\right) \\ \Rightarrow {\sin }^x\left(a\right)-{\sin }^2\left(a\right)+{\cos }^x\left(a\right)-{\cos }^2\left(a\right)\ge 1 \\ \Rightarrow {\sin }^2\left(a\right)\left({\sin }^{x-2}\left(a\right)-1\right)+{\cos }^2\left(a\right)\left({\cos }^{x-2}\left(a\right)-1\right)\ge 1 \end{gathered}$$

Let us call this LHS of the inequality be $Y$ then $Y\ge 1$.

Step 2: Analysis for the range of values of $x$

Since $Y\ge 1$, then we know that

$\left({\sin }^{x-2}\left(a\right)-1\right)>0$ and $\left({\cos }^{x-2}\left(a\right)-1\right)>0$

This is only possible when $x-2<0$, from this we have,

$x<2$

This shows that $x\in \left(-\infty ,2\right)$.

Hence, the correct option is (B).