\(f(x)=|x-1|+|x-3|+\sin x\)
\(\text { Since } \sin x \text { is differentiable everywhere }\)

At x = 1 and x = 3. The graph admit cups.
\(\therefore\) The derivative is not exist.
\(\therefore\) The number of points in R is 2.

Given f is differentiable
\(\therefore f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right)=1 \)
\(f^{\prime}\left(1^{-}\right) =\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x+1}=\lim _{x \rightarrow 1^{-}} \frac{\left(a x^{2}-b\right)-(a-b)}{x+1} \)
\(=\lim _{x \rightarrow 1^{-}} \frac{a x^{2}-b-a+b}{x+1}=\lim _{x \rightarrow 1^{-}} \frac{a\left(x^{2}-1\right)}{x+1} \)
\(=\lim _{x \rightarrow 1^{-}}-a(x+1) \)
\(=a(1+1)=2 a \)
\(\therefore f^{\prime}\left(1^{+}\right) =\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{\frac{1}{x}-1}{x-1} \)
\(=\lim _{x \rightarrow 1^{+}} \frac{1-x}{x(x-1)}=\lim _{x \rightarrow 1^{+}} \frac{-1}{x}=-1\)
\(\therefore 2 a =-1 \)
\(a =\frac{-1}{2} \)
\(\text { and } f(1)=1\)
\(a-b=1 \)
\(-1 / 2-1=b \)
\(b=-3 / 2 \)

\(f^{\prime}\left(a^{-}\right) =\lim _{x \rightarrow a^{-}} \frac{f(x)-f(a)}{x-a}=\lim _{x \rightarrow a^{-}} \frac{(2 a-x)-(2 a-a)}{x-a} \)
\(=\lim _{x \rightarrow a^{-}} \frac{-x+a}{x-a}=-1 \)
\(f^{\prime}\left(a^{+}\right) =\lim _{x \rightarrow a^{+}} \frac{f(x)-f(a)}{x-a} \)
\(=\lim _{x \rightarrow a^{-}} \frac{(3 x-2 a)-(3 a-2 a)}{x-a} \)
\(=\lim _{x \rightarrow a^{-}} \frac{3 x-2 a-a}{x-a}=\lim _{x \rightarrow a^{-}} \frac{3 x-3 a}{x-a}=3 \)
\(f^{\prime}\left(a^{-}\right) \neq f^{\prime}\left(a^{+}\right) \)
\(\therefore \text { It is not differentiable at } x=a\)
\(\therefore f^{\prime}(x) \text { does not exist }\)

\(f^{\prime}\left(3^{-}\right)=\lim _{x \rightarrow 3^{-}} \frac{f(x)-f(3)}{x-3}=\lim _{x \rightarrow 3^{-}} \frac{x+2-5}{x-3}\)
\(=\lim _{x \rightarrow 3^{-}} \frac{x-3}{x-3}=1\)
\(f^{\prime}\left(3^{+}\right)=\lim _{x \rightarrow 3^{+}} \frac{f(x)-f(3)}{x-3}=\lim _{x \rightarrow 3^{+}} \frac{8-x-5}{x-3}\)
\(=\lim _{x \rightarrow 3^{+}} \frac{3-x}{x-3} \)
\(=\lim _{x \rightarrow 3^{+}} \frac{-(x-3)}{(x-3)}=-1 \)
\(\text { by }(1) \text { and }(2), f^{\prime}\left(3^{-}\right) \neq f^{\prime}\left(3^{+}\right)\)
\(\text { It is not differentiable }\)
\(\therefore f^{\prime}(x) \text { does not exist }\)

\(\text { Given } f(0)=5\)
\(\lim _{x \rightarrow 0} \frac{f(x)-5}{x} =4 \)
\(\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0} =4 \)
\(f^{\prime}(0) =4 \)
\(g(x) =\left(x^{2}+2 x+1\right) f(x) \)
\(g^{\prime}(x) =\left(x^{2}+2 x+1\right) f^{\prime}(x)+(2 x+2) f(x) \)
\(g^{\prime}(0) =(1) f^{\prime}(0)+2 f(0) \)
\(=1(4)+2(5) \)
\(=4+10=14 \)

\(f^{\prime}\left(2^{-}\right)=\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}=\lim _{x \rightarrow 2^{-}} \frac{x+1-(2+1)}{x-2}\)
\(=\lim _{x \rightarrow 2^{-}} \frac{x+1-3}{x-2}=\lim _{x \rightarrow 2^{-}} \frac{x-2}{x-2}=1\)
\(f^{\prime}\left(2^{+}\right)=\lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}=\lim _{x \rightarrow 2^{+}} \frac{(2 x-1)-(4-1)}{x-2}\)
\(=\lim _{x \rightarrow 2^{+}} \frac{2 x-1-3}{x-2}=\lim _{x \rightarrow 2^{+}} \frac{2 x-4}{x-2}\)
\(=\lim _{x \rightarrow 2^{+}} \frac{2(x-2)}{(x-2)}=2\)
\(f^{\prime}\left(2^{-}\right) \neq f^{\prime}\left(2^{+}\right)\)
\(\therefore f^{\prime}(2) \text { does not exist. }\)

\(\lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a} \)
\(=\lim _{x \rightarrow a} \frac{x f(a)-a f(x)+a f(a)-a f(a)}{x-a} \)
\(=\lim _{x \rightarrow a} \frac{f(a)(x-a)-a[f(x)-f(a)]}{x-a} \)
\(=f(a) \lim _{x \rightarrow a} \frac{x-a}{x-a}-a \lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \)
\(=f(a)-a f^{\prime}(a) \)

\(\text { Given } f(x)=x+2\)
\(f^{\prime}(x) =1 \)
\(f^{\prime}(f(x)) =f^{\prime}(x+2)=1 \)

\( y=\log _{10} x=\frac{1}{\log _x 10}\)
Diff  w. r. to \(\log _x 10\)
\(y^{\prime}=\frac{-1}{\left(\log _x 10\right)^2}=-\left(\log _{10} x\right)^2 \)