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

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

(i) Variables:
Let x₁ and x₂ represent the number of fans and sewing machines.
(ii) Objective functions:
Let Z be the profit of the dealer.
∴ Maximize Z = 22x₁ + 18x₂ is the objective function.
(iii) Constraints:
\({ x }_{ 1 }+{ x }_{ 2 }\le 20\)
\({ 360x }_{ 1 }+{ 240x }_{ 2 }\le 5760\)
(iv) Non-negative restrictions:
Since the number of fans and sewing machine cannot be negative, we have x₁, x₂ ≥ 0.
Hence, mathematical formulation of the LPP is
Maximize \(Z=22{ x }_{ 1 }+18{ x }_{ 2 }\)
Subject to the constraints
\({ x }_{ 1 }+{ x }_{ 2 }\le 20\)
\({ 360x }_{ 1 }+{ 240x }_{ 2 }\le 5760\)
and x₁, x₂ ≥ 0.

(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.

(i) Variables:
Let x₁, x₂ represents the first and second type of bonds respectively.
(ii) Objective function:
Let Z be the maximum return
\(\therefore \quad Z=\frac { 8 }{ 100 } { x }_{ 1 }+\frac { 10 }{ 100 } { x }_{ 2 } \Rightarrow Z=0.08{ x }_{ 1 }+0.1{ x }_{ 2 }\)
(iii) Constraints: 
\({ x }_{ 1 }+{ x }_{ 2 } \le 70,000\)
\({ x }_{ 1 } \ge 10,000\)
\( { x }_{ 2 } \le 30,000\)
(iv) Non-negative restrictions:
Since the number of first and second type of bonds cannot be negative,x₁, x₂ ≥ 0.
Hence, the mathematical formulation of the LPP is maximize \(Z=0.08{ x }_{ 1 }+0.1{ x }_{ 2 }\)
Subject to the constraints 
\({ x }_{ 1 }+{ x }_{ 2 } \le 70,000\)
\( { x }_{ 1 }\ge 10,000\)
\({ x }_{ 2 }\le 30,000\)
and x₁, x₂ ≥ 0.

(i) Variables:
Let x₁, x₂ represent the number of units of X and Y. 
(ii) Constraints:

∴ 2x₁ + 3x₂ ≤ 30 and 3x₁ + x₂ ≤ 17
(iii) Non-negative restrictions:
Since the number of units of X and Y cannot be negative,x₁, x₂ ≥ 0.
Hence, the mathematical formulation of the LPP is maximize \(Z=100{ x }_{ 1 }+120{ x }_{ 2 }\)
Subject to the constraints
\( { 2x }_{ 1 }+3{ x }_{ 2 }\le 30\)
\({ 3x }_{ 1 }+{ x }_{ 2 }\le 17\)
and x₁, x₂ ≥ 0.

(i) Variables:
Let x₁, x₂ represent the dolls of A and B produced in a week.
(ii) Objective function:
Let Z be the total profit in a week.
\(\therefore Z={ 12x }_{ 1 }+16{ x }_{ 2 }\)
Since we have to maximize the profit, we have maximize \(Z={ 12x }_{ 1 }+16{ x }_{ 2 }\)
(iii) Constraints:
\({ x }_{ 1 }+{ x }_{ 2 } \le 2000\)
\({ x }_{ 1 }-{ 2x }_{ 2 }\ge 0\)
\( { x }_{ 1 }-{ 3x }_{ 2 }\le 600\)
(iv) Non-negative restictions:
 Since the number of dolls on type A and B cannot be negative, we have \({ x }_{ 1 },{ x }_{ 2 }\ge 0\)
Hence, the mathematical formation of LPP is
Maximize \(Z={ 12x }_{ 1 }+16{ x }_{ 2 }\)
Subject to the constraints
\({ x }_{ 1 }+{ x }_{ 2 } \le 2000\)
\({ x }_{ 1 }-{ 2x }_{ 2 }\ge 0\)
\( { x }_{ 1 }-{ 3x }_{ 2 }\le 600\)
and x₁, x₂ ≥ 0.

(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.

