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CAT 2020 Slot 2 — QA Question 26

Composite FunctionsEasy

Let f(x) = x² + ax + b and g(x) = f(x + 1) – f(x – 1). If f(x) ≥ 0 for all real x, and g(20) = 72, then the smallest possible value of b is

Answer & solution

  • A

    16

  • B

    1

  • 4

  • D

    0

Solution

Easy

Expand g(x)=f(x+1)−f(x−1)g(x)=f(x+1)-f(x-1); the quadratic terms cancel, leaving a linear function whose value at 2020 pins down aa. Then f(x)≥0f(x)\ge 0 for all real xx forces a non-positive discriminant, which bounds bb from below.

1

Simplify g(x)g(x). With f(x)=x2+ax+bf(x)=x^2+ax+b:

g(x)=[(x+1)2+a(x+1)+b]−[(x−1)2+a(x−1)+b]⇒ g(x)=[(x+1)2−(x−1)2]+a[(x+1)−(x−1)]⇒ g(x)=4x+2a(b and x2 cancel)\begin{aligned} &g(x)=\big[(x+1)^2+a(x+1)+b\big]-\big[(x-1)^2+a(x-1)+b\big]\\ &\Rightarrow\ g(x)=\big[(x+1)^2-(x-1)^2\big]+a\big[(x+1)-(x-1)\big]\\ &\Rightarrow\ g(x)=4x+2a \quad\text{(}b\text{ and }x^2\text{ cancel)} \end{aligned}
2

Use g(20)=72g(20)=72 to get aa. Substitute x=20x=20 into step 1:

4(20)+2a=72⇒ 80+2a=72⇒ a=−4\begin{aligned} &4(20)+2a=72\\ &\Rightarrow\ 80+2a=72\\ &\Rightarrow\ a=-4 \end{aligned}
3

Apply f(x)≥0f(x)\ge 0 for all real xx. With a=−4a=-4, f(x)=x2−4x+bf(x)=x^2-4x+b. An upward parabola is non-negative everywhere exactly when its discriminant is ≤0\le 0:

(−4)2−4(1)(b)≤0⇒ 16−4b≤0⇒ b≥4\begin{aligned} &(-4)^2-4(1)(b)\le 0\\ &\Rightarrow\ 16-4b\le 0\\ &\Rightarrow\ b\ge 4 \end{aligned}
4

Smallest bb. The lower bound from step 3 is attainable, so the least value is b=4b=4.

bmin⁡=4\begin{aligned} &b_{\min}=4 \end{aligned}
b=4b=4
CAT 2020 Slot 2 QA Q26: Let f(x) = x² + ax + b and g(x) = f(x + 1) – f(x – 1). If f(x) ≥ 0 for all real x, and g(2 — Solution | TheCATExam