Question

The number of solutions of the equation \(\sin(e^x)=5+x-5^x\), is

• A. 0
• B. 1
• C. 2
• D. infinitely many

Hint:

Always compare ranges: \(\sin(\cdot)\) lies in \([-1,1]\). If RHS does not lie in this range for any \(x\), there are no real solutions.

Answer 1

The Correct Option is A

Step 1: Range of LHS.
\[ \sin(e^x) \] Since sine function always satisfies: 
\[ -1 \le \sin(e^x) \le 1 \] Step 2: Analyze RHS: \(5+x-5^x\). 
Let: 
\[ f(x)=5+x-5^x \] We check its values. 
Step 3: Show RHS is always greater than 1 or less than -1. 
At \(x=0\): 
\[ f(0)=5+0-1=4 \] which is \(>1\). 
At \(x=1\): 
\[ f(1)=5+1-5=1 \] But LHS becomes \(\sin(e)\), and \(\sin(e)\ne 1\). 
For \(x>1\), \(5^x\) grows very rapidly, so \(f(x)\) becomes negative large. 
For \(x<0\), \(5^x\) becomes small but \(5+x\) stays near 5, so \(f(x)\) stays \(>1\). 
Thus, the equation cannot satisfy the bounded LHS in \([-1,1]\) for any real \(x\). 
Final Answer: \[ \boxed{0} \]