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
- A no real solution
- B exactly one real solution
- C exactly two real solutions
- D infinite number of real solutions
Answer & Solution
Correct Answer
(B) exactly one real solution
Step-by-step Solution
Detailed explanation
Given, \(f(x)=\frac{x}{1 !}+\frac{3}{2 !} x^{2}+\frac{7}{3 !} x^{3}+\frac{15}{4 !} x^{4}+\ldots\) \(=\frac{\left(2^{1}-1\right) x}{1 !}+\frac{\left(2^{2}-1\right) x^{2}}{2 !}+\frac{\left(2^{3}-1\right) x^{3}}{3 !}\) \(+\frac{\left(2^{4}-1\right)}{4 !} x^{4}+\ldots\)…
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