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CAT 2022 Slot 3 — QA Question 1

Inequality Maximization / MinimizationEasy

The minimum possibe value of x2-6x+103-x, for x < 3, is

Answer & solution

  • A

    -2

  • B

    12

  • 2

  • D

    -12

Solution

Easy

The numerator x2−6x+10x^2-6x+10 is just (x−3)2+1(x-3)^2+1. Rewriting in terms of (3−x)(3-x) turns the whole expression into "a positive number plus its reciprocal", which is always at least 22.

1

Complete the square in the numerator:

x2−6x+10=(x−3)2+1x^2-6x+10=(x-3)^2+1
2

Split the fraction over 3−x3-x:

(x−3)2+13−x=(x−3)23−x+13−x=(3−x)+13−xsince (x−3)2=(3−x)2\begin{aligned} &\frac{(x-3)^2+1}{3-x}=\frac{(x-3)^2}{3-x}+\frac{1}{3-x}\\ &=(3-x)+\frac{1}{3-x} \quad\text{since }(x-3)^2=(3-x)^2 \end{aligned}
3

Apply AM–GM. For x<3x<3 we have 3−x>03-x>0, so it is a positive number plus its reciprocal:

(3−x)+13−x≥2(3−x)⋅13−x=2(3-x)+\frac{1}{3-x}\ge 2\sqrt{(3-x)\cdot\frac{1}{3-x}}=2

Equality holds when 3−x=13-x=1, i.e. x=2x=2.

Minimum value=2\text{Minimum value}=\mathbf{2}

Any time you see t+1tt+\dfrac1t with t>0t>0, the minimum is 22 — no calculus needed.

CAT 2022 Slot 3 QA Q1: The minimum possibe value of x 2 - 6 x + 10 3 - x , for x < 3, is — Solution | TheCATExam