Question

If $|{\sin }^{2}x+17-x^2|=|16-x^2|+2{\sin }^{2}x+{\cos }^{2}x$ then subsets of solution are?

• A. $\left\{0\right\}$
• B. $[-4,4]$
• C. $[-8,8]$
• D. $[-\sqrt{17},\sqrt{17}]$

Answer 1

The correct option is B $[-4,4]$

Given:

$|{\sin }^{2}x+17-x^2|=|16-x^2|+2{\sin }^{2}x+{\cos }^{2}x$

$\Rightarrow |{\sin }^{2}x+17-x^2|=|16-x^2|+{\sin }^{2}x+{\sin }^{2}x+{\cos }^{2}x$

$\Rightarrow |{\sin }^{2}x+17-x^2|=|16-x^2|+{\sin }^{2}x+1$

We know that $-1\le \sin x\le 1$

$\Rightarrow 0\le {\sin }^{2}x\le 1$

If ${\sin }^{2}x=1\Rightarrow |{\sin }^{2}x+17-x^2|=|16-x^2|+{\sin }^{2}x+1$

$\Rightarrow |{\sin }^{2}x+17-x^2|=16-x^2+1+1$

$\Rightarrow |{\sin }^{2}x+17-x^2|=|18-x^2|$

Case$:1$If $x<-\sqrt{17}$

$\Rightarrow -{\sin }^{2}x-17+x^2=-16+x^2+1+{\sin }^{2}x$

$\Rightarrow -2{\sin }^{2}x=-15+17=2$

$\Rightarrow {\sin }^{2}x=-1$ is not possible since $0\le {\sin }^{2}x\le 1$

Case$:2$Let $-\sqrt{17}\le x<-4$

$\Rightarrow {\sin }^{2}x+17-x^2=-16+x^2+1+{\sin }^{2}x$

$\Rightarrow -2x^2=-15-17=-32$

$\Rightarrow x^2=16$

$\therefore x=\pm 4$ is not a valid solution since $-\sqrt{17}\le x<-4$

Case$:3$Let $-4\le x\le 4$

$\Rightarrow {\sin }^{2}x+17-x^2=16-x^2+1+{\sin }^{2}x$

$\Rightarrow 17=17$ is a solution

$\therefore -4\le x\le 4$ is a solution set.

Case$:4$Let $\sqrt{17}<x>4$

$\Rightarrow {\sin }^{2}x+17-x^2=-16+x^2+1+{\sin }^{2}x$

$\Rightarrow -2x^2=-15-17=-32$

$\Rightarrow x^2=16$

$\Rightarrow x=\pm 4$ is not a valid solution $\because \sqrt{17}<x>4$

Case$:5$Let $x\ge \sqrt{17}$

$\Rightarrow -{\sin }^{2}x-17+x^2=-16+x^2+1+{\sin }^{2}x$

$\Rightarrow -2{\sin }^{2}x-17=-16+1$

$\Rightarrow -2{\sin }^{2}x=-16+1+17=2$

$\Rightarrow {\sin }^{2}x=-1$ is not a valid solution $\because 0\le {\sin }^{2}x\le 1$

$\therefore$ the only possible solution is $x\in [-4,4]$