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 Concavity: The concavity of a function, \(f(x)\), is determined by the sign of its second derivative, \(f''(x)\). A positive second derivative (\(f''(x) > 0\)) indicates upward concavity (convexity), while a negative second derivative (\(f''(x) < 0\)) indicates downward concavity. The objective is to compute \(f''(x)\) for each given function and analyze its sign over specified intervals. Step 2: Detailed Analysis: (A) \(f(x) = e^{-x^2}\): \(f'(x) = -2xe^{-x^2}\). \(f''(x) = -2e^{-x^2} + (-2x)(-2xe^{-x^2}) = e^{-x^2}(-2+4x^2) = 2e^{-x^2}(2x^2-1)\). Concave downward when \(2x^2-1 < 0\), which implies \(x^2 < 1/2\), or \(-1/\sqrt{2} < x < 1/\sqrt{2}\). Thus, (A) is concave downward on \((-1/\sqrt{2}, 1/\sqrt{2})\). (A) matches (III). (B) \(f(x) = (1+x^2)e^{-x}\): \(f'(x) = 2xe^{-x} - (1+x^2)e^{-x} = e^{-x}(-x^2+2x-1) = -e^{-x}(x-1)^2\). \(f''(x) = e^{-x}(x-1)^2 - e^{-x}(2(x-1)) = e^{-x}(x-1)[(x-1)-2] = e^{-x}(x-1)(x-3)\). Concave upward when \((x-1)(x-3) > 0\), which occurs for \(x < 1\) or \(x > 3\). Thus, (B) is concave upward on \((-\infty, 1)\). (B) matches (II). (C) \(f(x) = 3x^4+4x^3-6x^2+12x+12\): \(f'(x) = 12x^3+12x^2-12x+12\). \(f''(x) = 36x^2+24x-12 = 12(3x^2+2x-1) = 12(3x-1)(x+1)\). The function is concave downward when \((3x-1)(x+1) < 0\), which is for \(-1 < x < 1/3\). For the given interval \((-\infty, -1)\), \(x+1 < 0\) and \(3x-1 < 0\), resulting in \(f''(x) > 0\). Therefore, this function is concave upward on \((-\infty, -1)\). There appears to be a discrepancy. Assuming the intended interval was \((-1, 1/3)\), but by elimination, (C) corresponds to (I). (D) \(f(x) = (x+1)^{1/3}\): \(f'(x) = \frac{1}{3}(x+1)^{-2/3}\). \(f''(x) = \frac{1}{3}\left(-\frac{2}{3}\right)(x+1)^{-5/3} = -\frac{2}{9}(x+1)^{-5/3}\). Concave upward when \((x+1)^{-5/3} < 0\), which implies \(x+1 < 0\), or \(x < -1\). Thus, (D) is concave upward on \((-\infty, -1)\). (D) matches (IV). Given the matches (A)-(III), (B)-(II), and (D)-(IV), (C) must correspond to (I) by elimination. Step 3: Final Answer: The pairings are: (A)-(III), (B)-(II), (C)-(I), (D)-(IV). Despite the apparent issue in part (C), the unique correct option is (B).