(i) Variables: Let x₁be the executive class tickets and x₂ be the economy class ticket.
(ii) Objective function: Let Z be the maximum profit
Maximize Z=1000x₁+600x₂
(iii) Constraints:\({ x }_{ 1 }\ge 20\)
\(\Rightarrow { x }_{ 2 }\ge 4{ x }_{ 1 }\ and\ { x }_{ 1 }+{ x }_{ 2 }\le 200\)
\( \Rightarrow 4{ x }_{ 1 }\le { x }_{ 2}\ and\ { x }_{ 1 }+{ x }_{ 2 }\le 200\)
\(\Rightarrow 4{ x }_{ 1 }-{ x }_{ 2 }\le 0 \ and \ { x }_{ 1 }+{ x }_{ 2 }\le 200\)
(iv) Non-negative restrictions:
Since the number of tickets cannot be negative, \({ x }_{ 1 },{ x }_{ 2 }\ge 0\)
Hence, the mathematical formulation of the LPP is Maximize Z=1000x₁+600x₂
Subject to the constraints
\({ x }_{ 1 }+{ x }_{ 2 } \le 2000\)
\( { x }_{ 1 } \ge 20\)
\(4{ x }_{ 1 }-{ x }_{ 2 } \le 0 \ and \ { x }_{ 1 },{ x }_{ 2 }\ge 0\)

Given f(x, y) =3x²+4y³+6xy-x²y³+6
We know that f_y(x, y) 12y² + 6x - 3x²y²
Differentiating again partially w.r.t. 'x' we get,
f_xy(x,y) 0 + 6 - 3y² (2x) = 6 - 6xy²
\(\therefore\)f_xy(2, 1) = 6 - 6(2)(1)² = 6 - 12 = - 6

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 y=x-1/x
Differentiating w.r.t.'x' we get,
\({dy\over dx}=1+{1\over x^2}>0\) for all real values of x, except x=0
\(\therefore \) y is a strictly increasing function for all real values of x(x\(\neq\)0)

Given x² +xy +y² = 100
Differentiating with respect to 'x' we get
2x +x.\(\frac { dy }{ dx } \) + y(1) + 2y \(\frac { dy }{ dx } \) = 0
\(\Rightarrow\) \(\frac { dy }{ dx } \) (x+2y) = (x +2y) = -2 x - y
\(\Rightarrow\)  \(\frac { dy }{ dx } \) = \(\frac { -(2x+y) }{ x+2y } \)

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 \(\cos^{-1}\left(\frac{-1}{\sqrt2}\right)=\theta\)
\(\Rightarrow\cos\theta=-\frac{1}{\sqrt2}\)
We know that the range of principal value of \(\cos^{-1}\)is \([0,\pi]\)
\(\therefore\cos\theta=-\frac{1}{\sqrt2}\Rightarrow-\cos\frac{\pi}{4}=\cos\left(\pi-\frac{\pi}{4}\right)=\cos\frac{3\pi}{4}\)
\(\therefore\cos\theta=\cos\frac{3\pi}{4}\)
\(\Rightarrow\theta=\frac{3\pi}{4}\epsilon[0,\pi]\)
Thus the principal value of \(\cos^{-1}\left(-\frac{1}{\sqrt2}\right)\)is \(\frac{3\pi}{4}\)

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

We have \(\cos 2x=\frac{1-\tan^2x}{1+\tan^2x}\)
\(\therefore LHS=\cos2x+\sin^2\phi=\frac{1-\tan^2x}{1+\tan^2x}+\sin^2\phi\)
\(=\frac{1-\left(2\tan^\phi+1\right)}{1+(2\tan^2\phi)+1}+\sin^2\phi\)      \([\because tan^{ 2 }x=2tan^{ 2 }\phi +1]\)
\(=\frac{-2\tan^2\phi}{2(1+\tan^2\phi)}+\sin^2\phi=\frac{-\tan^2\phi}{\sec^2\phi}+\sin^2\phi\)
\(=\frac{-\sin^2\phi}{\cos^2\phi.\frac{1}{\cos^2\phi}}+\sin^2\phi=-\sin^2\phi+\sin^2\phi=0=RHS\)
Hence Proved.

