SATHEE: Limit Continuity and Differentiability 7 Question 25

Limit Continuity and Differentiability 7 Question 25

25. For a real number $y$ , let $[y]$ denotes the greatest integer less than or equal to $y$ . Then, the function $f (x) = \frac{\tan π [(x - π)]}{1 + [x]^{2}}$ is

$(1981, 2 M)$

(a) discontinuous at some $x$

(b) continuous at all $x$ , but the derivative $f^{'} (x)$ does not exist for some $x$

(d) $f^{'} (x)$ exists for all $x$

Objective Questions II

(One or more than one correct option)

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Answer:

Correct Answer: 25. $(a, b)$

Solution:

Given, $f (x) = \log_{x} (\log x)$

$∴ f (x) = \frac{\log (\log x)}{\log x}$

On differentiating both sides, we get

$\begin{aligned} f^{'} (x) & = \frac{(\log x) \frac{1}{\log x} \cdot \frac{1}{x} - \log (\log x) \cdot \frac{1}{x}}{(\log x)^{2}} \\ ∴ f^{'} (e) & = \frac{1 \cdot \frac{1}{1} \cdot \frac{1}{e} - \log (1) \cdot \frac{1}{e}}{(1)^{2}} \\ \Rightarrow f^{'} (e) & = \frac{1}{e} \end{aligned}$

Limit Continuity and Differentiability 7 Question 25

25. For a real number $y$ , let $[y]$ denotes the greatest integer less than or equal to $y$ . Then, the function $f (x) = \frac{\tan π [(x - π)]}{1 + [x]^{2}}$ is

Objective Questions II

Answer:

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Limit Continuity and Differentiability 7 Question 25

25. For a real number y, let [y] denotes the greatest integer less than or equal to y. Then, the function f(x)=tan⁡π[(x−π)]1+[x]2 is

Objective Questions II

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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25. For a real number $y$ , let $[y]$ denotes the greatest integer less than or equal to $y$ . Then, the function $f (x) = \frac{\tan π [(x - π)]}{1 + [x]^{2}}$ is