JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Application of Derivatives

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

Answer: (A), (B), (C)

Solution

First, compute $\alpha$:

$$\alpha ={\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^4+{\left(\frac{1}{2}\right)}^6+\dots =\frac{1/4}{1-1/4}=\frac{1}{3}$$

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

At the endpoints:

$$g(0)=1+2^{1/3},g(1)=2^{1/3}+1$$

For critical points, $g^{\prime }(x)=0$:

$$\frac{1}{3}\ln 2\cdot 2^{x/3}-\frac{1}{3}\ln 2\cdot \frac{2^{1/3}}{2^{x/3}}=0$$

$$2^{2x/3}=2^{1/3}\Rightarrow x=\frac{1}{2}$$

At $x=1/2$:

$$g(1/2)=2^{1/6}+2^{1/6}=2\cdot 2^{1/6}=2^{7/6}$$

Since $g^{\prime \prime }(1/2)>0$, this is a minimum. The maximum occurs at both endpoints $x=0$ and $x=1$.

Therefore, (A), (B), and (C) are correct.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · One or more correct options