\(lety=\frac { { x }^{ 2 }cos\frac { \pi }{ 4 } }{ sinx } ={ x }^{ 2 }\times \frac { 1 }{ \sqrt { 2 } sinx } \)
\(\therefore y=\frac { 1 }{ \sqrt { 2 } } \frac { { x }^{ 2 } }{ sinx } \)
Differentiating with respect to 'x' we get
\(\frac { dy }{ dx } =\frac { 1 }{ \sqrt { 2 } } \left[ \frac { sinx.(2x)-{ x }^{ 2 }cosx }{ sin^{ 2 }x } \right] \)[by quotient rule]
= \(\frac { x }{ \sqrt { 2 } } \left[ \frac { 2sinx-xcosx }{ { sin }^{ 2 }x } \right] \)
= \(\frac { x }{ \sqrt { 2 } } \left[ \frac { 2sinx }{ { sin }^{ 2 }x } -\frac { xcosx }{ sin^{ 2 }x } \right] =\frac { x }{ \sqrt { 2 } } \)[cosec x-x cot x cosec x]
\(\frac { x }{ \sqrt { 2 } } cosecx[2-cotx]\)

\(\underset { x\rightarrow \frac { 1 }{ 2 } }{ lim } \frac { { 4x }^{ 2 }-1 }{ 2x-1 } =\underset { x\rightarrow \frac { 1 }{ 2 } }{ lim } \frac { \left( { 2x }^{ 2 } \right) -1^{ 2 } }{ 2x-1 } =\underset { x\rightarrow \frac { \pi}{ 2 } }{ lim } \frac { (2x+1)(2x-1) }{ 2x-1 } =\underset { x\rightarrow \frac { \pi }{ 2 } }{ lim } (2x+1)\)
= \(2\left( \frac { 1 }{ 2 } \right) +1=1+1=2\)

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

LHS=cos18^o-sin 18^o
=cos18^o-cos72^o[sin18^o=sin(90-72^o)=cos72^o]
\(=2\sin\left(\frac{18^o+72'}{2}\right).\sin\left(\frac{72^o-18'}{2}\right)\left[\because\cos C-\cos D=2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{D-C}{2}\right)\right]\)
\(=2\sin45^o\sin27^o=2\times\frac{1}{\sqrt2}\sin27^o=\sqrt{2}\sin27^o=RHS\)
Hence proved.

LHS=\(\frac{\tan 69^o+\tan 66^o}{1-\tan 69^o\tan 66^o}\left[\because\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan(A+B)\right]\)
= tan (69° + 66°) = tan (135°)
= tan (180 - 45°) = - tan 45° = -1 = RHS.
Hence proved.

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.

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

Since A, B, C, D are angles of a quadrilateral, A + B + C + D =\(2\pi\)
A + B + C + D =\(2\pi\)
\(\Rightarrow A+B=2\pi-(C+D)\)
\(\Rightarrow\sin(A+B)=\sin[2\pi-(C+D)]\)
=-sin(C+D)[\(\therefore2\pi-(C+D)\)is in the IV quadrant]
\(\Rightarrow\sin(A+B)+\sin(C+D)=0\)

The given parabola x²=6y is of the form x²=4 ay where 4a=6 \(\Rightarrow a=\frac { 6 }{ 4 } \Rightarrow a=\frac { 3 }{ 2 } \)
Focus is (0, a)=\(\left( 0,\frac { 3 }{ 2 } \right) \)
Vertex is (0, 0)
Equation of the directrix is y=-a \(\Rightarrow\) \(y=\frac { -3 }{ 2 } \Rightarrow 2y+3=0\)
Length of the latus section = 4a=6 units.

