\(y_1=\frac{1}{x}, y_2=\frac{-1}{x^2}\)

f(5) = 5² - 4(5) = 5

\(\tan (\pi / 4-x)=\frac{\tan \pi / 4-\tan x}{1+\tan \pi / 4 \tan x}=\frac{1-\tan x}{1+\tan x}\)

4cos³40^o - 3cos40^o = cos 3(40^o)
= cos 120^o
=  cos (180^o - 60^o)
= -cos 60^o
\(= \frac{-1}{2}\)

\(\sec \theta =\sqrt{1+\tan ^2 \theta}=\sqrt{1+\frac{1}{5}}=\sqrt{\frac{6}{5}} \)

\(4 a=1 \Rightarrow a=\frac{1}{4}\)
Equation x = a
\(x=\frac{1}{4}\)

\(2 n r=8 \pi\)
r = 4

(parallel line)

Since sum of all binomial coefficients is 2n
5C₀ + 5C₁ + 5C₂ + 5C₃ + 5C₄ + 5C₅ + (5C₁ + 5C₂ + 5C₃ + 5C₄)
= 2⁵ + (2⁵ - (5C₀ + 5C₅)
= 32 + 32 - (1 + 1)
= 64 - 2 = 2⁶ - 2

Equality coefficient of x in the numerator
kx = 4 (2n) + 1 (x)
kx = 9x

Since if n = 1 then (1) (2) (3) (4) = 24 is divisible by = 24

No. of ways = 5C₄ = 5C₁ = 5

Since \(R_2 \sim R_3\)

\(A=\left(\begin{array}{cc} \frac{4}{5} & \frac{-5}{12} \\ \frac{-2}{5} & \frac{1}{2} \end{array}\right)\)
\(|A|=\frac{2}{5}-\frac{1}{6}=\frac{12-5}{30}=\frac{7}{30}\)
\(A^{-1}=\frac{1}{|A|} \text { adjA }=\frac{30}{7}\left(\begin{array}{ll} \frac{1}{2} & \frac{5}{12} \\ \frac{2}{5} & \frac{4}{5} \end{array}\right)\)

A lies in I quadrant and B lies in the IlI quadrant

\(\cos A=\sqrt{1-\sin ^2 A}\)
\(=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
\(\cos B=-\frac{12}{13}\)
\(\sin B=-\sqrt{1-\cos ^2 B}\)
\(=-\sqrt{1-\frac{144}{169}}=-\sqrt{\frac{25}{169}}=\frac{-5}{13}\)
tan (A - B)
\(=\frac{\sin (A-B)}{\cos (A-B)} \)
\(=\frac{\frac{-16}{\frac{35}{63}}}{\frac{-63}{63}}=\frac{16}{63} \)

 cot 75°
tan 75° = tan (45° + 30°)
\(=\frac { 1+\frac { 1 }{ \sqrt { 3 } } }{ 1-1\left( \frac { 1 }{ \sqrt { 3 } } \right) } =\frac { \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } } }{ \sqrt { 3 } -\frac { 1 }{ \sqrt { 3 } } } =\frac { \sqrt { 3 } +1 }{ \sqrt { 3 } -1 } \left[ \tan { { 45 }^{ o } } =1,\ \tan { 30^{ o }=\frac { 1 }{ \sqrt { 3 } } } \right] \)
\(\cot { { 75 }^{ o } } =\frac { 1 }{ \tan { { 75 }^{ o } } } =\frac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } \)

