Question

Match List I with List - II. List - I & List - II
A. $x$ where $f(x)=9x(x-1)^{2}$ attains maximum & I. $e$

B. $x$ where $f(x)=\frac{1}{x}e^{-\frac{1}{2}(\log_{e}x-2)^{2}}$ attains maximum & II. $\frac{2}{3}$

C. $x$ where $f(x)=x^{2}(1-x)^{6}$ attains maximum & III. $\frac{1}{3}$

D. $x$ where $f(x)=x^{2}e^{-3x}$ attains maximum & IV. $\frac{1}{4}$

• A. A-III, B-I, C-IV, D-II
• B. A-III, B-I, C-II, D-IV
• C. A-IV, B-I, C-III, D-II
• D. A-II, B-III, C-I, D-IV

Hint:

For $x^m e^{-ax}$, the maximum is always at $x = m/a$. For $x^m (1-x)^n$, it is always at $x = m/(m+n)$. These patterns are very common in competitive exams.

Answer 1

The Correct Option is A

We evaluate each function to find the value of $x$ for which the maximum is reached.
Step 1: $f(x) = 9x(x-1)^2$. $f'(x) = 9[(x-1)^2 + 2x(x-1)] = 9(x-1)(3x-1)$. Critical points are $x=1$ and $x=1/3$. At $x=1, f=0$. At $x=1/3, f > 0$. So max is at $x=1/3$ (III).
Step 2: Let $u = \log x$. Then $f = e^{-u} e^{-\frac{1}{2}(u-2)^2}$. To maximize $f$, maximize the exponent: $-u - \frac{1}{2}(u^2 - 4u + 4) = -\frac{1}{2}u^2 + u - 2$. Derivative: $-u + 1 = 0 \implies u = 1 \implies \log x = 1 \implies x = e$ (I).
Step 3: Function of form $x^m(1-x)^n$. Max occurs at $x = \frac{m}{m+n}$. $x = \frac{2}{2+6} = \frac{2}{8} = \frac{1}{4}$ (IV).
Step 4: $f(x) = x^2 e^{-3x}$. $f'(x) = 2x e^{-3x} - 3x^2 e^{-3x} = x e^{-3x}(2 - 3x)$. Max at $x = 2/3$ (II). Final matching: A-III, B-I, C-IV, D-II. Matches Option (1).