Question:
For the function
\(f(x) = 4\log_e(x-1) - 2x^2 + 4x + 5,\ x \gt 1\)
, which one of the following is NOT correct?
For the function
\(f(x) = 4\log_e(x-1) - 2x^2 + 4x + 5,\ x \gt 1\)
, which one of the following is NOT correct?
\(f(x) = 4\log_e(x-1) - 2x^2 + 4x + 5,\ x \gt 1\)
, which one of the following is NOT correct?
Show Hint
Find \(f'\), \(f''\), the maximum value, and check signs of f at e and e + 1.
Updated On: Nov 4, 2026
- f is increasing in (1, 2) and decreasing in \((2, \infty)\)
- f(x) = \(-1\) has exactly two solutions
- \(f'(e) - f''(2) \lt 0\)
- f(x) = 0 has a root in the interval (e, e + 1)
Show Solution
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The Correct Option is C
Solution and Explanation
\(f'(x) = \frac{4}{x-1} - 4x + 4 = \frac{4 - 4(x-1)^2}{x-1} = \frac{4x(2-x)}{x-1}\); \(f''(x) = -\frac{4}{(x-1)^2} - 4\).
(A) \(f' \gt 0\) on (1, 2) and \(f' \lt 0\) on \((2,\infty)\): correct.
(B) \(f(2) = 0 - 8 + 8 + 5 = 5\) is the maximum, and \(f \to -\infty\) as \(x \to 1^+\) and as \(x \to \infty\). Since \(-1 \lt 5\), f(x) = -1 has exactly two solutions: correct.
(C) \(f'(e) = \frac{4e(2-e)}{e-1} \approx -4.54\), \(f''(2) = -4 - 4 = -8\). So \(f'(e) - f''(2) \approx -4.54 + 8 = 3.46 \gt 0\): the given statement \(\lt 0\) is NOT correct.
(D) \(f(e) = 4\ln(e-1) - 2e^2 + 4e + 5 \approx 3.26 \gt 0\) and \(f(e+1) = 4 - 2(e+1)^2 + 4(e+1) + 5 \approx -3.78 \lt 0\), so a root lies in (e, e + 1): correct.
Answer: (3) \(f'(e) - f''(2) \lt 0\)
(A) \(f' \gt 0\) on (1, 2) and \(f' \lt 0\) on \((2,\infty)\): correct.
(B) \(f(2) = 0 - 8 + 8 + 5 = 5\) is the maximum, and \(f \to -\infty\) as \(x \to 1^+\) and as \(x \to \infty\). Since \(-1 \lt 5\), f(x) = -1 has exactly two solutions: correct.
(C) \(f'(e) = \frac{4e(2-e)}{e-1} \approx -4.54\), \(f''(2) = -4 - 4 = -8\). So \(f'(e) - f''(2) \approx -4.54 + 8 = 3.46 \gt 0\): the given statement \(\lt 0\) is NOT correct.
(D) \(f(e) = 4\ln(e-1) - 2e^2 + 4e + 5 \approx 3.26 \gt 0\) and \(f(e+1) = 4 - 2(e+1)^2 + 4(e+1) + 5 \approx -3.78 \lt 0\), so a root lies in (e, e + 1): correct.
Answer: (3) \(f'(e) - f''(2) \lt 0\)
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