Question

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

• A. one real solution
• B. no solution
• C. more than one real solution
• D. Two solutions

Answer 1

The correct option is B no solution

$2{\cos }^2\left(\frac{x}{2}\right){\sin }^{2}x=2{\cos }^2\left(\frac{x}{2}\right){\left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)}^2=2{\cos }^2\left(\frac{x}{2}\right)\cdot 4{\sin }^2\frac{x}{2}{\cos }^2\left(\frac{x}{2}\right)=8{\cos }^4\left(\frac{x}{2}\right)\cdot {\sin }^2\frac{x}{2}Now,takex=\frac{\pi }{2}Then,8{\cos }^4\left(\frac{x}{2}\right)\cdot {\sin }^2\frac{x}{2}=8{\cos }^4\left(\frac{90°}{2}\right)\cdot {\sin }^2\frac{90°}{2}=8{\cos }^{4}45°\cdot {\sin }^{2}45°=8{\left(\frac{1}{\sqrt{2}}\right)}^4\cdot {\left(\frac{1}{\sqrt{2}}\right)}^2=8\times \frac{1}{2\times 2}\times \frac{1}{2}=1$

$\therefore 0\le x\le \frac{\pi }{2}$ which means it has no solution.