SATHEE: Limit Continuity and Differentiability 2 Question 4

Limit Continuity and Differentiability 2 Question 4

4. For each $t \in R$ , let $[t]$ be the greatest integer less than or equal to $t$ . Then,

$lim_{x \to 1 +} \frac{(1 - | x | + \sin | 1 - x |) \sin \frac{π}{2} [1 - x]}{| 1 - x | [1 - x]}$

(2019 Main, 10 Jan I)

(a) equals 0

(b) does not exist

(d) equals 1

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Answer:

Correct Answer: 4. (d)

Solution:

Here,

$f (x) = lim_{n \to \infty} \frac{n^{n} (x + n) x + \frac{n}{2} \dots x + \frac{n}{n}}{n! (x^{2} + n^{2}) x^{2} + \frac{n^{2}}{4} \dots x^{2} + \frac{n^{2}}{n^{2}}}, x > 0$

Taking log on both sides, we get

$\log_{e} f (x) = lim_{n \to \infty} \log \frac{n^{n} (x + n) x + \frac{n}{2} \dots x + \frac{n}{n}}{n! (x^{2} + n^{2}) x^{2} + \frac{n^{2}}{4} \dots x^{2} + \frac{n^{2}}{n^{2}}}$

$\begin{aligned} = lim_{n \to \infty} \frac{x}{n} \cdot \log \frac{\prod_{r = 1}^{n} x + \frac{1}{r / n}}{\prod_{r = 1}^{n} x^{2} + \frac{1}{(r / n)^{2}} \prod_{r = 1}^{n} (r / n)} \\ = x lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \log \frac{x + \frac{n}{r}}{x^{2} + \frac{n^{2}}{r^{2}} \frac{r}{n}} \\ = x lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \log \frac{\frac{r}{n} \cdot x + 1}{\frac{r^{2}}{n^{2}} \cdot x^{2} + 1} \end{aligned}$

Converting summation into definite integration, we get

$\begin{aligned} \log_{e} f (x) = x \int_{0}^{1} \log \frac{x t + 1}{x^{2} t^{2} + 1} d t \\ Put, t x = z \\ \Rightarrow x d t = d z \\ ∴ \log_{e} f (x) = x \int_{0}^{x} \log \frac{1 + z}{1 + z^{2}} \frac{d z}{x} \\ \Rightarrow \log_{e} f (x) = \int_{0}^{x} \log \frac{1 + z}{1 + z^{2}} d z \end{aligned}$

Using Newton-Leibnitz formula, we get

$\frac{1}{f (x)} \cdot f^{'} (x) = \log \frac{1 + x}{1 + x^{2}}$

Here, at $x = 1$ ,

$\begin{aligned} \frac{f^{'} (1)}{f (1)} & = \log (1) = 0 \\ ∴ & f^{'} (1) & = 0 \end{aligned}$

Now, sign scheme of $f^{'} (x)$ is shown below

$∴$ At $x = 1$ , function attains maximum.

Since, $f (x)$ increases on $(0, 1)$ .

$∴ f (1) > f (1 / 2)$

$∴$ Option (a) is incorrect.

$f (1 / 3) < f (2 / 3)$

$∴$ Option (b) is correct.

Also, $f^{'} (x) < 0$ , when $x > 1$

$\Rightarrow f^{'} (2) < 0$

$∴$ Option (c) is correct.

Also, $\frac{f^{'} (x)}{f (x)} = \log \frac{1 + x}{1 + x^{2}}$

$∴ \frac{f^{'} (3)}{f (3)} - \frac{f^{'} (2)}{f (2)} = \log \frac{4}{10} - \log \frac{3}{5}$

$\Rightarrow \frac{f^{'} (3)}{f (3)} < \frac{f^{'} (2)}{f (2)}$

$∴$ Option (d) is incorrect.

Limit Continuity and Differentiability 2 Question 4

4. For each $t \in R$ , let $[t]$ be the greatest integer less than or equal to $t$ . Then,

Answer:

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Limit Continuity and Differentiability 2 Question 4

4. For each t∈R, let [t] be the greatest integer less than or equal to t. Then,

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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4. For each $t \in R$ , let $[t]$ be the greatest integer less than or equal to $t$ . Then,