Question

The equation $\sin x(\sin x+\cos x)=k$ has real solutions, where $k$ is a real number. Then

• A. $0\le k\le \frac{1+\sqrt{2}}{2}$
• B. $2-\sqrt{3}\le k\le 2+\sqrt{3}$
• C. $0\le k\le 2-\sqrt{3}$
• D. $\frac{1-\sqrt{2}}{2}\le k\le \frac{1+\sqrt{2}}{2}$

Answer 1

Answer: (D)

Let, $f(x)=\sin x(\sin x+\cos x)$
$={\sin }^{2}x+\sin x\cos x$
$=\frac{1-\cos 2x}{2}+\frac{2\sin x\cos x}{2}$
$=\frac{1}{2}-\frac{1}{2}\cos .2x+\frac{1}{2}\sin 2x$
$=\frac{1}{2}+\frac{1}{2}(\sin 2x-\cos 2x)$
Now, $-\sqrt{(1{)}^2+(1{)}^2}\le \sin 2x-\cos 2x$
$\le \sqrt{(1{)}^2+(1{)}^2}$
$\left[\because \sqrt{a^2+b^2}\le a\sin x+b\cos x\le \sqrt{a^2+b^2}\right]$
$\Rightarrow -\sqrt{2}\le \sin 2x-\cos 2x\le \sqrt{2}$
$\Rightarrow -\frac{\sqrt{2}}{2}\le \frac{\sin 2x-\cos 2x}{2}\le \frac{\sqrt{2}}{2}$
$\Rightarrow \frac{1}{2}-\frac{\sqrt{2}}{2}\le \frac{1}{2}+\frac{\sin 2x-\cos 2x}{2}\le \frac{1}{2}+\frac{\sqrt{2}}{2}$
$\Rightarrow \frac{1-\sqrt{2}}{2}\le f(x)\le \frac{1+\sqrt{2}}{2}$
$\therefore \frac{1-\sqrt{2}}{2}\le k\le \frac{1+\sqrt{2}}{2}$