Q. The number of real solutions of $2\sin{\left( e^{x} \right) = 5^{x} + 5^{- x}}$ in [0, 1] is /are
A) 0
B) 1
C) 2
D) 4
Explanation
(a)
Since, $\frac{5^{x} + 5^{- x}}{2} \geq \sqrt{5^{x}.5^{- x}}$
$\Rightarrow \ 5^{x} + 5^{- x} \geq 2$
But $\sin{\left( e^{x} \right) \leq 1\ \Rightarrow 2\sin{(e^{x}) \leq 2}\ \ }$
At $x = 0,\ \ \ 5^{x} + 5^{- x} = 2$
Bit $2\sin{(e^{0}) \neq 2}$
Hence, no solution exist
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.