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CAT 2017 Slot 2 — QA Question 32

Basics (Functions)Easy

Let f(x) = 2x – 5 and g(x) = 7 – 2x. Then |f(x) + g(x)| = |f(x)| + |g(x)| if and only if

Answer & solution

Correct answer: 5 2 ≤ x ≤ 7 2

  • A

    52<x<72

  • B

    x≤52 or x≥72

  • C

    x<52 or x≥72

  • 52≤x≤72

Solution

Easy

The identity ∣p+q∣=∣p∣+∣q∣|p+q|=|p|+|q| holds exactly when pp and qq have the same sign (or either is zero), i.e. when p q≥0p\,q\ge 0. Apply this to f(x)f(x) and g(x)g(x).

1

State the condition. ∣f(x)+g(x)∣=∣f(x)∣+∣g(x)∣|f(x)+g(x)|=|f(x)|+|g(x)| holds if and only if f(x)f(x) and g(x)g(x) do not have opposite signs:

f(x) g(x)≥0\begin{aligned} &f(x)\,g(x)\ge 0 \end{aligned}
2

Case both ≥0\ge 0. With f(x)=2x−5f(x)=2x-5 and g(x)=7−2xg(x)=7-2x:

2x−5≥0 ⇒ x≥527−2x≥0 ⇒ x≤72⇒ 52≤x≤72\begin{aligned} &2x-5\ge 0\ \Rightarrow\ x\ge \tfrac52\\ &7-2x\ge 0\ \Rightarrow\ x\le \tfrac72\\ &\Rightarrow\ \tfrac52\le x\le \tfrac72 \end{aligned}
3

Case both ≤0\le 0.

2x−5≤0 ⇒ x≤527−2x≤0 ⇒ x≥72\begin{aligned} &2x-5\le 0\ \Rightarrow\ x\le \tfrac52\\ &7-2x\le 0\ \Rightarrow\ x\ge \tfrac72 \end{aligned}

This requires x≥72x\ge\tfrac72 and x≤52x\le\tfrac52 simultaneously, which is impossible — no solution here.

4

Combine. Only the first case contributes, so the condition is 52≤x≤72\dfrac52\le x\le\dfrac72, which is option (d).

52≤x≤72\dfrac{5}{2}\le x\le \dfrac{7}{2}

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CAT 2017 Slot 2 QA Q32 — Basics (Functions) solution | TheCATExam