JEE Advanced 2022 — Paper 2 — MATH-II solution

JEE Advanced2022Paper 2MATH-II

Let $α = k = 1 \sum \infty sin^{2 k} (\frac{π}{6})$ .

Let $g : [0, 1] \to R$ be the function defined by

$g (x) = 2^{α x} + 2^{α (1 - x)}$

Then, which of the following statements is/are TRUE?

A
The minimum value of $g (x)$ is $2^{7/6}$
B
The maximum value of $g (x)$ is $1 + 2^{1/3}$
C
The function $g (x)$ attains its maximum at more than one point
D
The function $g (x)$ attains its minimum at more than one point

Solution

Sum the geometric series: $α = k = 1 \sum \infty (\frac{1}{4})^{k} = \frac{1/4}{1 - 1/4} = \frac{1}{3}$ .

So $g (x) = 2^{x /3} + 2^{(1 - x) /3}$ .

By the AM-GM inequality, $g (x) \geq 2 2^{x /3} \cdot 2^{(1 - x) /3} = 2 2^{1/3} = 2 \cdot 2^{1/6} = 2^{7/6}$ , with equality when $2^{x /3} = 2^{(1 - x) /3}$ , i.e. $x = \frac{1}{2}$ . So statement (A) is true; the minimum is attained at exactly one point, so (D) is false.

For the maximum on $[0, 1]$ : since $g$ is convex (sum of two convex exponentials), the maximum occurs at an endpoint. $g (0) = 1 + 2^{1/3}$ and $g (1) = 2^{1/3} + 1$ , equal at both endpoints. Hence (B) and (C) are true.

The correct options are (A), (B), (C).

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