GATE 2020 IN GA Q.9: solution and answer

GATE 2020 · INGA Q.9 · 2 marks · MCQ

Select the graph that schematically represents BOTH $y = x^{m}$ and $y = x^{1/ m}$ properly in the interval $0 \leq x \leq 1$ , for integer values of $m$ , where $m > 1$ .

GATE 2020 IN GA question 9: Every option is a graph drawn the same way. — Figure from the GATE 2020 IN paper. It is part of the question.

(A)
Figure — upper concave curve labelled $x^{1/ m}$ (steep at the origin, levelling off at $(1, 1)$ ); lower convex curve labelled $x^{m}$ (flat at the origin, steep at $(1, 1)$ ); the two are mirror images about $y = x$
(B)
Figure — the same pair of curves as (A), with the labels swapped: upper concave curve labelled $x^{m}$ , lower convex curve labelled $x^{1/ m}$
(C)
Figure — upper curve, nearly straight and turning up sharply near $x = 1$ , labelled $x^{m}$ ; lower convex curve labelled $x^{1/ m}$ ; both on or below the diagonal
(D)
Figure — two concave curves above the diagonal: the steeper upper one labelled $x^{1/ m}$ , the flatter lower one labelled $x^{m}$

Answer and solution

ANSWER

(A): the curve labelled $x^{1/ m}$ is concave and lies above $y = x$ , the curve labelled $x^{m}$ is convex and lies below it, and both join $(0, 0)$ to $(1, 1)$ .

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

Endpoints: $0^{m} = 0^{1/ m} = 0$ and $1^{m} = 1^{1/ m} = 1$ , so both curves join $(0, 0)$ to $(1, 1)$ . All four options do this, so the endpoints alone cannot decide the answer.
Order between the endpoints: take $0 < x < 1$ . Since $m > 1$ , $x^{m - 1} < 1$ , so $x^{m} = x \cdot x^{m - 1} < x$ . Now put $t = x^{1/ m}$ , so $0 < t < 1$ . By the same argument $x = t^{m} < t$ , which means $x < x^{1/ m}$ . Together, $x^{m} < x < x^{1/ m} for 0 < x < 1.$ Check with $x = \frac{1}{4}$ and $m = 2$ : $x^{2} = \frac{1}{16} < \frac{1}{4} < \frac{1}{2} = x^{1/2}$ .
So the curve labelled $x^{1/ m}$ must lie above the line $y = x$ and the curve labelled $x^{m}$ must lie below it. That rules out (B), where the labels are reversed, and (C), where $x^{1/ m}$ is drawn below $x^{m}$ .
Shape of $y = x^{m}$ : $y^{'} = m x^{m - 1}$ and $y^{''} = m (m - 1) x^{m - 2} > 0$ for $0 < x < 1$ , so the curve is convex (it bends upward). Because $m - 1 \geq 1$ , the slope $m x^{m - 1}$ is $0$ at $x = 0$ and $m$ at $x = 1$ : flat at the start, steep at the end.
Shape of $y = x^{1/ m}$ : $y^{'} = \frac{1}{m} x^{\frac{1}{m} - 1}$ and $y^{''} = \frac{1}{m} (\frac{1}{m} - 1) x^{\frac{1}{m} - 2} < 0$ for $0 < x < 1$ , so the curve is concave. Since $\frac{1}{m} - 1 < 0$ , the slope grows without bound as $x \to 0^{+}$ , and it equals $\frac{1}{m}$ at $x = 1$ : steep at the start, levelling off at the end.
The two functions are inverses of each other on $[0, 1]$ ( $y = x^{m} ⟺ x = y^{1/ m}$ ), so their graphs are mirror images in the line $y = x$ . In (D) the curve labelled $x^{m}$ is drawn concave and above the diagonal, which contradicts steps 2 and 4. So (D) is wrong.
(A) draws $x^{1/ m}$ concave and above the diagonal, steep at the origin, and $x^{m}$ convex and below it, flat at the origin, as mirror images of each other. This matches every property found above. The answer is (A).

KEY IDEA

On $0 < x < 1$ , raising a fraction to a power greater than 1 makes it smaller and raising it to a power less than 1 makes it larger, so $x^{m} < x < x^{1/ m}$ . $x^{m}$ is convex with slope 0 at the origin, $x^{1/ m}$ is concave with a vertical tangent there, and as inverse functions their graphs are mirror images in $y = x$ .

WATCH OUT FOR

Carrying over the intuition from $x > 1$ , where $x^{m}$ is the larger of the two, and so choosing (B), which has the right shapes but swapped labels. Another mistake is to check only the endpoints $(0, 0)$ and $(1, 1)$ , which every option satisfies, and never test a point in between such as $x = \frac{1}{4}$ .

From the GATE 2020 IN paper. Every question of this paper · Download the full paper · All 58 Numerical questions, with notes

GATE 2020 IN GA Q.9: solution and answer

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