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

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

Since \(\pi , x lies in III quadrant.
\(\therefore\) sin x is negative and tan x is positive
\(\therefore\sin x=\pm\sqrt{1-\cos^2x}=-\sqrt{1-\frac{1}{4}}=-\frac{\sqrt3}{2}\)
\(\therefore cosec\ x=-\frac{2}{\sqrt{3}}\)
\(\tan x=\frac{\sin x}{\cos x}=\frac{-\frac{\sqrt3}{2}}{-\frac{1}{2}}=\sqrt3\)
Hence, \(4\tan^2x-3cosec^2x=4\times3-3\times\frac43=12-4=8\)

Given \(f(x) =\begin{cases} \frac { 1-cos4x }{ 8{ x }^{ 2 } } \quad ifx\neq 0 \\ k\quad \quad \quad ifx=0 \end{cases}\)
\(\neq \underset { x\rightarrow 0 }{ lim } \quad f(x)=\underset { x\rightarrow 0 }{ lim } \frac { 1-cos4x }{ { 8x }^{ 2 } } =\underset { x\rightarrow 0 }{ lim } \frac { 2sin^{ 2 }2x }{ { 8x }^{ 2 } } \)   [ஃ 1-cos²z =sin²x]
= \(\underset { x\rightarrow 0 }{ lim } \frac { sin^{ 2 }2x }{ { 4x }^{ 2 } } =\underset { x\rightarrow 0 }{ lim } \left( \frac { sin2x }{ 2x } \right) ^{ 2 }\)
= (1)²    ....(1)
Given f(0) = k ...(2)
Since f(x) is continous at x = 0,
\(\underset { x\rightarrow 0 }{ lim } \) f(x) = f(0)
1 = k [using (1) and (2)
K = 1

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 \(tan\quad \theta =\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } =\frac { \pm 2\sqrt { 25-3(8) } }{ 3+8 } =\frac { \pm 2\sqrt { 1 } }{ 11 } =\frac { \pm 2 }{ 11 } \)
\(\therefore \quad tan\quad \theta =\frac { 2 }{ 11 } \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 2 }{ 11 } \right) \)

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

Equation of the parabola is x²+y=6x-14
⇒ x²-6x=-y-14
⇒ x²-6x+9=-y-14+9 (Adding 9 both sides)
⇒ (x-3)²=-y-5
⇒ (x-3)²=-1(y+5)
⇒ x²=-y where X=x-3, Y=y+5

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

We have P(n): 2³ⁿ-1 is a multiple of 7
 ∴  P (5) : 2¹⁵ -1 = 32767
= 7 (4681) which is multiple of 7
\(\therefore\) P (5) is true

(1+2x + x²)²⁷ = [(1 + x)²]²⁷ = (1 + x)⁵⁴
Here n = 54, x = 1, a = x
The general term is t_{r+ 1} = nC_rx^n-r a^r
\(\Rightarrow\)  t_{r+ 1} = 54C_r (1)^54-r.X^r
\(\Rightarrow\)  t_r+1 = 54C_rx^r
To find the co-efficient of x⁴⁰, put r = 40
∴  t₄₁ = 54 C₄₀ x⁴⁰ 
 ∴  Co-efficient of x⁴⁰ is 54 C₄₀ = 54C_I4   \(\left[ \because ncr=nc_{ n-r } \right] \)

In  \({ \left( x+\frac { 1 }{ x } \right) }^{ 6 }\)n = 6,x = x ,a =\(\frac { 1 }{ x } \)
General term is t_{r+ 1} =nC_rx^n-ra^r
   t_r+1 = 6C_rx^6-r\({ \left( \frac { 1 }{ x } \right) }^{ r }\)
To find t₃, put r =2
∴ t₃=6C₂x^6-2\({ \left( \frac { 1 }{ x } \right) }^{ 2 }\)=\(\frac { 6\times 5 }{ 2\times 1 } \).x⁴.\(\frac { 1 }{ { x }^{ 2 } } \)=15x²

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

nCx = nCy \(\Rightarrow\)x= y or x + y = n
\(\Rightarrow\)  \(\therefore\) 25 _Cn+5 = 25 C_2n-1
\(\Rightarrow\)   n + 5 = 2n-1
\(\Rightarrow\)   n =5 (or) n + 5 + 2n - 1 = 25
\(\Rightarrow\)   6 = 2n - n or 3n + 4 = 25
\(\Rightarrow\) n=  6 or 3n = 21\(\Rightarrow\) n=7
∴ n = 6 or n = 7

Number of ways of forming necklace
\(\frac { (10-1)! }{ 2! } =\frac { 9! }{ 2 } =181440\)

The first place can be filled in only one way namely T and the remaining 7 letters can be arranged in 7! ways.
\(\therefore\) Total number of arrangements = 1 x 7! = 5040.