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

The letter of the word CHAT in alphabetical order are A, C, H, T.
(i) Number of words starting with A = 3! = 6 C begins
(ii) Number of words starting with CA are 2! = 2
Now the words CH begins
(iii) After that we get the word CHAT = 1
\(\therefore\) Rank of CHAT is = 6 + 2 + 1 = 9

x=acos³θ, y=asin³θ
\(\frac { dx }{ d\theta } =a.3cos^{ 2 }\theta \left( \frac { d }{ d\theta } (cos\theta ) \right) ;\quad \frac { dy }{ d\theta } =a.3cos^{ 2 }\theta \left( \frac { d }{ d\theta } (sin\theta ) \right) \)
\(\frac { dy }{ dx } =\frac { \frac { dy }{ d\theta } }{ \frac { dx }{ d\theta } } =\frac { 3asin^{ 2 }\theta cos\theta }{ -3acos^{ 2 }\theta sin\theta } =-\frac { sin\theta }{ cos\theta } =-tan\theta \)

Let cosec^-1(2) = y
\(\sin ^{-1}\left(\frac{1}{2}\right)=y\) where \(\frac{-\pi}{2}\le y\le \frac{\pi}{2}\)
\(\sin y=1 / 2=\sin (\pi / 6)\)
\(\Rightarrow y=\frac{\pi}{6}\)

f(x) = sin (x) + cos(x)
f(-x) = sin (-x) + cos(-x)
= -sin x + cos x
= cos x-sin x
\(\therefore\) f(-x) ≠ f(x) and f(-x) ≠ -f(x)
\(\therefore\) f is neither odd nor even function.

Let A = \(\begin{vmatrix}5&20\\ 0&-1 \end{vmatrix}\)
Minor of 5 = M₁₁ = -1
Minor of 20 = M₁₂ = 0
Minor of 0 = M₂₁ = 20
Minor of -1 = M₂₂ = 5
Co-factor of 5 = A₁₁ = -1
Co-factor of 20 = A₁₂ = 0
Co-factor of 0 = A₂₁ = -20
Co-factor of -1 = A₂₂ = 5

B = \(\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}\) 
I - B = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}=\begin{bmatrix} 0.4 & -0.9 \\ -0.20 & 0.20 \end{bmatrix}\)
|I - B| =\(\begin{bmatrix} 0.4 & -0.9 \\ -0.20 & 0.20 \end{bmatrix}\)
= 0.08 - 0.18 = - 0.1 < 0
Since |I - B| is negative, Hawkins - Simon conditions are not satisfied.
Therefore the given system is not viable.

\(\text {LHS }=2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}\)
\(=\cos \left(\frac{9 \pi}{13}+\frac{\pi}{13}\right)+\cos \left(\frac{9 \pi}{13}-\frac{\pi}{13}\right) +\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13} \)
\(=\cos \frac{10 \pi}{13}+\cos \frac{8 \pi}{13} +\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13} \)
\(=\cos \left(\pi-\frac{3 \pi}{13}\right)+\cos \left(\pi-\frac{5 \pi}{13}\right) +\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13} \)
\(=-\cos \frac{3 \pi}{13} -\cos \frac{5 \pi}{13} +\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13} \)
= 0 = RHS
Hence proved.

let p(x₁,y₁) be any point on the locus ​\(\therefore\)​  \(\sqrt { { \left( { x }_{ 1 }-4 \right) }^{ 2 }+{ \left( { y }_{ 1 }-0 \right) }^{ 2 } } =\frac { 1 }{ 2 } \left| \frac { { x }_{ 1 }-16 }{ \sqrt { { 1 }^{ 2 }+{ 0 }^{ 2 } } } \right| \)
\(\Rightarrow\)\(\sqrt { { \left( { x }_{ 1 }-4 \right) }^{ 2 }+{ y }_{ 1 }^{ 2 } } =\frac { 1 }{ 2 } \left| \frac { { x }_{ 1 }-16 }{ \sqrt { { 1 }^{ 2 }+{ 0 }^{ 2 } } } \right| \)
Squaring  both sides, (x₁ - 4)² + y₁² = \(\frac{1}{4}\)(x₁ - 16)²
⇒ 4[x₁² + 16 - 8 x₁ + y₁²] = x₁² - 32x₁ + 256
⇒ 3x₁² + 4y₁² - 192 = 0
Locus of (x₁, y₁) is 3x² + 4y² = 192

y = 2 sin x + 3cos x
y₁ = 2 cos x - 3 sin x
y₂ = 2 (-sin x) -3 cos x
y₂ = -2 sin x -3 cos x = -y
y₂ + y = 0.

