GATE 2020 EE Q.26: solution and answer

GATE 2020 · EEQ.26 · 2 marks · MCQ

For real numbers, $x$ and $y$ , with $y = 3 x^{2} + 3 x + 1$ , the maximum and minimum value of $y$ for $x \in [- 2, 0]$ are respectively, ______.

(A)
7 and 1/4.
(B)
7 and 1.
(C)
$- 2$ and $- 1/2$ .
(D)
1 and 1/4.

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Answer and solution

ANSWER

The maximum value of $y$ on $[- 2, 0]$ is $7$ (at $x = - 2$ ) and the minimum is $\frac{1}{4}$ (at $x = - \frac{1}{2}$ ), so the answer is (A) 7 and 1/4.

Official answer: ASource: Official answer key, GATE 2020 EE (IIT Delhi)

On a closed interval, a continuous function reaches its absolute maximum and minimum either at a critical point inside the interval or at an endpoint (closed-interval method). So we need $y$ at every critical point in $[- 2, 0]$ and at $x = - 2$ and $x = 0$ .
Critical point: $\frac{d y}{d x} = 6 x + 3 = 0 \Rightarrow x = - \frac{1}{2}$ , and $- \frac{1}{2} \in [- 2, 0]$ . Since $\frac{d ^{2} y}{d x ^{2}} = 6 > 0$ , this point is a local minimum.
Value at the critical point: $y (- \frac{1}{2}) = 3 \cdot \frac{1}{4} - \frac{3}{2} + 1 = \frac{3}{4} - \frac{6}{4} + \frac{4}{4} = \frac{1}{4}$ .
Values at the endpoints: $y (- 2) = 3 (4) + 3 (- 2) + 1 = 12 - 6 + 1 = 7$ , and $y (0) = 0 + 0 + 1 = 1$ .
Comparing ${7, 1, \frac{1}{4}}$ : the maximum is $7$ (at $x = - 2$ ) and the minimum is $\frac{1}{4}$ (at $x = - \frac{1}{2}$ ).
Check by completing the square: $y = 3 (x^{2} + x) + 1 = 3 (x + \frac{1}{2})^{2} - \frac{3}{4} + 1 = 3 (x + \frac{1}{2})^{2} + \frac{1}{4}$ . The minimum is $\frac{1}{4}$ at the vertex $x = - \frac{1}{2}$ , and $y$ is largest at the point of the interval farthest from $x = - \frac{1}{2}$ , which is $x = - 2$ (distance $\frac{3}{2}$ , against $\frac{1}{2}$ for $x = 0$ ). So the maximum and minimum are $7$ and $\frac{1}{4}$ : option (A).

KEY IDEA

For a continuous function on a closed interval, the absolute extrema can only occur at interior critical points or at the endpoints, so evaluate $y$ at all of them and compare. Here $y = 3 (x + \frac{1}{2})^{2} + \frac{1}{4}$ is an upward parabola whose vertex $x = - \frac{1}{2}$ lies inside $[- 2, 0]$ . The vertex gives the minimum, and the endpoint farther from the vertex gives the maximum.

WATCH OUT FOR

Checking only the endpoints gives $y (- 2) = 7$ and $y (0) = 1$ , which leads to option (B) "7 and 1". The interior critical point $x = - \frac{1}{2}$ , where $y = \frac{1}{4}$ , gets missed. Option (C) comes from confusing the $x$ -values (the endpoint $- 2$ , the vertex $- \frac{1}{2}$ ) with the values of $y$ .

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