For the function f(x) = 4 loge(x – 1) – 2x2 + 4x + 5, x &gt; 1, which one of the following is NOT correct?

f’(e) – f’’ (2) < 0

f is increasing in (1, 2) and decreasing in (2, ¥)

f (x) = -1 has exactly two solutions

f (x) = 0 has a root in the interval (e, e + 1)

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Application of Derivatives Ncert Solutions Maths class 12th Maths Class 12th

For the function f(x) = 4 log_e(x – 1) – 2x² + 4x + 5, x > 1, which one of the following is NOT correct?

Option 1 -

f is increasing in (1, 2) and decreasing in (2, ¥)

Option 2 -

f (x) = -1 has exactly two solutions

Option 3 -

f'(e) – f'' (2) < 0

Option 4 -

f (x) = 0 has a root in the interval (e, e + 1)

9 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

f (x) = 4log_e (x – 1) -2x² + 4x + 5, x > 1

$(A) f' (x) = \frac{4}{x - 1} - 4 x + 4$

$\Rightarrow f' (x) > 0 \Rightarrow \frac{x (- x + 2)}{x - 1} > 0$

option (A) is correct.

(B) f (x) = -1, has two solution

option (B) is correct

$(C) f " (x) = - \frac{4}{{(x - 1)}^{2}} - 4$

$f' (e) - f " (2) = \frac{4 e (2 - e)}{e - 1} + 8 > 0$

option (C) is not correct

(D) f (e) = 4log_e (e – 1) -2 (e² – 2e + 1) + 7 > 0

f (e + 1) = 4 – 2 (e + 1)² + 4 (e + 1) +5+ < 0

option (D) is correct

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