\(y=\frac { 1 }{ 5 } { x }^{ 2 }-6x+100\)
\(y=\frac{x^2-30 x+500}{5}\)
\(\Rightarrow\)  5y = x² - 30x + 500
\(\Rightarrow\) x² - 30x = 5y - 500
\(\Rightarrow\) (x - 15)² = 5y - 500 + 225
\(\Rightarrow\) (x - 15)² = 5y - 275
\(\Rightarrow\) (x - 15)² = 5(y - 55)
\(\Rightarrow\) X² = 5Y is a parabola
where X = x - 15, Y = y - 55.
The vertex of the parabola is (15, 55) with respect to (x, y) axis. The output and the average cost at the vertex are 15 kg and Rs. 55.

y = 500 e^7x + 600 e^-7x
y₁ =  500 (7)e^7x + 600 (-7)e^-7x
y₂ = 500 (49)e^7x + 600 (49)e^-7x
= 49 (500 e^7x + 600 e^-7x) = 49y
\(\therefore\) y₂ - 49y = 0

\(\lim _{ x\rightarrow 2 }{ \frac { { x }^{ 3 }+2 }{ x+1 } } \)
\(\lim _{ x\rightarrow 2 }{ \frac { { x }^{ 3 }+2 }{ x+1 } } \)=\(\frac { 2^{ 3 }+2 }{ 2+1 } =\frac { 10 }{ 3 } \)

A hexagon has 6 vertices and to draw a triangle we need 3 points
Number of triangles \(=6 C_3=\frac{6 \times 5 \times 4}{3 \times 2 \times 1}=20\)

Let B = \(\begin{bmatrix}3 & 1 \\ -1 & 3\end{bmatrix}\)
\(A=\left(\begin{array}{cc} 3 & 1 \\ -1 & 3 \end{array}\right)\)
\(|A|=9+1=10 \neq 0\)
\(\therefore \mathrm{A}^{-1} \text { exists }\)
Now, \({A}^{-1}={{1}\over{|A|}}adj\ A={{1}\over{10}}\begin{bmatrix} 3 & -1 \\ 1 & 3\end{bmatrix}\)

\(\left(2 x^2+\frac{1}{x}\right)^{12}\)
\(t_{r+1}=n C_r x^{n-r} \text { ar }\)
\(=12 \mathrm{C}_{\mathrm{r}}\left(2 x^2\right)^{12-\mathrm{r}}\left(\frac{1}{x}\right)^{\mathrm{r}}\)
\(=12 \mathrm{C}_{\mathrm{r}} 2^{12-r} x^{24-2 r}\left(\frac{1}{x^r}\right)\)
\(=12 \mathrm{C}_{\mathrm{r}} 2^{12-r} x^{24-2 r}-r\)
\(=12 \mathrm{C}_{\mathrm{r}} 2^{12-r} x^{24-3 r}\)
To find the term independent of x, equate the power of x as 0
24 - 3r = 0
\(3 r=24 \Rightarrow r=8\)
\(t_{r+1}=12 C_8 2^{12-8}\)
\(=12 C_4 2^4=\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} \times 16\)
= 7920

Length of perpendicular from P (l, m) to the line ax + by + c = 0 is \(\left| \frac { { al }+{ bm }+c }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \right| \)
Length of perpendicular from P (4, 1) to the line 3x - 4y + 12 =- 0 is
d = \(\frac { 3(4)-4(1)+12 }{ \sqrt { { 3 }^{ 2 }+(-4)^{ 2 } } } \)
\(\Rightarrow\) d = \(\left| \frac { 12-4+12 }{ \sqrt { 9+16 } } \right| =\left| \frac { 20 }{ 5 } \right| =4\)
\(\Rightarrow\)  d = 4 units

