Question

Let f(x)=\(\frac{x^{100}}{100}+\frac{x^{101}}{101}+…\infty\). Find a point c ∈ (- ∞, ∞) such that f'(c) = 0

(a) 1

(b) 2

(c) 0

(d) -1

I got this question in an interview.

Question is taken from Lagrange’s Mean Value Theorem topic in division Differential Calculus of Engineering Mathematics

Answer 1

Right answer is (c) 0

Easiest explanation: f'(x) = x^99+x^100+x^101+…∞

Using geometric series we have

f'(x)=\(\frac{x^{99}}{1-x}\)

Equating to 0 we have

x^99 = 0

Observe that x = 0 satisfies the equation. Neglect other roots as they are complex roots.