Let \(\left(\cos ^{-1} \frac{8}{17}\right)=\theta\) 
\(\cos \theta=\frac{8}{17}\) 
\(\sin\theta =\sqrt { 1-{ \cos }^{ 2 }\theta } \)
= \(\sqrt { 1-\cfrac { 64 }{ 289 } } \) 
= \(\cfrac { 15 }{ 17 } \)   ...(2)
From (1) and (2), we get
Now \(\tan\left( \cos\cfrac { 8 }{ 17 } \right) =\tan\theta =\cfrac { \sin\theta }{ \cos\theta } \)
\(=\cfrac { \frac { 15 }{ 17 } }{ \frac { 8 }{ 17 } } =\cfrac { 15 }{ 8 } \)

Let \(\sin ^{-1}(1 / 2)=y\ \text {where }-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)
\(\therefore \sin y=(1 / 2)=\sin \left(\frac{\pi}{6}\right)\) 
\(y=\frac{\pi}{6}\) 
The principal value of \(\sin ^{-1}(1 / 2) \text { is } \frac{\pi}{6}\)

Equation of a circle when the end points of the diameter are given is
\(\left( x-{ x }_{ 1 } \right) \left( x-{ x }_{ 2 } \right) +\left( y-{ y }_{ 1 } \right) \left( y-{ y }_{ 2 } \right) =0\) 
Here \(\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 2,4 \right) \) and
\(\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 3,-2 \right) \)
\(\therefore \left( x-2 \right) \left( x-3 \right) +\left( y-4 \right) \left( y+2 \right) =0\)
\({ x }^{ 2 }+{ y }^{ 2 }-5x-2y-2=0\)

Let A(2, –3) and B (3, –4) be the given points
Let P(x₁,y₁) be any point on the locus.
Given that PA = PB.
PA² = PB²
\(\left( { x }_{ 1 }-2 \right) ^{ 2 }+\left( { y }_{ 1 }+3 \right) ^{ 2 }=\left( { x }_{ 1 }-3 \right) ^{ 2 }+\left( { y }_{ 1 }+4 \right) ^{ 2 }\)
\({ x }_{ 1 }^{ 2 }-4{ x }_{ 1 }+4+y_{ 1 }^{ 2 }+6{ y }_{ 1 }+9={ x }_{ 1 }^{ 2 }-6{ x }_{ 1 }+9+{ y }_{ 1 }^{ 2 }+8{ y }_{ 1 }+16\)
\(i.e,{ 2x }_{ 1 }-{ 2y }_{ 1 }-12=0\)
\( i.e.,{ x }_{ 1 }-{ y }_{ 1 }-6=0\) 
The locus of \(P({ x }_{ 1 },{ y }_{ 1 })\ is \ x-y-6=0\)

Let A = \(\left[ \begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix} \right] \)
|A| = \(\left| \begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix} \right| \)
= 4 – 4 = 0
\(\therefore\) A is a singular matrix

(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 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.

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

\(r =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{\sqrt{N \Sigma X^2-(\Sigma X)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{10(-115)-(50)(-30)}{\sqrt{10(290)-(50)^2} \sqrt{10(300)-(-30)^2}} \)
\(=\frac{-1150+1500}{\sqrt{400 \times 2100}}=\frac{350}{916.52}=0.382\)

Since we are given rate per rupee, harmonic mean will give the correct answer.
\(HM=\frac { n }{ \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } +\frac { 1 }{ d } } \)
\(=\frac { 4 }{ \frac { 1 }{ 1 } +\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +\frac { 1 }{ 4 } } \)
\(=\frac { 4\times 12 }{ 25 } \)
= 1.92 kg per rupee.

Here a = 9,000 and i = 0.15
p = \(\frac{a}{i}\)
= \(\frac{9000}{0.15}\)
= \(\frac{9,00,000}{15}\)
= 60,000
Therefore the amount to be deposited is Rs. 60,000.

Here a = 64,000, n = 12 and i = \(\frac{10}{100}=0.1\)
Amount of Ordinary annuity (A) = \(\frac{a}{i}[(1+i)^n-1]\)
= \(\frac{64000}{0.1}[(1+0.1)^{12}-1]\)
= 6,40,000 [(1.1)¹² – 1]
= 6,40,000[3.3184 – 1]
= 6,40,000 [2.3184]
= 64 x 23184
∴ A  = Rs. 14,83,776.

Here n = 11 observations are arranged into ascending order
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
D₂ = size of 2 \(\left( \frac { n+1 }{ 10 } \right) \)^th value
D₆ = size of 6  \(\left( \frac { n+1 }{ 10 } \right) \)^th value
D₂ = size of 2.4^th value ≈ size of 2^nd value = 4
D₆ = size of 7.2^th value ≈ size of 7^th value = 14

Given Σxy = 120, Σx² = 90, Σy² = 640
Then r = \(\frac { \Sigma xy }{ \sqrt { \Sigma { x }^{ 2 }\Sigma { y }^{ 2 } } } =\frac { 120 }{ \sqrt { 90(640) } } =\frac { 120 }{ \sqrt { 57600 } } =\frac { 120 }{ 240 } \) = 0.5

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

xpⁿ = k ⇒ x = kp^-n
\({dx\over dp}=-nkp^{-n-1}\)
Elasticity of demand: \(η_d=-{p\over x}.{dx\over dp}\)
\(=-{p\over kp^{-n}}(-nkp^{-n-1})\) = n, which is a constant.

