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CAT 2018 Slot 2 — QA Question 31

Basics (Functions)Easy

Let f(x) = max {5x, 52 – 2x2}, where x is any positive real number. Then the minimum possible value of f(x) is

Answer & solution

Correct answer: 20

Answer: 20

Solution

Easy

f(x)f(x) takes the larger of an increasing line 5x5x and a decreasing (for x>0x>0) downward parabola 52−2x252-2x^2. The pointwise maximum of a rising and a falling curve is minimized exactly where they cross. Find that intersection and evaluate.

1

Why the minimum is at the crossing. For small xx, 52−2x252-2x^2 dominates and ff decreases; for large xx, 5x5x dominates and ff increases. So ff bottoms out where the two pieces are equal.

5x=52−2x2\begin{aligned} &5x = 52 - 2x^2 \end{aligned}
2

Solve for the positive xx. Rearrange into a quadratic and factor.

2x2+5x−52=0⇒ (x−4)(2x+13)=0⇒ x=4(reject x=−132, need x>0)\begin{aligned} &2x^2 + 5x - 52 = 0\\ &\Rightarrow\ (x-4)(2x+13) = 0\\ &\Rightarrow\ x = 4 \quad\text{(reject } x=-\tfrac{13}{2}\text{, need }x>0\text{)} \end{aligned}
3

Evaluate ff at x=4x=4. Both pieces agree there (from step 1).

f(4)=5(4)=20⇒ check: 52−2(4)2=52−32=20 ✓\begin{aligned} &f(4) = 5(4) = 20\\ &\Rightarrow\ \text{check: } 52 - 2(4)^2 = 52 - 32 = 20\ \checkmark \end{aligned}
min⁡f(x)=20\min f(x) = \textbf{20}

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CAT 2018 Slot 2 QA Q31: Let f(x) = max {5x, 52 – 2x 2 }, where x is any positive real number. Then the minimum possible value of — Solution | TheCATExam