GM = Anti \(\log { \left( \frac { \sum { flog \ m } }{ N } \right) } \)
= Anti \(\log { \left( \frac { 67.1394 }{ 52 } \right) } \)
= Anti log(1.2911)
GM = 19.54

Here n = 10, a = 12,000 and i = 0.15
Now P = \(\frac{a}{i}[1-\frac{1}{(1+i)^n}]\)
= \(\frac{12,000}{0.15}[1-\frac{1}{(1+0.15)^{10}}]\)
= \(\frac{12,000}{0.15}[1-\frac{1}{(1.15)^{10}}]\)
= \(\frac{12,00,000}{15}[1-\frac{1}{4.016}]\)
= 80,000 \([\frac{4,016-1}{4,016}]\)
= 80,000 \([\frac{3,016}{4,016}]≈ 60,080\) 
\(\therefore \) P = Rs. 60,080

Correlation Co-efficient r =\(\frac { N\sum { xy-(\sum { x)(\sum { y) } } } }{ \sqrt { N\sum { { x }^{ 2 }-({ \sum { x) } }^{ 2 }\times \sqrt { N{ \sum { y } }^{ 2 }-\left( { \sum { y } }^{ 2 } \right) } } } } \)
= \(\frac { 10(288)-(60)(40) }{ \sqrt { 10(494)-{ (60) }^{ 2 }\sqrt { 10(211)-{ (40) }^{ 2 } } } } \)
= \(\frac { 2880-2400 }{ \sqrt { 1340 } .\sqrt { 520 } } \)
= \(\frac { 480 }{ (36.61)(22.80) } =\frac { 480 }{ 834.71 } \)
r = 0.575

(i) Variables: Let x₁ be the number of days required to run plant C₁ and x₂ be the number of days required to run plant C₂
Objective function: Minimize Z = 600 x₁ + 400 x₂
(ii) Constraints: 3000 x₁ + 1000 x₂ ≥ 24000 (since there is a demand of 24000 bottles of drink A, production should not be less than 24000)
1000 x₁ + 1000 x₂ ≥ 16000
2000 x₁ + 6000 x₂ ≥ 48000
(iii) Non-negative restrictions: Since be the number of days required of a firm are non-negative, we have x₁, x₂ ≥ 0
Thus we have the following LP model.
Minimize Z = 600 x₁ + 400 x₂
subject to 3000 x₁ + 1000 x₂ ≥ 24000
1000 x₁ + 1000 x₂ ≥ 16000
2000 x₁ + 6000 x₂ ≥ 48000 and x₁, x₂ ≥ 0

\(x={20\over p+1}\)
\({dx\over dp}={-20\over (p+1)^2}\)
Elasticity of demand: \(η_d=-{P\over x}.{dx\over dP}\)
\(=-{P\over \left(20\over (p+1)\right)}.{-230\over (p+1)}\)\(={p\over p+1}\)
When p = 3, \(η_d={3\over 4}\) (or) 0.75
Here |η_d |<1
∴ demand is inelastic.

LHS \(=\sqrt{3}\ cosec{20^o}-\sec20^o=\sqrt{3}.\frac{1}{\sin20^o}-\frac{1}{\cos20^o}\)
\(=\frac{\sqrt3\cos20^o-\sin20^o}{\sin20^o\cos20^o}=2\left[\frac{\frac{\sqrt3}{2}\cos20^o-\frac{1}{2}\sin20^o}{\sin20^o\cos20^o}\right]\)
\(=2\frac{(\sin60^o\cos20^o-\cos60^o\sin20^o)}{\sin20^o\cos20^o}\left[\because\sin60^o=\frac{\sqrt3}{2}\ and\cos60^o=\frac{1}{2}\right]\)
\(=2\frac{\sin(60^o-20^o)}{\sin20^o\cos20^o}=\frac{4\sin40^o}{2\sin20^o\cos20^o}=\frac{4\sin40^o}{\sin40^o}[\because2\sin A\cos A=\sin2A]\)
= 4 = RHS
Hence proved

u = sin³x ; v = cos³x
\(\frac { du }{ dx } =3sin^{ 2 }x; \frac { d }{ dx } (sinx)=3{ sin }^{ 2 }xcosx\)
\(\therefore \frac { du }{ dv } =\frac { \frac { du }{ dx } }{ \frac { dv }{ dx } } =\frac { 3{ sin }^{ 2 }x\cos x }{ -3{ cos }^{ 2 }x\sin x } =-\frac { \sin x }{ \cos x } =-\tan x\)

m₁ = \(\frac { 1 }{ 2 } \), m₂ = 3
tan \(\theta\) = \(\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| \)
\(\Rightarrow \)  tan \(\theta\) \(=\left| \frac { \frac { 1 }{ 2 } -3 }{ 1+\frac { 1 }{ 2 } \left( 3 \right) } \right| \)
\(\Rightarrow \) tan \(\theta\) = \(\left| \frac { \frac { -5 }{ 2 } }{ \frac { 5 }{ 2 } } \right| =1\)
\(\Rightarrow \)  tan \(\theta\)  = 1 \(\Rightarrow \)  \(\theta\) = 45^°
Angle between the line is 45^°

Expanding the given determinant along R₁ we
\(\left|\begin{array}{ccc} 7 & 4 & 11 \\ -3 & 5 & x \\ -x & 3 & 1 \end{array}\right|=0\)
⇒ 7 (5 - 3x) - 4 (-3 + x²) + 11 (-9 + 5x) = 0
⇒ 35 - 21x + 12 - 4x² - 99 + 55x =  0
⇒ - 4x² + 34x - 52 = 0
\(\div\)  by - 2
\(2 x^2-17 x+26 =0\)
\(2 x^2-4 x-13 x+26 =0 \)
\(2 x(x-2)-13(x-2) =0 \)
\((2 x-13)(x-2) =0\)
\(x =2, \frac{13}{2}\)

\((x-2 y)^{13}\)
\(T_{r+1}=n C_r x^{n-r} a^r\)
\(n=13, r=4\)
\(t_{r+1}=13 C_r x^{13-r}(-2 y)^r\)
\(t_5=13 C_4 x^{13-4}(-2 y)^4\)
\(=\frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} x^9(16) y^4\)
\(=11440 x^9 y^4\)