\(\left(2 x^2-\frac{3}{x^3}\right)^{10}\)
n = 10
middle term is \(t_{\frac{n}{2}+1}=t_{\frac{10}{2}+1}=t_6\)
r = 5
\(t_{r+1}=n C_{r x}^{n-r} a^r\)
\(t_6=10 C_5\left(2 x^2\right)^5\left(\frac{-3}{x^3}\right)^5\)
\(=-10 C_5 \frac{x^{10}}{x^{15}} 2^5 \cdot 3^5\)
\(=-10 C_5 \frac{(6)^5}{x^5}\)

On x axis y = 0
Let P (x₁, 0) be any point on the locus and A (7, -6) and B (3 , 4) are the two given points
PA = PB
\(\Rightarrow\) PA² = PB²
\(\Rightarrow\)  (x₁- 7)² + (0+ 6)^{2  =} (x₁ - 3)² + ( 0 - 4)²
\(\Rightarrow\)   \({ x }_{ 1 }^{ 2 }\) - 14x₁ + 46 + 36    = \({ x }_{ 1 }^{ 2 }\) - 6x₁ +9 +16
\(\Rightarrow\)  -8x₁ = - 60
\(\Rightarrow\)   x₁ \( =\frac { 15 }{ 2 } \)
Required point is \(\left(\frac{15}{2}, 0\right)\)

\(X=\left(\begin{array}{ccc} 8 & -1 & -3 \\ -5 & 1 & 2 \\ 10 & -1 & -4 \end{array}\right)\)
\(|X|=8(-4+2)+1(20-20)-3(5-10)\)
\(=-16+15=-1 \neq 0\)
\(X^{-1} \text { exists }\)
\(A_{11}=\text {Co-factor of } 8=-4+2=-2 \)
\(A_{12}=\text {Co-factor of }-1=-(20-20)=0\)
\(A_{13}=\text {Co-factor of }-3=5-10=-5\)
\(\mathrm{A}_{21}=\text {Co-factor of }-5=-(4-3)=-1 \)
\(\mathrm{A}_{22}=\text {Co-factor of } 1=-32+30=-2 \)
\(\mathrm{A}_{23}=\text {Co-factor of } 2=-(-8+10)=-2\)
\(\mathrm{A}_{31}=\text {Co-factor of } 10=-2+3=1 \)
\(\mathrm{A}_{32}=\text {Co-factor of }-1=-(16-15)=-1 \)
\(\mathrm{A}_{33}=\text {Co-factor of }-4=8-5=+3\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} -2 & 0 & -5 \\ -1 & -2 & -2 \\ 1 & -1 & +3 \end{array}\right)\)
\(\operatorname{Adj} X=\left(\begin{array}{ccc} -2 & -1 & 1 \\ 0 & -2 & -1 \\ -5 & -2 & +3 \end{array}\right)\)
\(X^{-1}=\frac{1}{|X|} \operatorname{adj} X\)
\(=\frac{-1}{1}\left(\begin{array}{ccc} -2 & -1 & 1 \\ 0 & -2 & -1 \\ -5 & -2 & +3 \end{array}\right)\)
\(\text {Given } Y=X^{-1}\)
\(\left(\begin{array}{ccc} 2 & 1 & -1 \\ 0 & 2 & 1 \\ 5 & p & q \end{array}\right)=\left(\begin{array}{ccc} 2 & 1 & -1 \\ 0 & 2 & 1 \\ 5 & 2 & -3 \end{array}\right)\)
\(\mathrm{p}=2 ; \mathrm{q}=-3\)