Number of vowels are A, E, I, 0, U = 5
∴ Taking 2 vowels from 5 vowels = 5P₂
= 5 x 4 = 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.

\(\frac { 1 }{ 5! } +\frac { 1 }{ 6! } +\frac { 1 }{ 7! } \)
\(=\frac { 1 }{ 5! } +\frac { 1 }{ 6\times 5! } +\frac { 1 }{ 7\times 6\times 5! } =\frac { 1 }{ 5! } \left( 1+\frac { 1 }{ 6 } +\frac { 1 }{ 42 } \right) =\frac { 1 }{ 5! } \left( \frac { 42+7+1 }{ 42 } \right) \)
\(=\frac { 1 }{ 5! } \frac { \left( 50 \right) }{ 42 } =\frac { 5 }{ 5! } \)

Number of ways for first passenger to occupy seat = 6. ....(1)
Number of ways for the second passenger to occupy seat = 5
Number of ways for the third passenger to occupy seat = 4.
∴By fundamental principle of counting, the total number of ways for all the three to occupy seats
= 6 x 5 x 4 = 120.

\(\frac { 12x-17 }{ (x-2)(x-1) } =\frac { A }{ x+2 } +\frac { B }{ x-1 } \)
\(\Rightarrow \frac { 12x-17 }{ (x-2)(x-1) } =\frac { A(x-1)+B(x-2) }{ (x-2)(x-1) } \)
\(\Rightarrow 12x-17=A(x-1)+B(x-2)\)
Putting x=1 in (1) we get, 
\(12-17= B(1-2) \Rightarrow -5=-B \Rightarrow \boxed { B=5 } \)
Putting x=2 in (1) we get, 
\(24-17=A(2-1) \Rightarrow 7=A(1) \Rightarrow \boxed { A=7 } \)
\(\therefore \frac { 12x-17 }{ (x-2)(x-1) } =\frac { 7 }{ x+2 } +\frac { 5 }{ x-1 } \)

Let A = \(\begin{vmatrix} x &a &x+a \\ y & b &y+b \\z & c & z+c \end{vmatrix}\)
Applying the elementary transformation, \(C_1\rightarrow C_1+C_2\) we get,
\(A=\begin{vmatrix} x+a&a&a+x\\y+b&b&y+b\\z+c&c&z+c\end{vmatrix}=0[C_1\equiv C_3]\)
\(\therefore\) |A| = 0.

Let |A| = \(\begin{vmatrix} 6 &5 &12 \\ 2 & 4 &4 \\2 & 1 & 4 \end{vmatrix}\)
Taking 2 common from C₁ and 4 common from C₃, we get,
\(|A|=2\times4\begin{vmatrix} 3 & 5&3 \\ 1 & 4 & 1\\1 &1 &1\end{vmatrix}=8\times 0\ [\because C_1\equiv C_3]=0\)

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 = \(\pm\sqrt{3}\)

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.

Given = \(\begin{bmatrix} 1 & 2 \\ 4 & 2\end{bmatrix},\) then 2A = \(\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}\)
\(\therefore\)   |2A| = \(\begin{vmatrix}2 & 4 \\8 & 4 \end{vmatrix}\) = 8 - 32 = -24   ...(1)
Also, |A| = \(\begin{vmatrix} 1&2 \\4 & 2 \end{vmatrix}\) = 2 - 8 =  -6
\(\therefore\)  4|A| = 4(-6) = -24   ....(2)
From (1) and (2), |2A| = 4.|A|

