Question

The equation $2{\cos ^2}\dfrac{x}{2}{\sin ^2}x = {x^2} + {x^{ - 2}}$, $0 < x \leqslant \dfrac{\pi }{2}$ has
A) No real solution
B) One real solution
C) More than one solution
D) None of these

Answer 1

In the given problem, the left hand side can be simplified by using properties of sine and cosine functions, in such a way that the range of the left hand side within the given interval can be easily determined. And in the right hand side, the basic properties of algebra can be applied to get a range of the function on the right hand side. Then, by comparing the range of the left hand side and of the right hand side, we can determine how many solutions the following equation has.

Complete step by step answer:
Given equation is, $2{\cos ^2}\dfrac{x}{2}{\sin ^2}x = {x^2} + {x^{ - 2}}$, $0 < x \leqslant \dfrac{\pi }{2}$.
Now, we know, $\cos \theta = 2{\cos ^2}\dfrac{\theta }{2} - 1$
$\Rightarrow \cos \theta + 1 = 2{\cos ^2}\dfrac{\theta }{2}$
Now, using this property we get,
$\left( {1 + \cos x} \right){\sin ^2}x = {x^2} + \dfrac{1}{{{x^2}}} - - - \left( 1 \right)$
Now, working over the left hand side,
Now, given, $0 < x \leqslant \dfrac{\pi }{2}$
Since, $\cos x$ is decreasing in the interval $\left[ {0,\dfrac{\pi }{2}} \right]$.
Therefore, inequality sign changes it’s direction on taking cosine over the inequality.
i.e., $\cos \dfrac{\pi }{2} \leqslant \cos x < \cos 0$
$\Rightarrow 0 \leqslant \cos x < 1$
Now, adding $1$ over the inequality, we get,
$\Rightarrow 1 \leqslant \cos x + 1 < 2 - - - \left( 2 \right)$
We know, $- 1 \leqslant \sin x \leqslant 1$
Now, squaring over the inequality, we get,
$0 \leqslant {\sin ^2}x \leqslant 1 - - - \left( 3 \right)$
Now, multiplying $\left( 2 \right)$ and $\left( 3 \right)$, we get,
$0 \leqslant \left( {\cos x + 1} \right){\sin ^2}x \leqslant 2 - - - \left( 4 \right)$
Now, working over the right hand side,
${x^2} + \dfrac{1}{{{x^2}}}$
$= {x^2} + \dfrac{1}{{{x^2}}} - 2.x.\dfrac{1}{x} + 2.x.\dfrac{1}{x}$
$= {\left( {x - \dfrac{1}{x}} \right)^2} + 2$
Clearly, $0 \leqslant {\left( {x - \dfrac{1}{x}} \right)^2} \leqslant \infty$
Now, adding $2$ over the inequality, we get,
$\Rightarrow 2 \leqslant {\left( {x - \dfrac{1}{x}} \right)^2} + 2 \leqslant \infty + 2$
[Since, $\infty + 2 = \infty$]
$\Rightarrow 2 \leqslant {x^2} + \dfrac{1}{{{x^2}}} \leqslant \infty - - - \left( 5 \right)$
From $\left( 4 \right)$ and $\left( 5 \right)$, we can clearly see that the only real solution is $2$.
Therefore, the given equation has only one real solution, i.e., option (B).

Note:
Comparing equations over inequality is a very basic and effective way of solving problems. As done over equal to, every operation can be worked over with inequality, but only one thing is to be kept in mind, the sign and direction of the inequality changes when the reciprocal of the inequality is taken or it is multiplied with some negative value.