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CAT 2018 Slot 1 — QA Question 30

Basics (Functions)Easy

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

Answer & solution

Answer: 32

Solution

Easy

f(x)f(x) is the lower of two curves: 2x22x^2 (rising) and 52−5x52-5x (falling). The "min" follows the rising curve, then switches to the falling one; the peak of the lower envelope sits exactly where the two curves cross.

(4, 32) 2x² 52 − 5x
1

Locate where the two curves meet. For x>0x>0, 2x22x^2 increases and 52−5x52-5x decreases, so the minimum of the two is largest at their intersection.

2x2=52−5x⇒ 2x2+5x−52=0⇒ (x−4)(2x+13)=0⇒ x=4(reject x=−132, not positive)\begin{aligned} &2x^2 = 52-5x\\ &\Rightarrow\ 2x^2+5x-52 = 0\\ &\Rightarrow\ (x-4)(2x+13)=0\\ &\Rightarrow\ x = 4 \quad\text{(reject }x=-\tfrac{13}{2}\text{, not positive)} \end{aligned}
2

Evaluate ff at the crossing point x=4x=4.

f(4)=min⁡{ 2(4)2, 52−5(4) }⇒ f(4)=min⁡{32, 32}=32\begin{aligned} &f(4)=\min\{\,2(4)^2,\ 52-5(4)\,\}\\ &\Rightarrow\ f(4)=\min\{32,\ 32\}=32 \end{aligned}
max⁡f(x)=32\max f(x) = 32
CAT 2018 Slot 1 QA Q30: Let f(x) = min {2x 2 , 52 − 5x}, where x is any positive real number. Then the maximum possible value of — Solution | TheCATExam