GATE 2023 IN GA Q.9: solution and answer

GATE 2023 · INGA Q.9 · 2 marks · MCQ

Which one of the given figures P, Q, R and S represents the graph of the following function?

f (x) = ∣ ∣ x + 2∣ - ∣ x - 1∣ ∣

GATE 2023 IN GA question 9: Four graphs arranged in a 2 x 2 grid, each a thick blue piecewise-linear curve on axes labelled f(x) (vertical) and x… — Figure from the GATE 2023 IN paper. It is part of the question.

(A)
P
(B)
Q
(C)
R
(D)
S

Answer and solution

ANSWER

The graph of $f (x) = ∣ ∣ x + 2∣ - ∣ x - 1∣ ∣$ is figure P: flat at 3 for $x \leq - 2$ and for $x \geq 1$ , and $∣2 x + 1∣$ in between, touching 0 at $x = - \frac{1}{2}$ . The answer is option (A).

Official answer: ASource: Official answer key, GATE 2023 IN (IIT Kanpur)

The inner moduli change sign at $x = - 2$ (from $∣ x + 2∣$ ) and at $x = 1$ (from $∣ x - 1∣$ ). Work out $g (x) = ∣ x + 2∣ - ∣ x - 1∣$ on the three intervals these points create, then apply the outer modulus: $f (x) = ∣ g (x) ∣$ .
For $x \leq - 2$ : $x + 2 \leq 0$ and $x - 1 < 0$ , so $∣ x + 2∣ = - (x + 2)$ and $∣ x - 1∣ = 1 - x$ . Then $g (x) = - (x + 2) - (1 - x) = - 3$ , so $f (x) = ∣ - 3∣ = 3$ . On the left the graph is flat at height 3.
For $x \geq 1$ : both $x + 2$ and $x - 1$ are non-negative, so $g (x) = (x + 2) - (x - 1) = 3$ and $f (x) = 3$ . On the right the graph is also flat at height 3.
For $- 2 \leq x \leq 1$ : $x + 2 \geq 0$ and $x - 1 \leq 0$ , so $g (x) = (x + 2) - (1 - x) = 2 x + 1$ and $f (x) = ∣2 x + 1∣$ . It equals $∣ - 3∣ = 3$ at $x = - 2$ , drops to 0 at $x = - \frac{1}{2}$ and rises back to 3 at $x = 1$ .
Putting it together: $f (x) = ⎩ ⎨ ⎧ 3, ∣2 x + 1∣, 3, x \leq - 2 - 2 \leq x \leq 1 x \geq 1$ The graph never goes below 0, never goes above 3, and has a V touching 0 at $x = - \frac{1}{2}$ between two flat arms at height 3.
Now check the figures. Q goes down to $- 3$ on the left. That is $g (x)$ before the outer modulus is applied, and an absolute value can never be negative. R and S both reach about 9 at $x \approx - 5$ , but $f (- 5) = 3$ . Those left arms come from continuing the middle-interval formula past $x = - 2$ , and S also never dips to 0 at $x = - \frac{1}{2}$ . Only P has flat arms at 3 on both sides and a V touching 0 at $x = - 0.5$ . The answer is P, option (A).

KEY IDEA

Break a sum of moduli at the points where each inner expression is zero ( $x = - 2$ and $x = 1$ ), write the function piece by piece, and apply the outer modulus last. A quick check: $∣ x + 2∣ - ∣ x - 1∣$ is (distance from $x$ to $- 2$ ) minus (distance from $x$ to $1$ ). The triangle inequality keeps that difference between $- 3$ and $3$ , so $0 \leq f (x) \leq 3$ for every $x$ . That alone rules out Q (it goes negative) and R and S (they reach about 9), leaving only P.

WATCH OUT FOR

Forgetting the outer modulus gives the ramp from $- 3$ to $3$ , which is figure Q. Using the middle-interval formula $∣2 x + 1∣$ (or $- (2 x + 1)$ ) for all $x < - 2$ instead of the constant 3 produces the rising left arm of R or S. Always evaluate each piece only on its own interval.

From the GATE 2023 IN paper. Every question of this paper · Download the full paper · All 58 Numerical questions, with notes

GATE 2023 IN GA Q.9: solution and answer

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