Question

If $y={\cos }^2({45}^{\circ }+x)+(\sin x-\cos x{)}^2$, where $x\in (0,\frac{\pi }{2}]$, then the possible value(s) of $y$ is/are

• A. $0$
• B. $\frac{3}{2}$
• C. $3$
• D. $1$

Answer 1

Answer: (A), (B), (D)

$1$

${\cos }^2({45}^{\circ }+x)+(\sin x-\cos x{)}^2=(\cos {45}^{\circ }\cos x-\sin {45}^{\circ }\sin x{)}^2+(1-\sin 2x)$
$={[\frac{\cos x}{\sqrt{2}}-\frac{\sin x}{\sqrt{2}}]}^2+(1-\sin 2x)=\frac{(1-\sin 2x)}{2}+(1-\sin 2x)$
$=\frac{3}{2}(1-\sin 2x)$
As,
$0<x\le \frac{\pi }{2}\Rightarrow 0<2x\le \pi \Rightarrow 0\le \sin 2x\le 1\Rightarrow 0\ge -\sin 2x\ge -1\Rightarrow -1\le -\sin 2x\le 0\Rightarrow 0\le 1-\sin 2x\le 1\Rightarrow 0\le \frac{3}{2}(1-\sin 2x)\le \frac{3}{2}\therefore y\in [0,\frac{3}{2}]$