\(x=a\theta \) 
\(\cfrac {dx }{ d\theta } =a\) 
\(y=\cfrac { a }{ \theta } \) 
\(\cfrac {dy }{ d\theta } =\cfrac { -a }{ { \theta }^{ 2 } } \) 
\(\cfrac { dy }{ dx } =\cfrac { \frac { dy }{ d\theta } }{ \frac { dx }{ d\theta } } \) 
\(=\cfrac { \left( \frac { -\alpha }{ { \theta }^{ 2 } } \right) }{ a } \) 
\(=\cfrac { 1 }{ { \theta }^{ 2 } } \) 
\(-\cfrac { y }{ x } \) 
i.e.,\(\cfrac { dy }{ dx } +\cfrac { y }{ x } =0\) 
Aliter :
Take \(xy=a\theta .\cfrac { a }{ \theta } \) 
xy = a²
Differentiating with respect to x,
\(x\cfrac { dy }{ dx } +y=0\)
\( \cfrac { dy }{ dx } +\cfrac { y }{ x } =0\)

\(\frac{\sin A-\sin C}{\cos C-\cos A} =\frac{2 \sin \left(\frac{A-C}{2}\right) \cos \left(\frac{A+C}{2}\right)}{2 \sin \left(\frac{A+C}{2}\right) \sin \left(\frac{A-C}{2}\right)}\)
\(=\frac{\cos \left(\frac{A+C}{2}\right)}{\sin \left(\frac{A+C}{2}\right)}=\cot \left(\frac{A+C}{2}\right)=\cot B\text { (since } A, B, C \text { are in } A . P ., B=\frac{A+C}{2} \text { ) }\)

\(\tan2A=\cfrac { 2\tan A }{ 1-{ \tan }^{ 2 }A } \) 
Put A = 30^o in the above identity, we get
\({ \tan60 }^{ o }=\cfrac { 2\tan{ 60 }^{ o } }{ 1-{ \tan }^{ 2 }{ 30 }^{ o } } \) 
= \(\cfrac { 2-\frac { 1 }{ \sqrt { 3 } } }{ 1-\frac { 1 }{ 3 } } =\cfrac { \frac { 2 }{ \sqrt { 3 } } }{ \frac { 2 }{ 3 } } =\cfrac { 2 }{ \sqrt { 3 } } \times \cfrac { 3 }{ 2 } =\sqrt { 3 } \)

Let \(\frac { 1 }{ \left( { x }^{ 2 }-1 \right) } =\frac { A }{ x-1 } +\frac { B }{ x+1 } \)
Multiplying both sides by (x - 1)(x + 1), we get
1 = A(x +1) + ( Bx -1) ... (1)
Put x =1 in (1) we get, 1 = A(2)
 \(\therefore\) A = \(\frac{1}{2}\)
Put x = -1 in (1) we get, 1 = A(0) + B(-2)
\(\therefore\)  B = - \(\frac{1}{2}\)

\(\left| \begin{matrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{matrix} \right| \) = 1 (Minor of 1) –2 (Minor of 2) + 4 (Minor of 4)
\(=1\left| \begin{matrix} 3 & 0 \\ 1 & 0 \end{matrix} \right| -2\left| \begin{matrix} -1 & 0 \\ 4 & 0 \end{matrix} \right| +4\left| \begin{matrix} -1 & 3 \\ 4 & 1 \end{matrix} \right| \)
= 0 – 0 – 52 = –52.

(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 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.

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

Arranging in ascending order
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
n = 11
 Q₁ = size of \((\frac { { n+1} }{ 4 } )^{ th }\)  value
= size of \({ \left( \frac { 11+1 }{ 4 } \right) }^{ th }\) value
= size of 3^rd value = 6
Q₃ = Size of \({ \left( \frac { 3(n+1) }{ 4 } \right) }^{ th }\) value
= size of \({ \left( 3\left( \frac { 11+1 }{ 4 } \right) \right) }^{ th }\) value
=size of 3(3)^th value
= size of 9^th value = 18

Given C = 10- 4x³ + 3x⁴
(i) Average Cost (AC) = \({C\over x}=\frac { 10 }{ x } \) - 4x² + 3x³
(ii) Marginal Cost (MC) = \({dC\over dx}\)= -12x² + 12x³
(iii) Marginal Average Cost (MAC) = \({d\over dx }(AC)=-\frac { 10 }{ { x }^{ 2 } } \) - 8x + 9x²

