Using the precedence relationship and following the rules of network construction, the required network is shown in the following figure.

(i) Variables:
Let x₁, x₂ represent bags of brand P and brand Q. 
(ii) Objective function:
Let Z be the amount of nitrogen. Since the amount of nitrogen is to be minimized, we have minimize Z = 3x + 3.5y
(iii) Constraints:
For Phosphoric acid,\({ x }_{ 1 }+2{ x }_{ 2 }\ge 240\)
For Potash, \({ 3x }_{ 1 }+{ 1.5x }_{ 2 }\ge 270\)
For Chlorine, \({ 1.5x }_{ 1 }+2{ x }_{ 2 }\ge 310\)
(iv) Non-negative restrictions:
Since the number of bags of brand P and Q, cannot be negative,\({ x }_{ 1 },{ x }_{ 2 }\ge 0\).
Here Mathematical form of the LPP is Minimize Z = 3x₁ + 3.5x₂ 
Subject to the constraints
\({ x }_{ 1 }+2{ x }_{ 2 }\ge 240\)
\({ 3x }_{ 1 }+{ 1.5x }_{ 2 }\ge 270\)
\( { 1.5x }_{ 1 }+2{ x }_{ 2 }\ge 310\)
and x₁, x₂ ≥ 0.

Given f(x, y) =3x²+4y³+6xy-x²y³+6
Differentiating partially w.r.t. 'x' we get,
fx(x, y)=6x + 0 + 6y(1) -y³(2x) + 0
=6x + 6y- 2xy³
\(\therefore f_x(1,-1)=\) 6(1) +6(-1)-2(1)(-1)³
= 6-6+2
= 2

Let f(x) = \({ x }^{ \frac { 2 }{ 3 } }\)
f(x+h) = (x+h)\(^{ \frac { 2 }{ 3 } }\)
\(\frac { d }{ dx } (f(x))=\underset { h\rightarrow 0 }{ lim } \frac { f(x+h)-f(x) }{ h } \)
= \(\underset { h\rightarrow 0 }{ lim } \frac { (x+h)^{ \frac { 2 }{ 3 } }-x^{ \frac { 2 }{ 3 } } }{ x+h-x } \) [adding and subtracting x in the denominator]
= \(\frac { 2 }{ 3 } .x^{ \frac { 2 }{ 3 } -1 }\)
\(\left[ \therefore \underset { x\rightarrow a }{ Lt } \frac { { x }^{ n }-{ a }^{ n } }{ x-a } =n-a^{ n-1 } \right] =\frac { 2 }{ 3 } { x }^{ \frac { -1 }{ 3 } }\)
\(\therefore \frac { d }{ dx } \left( x^{ \frac { 2 }{ 3 } } \right) -\frac { 2 }{ 3 } .x^{ \frac { -1 }{ 3 } }\)

Let \(\cos^{-1}(\frac12)=\theta\)
\(\Rightarrow\frac12=\cos\theta\Rightarrow\cos=\frac{\pi}{3}\cos\theta\)
\(\Rightarrow\theta=\frac{\pi}{3}\)
\(\therefore\cos\left[\frac{\pi}{3}-\cos^{-1}(\frac{1}{2})\right]=\cos\left[\frac{\pi}{3}-\frac{\pi}{3}\right]=\cos(0)=1.\)

\(\frac{15\pi}{4}=15\times45^o=675^o\)
\(\cot\left(\frac{-15\pi}{4}\right)=\)cot (-675°)= - cot 675° = - cot (720 -45°) = -cot (2 x 360° - 45°)
= -(-cot 45°)(\(\therefore\) 675° is in the IV quadrant) =-(-1) = 1.

Given f(x) = 5x - \(\left| x \right| \)
\(\therefore f(x)=\begin{cases} 5x-x\quad if\quad x\ge 0 \\ 5x-(-x)\quad if\quad x>0 \end{cases}=\begin{cases} 4x\quad ifx\ge 0 \\ 6x\quad if\quad x<0 \end{cases}\)
\(L\left[ f\left( x \right) \right] _{ x=0 }=\underset { x\rightarrow 0 }{ lim }f\left( x \right) =\underset { h\rightarrow 0- }{ lim } \quad f(o-h)\)
\(=\underset { h\rightarrow 0 }{ lim } f(-h)=\underset { h\rightarrow 0 }{ lim } 6(-h)\quad \quad \left[ \therefore f(x)=6x\quad ifx\ge 0 \right] \)
= 0
\(R\left[ f\left( x \right) \right] _{ x=0 }=\underset { x\rightarrow 0 }{ lim } lif\left( x \right) =\underset { h\rightarrow 0 }{ lim } \quad f(o+h)\)
\(\underset { h\rightarrow 0 }{ lim } f(h)=\underset { h\rightarrow 0 }{ lim } 4(h)\quad \left[ \therefore f(x)=4xi\quad fx\ge 0 \right] \)
=0
∴ \(L\left[ f\left( x \right) \right] _{ x=0 }=R\left[ f\left( x \right) \right] _{ x=0 }\)
∴ f(x) is continuous at x = 0

The given equation is
y²=4x+4y
⇒ y²-4y=4x
⇒ y²-4y=4x
⇒ y²-4y+4=4x+4 (Adding 4 on both sides)
⇒ (y-2)²=4(x+1)
⇒ y²=4 where X=x+1 ⇒ Y=y-2

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.

Let \(|A|=\begin{vmatrix} 2&-1&-2 \\ 0 &2&-1\\3&-5&0 \end{vmatrix}\)
Expanding along R₁ we get, 
\(|A|=2\begin{vmatrix} 2 & -1\\ -5 & 0\end{vmatrix}+1\begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix}-2\begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix}\)
= 2 ( 0 - 5 ) + 1 ( 0 + 3 )  - 2 ( 0 - 6 )
= - 10 + 3 + 12 = 5
\(\therefore\) |A| = 5.