Limit Continuity and Differentiability 6 Question 1
1. For the function $f(x)=x \cos \frac{1}{x}, x \geq 1$,
(a) for atleast one $x$ in the interval
(2009)
$$ [1, \infty), f(x+2)-f(x)<2 $$
(b) $\lim _{x \rightarrow \infty} f^{\prime}(x)=1$
(c) for all $x$ in the interval $[1, \infty), f(x+2)-f(x)>2$
(d) $f^{\prime}(x)$ is strictly decreasing in the interval $[1, \infty)$
Analytical & Descriptive Questions
$$ \begin{aligned} & \text { 2. Let } \quad f(x)=\begin{array}{cc} x+a, & \text { if } x<0 \\ |x-1|, & \text { if } x \geq 0 \end{array} \\ & g(x)=\begin{array}{cl} x+1, & \text { if } x<0 \\ (x-1)^{2}+b, & \text { if } x \geq 0 \end{array} \end{aligned} $$
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Answer:
Correct Answer: 1. (b, c, d)
Solution:
- Given function, $g(x)=|f(x)|$
where $f: R \rightarrow R$ be differentiable at $c \in R$ and $f(c)=0$, then for function ’ $g$ ’ at $x=c$
$$ \begin{aligned} g^{\prime}(c) & =\lim _{h \rightarrow 0} \frac{g(c+h)-g(c)}{h} \quad[\text { where } h>0] \\ & =\lim _{h \rightarrow 0} \frac{|f(c+h)|-|f(c)|}{h}=\lim _{h \rightarrow 0} \frac{|f(c+h)|}{h} \quad[\operatorname{as} f(c)=0 \text { (given)] } \\ & =\lim _{h \rightarrow 0}\left|\frac{f(c+h)-f(c)}{h}\right| \\ & =\left|\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}\right| \\ & =\left|f^{\prime}(c)\right| \quad[\because f \text { is differentiable at } x=c] \end{aligned} $$
Now, if $f^{\prime}(c)=0$, then $g(x)$ is differentiable at $x=c$, otherwise LHD (at $x=c$ ) and RHD (at $x=c$ ) is different.
