Question

The equation
$2{\cos }^2\frac{x}{2}{\sin }^{2}x=x^2+x^{-2};0<x\le \frac{\pi }{2}$
has

• A. No real solution
• B. One real solution
• C. more than one solution
• D. none of these

Answer 1

The correct option is A No real solution

The given equation is
$2{\cos }^2\left(\frac{x}{2}\right){\sin }^{2}x=x^2+\frac{1}{x^2}$
where $0<x\le \frac{\pi }{2}$
L.H.S.=$2{\cos }^2\left(\frac{x}{2}\right){\sin }^{2}x=(1+\cos x){\sin }^{2}x\because 1+\cos x<2\text{and}{\sin }^{2}x\le 1\text{for}0<x\le \frac{\pi }{2}\therefore (1+\cos x){\sin }^{2}x<2$
and R.H.S.=$x^2+\frac{1}{x^2}\ge 2$
Therefore, for $0<x\le \frac{\pi }{2}$, the given equation is not possible for any real value of x.