Question

Assertion (A) : The number of real roots of the equation $\sin 2^{x} \cos 2^{x}=\frac{2^{x}+2^{-x}}{4}$ is 2
Reason (R) : $ A . M . \geq G$. $M$

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A)
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A)
• C. (A) is true but (R) is false
• D. (A) is false but (R) is true

Answer 1

Answer: (D)

Given equation is $4 \sin 2^{x} \cos 2^{x}=2^{x}+2^{-x}$
$\Rightarrow 2\left(2 \sin 2^{x} \cdot \cos 2^{x}\right)=2^{x}+2^{-x}$
$\Rightarrow 2 \sin \left(2^{x+1}\right)=2^{x}+2^{-x} \,\,...(A)$
(i) For $x=0,$ we have $2 \sin 2=2$
$\Rightarrow \sin 2=1,$ which is not possible
$\therefore x=0$ is not the solution of the equation
(ii) If $x \neq 0$ $\because \mathrm{A.M.} \geq \mathrm{G.M.}$
$\therefore \quad \frac{2^{x}+2^{-x}}{2} > \sqrt{2^{x} \cdot 2^{-x}} \,\,\,(\because x \ne 0)$
$\Rightarrow \frac{2^{x}+2^{-x}}{2}>1$
$\Rightarrow \sin 2^{x+1} >1 \,\,\,$ (From $(A))$
Which is not possible as $-1 \leq \sin x \leq 1$
Hence, the equation $\sin 2^{x} \cos 2^{x}=\frac{2^{x}+2^{-x}}{4}$ has no real
solution. So Assertion (A) is false but Reason (R) is true.