cosec A + sec A = cosec B + sec B
cosec A - cosec B  = sec B - sec A
\(\frac{1}{\sin A}-\frac{1}{\sin B}=\frac{1}{\cos B}-\frac{1}{\cos A}\)
\(\frac{\sin B-\sin A}{\sin A \sin B}=\frac{\cos A-\cos B}{\cos B \cos A}\)
\(\frac{\sin B-\sin A}{\cos A-\cos B}=\frac{\sin A \sin B}{\cos A \cdot \cos B}\)
\(\frac{2 \sin \frac{B-A}{2} \cos \frac{B+A}{2}}{-2 \sin \frac{A-B}{2} \sin \frac{A+B}{2}}=\tan A \tan B\)
\(\cot \left(\frac{A+B}{2}\right)=\tan A \tan B\)

Let x represent the number of units and y its cost.
By the given data,
x₁(6000) y₁(33,000)
x₂(8000) y₂(40,000)
Using two point form, \(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
⇒ \(\frac { y-33000 }{ 40,000-33000 } =\frac { x-6000 }{ 8000-6000 } \)
⇒ \(\frac { y-33000 }{ 7000 } =\frac { x-6000 }{ 2000 } \)
⇒ \(\frac { y-33000 }{ 7 } =\frac { x-6000 }{ 2 } \)
⇒ 2y - 66000 = 7x - 42000
⇒ 2y = 7x - 42000 + 66000
2y = 7x + 24000, which is the required linear function.

\(\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}=\frac{1}{8}\)
\(\text {LHS }=\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}\)
\(=\cos 20^{\circ} {\left[\cos \left(60^{\circ}-20^{\circ}\right)\right.} \left.\cos \left(60^{\circ} \div 20^{\circ}\right)\right]\)
\(=\cos 20^{\circ}\left[\cos ^2 60^{\circ}-\sin ^2 20^{\circ}\right]\)
\(=\cos 20^{\circ}\left[\left(\frac{1}{2}\right)^2-\left(1-\cos ^2 20^{\circ}\right)\right]\)
\(=\cos 20\left[\frac{1-4+4 \cos ^2 20^{\circ}}{4}\right]\)
\(=\frac{1}{4}\left[4 \cos ^3 20^{\circ}-3 \cos 20^{\circ}\right]\)
\(=\frac{1}{4} \cos 3(20)=\frac{1}{4} \cos 60^{\circ}\)
\(=\frac{1}{4}\left(\frac{1}{2}\right)=1 / 8=\mathrm{RHS}\)
Hence proved.

\(L H L=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} 1-x=0\)
\(R H L=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(1-x)(2-x)=0\)
LHL = RHL
\(\therefore\) f is continuous at x = 1
\(L H D=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{1-h-1}\)
\(=\lim _{h \rightarrow 0} \frac{(1-1-h)(2-1-h)}{h}\)
\(R H D=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{1+h-1}\)
\(=\lim _{h \rightarrow 0} \frac{-h(1-h)}{h}=-1\)
LHD = RHD
\(\therefore\) f is differentiable at x = 1
\(L H L=\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}(1-x)(2-x)=0\)
\(R H L=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2} 3-2=1\)
\(L H L \neq R H L\)
\(\therefore\) f is not continuous at x = 2 and hence not differentiable at x = 2

y = \({ \left( x+\sqrt { 1+{ x }^{ 2 } } \right) }^{ m }\)  
\(\frac{d y}{d x}=m\left(x+\sqrt{1+x^2}\right)^{m-1}\left(1+\frac{1}{2 \sqrt{1+x^2}}(2 x)\right) \)
\(y_1=m\left(x+\sqrt{1+x^2}\right)^{m-1}\left(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\right) \)
\(\sqrt { 1+{ x }^{ 2 } } y_{ 1 } = { m{ \left( x+\sqrt { 1+{ x }^{ 2 } } \right) }^{ m } }\)
\(\Rightarrow \sqrt { 1+{ x }^{ 2 } } y_{ 1 } =my\)
Squaring both sides
\(\left(1+x^2\right) y_1^2=m^2 y^2\)
\(\left(1+x^2\right) 2 y_1 y_2+y_1^2(2 x)=2 m^2 y y_1\)
\(\div 2 y_1 \ \left(1+x^2\right) y_2+x y_1-m^2 y=0\)

