\(f(x)=|x|=\left\{\begin{array}{cll} x & \text { if } & x \geq 0 \\ -x & \text { if } & x<0 \end{array}\right.\)
\(\mathrm{L}\left[f^{\prime}(0)\right]=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\)
\(=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0}, x=0-h\)
\(=\lim _{h \rightarrow 0} \frac{f(-h)-f(0)}{-h}\)
\(=\lim _{h \rightarrow 0} \frac{|-h|-|0|}{-h}\)
\(=\lim _{h \rightarrow 0} \frac{|-h|}{-h}\)
\(=\lim _{h \rightarrow 0} \frac{h}{-h}\)
\(=\lim _{h \rightarrow 0}(-1)=-1\)
\(\mathrm{R}\left[f^{\prime}(0)\right]=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}\)
\(=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{0+h-0}, x=0+h\)
\(=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{|h|-|0|}{h}\)
\(=\lim _{h \rightarrow 0} \frac{|h|}{h}\)
\(=\lim _{h \rightarrow 0} \frac{h}{h}\)
\(=\lim _{h \rightarrow 0} 1=1\)
\(\text { Here, } \mathrm{L}\left[f^{\prime}(0)\right] \neq \mathrm{R}\left[f^{\prime}(0)\right]\)
\(\therefore\) f(x) is not differentiable at x = 0

Since \(\cfrac { 3\pi }{ 2 } <\left( A,B \right) <2\pi \) ,both A and B lie in the fourth quadrant,
 \(\therefore \) sinA and sinB are negative
Given \(\cos A=\frac{4}{5} \text { and } \cos B=\frac{12}{13}\)
Therefore, \(\sin A=-\sqrt{1-\cos ^2 A}\)
\(=-\sqrt{1-\frac{16}{25}}\)
\(=-\sqrt{\frac{25-16}{25}}\)
\(=-\frac{3}{5}\)
\(\operatorname{Sin} \mathrm{B}=-\sqrt{1-\cos ^2 B}\)
\(=-\sqrt{1-\frac{144}{169}}\)
\(=-\sqrt{\frac{169-144}{169}}\)
\(=-\frac{5}{13}\)
\( \cos (A+B)= \cos A \cos B- \sin A \sin B \)
\(=\frac{4}{5} \times \frac{12}{13}-\left(\frac{-3}{5}\right) \times\left(\frac{-5}{13}\right)\)
\(=\frac{48}{65}-\frac{15}{65}=\frac{33}{65}\)

The equation of the circle is \({ x }^{ 2 }+{ y }^{ 2 }-4x+4y-8=0\) 
\({ PT }^{ 2 }={ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }-{ 4x }_{ 1 }+{ 4y }_{ 1 }-8\) 
At \(P(0,1)\) 
\({ PT }^{ 2 }=0+1+0+4-8=-3<0\)
\(At \ Q(5,9)\)
\({ QT }^{ 2 }=25+81-20+36-8=114>0\)
\(At \ R(-2,3)\)
\({ RT }^{ 2 }=4+9+12-8=25>0\)
\(At \ S(2,2)\)
\(ST^{ 2 }=4+4-8+8-8=0\) 
\(\therefore \) The point P lies inside the circle. The points Q and R lie outside the circle and the point S lies on the circle.

Let E₁, E₂, E₃ and A be the events defined as
E₁ = Screw is manufactured by machine A
E₂ = Screw is manufactured by machine B
E₃ = Screw is manufactured by machine C
A → screws drawn is defective
\(P({ E }_{ 1 })=\frac { 30 }{ 100 } \), \(P({ E }_{ 2 })=\frac { 40 }{ 100 } \), \(P({ E }_{ 3 })=\frac { 30 }{ 100 } \)
P(A/E₁) =\(\frac { 20 }{ 100 } \), P(A/E₂) = \(\frac { 4 }{ 100 } \), P(A/E₃) \(=\frac { 6 }{ 100 } \)
\(P({ E }_{ 3 }/A)=\frac { P({ E }_{ 3 }).P(A/{ E }_{ 3 }) }{ P({ E }_{ 1 }).P(A/E_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 })+P({ E }_{ 3 }).P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 30 }{ 100 } \times \frac { 6 }{ 100 } }{ \frac { 30 }{ 100 } \times \frac { 2 }{ 100 } +\frac { 40 }{ 100 } \times \frac { 4 }{ 100 } +\frac { 30 }{ 100 } \times \frac { 6 }{ 100 } } \)
\(=\frac{180}{60+160+180}\)
\(=\frac { 180 }{ 400 } \)
= 0.45

\(\text {(i) Number of shares }=\frac{\text { Investment }}{M \cdot V}\)
\(=\frac{8000}{80}=100\)
\(\text {Amount of dividend }=100 \times 100 \times \frac{7}{100}= 700\)
\(\text {(ii) SP of } 100 \text { shares }=75 \times 100= 7500\)
\(\text {Investment }=\mathrm{SP}+\text { Dividend }\)
\(=7500+700= 8200\)
\(\text {Income }=\frac{\text { Investment }}{\mathrm{M} . \mathrm{V}} \times \mathrm{FV} \times \text { Rate of dividend }\)
\(=\frac{8200}{41} \times 25 \times \frac{18}{100}= 900 \text {. }\)
\(\text {(iii) Increase in his return }=900-700\)
\(\text { = } 200\)
\(\% \text { of increase in return }=\frac{\text { Increase in return }}{\text { Investment }} \times 100\)
\(=\frac{200}{8000} \times 100=2.5 \% \)

