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) : $A.M.\ge G$. $M$

• 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 $x\ne 0$ $\because A.M.\ge G.M.$
$\therefore \frac{2^x+2^{-x}}{2}>\sqrt{2^x\cdot 2^{-x}}(\because x\ne 0)$
$\Rightarrow \frac{2^x+2^{-x}}{2}>1$
$\Rightarrow \sin 2^{x+1}>1$ (From $(A))$
Which is not possible as $-1\le \sin x\le 1$
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.