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PEAMCQEasy-Moderate

2024 PEA Q4

Let f:R→Rf:\mathbb{R}\to\mathbb{R} be defined as

f(x)={cx2+ax+b,x<0,bx2+cx+a,0≤x<2,ax2+bx+c,x≥2, f(x)=\begin{cases} cx^{2}+ax+b, & x<0,\\ bx^{2}+cx+a, & 0\le x<2,\\ ax^{2}+bx+c, & x\ge 2, \end{cases}

where a,b,ca,b,c are positive real numbers. Which of the following statements is correct, under the assumption that ff is continuous?

Reveal answer and solution

Answer

D

Solution

  1. 1

    At x=0x=0: left-hand value =b=b; right-hand value =a=a. Hence a=ba=b.

  2. 2

    At x=2x=2: left-hand value =4b+2c+a= 4b+2c+a; right-hand value =4a+2b+c=4a+2b+c. Using a=ba=b:

  3. 3
    4a+2c+a=5a+2c,4a+2a+c=6a+c. 4a+2c+a=5a+2c,\qquad 4a+2a+c=6a+c.
  4. 4

    Equating: 5a+2c=6a+c⇒c=a5a+2c=6a+c \Rightarrow c=a. Therefore a=b=ca=b=c.

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.