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,
$\lim _{x \rightarrow 1+} \frac{(1-|x|+\sin |1-x|) \sin \frac{\pi}{2}[1-x]}{|1-x|[1-x]}$
(2019 Main, 10 Jan I)
(a) equals 0
(b) does not exist
(c) equals -1
(d) equals 1
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Answer:
Correct Answer: 4. (d)
Solution:
- Here,
$$ f(x)=\lim _{n \rightarrow \infty} \frac{n^{n}(x+n) x+\frac{n}{2} \ldots x+\frac{n}{n}}{n !\left(x^{2}+n^{2}\right) x^{2}+\frac{n^{2}}{4} \ldots x^{2}+\frac{n^{2}}{n^{2}}} \quad, x>0 $$
Taking log on both sides, we get
$$ \log _e{f(x)}=\lim _{n \rightarrow \infty} \log \frac{n^{n}(x+n) x+\frac{n}{2} \ldots x+\frac{n}{n}}{n !\left(x^{2}+n^{2}\right) x^{2}+\frac{n^{2}}{4} \ldots x^{2}+\frac{n^{2}}{n^{2}}} $$
$$ \begin{aligned} & =\lim _{n \rightarrow \infty} \frac{x}{n} \cdot \log \frac{\prod _{r=1}^{n} x+\frac{1}{r / n}}{\prod _{r=1}^{n} x^{2}+\frac{1}{(r / n)^{2}} \prod _{r=1}^{n}(r / n)} \\ & =x \lim _{n \rightarrow \infty} \frac{1}{n} \sum _{r=1}^{n} \log \frac{x+\frac{n}{r}}{x^{2}+\frac{n^{2}}{r^{2}} \frac{r}{n}} \\ & =x \lim _{n \rightarrow \infty} \frac{1}{n} \sum _{r=1}^{n} \log \frac{\frac{r}{n} \cdot x+1}{\frac{r^{2}}{n^{2}} \cdot x^{2}+1} \end{aligned} $$
Converting summation into definite integration, we get
$$ \begin{aligned} & \log _e{f(x)}=x \int _0^{1} \log \frac{x t+1}{x^{2} t^{2}+1} d t \\ & \text { Put, } \quad t x=z \\ & \Rightarrow \quad x d t=d z \\ & \therefore \quad \log _e{f(x)}=x \int _0^{x} \log \frac{1+z}{1+z^{2}} \frac{d z}{x} \\ & \Rightarrow \quad \log _e{f(x)}=\int _0^{x} \log \frac{1+z}{1+z^{2}} d z \end{aligned} $$
Using Newton-Leibnitz formula, we get
$$ \frac{1}{f(x)} \cdot f^{\prime}(x)=\log \frac{1+x}{1+x^{2}} $$
Here, at $x=1$,
$$ \begin{aligned} & & \frac{f^{\prime}(1)}{f(1)} & =\log (1)=0 \\ \therefore & & f^{\prime}(1) & =0 \end{aligned} $$
Now, sign scheme of $f^{\prime}(x)$ is shown below
$\therefore$ At $x=1$, function attains maximum.
Since, $f(x)$ increases on $(0,1)$.
$$ \therefore \quad f(1)>f(1 / 2) $$
$\therefore$ Option (a) is incorrect.
$$ f(1 / 3)<f(2 / 3) $$
$\therefore$ Option (b) is correct.
Also, $\quad f^{\prime}(x)<0$, when $x>1$
$\Rightarrow \quad f^{\prime}(2)<0$
$\therefore$ Option (c) is correct.
Also, $\quad \frac{f^{\prime}(x)}{f(x)}=\log \frac{1+x}{1+x^{2}}$
$\therefore \quad \frac{f^{\prime}(3)}{f(3)}-\frac{f^{\prime}(2)}{f(2)}=\log \frac{4}{10}-\log \frac{3}{5}$
$$ \Rightarrow \quad \frac{f^{\prime}(3)}{f(3)}<\frac{f^{\prime}(2)}{f(2)} $$
$\therefore$ Option (d) is incorrect.