Let x, y, z be the salary of each type of staff in A, B and C categories
4x + 2y + 3z = 4900
3x + 3y + 2z = 4500
4x + 3y + 4z = 5800
In matrix form\(\begin{bmatrix} 4&2&3\\3&3&2\\4&3&4 \end{bmatrix}\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 4900\\4500\\5800 \end{bmatrix}\)
\(\Rightarrow\) \(A X=B \Rightarrow X=A^{-1} B\)
Where A  \(=\begin{bmatrix}4&2&3\\3&3&2\\4&3^4 \end{bmatrix},X=\begin{bmatrix} x\\y\\z \end{bmatrix},B=\begin{bmatrix} 4900 \\4500\\5800 \end{bmatrix}\)
= 4 (12 - 6) - 2(12 - 8) + 3 (9 - 12)
= 4 (6) - 2 (4) + 3 (-3) = 24 - 8 - 9 = 7 = 1
\(\therefore\) A^-1 exists.
\(Co-factor matrix=\begin{bmatrix} 6&-4&-3\\1&4&-4\\-5&1&6 \end{bmatrix}\)
\(\therefore\ {A}^{-1}={{1}\over{|A|}}adj\ A={{1}\over{7}}\begin{bmatrix} 6&1&-5\\-4&4&1\\-3&-4&6 \end{bmatrix}\)
\(\therefore\ X={A}^{-1}B={{1}\over{7}}\begin{bmatrix} 6&1&-5\\-4&4&1\\-3&-4&6 \end{bmatrix}\begin{bmatrix} 4900\\45000\\5800 \end{bmatrix}={{1}\over{7}}\begin{bmatrix} 29400&+45000&-29,000\\-19600&+18000&+5800\\-14700&-18,000&+34800 \end{bmatrix}\)
\(\left(\begin{array}{l} x \\ y \\ z \end{array}\right)={{1}\over{7}}\begin{bmatrix} 49000\\4200\\2100\end{bmatrix}=\begin{bmatrix} 700\\600\\300 \end{bmatrix}\)
Salary for each type of staff under category A, B and C are respectively Rs 700,Rs 2600 and Rs 7300

\(\left(x+\frac{2^{-}}{x^2}\right)^{17}\)
\(n=17\)
\(T_{r+1}=n C_r \cdot x^{n-r} a^{r h}\)
\(=17 C_r x^{17-r}\left(\frac{2}{x^2}\right)^{\mathrm{r}}\)
\(=17 C_r 2^r x^{17-r-2 r}=17 C_r 2^r x^{17-3 r}\)
To find co-efficient of \(x^{11}\) equate the power of x to 11
17 - 3r = 11
17 - 11 = 3r
6 = 3r \(\Rightarrow\) r = 2
Co-efficient of \(x^{11}=17 C_2 2^2\)
\(=\frac{17 \times 16}{2 \times 1} \times 2^2=544\)

(i) Number of ways of selecting 11 players out of 15 is 15C₁₁ = 15C₄
= \(\frac { 15\times 14\times 13\times 12 }{ 4\times 3\times 2\times 1 } \) = 1365
(ii) If a particular player is always chosen this means that we have to select 10 players out of the remaining 14 players
Number of ways = 14C₁₀ = 14C₄
\(=\frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1}=1001\)
(iii) If a particular player is never chosen this means that we have to select 11 players out of the remaining 14 players
Number of ways = 14C₁₁ = 14C₃
\(=\frac{14 \times 13 \times 12}{3 \times 2 \times 1}=364\)

