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2024 PEA Q1

If f:R→Rf:\mathbb{R}\to\mathbb{R} is a differentiable function at a∈Ra\in\mathbb{R} such that f′(a)=af(a)f'(a)=af(a), then what is

lim⁡x→axf(a)−af(x)x−a ? \lim_{x\to a}\frac{xf(a)-af(x)}{x-a}\, ?
Reveal answer and solution

Answer

C

Solution

  1. 1

    At x=ax=a, the numerator equals af(a)−af(a)=0af(a)-af(a)=0, so the limit is of the form 0/00/0. Write

  2. 2
    xf(a)−af(x)x−a=(x−a)f(a)+af(a)−af(x)x−a=f(a)−a⋅f(x)−f(a)x−a. \frac{xf(a)-af(x)}{x-a} = \frac{(x-a)f(a) + af(a) - af(x)}{x-a} = f(a) - a\cdot\frac{f(x)-f(a)}{x-a}.
  3. 3

    Taking x→ax\to a and using differentiability,

  4. 4
    lim⁡x→axf(a)−af(x)x−a=f(a)−a f′(a)=f(a)−a⋅af(a)=(1−a2)f(a). \lim_{x\to a}\frac{xf(a)-af(x)}{x-a} = f(a) - a\,f'(a) = f(a) - a\cdot af(a) = (1-a^{2})f(a).

Answer structure / marking notes

No major trap beyond standard calculation care.

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Content note

Imported from public/resources/isi/msqe/solutions/pea/2024/ISI_MSQE_PEA_2024_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.