Let f : be a twice differentiable function such that f(3)=18, f'(3) = 0 and f''(3) = 4. Then lim x 1 ( l og e ( f ( 2 + x ) f ( 3 ) ) 18 ( x - 1 ) 2 ) is equal to [2026]

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Solution

Q.
Let $f : ℝ \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18, f' (3) = 0$ and $f'' (3) = 4$ . Then $\lim_{x \to 1} (l {og}_{e} {(\frac{f (2 + x)}{f (3)})}^{\frac{18}{{(x - 1)}^{2}}})$ is equal to [2026]

1 2

2 9

3 1

4 18

Ans.
(1)

$Let T = \lim_{x \to 1} {(\frac{f (x + 2)}{f (3)})}^{\frac{18}{{(x - 1)}^{2}}}; 1^{\infty} form$

$\Rightarrow T = e^{\lim_{x \to 1} \frac{18}{{(x - 1)}^{2}} (\frac{f (x + 2) - f (3)}{f (3)})}$

$\Rightarrow T = e^{\lim_{x \to 1} \frac{18}{{(x - 1)}^{2}} (\frac{f (x + 2) - f (3)}{18})}$

$\Rightarrow T = e^{\lim_{x \to 1} (\frac{f (x + 2) - f (3)}{{(x - 1)}^{2}}) \frac{0}{0} form}$ apply L' opital

$\Rightarrow T = e^{\lim_{x \to 1} \frac{f' (x + 2)}{2 (x - 1)}; \frac{0}{0} form}$ apply L' opital

$\Rightarrow T = e^{\lim_{x \to 1} \frac{f'' (x + 2)}{2}}$ ; $= e^{\frac{4}{2}} = e^{2}$

$\Rightarrow \log_{e} (T) = 2$

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