Match List-I with List-II
| List-I (Function) | List-II (Largest open interval in which f is strictly decreasing) | ||
| (A) | \(f(x) = x^{2} - 4x + 1\) | (I) | \((e, \infty)\) |
| (B) | \(f(x) = 3 + 8x - 2x^{2}\) | (II) | \((-2, 2)\) |
| (C) | \(f(x) = x^{3} - 12x\) | (III) | \((2, \infty)\) |
| (D) | \(f(x) = \displaystyle\frac{\log x}{x},\ x \gt 0\) | (IV) | \((-\infty, 2)\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
In each case f is strictly decreasing where \(f'(x) \lt 0\).
(A) \(f(x) = x^{2} - 4x + 1\)
This matches (IV) \((-\infty, 2)\).
\( f'(x) = 2x - 4 \lt 0 \iff x \lt 2 \)
(B) \(f(x) = 3 + 8x - 2x^{2}\)
This matches (III) \((2, \infty)\).
\( f'(x) = 8 - 4x \lt 0 \iff x \gt 2 \)
(C) \(f(x) = x^{3} - 12x\)
This matches (II) (−2, 2).
\( f'(x) = 3x^{2} - 12 = 3(x - 2)(x + 2) \lt 0 \iff -2 \lt x \lt 2 \)
(D) \(f(x) = \displaystyle\frac{\log x}{x},\ x \gt 0\)
This matches (I) \((e, \infty)\).
\( \displaystyle f'(x) = \frac{x \cdot \frac{1}{x} - \log x}{x^{2}} = \frac{1 - \log x}{x^{2}} \)
The denominator is positive, and \(1 - \log x \lt 0 \iff \log x \gt 1 \iff x \gt e\).