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

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.

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

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

\(x =a \sin \theta, y=b \cos \theta \)
\(\frac{d x}{d \theta} =a \cos \theta, \frac{d y}{d \theta}=-b \sin \theta\)
\(\frac{d y}{d x} =\frac{d y / d \theta}{d x / d \theta}=\frac{-b \sin \theta}{a \cos \theta}=\frac{-b}{a} \tan \theta \)
\(\frac{d^{2} y}{d x^{2}} =\frac{-b}{a} \sec ^{2} \theta \frac{d \theta}{d x}\)
\(=\frac{-b}{a} \sec ^{2} \theta \cdot \frac{1}{a \cos \theta} \)
\(= \frac{-b}{a^{2}} \sec ^{3} \theta \)

\(y =m x+c \)
\(f(x) =m x+c \Rightarrow f(0)=c=1 \)
\(\therefore c =1 \)
\(f^{\prime}(x) =m \)
\(f^{\prime}(0) =1=m \)
\(\therefore m =1 \)
\(\therefore f(x) =x+1 \)
\(f(2) =2+1=3 \)

\(\text { Given } y=\frac{1}{a-z}\)
\(a-z =\frac{1}{y} \)
\(a-\frac{1}{y} =z \)
\(z =a-\frac{1}{y} \)
\(\frac{d z}{d y}=0+\frac{1}{y^{2}}=\frac{1}{y^{2}} =\frac{1}{1 /(a-z)^{2}} \)
\(=(a-z)^{2} \)

\(\overrightarrow{O A}=10 \hat{i}+3 \hat{j}, \overrightarrow{O B}=12 \hat{i}-5 \hat{j}, \overrightarrow{O C}=a \hat{i}+11 \hat{j} \)
\(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=2 \hat{i}-8 \hat{j} \)
\(\overrightarrow{B C} =\overrightarrow{O C}-\overrightarrow{O B} \)
\(=(a-12) \hat{i}+16 \hat{j} \)
Condition:
\(\overrightarrow{B C} =-2 \overrightarrow{A B} \)
\((a-12) \hat{i}+16 \hat{j} =2(2 \hat{i}-8 \hat{j}) \)
\((a-12) \hat{i}+16 \hat{j} =-4 \hat{i}+16 \hat{j} \)
\(a-12 =-4 \)
\(a =-4+12=8 \)
\(a =8 \)

\(A B=B C=2 \text { (ie) } \vec{a}=\vec{b}=2\)
\(\therefore A B C \text { is isosceles triangle }\)
\(\therefore \angle A+\angle B =180^{\circ}-120^{\circ} \)
\(2 \angle A =60^{\circ} \)
\(\angle A =30^{\circ} \)

\((\vec{a}+3 \vec{b}) \times(3 \vec{a}-\vec{b}) =3(\vec{a} \times \vec{a})-\vec{a} \times \vec{b}+9 \vec{b} \times \vec{a}-3 \vec{b} \times \vec{b} \)
\(=\overrightarrow{0}+\vec{b} \times \vec{a}+9(\vec{b} \times \vec{a})-\overrightarrow{0} \)
\(=10 \vec{b} \times \vec{a} \)
\(\frac{\sqrt{3}}{2} \times 2 =|\vec{a} \times \vec{b}| \)
\(|\vec{a} \times \vec{b}| =\sqrt{3} \)
\([(\vec{a}+3 \vec{b}) \times(3 \vec{a}-\vec{b})]^{2} =100(\vec{b} \times \vec{a})^{2} \)
\(=100(\vec{a} \times \vec{b})^{2}=100(\sqrt{3})^{2} \)
\(=100 \times 3=300 \)

\(\cos 60^{\circ} =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \)
\(\frac{1}{2} =\frac{\frac{1}{2}}{|\vec{a} \| \vec{a}|} \)
\(|\vec{a}|^{2} =1 \Rightarrow|a|=1 \)

\(\left(\frac{3+2+x_{1}}{3},\right. \left.\frac{4+3+x_{2}}{3}, \frac{-4+4+x_{3}}{3}\right)=(1,2,3) \)
\(3+2+x_{1} =3 ; 4+3+x_{2}=6 ;-4+4+x_{3}=9 \)
\(5+x_{1}=3 7+x_{2}=6 \)
\(x_{1}=-2 \quad x_{2}=-1 \)
\(\therefore \text { Third vertex is }(-2,-1,9)\)

\(\overrightarrow{A B}=-\overrightarrow{C D}\)
\(\overrightarrow{A D}=\overrightarrow{B C}=-\overrightarrow{C B}\)
\(\therefore \overrightarrow{A B}+\overrightarrow{A D}+\overrightarrow{C B}+\overrightarrow{C D}=-\overrightarrow{C D}-\overrightarrow{C B}+\overrightarrow{C B}+\overrightarrow{C D}=\overrightarrow{0}\)

\(\vec{a}+\vec{b} =2 \hat{i}-\hat{j} \)
\(|\vec{a}+\vec{b}| =\sqrt{4+1}=\sqrt{5} \)
\(\text { Unit vector }=\frac{2 \hat{i}-\hat{j}}{\sqrt{5}}\)

\(\frac{{ }^{2 n} C_{3}}{{ }^{n} C_{3}} =\frac{11}{1} \)
\(\frac{(2 n)(2 n-1)(2 n-2)}{n(n-1)(n-2)} =\frac{11}{1} \)
\(\frac{2 n(2 n-1) 2(n-1)}{n(n-1)(n-2)} =11 \)
\(4(2 n-1) =11(n-2) \)
\(8 n-4 =11 n-22 \)
\(18 =3 n \)
\(n = 6\)

\(\text { Number of triangles }={ }^{10} \mathrm{C}_{3}-{ }^{4} \mathrm{C}_{3}\)
\(=\frac{10 \times 9 \times 8}{1 \times 2 \times 3}-4 \)
\(=120-4 =116 \)

Total numbers = 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 24,
Sum of alt integers in tenth place
= 6 ( 2 + 4 + 5 + 7) = 108