GATE 2023 IN Q.57: solution and answer

GATE 2023 · INQ.57 · 2 marks · NAT

Consider the real-valued function $g (x) = max {(x - 2)^{2}, - 2 x + 7}$ , where $x \in (- \infty, \infty)$ . The minimum value attained by $g (x)$ is ________ (rounded off to one decimal place).

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Answer and solution

ANSWER

The minimum value of $g (x)$ is $1.0$ . It occurs at $x = 3$ , where $(x - 2)^{2}$ and $- 2 x + 7$ are equal.

Official answer: 0.9 to 1.1Source: Official answer key, GATE 2023 IN (IIT Kanpur)

Where the two pieces cross. Set them equal: $(x - 2)^{2} = - 2 x + 7 \Rightarrow x^{2} - 4 x + 4 = - 2 x + 7 \Rightarrow x^{2} - 2 x - 3 = 0 \Rightarrow (x - 3) (x + 1) = 0$ . So they cross at $x = - 1$ and $x = 3$ .
Which piece is larger where. Take the difference $d (x) = (x - 2)^{2} - (- 2 x + 7) = x^{2} - 2 x - 3 = (x - 3) (x + 1)$ . It is $\leq 0$ on $[- 1, 3]$ and $> 0$ outside that interval. So the line $- 2 x + 7$ is the larger piece on $[- 1, 3]$ , and the parabola is larger everywhere else. (Check at $x = 2$ : $(x - 2)^{2} = 0$ and $- 2 x + 7 = 3$ .)
Write $g$ piecewise: $g (x) = ⎩ ⎨ ⎧ (x - 2)^{2}, - 2 x + 7, (x - 2)^{2}, x \leq - 1 - 1 \leq x \leq 3 x \geq 3$
Find the minimum of each piece on its own interval. On $[- 1, 3]$ , $- 2 x + 7$ is strictly decreasing, so its smallest value is at $x = 3$ : $- 2 (3) + 7 = 1$ . For $x \geq 3$ , $(x - 2)^{2}$ is increasing because its vertex $x = 2$ is to the left, so $g (x) \geq (3 - 2)^{2} = 1$ . For $x \leq - 1$ , $(x - 2)^{2}$ is decreasing on that interval, so $g (x) \geq (- 1 - 2)^{2} = 9$ .
Conclusion. The smallest of these values is $1$ . It occurs at the corner $x = 3$ , where both pieces equal $1$ . So $g_{m i n} = 1.0$ .

KEY IDEA

The graph of $max {f, h}$ is the upper boundary of the two curves. Its minimum is either at the lowest point of a piece inside that piece's own interval, or at a crossing point where the upper boundary switches from one curve to the other. Here the line is on top on $[- 1, 3]$ and keeps falling up to $x = 3$ . After $x = 3$ the parabola is on top and keeps rising. So the minimum is at the crossing $x = 3$ .

WATCH OUT FOR

Taking the vertex of the parabola, $x = 2$ , $g = 0$ , as the answer. At $x = 2$ the parabola is not the larger piece: the line gives $3$ , so $g (2) = 3$ . The opposite slip is to minimise $min {f, h}$ instead of $max {f, h}$ . The vertex value $0$ is only the minimum of the parabola alone, and $g$ never gets that low.

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GATE 2023 IN Q.57: solution and answer

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