Question

Number of ordered pairs $(x,y)$ which satisfies $\frac{x^4+1}{8x^2}={\sin }^{2}y{\cos }^{2}y;$ where $y\in [0,2\pi ]$ is

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is D $8$

$\frac{x^4+1}{8x^2}={\sin }^{2}y{\cos }^{2}y\Rightarrow \frac{x^4+1}{x^2}=2(2\sin y\cos y{)}^2\Rightarrow x^2+\frac{1}{x^2}=2{\sin }^{2}2yL.H.S=x^2+\frac{1}{x^2}\ge 2(\because a+b\ge 2\sqrt{ab})R.H.S=2{\sin }^{2}2y\le 2$
For equation to have solution
$L.H.S=R.H.S=2$

$L.H.S=2\Rightarrow x^2+\frac{1}{x^2}=2\Rightarrow x^4-2x^2+1=0\Rightarrow (x^2-1{)}^2=0\Rightarrow x=\pm 1$

$R.H.S=2\Rightarrow {\sin }^{2}2y=1={\sin }^2\frac{\pi }{2}\Rightarrow 2y=n\pi \pm \frac{\pi }{2},n\in Z\Rightarrow y=\frac{n\pi }{2}\pm \frac{\pi }{4}$
In $y\in [0,2\pi ],y=\{\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}\}$

Hence, number of possible ordered pairs for $(x,y)=8$