SATHEE: Limit Continuity and Differentiability 7 Question 13

Limit Continuity and Differentiability 7 Question 13

13. Let $f (x) = x_{0}^{2} | \cos \frac{π}{x} | \begin{matrix} x \neq 0, x \in R, then f is x = 0 \end{matrix}$

(2012)

(a) differentiable both at $x = 0$ and at $x = 2$

(b) differentiable at $x = 0$ but not differentiable at $x = 2$

(d) differentiable neither at $x = 0$ nor at $x = 2$

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Answer:

Correct Answer: 13. $(b, c)$

Solution:

Since, $f (x) = e^{g (x)} \Rightarrow e^{g (x + 1)} = f (x + 1) = x f (x) = x e^{g (x)}$

and

$g (x + 1) = \log x + g (x)$

i.e.

$g (x + 1) - g (x) = \log x$

Replacing $x$ by $x - \frac{1}{2}$ , we get

$\begin{aligned} g & x + \frac{1}{2} - g x - \frac{1}{2} = \log x - \frac{1}{2} = \log (2 x - 1) - \log 2 \\ ∴ g^{''} & x + \frac{1}{2} - g^{''} x - \frac{1}{2} = \frac{- 4}{(2 x - 1)^{2}} \end{aligned}$

On substituting, $x = 1, 2, 3, \dots, N$ in Eq. (ii) and adding, we get

$g^{''} N + \frac{1}{2} - g^{''} \frac{1}{2} = - 4 1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{(2 N - 1)^{2}} .$

Limit Continuity and Differentiability 7 Question 13

13. Let $f (x) = x_{0}^{2} | \cos \frac{π}{x} | \begin{matrix} x \neq 0, x \in R, then f is x = 0 \end{matrix}$

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Limit Continuity and Differentiability 7 Question 13

13. Let f(x)=x02|cos⁡πx|x≠0,x∈R, then f is x=0

Answer:

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13. Let $f (x) = x_{0}^{2} | \cos \frac{π}{x} | \begin{matrix} x \neq 0, x \in R, then f is x = 0 \end{matrix}$