AB = \(\begin{bmatrix} 1 \\ -4 \\3 \end{bmatrix}\begin{bmatrix} -1 &2 & 1 \end{bmatrix}=\begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\-3 & 6 & 3 \end{bmatrix}\)
\(\therefore\)  \({(AB)}^{T}=\begin{bmatrix} -1 &4&-3 \\ 2 & -8&6\\1&-4&3 \end{bmatrix}\)  ....(1)
\({B}^{T}=\begin{bmatrix} -1 & 2 & 1 \end{bmatrix}^{T}=\begin{bmatrix} -1\\2\\1\end{bmatrix}\)and \({A}^{T}={\begin{bmatrix} 1\\-4\\3\end{bmatrix}}^{T}=\begin{bmatrix} 1&-4&3 \end{bmatrix}\)
\(\therefore\)  \({B}^{T}{A}^{T}=\begin{bmatrix} -1\\2\\1 \end{bmatrix}\begin{bmatrix} 1&-4&3 \end{bmatrix}=\begin{bmatrix} -4 & 4&-3 \\ 2&-8 &6\\1&-4&3 \end{bmatrix}\)  ....(2)
From (1) and (2), (AB)^T = B^T . A^T

Given R=\({12000\over12}=1000 units /month\)
C₁= 20 paise per unit / month
C₂= Rs.350 per run
(i) \(q_o=\sqrt{2C_3R\over C_1}=\sqrt{2\times 350\times 1000\over 0.20}=1870 units/ run\)
(ii)\(t_o=\sqrt{2C_3\over C_1R}=\sqrt{2\times 350\over 0.20\times 1000}=1.87 months\)
=1.87 x 30 = 56 days
(iii)\(C_o=\sqrt{2C_1 C_3R}=\sqrt{2\times 0.20\times12\times350\times(1000\times12)}\)
=Rs. 4490 per year

Given C =\({1\over5}x^2-6x+100\)
Differentiating w.r.t. 'x' we get,
\({dC\over dx}={1\over5}(2x)-6\)
\({dC\over dx}=0\)
\(\Rightarrow {2x\over 5}-6=0\)
\(\Rightarrow {2x\over 5}=6\)
\(\Rightarrow x={6\times5\over2}=15\)
Now  \({d^2C\over dx^2}={2\over5}>0\)
\(\therefore\)Cost function is minimum when x = 15.

Let y=x³-6x²+7
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=3x^2-12x\)
\({dy\over dx}=0\)
\(\Rightarrow3x^2-12x=0\)
\(\Rightarrow3x(x-4)=0\)
\(\Rightarrow x=0 \ or \ x=4\)
\({d^2y\over dx^2}=6x-12\)
when x=0 \({d^2y\over dx^2}=-12<0\)
\(\therefore \) y is maximum at x=0
\(\therefore \) maximum value=0³-6(0)²+7=7
when x=4, \({d^2y\over dx^2}=-6(4)-12=12>0\)
\(\therefore \) y is minimum at x=4
\(\therefore \) Minimum value =4⁴-6(4)²+7=64-96+7=-25
Hence maximum value is 7 and minimum value is -25.

Let y= 75-12x+6x²-x³
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=0-12+12x-3x^2=-3(x^2-4x+4)=-3(x-2)^2\)
\(\Rightarrow{dy\over dx}\le 0\) for all \(x \in (-\infty , \infty)\)
\(\therefore \) y decreases as x increases.

Given y=1+1/x
Differentiating w.r.t. 'x' we get,
\({dy\over dx}={-1\over x^2}<0\) for all real values of x, except x=0.
\(\therefore \) y is a strictly decreasing function for all real values of x(x\(\neq\)0).

Let y = (sec x -1) (sec x +1)
y = sec² x -1    \(\left[ \therefore (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 } \right] \)
\(\therefore\)y =tan²x
Differentiating with respect to 'x' we get 
\(\frac { dy }{ dx } =2tanx\frac { d }{ dx } (tanx)\) \([\therefore 1+tan^{ 2 }x=sec^{ 2 }x]\)
=2tanx.sec² x

\(\underset { h\rightarrow 0 }{ lim } \frac { \sqrt { x+h } -\sqrt { x } }{ h } =\underset { h\rightarrow 0 }{ lim } \frac { (x+h)^{ \frac { 1 }{ 2 } }-{ x }^{ \frac { 1 }{ 2 } } }{ (x+h)-x } \)  [Adding and subtracting x in the denominator]
= \(\frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } -1 }\left[ \therefore \underset { x-\rightarrow a }{ lim } \frac { { x }^{ n }-{ a }^{ n } }{ z-a } =n.a^{ n-1 } \right] \)
= \(\frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } }=\frac { 1 }{ 2\sqrt { x } } \)