Question

Match List-I with List-II

\[ \begin{array}{|l|l|} \hline \textbf{List-I (Functions)} & \textbf{List-II (Concavity and Convexity)} \\ \hline (A) \; f(x) = e^{-x^2} & (I) \; \text{Concave downward in } (-\infty, -1) \\ (B) \; f(x) = (1+x^2)e^{-x} & (II) \; \text{Concave upward in } (-\infty, 1) \\ (C) \; f(x) = 3x^4+4x^3-6x^2+12x+12 & (III) \; \text{Concave downward in } \left(-\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}}\right) \\ (D) \; f(x) = (x+1)^{1/3} & (IV) \; \text{Concave upward in } (-\infty, -1) \\ \hline \end{array} \]
Choose the correct answer from the options given below:

• (A) - (III), (B) - (II), (C) - (IV), (D) - (I)
• (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
• (A) - (III), (B) - (I), (C) - (II), (D) - (IV)
• (A) - (IV), (B) - (II), (C) - (I), (D) - (III)

Hint:

When doing concavity problems, find the second derivative \(f''(x)\) and then find its roots. These roots are the potential inflection points. Test the sign of \(f''(x)\) in the intervals between these roots to determine the concavity. A simple sign chart is very effective.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The concavity of a function is determined by the sign of its second derivative, \(f''(x)\).
If \(f''(x) > 0\) on an interval, the function is concave upward (convex).
If \(f''(x) < 0\) on an interval, the function is concave downward.
We need to find \(f''(x)\) for each function and determine its sign on the given intervals.
Step 2: Detailed Explanation:
(A) \(f(x) = e^{-x^2}\):
\(f'(x) = -2xe^{-x^2}\).
\(f''(x) = -2e^{-x^2} + (-2x)(-2xe^{-x^2}) = e^{-x^2}(-2+4x^2) = 2e^{-x^2}(2x^2-1)\).
\(f''(x) < 0\) when \(2x^2-1 < 0 \implies x^2 < 1/2 \implies -1/\sqrt{2} < x < 1/\sqrt{2}\).
So, it is concave downward in \((-1/\sqrt{2}, 1/\sqrt{2})\). (A) matches (III).

(B) \(f(x) = (1+x^2)e^{-x}\):
\(f'(x) = 2xe^{-x} - (1+x^2)e^{-x} = e^{-x}(-x^2+2x-1) = -e^{-x}(x-1)^2\).
\(f''(x) = e^{-x}(x-1)^2 - e^{-x}(2(x-1)) = e^{-x}(x-1)[(x-1)-2] = e^{-x}(x-1)(x-3)\).
\(f''(x) > 0\) when \((x-1)(x-3) > 0\), which occurs for \(x < 1\) or \(x > 3\).
So, it is concave upward in \((-\infty, 1)\). (B) matches (II).

(C) \(f(x) = 3x^4+4x^3-6x^2+12x+12\):
\(f'(x) = 12x^3+12x^2-12x+12\).
\(f''(x) = 36x^2+24x-12 = 12(3x^2+2x-1) = 12(3x-1)(x+1)\).
\(f''(x) < 0\) when \((3x-1)(x+1) < 0\), which occurs for \(-1 < x < 1/3\).
The provided interval is \((-\infty, -1)\). In this interval, \(x+1 < 0\) and \(3x-1 < 0\), so \(f''(x) > 0\).
This function is concave upward in \((-\infty, -1)\).
There seems to be a mismatch. Likely, the intended interval was \((-1, 1/3)\). Still, by elimination, (C) matches (I).

(D) \(f(x) = (x+1)^{1/3}\):
\(f'(x) = \frac{1}{3}(x+1)^{-2/3}\).
\(f''(x) = \frac{1}{3}\left(-\frac{2}{3}\right)(x+1)^{-5/3} = -\frac{2}{9}(x+1)^{-5/3}\).
\(f''(x) > 0\) when \((x+1)^{-5/3} < 0 \implies x+1 < 0 \implies x < -1\).
So, it is concave upward in \((-\infty, -1)\). (D) matches (IV).

With (A)-(III), (B)-(II), (D)-(IV), the only possibility for (C) is (I).
Step 3: Final Answer:
Final Matching: (A)-(III), (B)-(II), (C)-(I), (D)-(IV).
Despite the apparent error in part (C), the unique correct option is (B).