Question

The equation $2{\cos }^2\frac{x}{2}{\sin }^{2}x=x^2+\frac{1}{x^2},0\le x\le \frac{\pi }{2}$ has

• A. one real solution
• B. no real solution
• C. more than one real solution
• D. None of these

Answer 1

Answer: (B)

We have for any real $x,{\cos }^2\frac{x}{2}\le 1$ and ${\sin }^{2}x\le 1$
$\therefore 2{\cos }^2\frac{x}{2}{\sin }^{2}x\le 2$
$\Rightarrow$ LHS $\le 2$
Again for any real $x$
$x^2+\frac{1}{x^2}=x^2+\frac{1}{x^2}-2+2={\left(x-\frac{1}{x}\right)}^2+2\ge 2$
$\therefore RHS\ge 2$ Thus, the given equation can have solution only if $LHS=RHS=2$
$\Rightarrow 2{\cos }^2\frac{x}{2}{\sin }^{2}x=x^2+\frac{1}{x^2}=2$
Now, $x^2+\frac{1}{x^2}=2$
$\Rightarrow x=\pm 1$
but for $x=\pm 1,{\cos }^2\frac{x}{2}{\sin }^{2}x\ne 1$
That is, LHS and RHS cannot be simultaneously equal to 2 for any value of $x$.
$\therefore$ The equation has no real solution.