Question

If ${\sin }^x\alpha +{\cos }^x\alpha \ge 1,0<\alpha <\frac{\pi }{2},$ then

• A. $x\in [2,+\infty )$
• B. $x\in (-\infty ,2]$
• C. $x\in [-1,1]$
• D. none of these

Answer 1

The correct option is B $x\in (-\infty ,2]$

${\sin }^x\alpha +{\cos }^x\alpha \ge 1$ , $0<\alpha <\frac{\pi }{2}$

We know $-1\le \sin \alpha \le 1$ and $-1\le \cos \alpha \le 1$.

It is also known ${\sin }^{2}x+{\cos }^{2}x=1$

So, $x=2$ is either lower or upper bound.

In $0<\alpha <\frac{\pi }{2},1>\sin \alpha >0$ and $1>\cos \alpha >0$

So, if ${\sin }^x\alpha +{\cos }^x>1$, then

either ${\sin }^x\alpha >1$ or ${\cos }^x\alpha >1$

or both ${\sin }^x\alpha >1$ and ${\cos }^x\alpha >1$

So, $x$ must be negative, then ${\sin }^x\alpha$ and ${\cos }^x\alpha$ will be $\frac{1}{y}$ where $0<y<1$.

So, $\frac{1}{y}>1$

This is only possible for all values $<2$.

So, $x\in (-\infty ,2]$