Given \(TR=12x+{x^2\over2}-{x^3\over 3}\)
Average Revenue \(AR={TR\over x}={12x+x^2/2-x^3/3\over x}\)
\(AR=12+{x\over2}-{x^2\over3}\)     ...(1)
Let y \(=12+{x\over2}-{x^2\over3}\)
\({dy\over dx}={1\over2}-{2x\over 3}\)
Condition for maximum is
\({dy\over dx}=0\ and\ {d^2 y\over dx^2}<0\)
\(∴\ {1\over2}-{2x\over3}=0⇒{1\over2}={2x\over 3}\)
\(x={1\over2}\times{3\over 2}={3\over 4}\)
\({d^2y\over dx^2}={-2\over3}<0\)
∴ AR is maximum at x=3/4
when x=3/4, \(AR=12+{3/4\over2}-{\left(3/4\right)^2\over 3}\)  [From )1)]
\(=12+{3\over8}-{9\over 48}=12.1875\)     ....(2)
\(MR={{dR\over dx}}={d\over dx}\left(12x+{x^2\over 2}-{x^3\over 3}\right)=12+x-x^2\)
when x = 3/4, \(MR=12+{3\over 4}-\left(3\over4\right)^2=12+{3\over4}-{9\over 16}=12.1875 \)    ..(3)
From (2) and (3), at the highest point of AR,
AR = MR = 12.1875

Here n=5,
b_yx=\(\frac { n\sum { xy } -(\sum { x } )(\sum { y } ) }{ n\sum { { x }^{ 2 }-{ (\sum { x } ) }^{ 2 } } } \)
=\(\frac { 5(44)-(15)(15) }{ 5(49)-{ (15) }^{ 2 } } \)
=\(\frac { 220-225 }{ 245-225 } =\frac { -5 }{ 20 } =\frac { -1 }{ 4 } \)
b_xy=\(\frac { n\sum { xy } -(\sum { x } )(\sum { y } ) }{ n\sum { { y }^{ 2 }-{ (\sum { y } ) }^{ 2 } } } \)
=\(\frac { 5(44)-15(15) }{ 5(49)-{ (15) }^{ 2 } } \)
\(\frac { 220-225 }{ 245-225 } =\frac { -5 }{ 20 } =\frac { -1 }{ 4 } \)
b_xy and b_yx are negative, r is also negative.
\(\therefore\)Correlation co-efficient
r(x,y)=-\(\sqrt { { b }_{ xy }.{ b }_{ yx } } \)
=-\(\sqrt { \left( \frac { -1 }{ 4 } \right) \left( \frac { -1 }{ 4 } \right) } =\frac { -1 }{ 4 } \)
r=-0.25

EFT and LFT are same in the activity, 1 - 2, 2 - 4 and 4 - 6.
Hence the critical path is 1 - 2 - 4 - 6 and the project completion time is 22 Weeks.

Let A = {First ball drawn is green}
B = {Second ball drawn is green}
\(P(A)=\frac { n(A) }{ n(S) } =\frac { 6 }{ 10 } \)   [\(\because\) Total number of balls = 4+6 = 10]
Let C = {getting green ball after taking out first green ball}
\(P(C)=\frac { 5{ C }_{ 1 } }{ 9{ C }_{ 1 } } =\frac { 5 }{ 9 } \)   [\(\because\) First ball is not replaced]
\(\therefore\) P(getting green ball) = P(A).P(C) = \(\frac { 6 }{ 10 } \times \frac { 5 }{ 9 } \)
\(\therefore P(A\cap B)=\frac { 1 }{ 3 } \)
\(P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { \frac { 1 }{ 3 } }{ \frac { 6 }{ 10 } } \)
\(P(B/A)=\frac { 1 }{ 3 } \times \frac { 10 }{ 6 } =\frac { 5 }{ 9 } \)

Given A = 10,000 r = \(\cfrac { 8 }{ 100 } \times \cfrac { 1 }{ 4 } \) = 2% = 0.02, n = 5 X 4 =20
A = \(\cfrac { a }{ i } \left( 1+i \right) \left[ \left( 1+i \right) ^{ n }-1 \right] \)
10,000 = \(\cfrac { a }{ 0.02 } \) (1.02)[(1.02)²⁰-1]
\(\cfrac { 10,000\times 0.02 }{ 1.02 } \) = a[1.4859-1]
196.08 = a[0.4859]
a =\(\cfrac { 196.08 }{ 0.4859 } \) = Rs.403.53
a = Rs.404
(1.02)²⁰ =20 log(1.02)
= 20(0.0086)
= 0.172
Antilog of 0.172 is 1.4859

