JEE Advanced 2022 Paper 2 Q10 Mathematics Differentiation & Applications Maxima & Minima Medium

JEE Advanced 2022 Paper 2 · Q10 · Maxima & Minima

Let $\displaystyle\alpha = \sum_{k=1}^{\infty}\sin^{2k}\left(\dfrac{\pi}{6}\right)$. Let $g:[0,1]\to\mathbb{R}$ be the function defined by $g(x)=2^{\alpha x}+2^{\alpha(1-x)}$. Then, which of the following statements is/are TRUE?

  1. A. The minimum value of $g(x)$ is $2^{7/6}$
  2. B. The maximum value of $g(x)$ is $1+2^{1/3}$
  3. C. The function $g(x)$ attains its maximum at more than one point
  4. D. The function $g(x)$ attains its minimum at more than one point
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, C

Solution

$\sin^2(\pi/6)=1/4$. So $\alpha = \sum_{k=1}^\infty (1/4)^k = \dfrac{1/4}{1-1/4}=\dfrac{1}{3}$. Then $g(x)=2^{x/3}+2^{(1-x)/3}$. By AM-GM: $g(x)\ge 2\sqrt{2^{x/3}\cdot 2^{(1-x)/3}}=2\cdot 2^{1/6}=2^{7/6}$, with equality when $2^{x/3}=2^{(1-x)/3}$ i.e. $x=1/2$. So minimum $=2^{7/6}$ attained only at $x=1/2$ (one point), so (A) true, (D) false. On $[0,1]$, $g'(x)=\dfrac{\ln 2}{3}(2^{x/3}-2^{(1-x)/3})$, which is $<0$ for $x<1/2$ and $>0$ for $x>1/2$, so max is at endpoints $x=0$ and $x=1$: $g(0)=g(1)=1+2^{1/3}$. So (B) true (max value $=1+2^{1/3}$), and (C) true (attained at $x=0$ and $x=1$).

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