GATE 2022 EC Q.36: solution and answer

GATE 2022 · ECQ.36 · 2 marks · MCQ

The function $f (x) = 8 lo g_{e} x - x^{2} + 3$ attains its minimum over the interval $[1, e]$ at $x =$ ________.

(Here $lo g_{e} x$ is the natural logarithm of $x$ .)

(A)
$2$
(B)
$1$
(C)
$e$
(D)
$\frac{1 + e}{2}$

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Answer and solution

ANSWER

The minimum of $f (x) = 8 ln x - x^{2} + 3$ on $[1, e]$ is at $x = 1$ , where $f (1) = 2$ . The answer is option (B).

Official answer: BSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Closed-interval extremum rule: $f$ is continuous on $[1, e]$ , so its absolute minimum there is reached either at a critical point inside the interval or at an endpoint. We find the critical points, then compare values.
Critical point: $f^{'} (x) = \frac{8}{x} - 2 x = 0 \Rightarrow x^{2} = 4 \Rightarrow x = 2$ (we drop $x = - 2$ , which lies outside the domain $x > 0$ ). Since $e \approx 2.718$ , the point $x = 2$ lies inside $[1, e]$ .
Second-derivative test: $f^{''} (x) = - \frac{8}{x ^{2}} - 2 < 0$ for every $x > 0$ , so $x = 2$ is a local maximum, not a minimum. Also, $f$ is concave on the whole interval, so its minimum on $[1, e]$ has to be at an endpoint. (Equivalently: $f^{'} > 0$ on $[1, 2)$ and $f^{'} < 0$ on $(2, e]$ , so $f$ rises and then falls.)
Evaluate the endpoints: $f (1) = 8 (0) - 1 + 3 = 2$ and $f (e) = 8 (1) - e^{2} + 3 = 11 - e^{2} \approx 11 - 7.389 = 3.611$ .
Compare: $f (1) = 2 < f (e) \approx 3.611$ , and the interior value is $f (2) = 8 ln 2 - 1 \approx 4.545$ . So the absolute minimum on $[1, e]$ is at $x = 1$ , which is option (B).

KEY IDEA

On a closed interval, the absolute minimum is at a critical point or at an endpoint. Here the only critical point, $x = 2$ , is a maximum, because $f^{''} < 0$ everywhere. A concave function takes its minimum at an endpoint, so you only have to compare $f (1) = 2$ with $f (e) = 11 - e^{2} \approx 3.61$ .

WATCH OUT FOR

Setting $f^{'} (x) = 0$ , getting $x = 2$ and picking option (A) without checking which kind of extremum it is. $f^{''} (2) = - 4 < 0$ , so $x = 2$ is where $f$ is largest on the interval. The other trap is picking $x = e$ because it is the right-hand endpoint, without actually comparing $f (1)$ and $f (e)$ .

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