Let the sales be X and advertisement expenditure be Y
Given \(\bar { X } \) = 40, \(\bar { Y } \) = 6, σ_x = 10, σ_y = 1.5 and r = 0.9
Equation of line of regression x on y is
X -\(\bar { X } \) = r\(\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } (Y-\bar { Y } )\)
X - 40 = (0.9)\(\frac{10}{1.5}\)(Y - 6)
X - 40 = 6Y - 36
X = 6Y + 4
When advertisement expenditure is 10 crores i.e., Y = 10 then sales X = 6(10) + 4 = 64 which implies sales is 64.

The required network for the above information.

\(\mathrm{LHS}=\frac{\cos 2 A-\cos 3 A}{\sin 2 A+\sin 3 A}\)
\(=\frac{-2 \sin \frac{2 A-3 A}{2} \sin \frac{2 A+3 A}{2}}{2 \sin \frac{2 A+3 A}{2} \cos \frac{2 A-3 A}{2}}\)
\(=\frac{\sin A / 2}{\cos A / 2}=\tan A / 2\)
= RHS
Hence proved.

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

sin 76^o cos 16^o + cos 76^o sin 16^o
= sin (76 ° + 16°) = sin 92°

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

Let y = \(\frac { 5 }{ { x }^{ 4 } } -\frac { 2 }{ { x }^{ 3 } } +\frac { 5 }{ x } \) = 5x^-4 - 2x^-3 + 5x^-1
dy/dx = -20x^-5 + 6x^-4 - 5x^-2
= \(-\frac { 20 }{ { x }^{ 5 } } +\frac { 6 }{ { x }^{ 4 } } -\frac { 5 }{ { x }^{ 2 } } \)

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

x = 3 cos\(\theta\) , y = 3 sin\(\theta\)
\(\cos \theta=\frac{x}{3}\ \sin \theta=\frac{y}{3}\)
We know that \(\cos ^2 \theta+\sin ^2 \theta=1\)
\(\frac {x^2}{9} + \frac {y^2}{9} = 1\)
\(\Rightarrow\) x² + y² = 9

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

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 = \(\pm\sqrt{3}\)

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

\(f(x)=\frac { { a }^{ -x }-1 }{ { a }^{ -x }+1 } \) 
\(f(-x) =\frac { \frac { 1 }{ { a }^{ x } } -1 }{ \frac { 1 }{ { a }^{ x } } +1 } =\frac { \frac { 1-{ a }^{ x } }{ { a }^{ x } } }{ \frac { 1+{ a }^{ x } }{ { a }^{ x } } } \)
\(=-\left( \frac { { a }^{ x }-1 }{ { a }^{ x }-1 } \right) \) = -f(x)
\(\therefore\) f is an odd function.

B \(=\begin{bmatrix} 0.50 & 0.30 \\ 0.41 & 0.33 \end{bmatrix}\)
I - B = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.50 & 0.30 \\ 0.41 & 0.33 \end{bmatrix}=\begin{bmatrix} 0.50 & -0.30 \\ -0.41 & 0.67 \end{bmatrix}\)
= (0.50) (0.67) - (0.30) (0.41)
\(|I-B|\) = 0.335 - 0.123 = 0.212 >  0
Since the main diagonal elements of I - B are positive and |I-B| is positive. Hawkins Simon conditions are satisfied. Therefore given system is viable

\(\frac { 4x+1 }{ (x-2)(x+1) } =\frac { A }{ x-2 } +\frac { B }{ x+1 } \)
⇒ \(\frac { 4x+1 }{ (x-2)(x+1) } =\frac { A(x+1)+B(x-2) }{ (x-2)(x+1) } \)
⇒ 4x + 1 = A(x + 1) + B(x - 2)
If x = 2
9 = A(2 + 1)
\(\Rightarrow A=\frac{9}{3}=3\)
If x = -1
-4 + 1 = B(-1 - 2)
⇒ -3 = -3B  ⇒ \(\ B=1 \)
\(\frac { 4x+1 }{ (x-2)(x+1) } =\frac { 3 }{ x-2 } +\frac { 1 }{ x+1 } \)