Given equations are 2x - z = 0, 5x + y = 4 and y + 3z = 5·
Given equations can be written in matrix form as 
\(\begin{bmatrix} 2&0&-1\\ 5&1&0\\ 0&1&3 \end{bmatrix}\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 0\\4\\5 \end{bmatrix}\)
\(\Rightarrow\) AX  =  B
Where A \(=\begin{bmatrix} 2&0&-1\\5&1&0\\0&1&3 \end{bmatrix},X=\begin{bmatrix} x\\y\\z \end{bmatrix}\) and B \(=\begin{bmatrix} 0\\4\\ 5\end{bmatrix}\)
\(|A|\) = 2 ( 3 - 0 ) - 0 - 1 ( 5 - 0 ) = 6 - 5 = 1 \(\neq\) 0 
\(\therefore\)  A^-1 exists.
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2 \end{array}\right)\)
\({A}^{-1}={{1}\over{|A|}}adj\ A={{1}\over{1}}\begin{bmatrix} 3&-1&1\\ -15&6&-5\\ 5&-2&2 \end{bmatrix}\)
\(\therefore\)  \(X={A}^{-1}B=\begin{bmatrix} 3&-1&1\\-15&6&-5\\ 5&-2&2 \end{bmatrix}\begin{bmatrix} 0\\4\\5 \end{bmatrix}\)
\(\begin{bmatrix} x\\y\\z \end{bmatrix}= \begin{bmatrix} 0-4+5\\ 0+24-25\\ 0-8+10 \end{bmatrix}=\begin{bmatrix} 1\\-1\\2 \end{bmatrix}\)
\(x=1, \mathrm{y}=-1, \mathrm{z}=2\)

\(\tan ^{-1}(x+1)+\tan ^{-1}(x-1)=\tan ^{-1}\left(\frac{4}{7}\right)\)
\(\tan ^{-1}\left(\frac{x+1+x-1}{1-\left(x^2-1\right)}\right)=\tan ^{-1}\left(\frac{4}{7}\right)\)
\(\frac{2 x}{2-x^2}=\frac{4}{7} \ x^2-1<1\)
\(14 x=8-4 x^2 \ x^2<2\)
\((4 x^2+14 x-8=0\ -\sqrt{2})\)
\(\div 2 \ \ 2 x^2+7 x-4=0\)
\((2 x-1)(x+4)=0\)
\((x=\frac{1}{2} \ \ [\because-\sqrt{2}]\)

Compare the equation
12x² - 10xy + 2y² + 14x - 5y + 2 = 0 with
ax² + 2hxy + by² + 2gx + 2fy + c = 0
We get a = 12, 2h = -10, b = 2, 2g = 14, 2f = -5
\(h=-5\quad g=7\quad f=-\frac { 5 }{ 2 } ,c=2\)
\(\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=\left|\begin{array}{ccc} 12 & -5 & 7 \\ -5 & 2 & \frac{-5}{2} \\ 7 & \frac{-5}{2} & 2 \end{array}\right|\)
\(=12\left(4-\frac{25}{4}\right)+5\left(-10+\frac{35}{2}\right)+7\left(\frac{25}{2}-14\right)\)
\(=48-75-50+\frac{175}{2}+\frac{175}{2}-98\)
= -175 + 175 = 0
Hence the given equations represent a pair of straight lines.
To find separate equation
\(12 x^2-10 x y+2 y^2 =12 x^2-6 x y-4 x y+2 y^2 \)
\(=6 x(2 x-y)-2 y(2 x-y) \)
\(=(6 x-2 y)(2 x-y)\)
\(12 x^2-10 x y+2 y^2+ 14 x-5 y+2 =(6 x-2 y+l)(2 x-y+\mathrm{m})\)
Comparing the coefficient of x and y
14 = 6m + 2l
divided by 2
7 = 3m + l .........(1)
-5 = -2m - l .......(2)
Solving (1) and (2) we get m = 2, 1 = 1
The separate equations are
6x - 2y + 1 = 0
2x - y + 2 = 0