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WBJEE · Maths · Quadratic Equation

For every real number \(x\), let \(f(x)=\frac{x}{1 !}+\frac{3}{2 !} x^{2}+\frac{7}{3 !} x^{3}+\frac{15}{4 !} x^{4}+\ldots .\) Then the equation \(f(x)=0\) has

  1. A no real solution
  2. B exactly one real solution
  3. C exactly two real solutions
  4. D infinite number of real solutions
Verified Solution

Answer & Solution

Correct Answer

(B) exactly one real solution

Step-by-step Solution

Detailed explanation

Given, \(f(x)=\frac{x}{1 !}+\frac{3}{2 !} x^{2}+\frac{7}{3 !} x^{3}+\frac{15}{4 !} x^{4}+\ldots\) \(=\frac{\left(2^{1}-1\right) x}{1 !}+\frac{\left(2^{2}-1\right) x^{2}}{2 !}+\frac{\left(2^{3}-1\right) x^{3}}{3 !}\) \(+\frac{\left(2^{4}-1\right)}{4 !} x^{4}+\ldots\)…