(i) Variables:
Let x₁, x₂  represent the machines of type A and B.
(ii) Objective function:
Let Z be the maximum daily output \(\therefore Z=60{ x }_{ 1 }+40{ x }_{ 2 }\)
(iii) Constraints: 
\(1000{ x }_{ 1 }+2000{ x }_{ 2 }\le 9000\)
\({ 12x }_{ 1 }+8{ x }_{ 2 }\le 72\)
(iv) Non-negative restrictions:
Since the number of machines of type A and B cannot be negative, x₁, x₂ ≥ 0.
Hence, mathematical formulation of the LPP is  Maximize \(Z=60{ x }_{ 1 }+40{ x }_{ 2 }\)
Subject to the constraints
\(1000{ x }_{ 1 }+2000{ x }_{ 2 }\le 9000\)
\( { 12x }_{ 1 }+8{ x }_{ 2 }\le 72\)
and x₁, x₂ ≥ 0.

Given f(x, y) =3x²+4y³+6xy-x²y³+6
Differentiating 'f' partially w.r.t. 'y' we get
f_y(x,y) = 0+ 12y²+6x(1)-x²(3y²)+0
=12y²+ 6x - 3x²y²
Differentiating again partially w.r.t. 'y' we get,
fy(x, y) =24y + 0 - 3x²(2y) = 24y - 6x²y
\(\therefore\)f_yy(1, 1)= 24(1) - 6(1²)(1) = 24 - 6 = 18

Given f(x) = |x - 1|

\(f(x) = \begin{cases} x-1\quad if\quad x\ge 1 \\ 1-x\quad if\quad x<1 \end{cases}\)
\(L\left[ f\left( 1 \right) \right] =\underset { h\rightarrow 0 }{ lim } \frac { f(1-h)-f(1) }{ 1-h-1 } \left[ \therefore f(x)=1-xifx<1 \right] \)
\(=\underset { h\rightarrow 0 }{ lim } \frac { \left[ 1-(1-h)-[1-1] \right] }{ 1-h-1 } \)
\(=\underset { h\rightarrow 0 }{ lim } \frac { (1-1+h)-0 }{ -h } =\underset { h\rightarrow 0 }{ lim } \frac { h }{ -h } =-1\);...(1)
\(R\left[ f\left( 1 \right) \right] =\underset { h\rightarrow 0 }{ lim } \frac { f(1+h)-f(1) }{ 1+h-1 } [\therefore f(x)-x-1ifx\ge 1]\)
\(=\underset { h\rightarrow 0 }{ lim } \frac { (1+h)-1-[1-1] }{ 1+h-1 } \)
= \(\underset { h\rightarrow 0 }{ lim } \frac { h-0 }{ h } =1\) ......(2)
From (1) and (2),
L[f '(1)]\(\neq \) \(R\left[ { f }^{ ' }\left( 1 \right) \right] \)
f(x) is not differentiable at x=1

Using \(\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)\), we get
\(LHS=\sin^2\left(\frac{\pi}{8}+\frac{x}{2}\right)-\sin^2\left(\frac{\pi}{8}-\frac{x}{2}\right)=\sin\left(\frac{\pi}{8}+\frac x2+\frac{\pi}{8}-\frac x2\right).\sin\left(\frac{\pi}{8}+\frac x2-\frac{\pi}{8}+\frac x2\right)\)
\(=\sin\left(\frac{2\pi}{8}\right).\sin\left(\frac{2x}{2}\right)=\sin\left(\frac{\pi}{4}\right).\sin x=\frac{1}{\sqrt2}.\sin x=RHS\)
Hence proved.

\(\cos\left(\frac{5\pi}{12}\right)=\cos\left(\frac{\pi}{4}+\frac{\pi}{6}\right)\)
\(=\cos\frac{\pi}{4}\cos\frac{\pi}{6}-\sin\frac{\pi}{4}\sin\frac{\pi}{6}[\therefore \cos(A+B)=\cos A\cos B-\sin A\sin B]\)
\(=\frac{1}{\sqrt2}\times\frac{\sqrt3}{2}-\frac{1}{\sqrt2}\times\frac{1}{2}=\frac{\sqrt3-1}{2\sqrt2}\)

Given f(x) = 5x - 3
\(L\left[ f\left( x \right) \right] _{ x=3 }\) =\(\underset { x\rightarrow 3 }{ lim } f(x)=\underset { h\rightarrow 0 }{ lim } f(3-h)\)
= \(\underset { h\rightarrow 0 }{ lim } 5(3-h)-3=\underset { h\rightarrow 0 }{ lim } (15-5h-3)\)
= \(\underset { h\rightarrow 0 }{ lim } \) (12-5h) = 12-0=12
\(R\left[ f\left( x \right) \right] _{ x=3 }=\underset { x\rightarrow 3^{ + } }{ lim } f(x)=\underset { h\rightarrow 0 }{ lim } f(3+h)\)
= \(\underset { h\rightarrow 0 }{ lim } 5(3+h)-3=\underset { h\rightarrow 0 }{ lim } 15+5h-3\)
= \(\underset { h\rightarrow 0 }{ lim } 12+5h=12-0=12\)
\(L\left[ f\left( x \right) \right] _{ x=3 }=R\left[ f\left( x \right) \right] _{ x=3 }\) 
ஃ f(x) is continous at x =3

Given pair of lines is
3x²+10xy+8y²+14x+22y+15=0
2h=10
Here a=3, h=5, b=8,
Let \(\theta\) be the angle between the pair of lines
Then \(tan\quad \theta =\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } =\frac { \pm 2\sqrt { 25-3(8) } }{ 3+8 } =\frac { \pm 2\sqrt { 1 } }{ 11 } =\frac { \pm 2 }{ 11 } \)
\(\therefore \quad tan\quad \theta =\frac { 2 }{ 11 } \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 2 }{ 11 } \right) \)

An octagon has 8 angular points 
∴ Number of lines = 8 C₂ =\(\frac { 8\times 7 }{ 1\times 2 } =27\)
Number of sides = 8
∴ Number of diagonals 28 - 8 = 20

LHS 10P₃ = 10 x 9 x 8 = 720
RHS 9P₃ + 3. 9P₂ = 9 x 8 x 7 + 3 x 9 x 8
= 9 x 8 (7 + 3) = 72 (10) = 720
LHS= RHS  Hence proved.

Given \(\begin{vmatrix}2 & 4 \\5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
\(\Rightarrow\)   2 - 20 = 2x² - 24
\(\Rightarrow\)  -18 = 2x² - 24
\(\Rightarrow\)  -18 + 24 = 2x²
\(\Rightarrow\)  6 = 2x²
\(\Rightarrow\) x² = 3 
\(\Rightarrow\) x = \(\pm\sqrt{3}\)