Question

The number of real solution of $\sin e^x.\cos e^x=2^{x-2}+2^{-x-2}$is: -

• A. $0$
• B. $1$
• C. $2$
• D. infinite

Answer 1

The correct option is A $0$

$\sin e^x\cos e^x=\frac{2^x}{2^2}+\frac{2^{-x}}{2^2}$

$\Rightarrow 2^2\sin e^x\cos e^x=2^x+2^{-x}$

$\Rightarrow 2\sin 2e^x=2^x+2^{-x}$

Consider $2^x>0,2^{-x}>0$

Using AM-GM inequality,$AM\ge GM$

$\frac{2^x+2^{-x}}{2}\ge {\left(2^x{2}^{-x}\right)}^{\frac{1}{2}}$

$\Rightarrow \frac{2^x+2^{-x}}{2}\ge {\left(1\right)}^{\frac{1}{2}}$

$\Rightarrow \frac{2^x+2^{-x}}{2}\ge 1$

$\Rightarrow 2^x+2^{-x}\ge 2$

$\therefore$Minimum value of $2^x+2^{-x}$ is $2$

Consider $-1\le \sin 2e^x\le 1$

or $-2\le 2\sin 2e^x\le 2$

Equality holds when both the expression$=2$

$2^x+2^{-x}=2$ when $x=0$

$2\sin 2e^x=2$ when $\sin 2e^x=1$

When $x=0\Rightarrow \sin 2\ne 1$

When $x=1\Rightarrow 2^x+2^{-x}\ne 2$

and $2\sin 2e\ne 1$

When $x=2\Rightarrow 2^x+2^{-x}\ne 2$

and $2\sin 2e^2\ne 1$

When $x=\infty \Rightarrow$ both the LHS and RHS are not defined.

Hence number of solutions$=0$