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2024 PEA Q5

Consider f:[0,1]→[0,1]f:[0,1]\to[0,1] such that f(x)=x2−xf(x)=\dfrac{x}{2-x}. Which of the following statements is incorrect?

Reveal answer and solution

Answer

C

Solution

  1. 1

    A. f(0)=0f(0)=0 and f(1)=1/1=1f(1)=1/1=1. True.

  2. 2

    B. Compute 1−f(x)=1−x2−x=2−2x2−x1-f(x) = 1-\tfrac{x}{2-x}=\tfrac{2-2x}{2-x}. Then

  3. 3
    f ⁣(2−2x2−x)=2−2x2−x2−2−2x2−x=2−2x2(2−x)−(2−2x)=2−2x2=1−x. True. f\!\left(\tfrac{2-2x}{2-x}\right) =\frac{\frac{2-2x}{2-x}}{2-\frac{2-2x}{2-x}} =\frac{2-2x}{2(2-x)-(2-2x)} =\frac{2-2x}{2}=1-x. \text{ True.}
  4. 4

    C. Differentiate:

  5. 5
    f′(x)=(2−x)−x(−1)(2−x)2=2(2−x)2>0,f′′(x)=4(2−x)3>0  on  (0,1). f'(x)=\frac{(2-x)-x(-1)}{(2-x)^2}=\frac{2}{(2-x)^2}>0,\qquad f''(x)=\frac{4}{(2-x)^3}>0\;\text{on}\;(0,1).
  6. 6

    Hence ff is strictly convex, not concave. Incorrect statement.

  7. 7

    D. f′(x)>0f'(x)>0, so strictly increasing. True.

Answer structure / marking notes

Sign of second derivative; do not confuse convex with concave.

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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.