x = 2p²-5p+1
\({dx\over dp}=4p-5\)
Elasticity of supply: \(η_s={P\over x}.{dx\over dp}\)
\(={p\over 2p^2-5p+1}.(4p-5)\) \(={4p^2-5p\over 2p^2-5p+1}\)

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

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 \(f(x) =\begin{cases} \frac { 1-cos4x }{ 8{ x }^{ 2 } } \quad ifx\neq 0 \\ k\quad \quad \quad ifx=0 \end{cases}\)
\(\neq \underset { x\rightarrow 0 }{ lim } \quad f(x)=\underset { x\rightarrow 0 }{ lim } \frac { 1-cos4x }{ { 8x }^{ 2 } } =\underset { x\rightarrow 0 }{ lim } \frac { 2sin^{ 2 }2x }{ { 8x }^{ 2 } } \)   [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 \(tan\quad \theta =\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } =\frac { \pm 2\sqrt { 25-3(8) } }{ 3+8 } =\frac { \pm 2\sqrt { 1 } }{ 11 } =\frac { \pm 2 }{ 11 } \)
\(\therefore \quad tan\quad \theta =\frac { 2 }{ 11 } \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 2 }{ 11 } \right) \)

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 = (x² - 3x + 2)(x + 1)
\(\frac{d y}{d x}=\left(x^2-3 x+2\right)(1)+(x+1)(2 x-3)\)
= (x² - 3x + 2)(1) + (x + 1)(2x - 3) = x² - 3x + 2 + 2x² - 3x + 2x - 3
= 3x² - 4x - 1

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

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.

\(\frac { 7! }{ 6! } =\frac { 7\times 6! }{ 6! } =7\)

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 = \(\pm\sqrt{3}\)

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|

\(a_{ 11 }=15,\quad { a }_{ 12 }=10,\quad { x }_{ 1 }=35\)
\(a_{ 21 }=20,\quad { a }_{ 22 }=30,\quad { x }_{ 2 }=65\)
\(b_{ 11 }=\frac { { a }_{ 11 } }{ { x }_{ 1 } } =\frac { 15 }{ 35 } =\frac { 3 }{ 7 } ;\quad b_{ 12 }=\frac { { a }_{ 12 } }{ { x }_{ 2 } } =\frac { 10 }{ 65 } =\frac { 2 }{ 13 } \)
\(b_{ 21 }=\frac { { a }_{ 21 } }{ { x }_{ 1 } } =\frac { 20 }{ 35 } =\frac { 4 }{ 7 } ;\quad b_{ 22 }=\frac { { a }_{ 12 } }{ { x }_{ 2 } } =\frac { 30 }{ 65 } =\frac { 6 }{ 13 } \)
B = \(\begin{bmatrix} \frac { 3 }{ 7 } & \frac { 2 }{ 13 } \\ \frac { 4 }{ 7 } & \frac { 6 }{ 13 } \end{bmatrix}\)
\(I-B=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} \frac { 3 }{ 7 } & \frac { 2 }{ 13 } \\ \frac { 4 }{ 7 } & \frac { 6 }{ 13 } \end{bmatrix}=\begin{bmatrix} \frac { 4 }{ 7 } & \frac { -2 }{ 13 } \\ \frac { 4 }{ 7 } & \frac { 7 }{ 13 } \end{bmatrix}\)
\(|I-B|=\frac{4}{7} \times \frac{7}{13}-\left(\frac{2}{13} \times \frac{4}{7}\right)=\frac{28-8}{91}=\frac{20}{91}\)
Since the diagonal elements of I - B are positive and |I - B| is positive the system is available
 \((I-B)^{-1}=\frac{1}{|I-B|} \operatorname{adj}(\mathrm{I}-\mathrm{B})=\frac{91}{20}\left(\begin{array}{cc} \frac{7}{13} & \frac{2}{13} \\ \frac{4}{7} & \frac{4}{7} \end{array}\right)\)
Now, X = (I - B)^-1 D where D \(=\left[ \begin{matrix} 12 \\ 18 \end{matrix} \right] =\frac { 91 }{ 20 } \begin{bmatrix} \frac { 7 }{ 13 } & \frac { 2 }{ 13 } \\ \frac { 4 }{ 7 } & \frac { 4 }{ 7 } \end{bmatrix}\left[ \begin{matrix} 12 \\ 18 \end{matrix} \right]=\frac{91}{20}\left(\begin{array}{l} \frac{84+36}{13} \\ \frac{48+72}{7} \end{array}\right)\)
\(=\left(\begin{array}{l} 42 \\ 78 \end{array}\right)\)
The gross output for two sectors X and Y are 42 and 78 respectively

\(\frac { 3x+7 }{ { x }^{ 2 }-3x+2 } =\frac { 3x+7 }{ (x-1)(x-2) } =\frac { A }{ x-2 } +\frac { B }{ x-1 } \)
\(\frac{3 x+7}{(x-1)(x-2)}=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}\)
\(3 x+7=A(x-2)+B(x-1)\)  ...(1)
If x = 1
\(3+7=\mathrm{A}(1-2) \Rightarrow \mathrm{A}=-10\)
If x = 2
\(6+7=\mathrm{B}(2-1) \Rightarrow \mathrm{B}=13\) 
\(\therefore\)\(\frac { 3x+7 }{ { x }^{ 2 }-3x+2 } \)=\(\frac { -10 }{ x-1 } + \frac { 13 }{ x-2 } \)