2X = 8 - 3Y
Regression equation of X on Y is
\(X=4-\frac{3}{2} Y \)
\(b_{x y}=-\frac{3}{2}<1\)
Regression equation of Y on X be
2Y = 5 - X
\(Y=\frac{5}{2}-\frac{X}{2} \)
\(b_{y x}=-\frac{1}{2}<1\)
\(\because\) both are negative, r is - ve.
\(r =-\sqrt{b_{x y} \cdot b_{y x}} \)
\(=-\sqrt{\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)}=\frac{-\sqrt{3}}{2} \)
\(=-\frac{1.732}{2}=-0.866\)

\(p=100-6 x^2\)
\(R=p x=100 x-6 x^3\)
\(\text { LHS }=\text { Marginal Revenue }=\frac{d R}{d x}=100-18 x^2\)
\(\frac{d p}{d x}=-12 x\)
\(\text { Elasticity of demand }=\eta_d=\frac{-p}{x} \frac{d x}{d p}\)
\(=\frac{-\left(100-6 x^2\right)}{x} \cdot\left(\frac{-1}{12 x}\right)\)
\(=\frac{100-6 x^2}{12 x^2}\)
\(R H S=p\left[1-\frac{1}{\eta_d}\right]=\left(100-6 x^2\right)\left[1-\frac{12 x^2}{100-6 x^2}]\right.\)
\(=\left(100-6 x^2\right)\left[\frac{100-6 x^2-12 x^2}{100-6 x^2}\right]=100-18 x^2\)
\(\therefore \text { From (1) and (2), }\)
\(M R=p\left[1-\frac{1}{\eta_d}\right] \)

First we have to find the feasible region using the given conditions.
Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant write all the inequalities of the constraints in the form of equations.
\(\therefore \) We have the lines \(x_1+x_2 \leq 6 ; x_1 \leq 4; x_2 \leq 5\)
\({ x_{ 1 } }+{ x }_{ 2 }=6\) is a line passing through the points (0,6) and (6,0).
x₁ = 4 is a line parallel to x₂ axis at a distance of 4 units.
x₂ = 5 is a line parallel to x₁ axis at a distance of 5 units.
Now we draw the graph

The optimal solution is occurs at C(1,5)
x₁ = 1, x₂ = 5, Z_max = 28
Verification
\(If \ x_1=4, \quad x_2=2 \quad \mathrm{~B}(4,2) \)
\(If \ x_2=5, \quad x_1=1 \quad C(1,5)\)

(a) Since letters cannot be repeated, number of ways = nP_r
\(=6 P_5=6 \times 5 \times 4 \times 3 \times 2=720\)
(b) Since letters can be repeated Number of ways \(=n^r=65=7776\)
(c) Since the first place is filled with E the remaining 5 letters can be filled in 4 places in \(5 P_4 \text { ways }=5 \times 4 \times 3 \times 2=120\)
(d) Since the last 3 letters are already filled the remaining 3 letters can be filled in first 2 places in \(3 P_2 \text { ways }=3 \times 2=6\)

Let x, y, z represent the rate of commission payable on A, B and C respectively.
Then, 9x + 10y + 2z = 800
15x + 5y + 4z = 900
6x + 10y + 3z = 850
 \(\begin{bmatrix}9&10&2\\15&5&4\\6&10&3\end{bmatrix}\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 800\\900\\850 \end{bmatrix}\)
AX = B
\(X=A^{-1} B\)
Where A \(=\begin{bmatrix} 9&10&2\\15&5&4\\6&10&3 \end{bmatrix},X=\begin{bmatrix} x\\y\\z \end{bmatrix}\) and B = \(\begin{bmatrix} 800\\900\\850 \end{bmatrix}\)
= 9 (15 - 40) - 10 (45 - 24) + 2 (150 - 30)
= 9 (- 25) - 10 (21) + 2 (120)
= - 225 - 210 + 240 = - 195 \(\neq0\):
\(\therefore\) A^-1 exists.
Co-factor matrix \(=\begin{bmatrix} -25&-21&120\\-10& 15&-30\\30&-6&-105 \end{bmatrix}\)
\({A}^{-1}={{1}\over{|A|}}adj\ A={{-1}\over{195}}\begin{bmatrix} -25&-10 &30\\-21&15&-6\\120&-30&-105 \end{bmatrix}\)
X = A^-1 B
= \({{-1}\over{195}}\begin{bmatrix} -25&-10&30\\-21&15&-6\\120&-30&-105 \end{bmatrix}\begin{bmatrix}800\\900\\850 \end{bmatrix}={{-1}\over{195}}\begin{bmatrix} -20,000&-9,000&+25,5000\\ -16,8000&+13,500&-5,1000\\96,000&-27,000&-89,250 \end{bmatrix}\)
\(={{-1}\over{195}}\begin{bmatrix} -3500\\-8400\\-20,250 \end{bmatrix}=\begin{bmatrix} {{3500}\over{195}} \\{{8400}\over{195}}\\{{20,250}\over{195}} \end{bmatrix}\)
\(\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 17.95\\43.05\\103.85 \end{bmatrix}\)
\(x=17.95, y=43.08, z=103.85\)