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CAT 2022 Slot 2 — QA Question 11

Graph & Maximum or Minimum value of Quadratic functionEasy

Let f(x) be a quadratic ploynomial in x such that f(x) ≥ 0 for all real numbers x. If f(2) = 0 and f(4) = 6, then f(-2) is equal to

Answer & solution

Correct answer: 24

  • 24

  • B

    6

  • C

    36

  • D

    12

Solution

Easy

f(x)≥0f(x)\ge0 for all xx with f(2)=0f(2)=0 means x=2x=2 is a repeated root — the parabola just touches the axis there. So f(x)=a(x−2)2f(x)=a(x-2)^2 with a>0a>0. One more data point fixes aa, and then f(−2)f(-2) is immediate.

1

Double-root form. A non-negative quadratic touching zero at x=2x=2 is

f(x)=a(x−2)2,a>0.f(x)=a(x-2)^2,\qquad a>0.
2

Use f(4)=6f(4)=6 to find aa:

a(4−2)2=6  ⇒  4a=6  ⇒  a=32.a(4-2)^2=6 \;\Rightarrow\; 4a=6 \;\Rightarrow\; a=\frac32.
3

Evaluate f(−2)f(-2):

f(−2)=32(−2−2)2=32⋅16=24.f(-2)=\frac32(-2-2)^2=\frac32\cdot16=24.

f(−2)=24f(-2)=\mathbf{24} — option (a).

By symmetry about x=2x=2: f(−2)=f(6)f(-2)=f(6). Distances from the vertex are 2,4,62,4,6 at x=4,6,−2x=4,6,-2, and f∝(distance)2f\propto(\text{distance})^2, so f(6)=f(4)⋅4222=6⋅4=24f(6)=f(4)\cdot\frac{4^2}{2^2}=6\cdot4=24.

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CAT 2022 Slot 2 QA Q11 — Graph & Maximum or Minimum value of Quadratic function solution | TheCATExam