Question

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

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

Answer 1

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

$si{n}^x\alpha +co{s}^x\alpha ⩾1,0<\alpha <\frac{\pi }{2}$
$si{n}^x\alpha +co{s}^x\alpha ⩾si{n}^2\alpha +co{s}^2\alpha$
$si{n}^x\alpha -si{n}^2\alpha )+(co{s}^x\alpha -co{s}^2\alpha )⩾0$
$f\left(x\right)=si{n}^2\alpha (si{n}^{x-2}\alpha -1)+co{s}^2\alpha (co{s}^{x-2}\alpha -1)⩾0$
But both $si{n}^a\alpha ,co{s}^a\alpha <1$ for a>0 when $0<a<\frac{\pi }{2}$
Here if both $(si{n}^{x-2}\alpha -1)$ and $(co{s}^{x-2}\alpha -1)$ are negative , $f\left(x\right)<0$
So,$(si{n}^{x-2}\alpha -1)$ and $(co{s}^{x-2}\alpha -1)$ are >0
only possible if $x-2<0$
=>$x<2$
$xϵ(-\infty ,2)$
Hence option B is correct