We have f (x) = $\frac{x^{100}}{100}+\frac{x^{99}}{99}$ + ... + $\frac{x^{2}}{2}$ + x + 1. ... (i) Differentiating (i) with respect to x we get f' (x) = $\frac{100x^{99}}{100}+\frac{99x^{98}}{99}$ + ... + $\frac{2x}{2}$ + 1 ⇒ f' (x) = $1^{99}+1^{98}$ + ... + x + 1 At x = 1 f' (1) = $1^{99}+1^{98}$ + ... + 1 + 1 = 100 and f' (0) = 0 + 0 + ... + 0 + 1 = 1 Hence f ′(1) = 100 f ′(0).