LHS = cos 6x
= cos 3(2x) = 4 cos³2x- 3cos2x    [∴ cos 3\(\theta\) = 4 cos³\(\theta\) - 3 cos\(\theta\)]
= 4 (2cos²x - 1)³- 3(2cos²x - 1)  [∴ cos2x = 2cos²x - 1]
= 4 (8 cos⁶x-12 cos⁴x + 6 cos²x-1) - 6 cos²x + 3
= 32 cos⁶x - 48 cos⁴x + 18 cos²x - 1
= RHS. Hence proved.

Taking vertex of the parabola as reflector at origin, x-axis along the axis of the parabola, equation of the parabola is y² = 4ax
Given depth = 5 cm, diameter = 20 cm
\(\therefore\) (5, 10) lies on the parabola
\(\therefore\) 10² = 4a(5) ⇒ 100 = 20a ⇒ a = \(\frac{100}{20}\) = 5

\(\therefore\) Focus is (a, 0) = (5, 0) which is the mid-point of the given diameter

Given y = e^a cos^-1 x    ...(1)
Differentiating with respect to 'x' we get,
\({{dy}\over{dx}}={e}^{a\ {cos}^{-1}x}.{{d}\over{dx}}\left( a\ {\cos}^{-1}x \right)={e}^{a\ {cos}^{-1}x}.\left( {{-a}\over{\sqrt{1-{x}^{2}}}} \right)\)
\(={{-ay}\over{\sqrt{1-{x}^{2}}}}\)  [ using (1) ] 
\(\Rightarrow\sqrt{1-x^2}.{{dy}\over{dx}}=-ay\)
Squaring both sides we get,
\((1-x^2).{\left( {{dy}\over{dx}} \right)}^{2}=a^2y^2\)
Differentiating again with respect to 'x' we get,
\(\left( (1-x^2).2\left( {{dy}\over{dx}} \right)\left({{d^2y}\over{dx^2}} \right) +{\left({{dy}\over{dx}} \right)}^{2}(-2x)=a^2(2y)\left( {{dy}\over{dx}} \right) \right)\)
Dividing throughout by 2 \(\left({{dy}\over{dx}} \right)\) we get,
\((1-x^2).\left({{d^2y}\over{dx^2}} \right)-x\left( {{dy}\over{dx}} \right)=a^2 y\Rightarrow(1-x^2)\left( {{d^2y}\over{dx^2}} \right)-x\left( {{dy}\over{dx}} \right)-a^2y=0.\)
Hence proved.

Given \(({\sqrt{2}+1})^{5}+{(\sqrt{2}-1)}^{5}\)
=[\({ \left( \sqrt { 2 } \right) }^{ 5 }\)+ 5C_I \({ \left( \sqrt { 2 } \right) }^{ 4}\) (1)¹ + 5C₂ \({ \left( \sqrt { 2 } \right) }^{ 3 }\) .(1)² + 5C₃ \({ \left( \sqrt { 2 } \right) }^{ 2 }\) . (1)³+ 5C₄ \({ \left( \sqrt { 2 } \right) }^{ 1}\) .(1)⁴ + (1)⁵] +[\({ \left( \sqrt { 2 } \right) }^{ 5 }\) - 5C_I \({ \left( \sqrt { 2 } \right) }^{ 4}\) (l)¹ + 5C₂ \({ \left( \sqrt { 2 } \right) }^{ 3 }\) (1)² - 5C₃ \({ \left( \sqrt { 2 } \right) }^{ 2 }\) (1)³+ 5C₄ \({ \left( \sqrt { 2 } \right) }^{ }\) (1)⁴ -1⁵ ]
=2[\({ \left( \sqrt { 2 } \right) }^{ 5 }\) + 10\({ \left( \sqrt { 2 } \right) }^{ 3 }\) +5 \({ \left( \sqrt { 2 } \right) }^{ }\)] = 2[ 4\(\sqrt { 2 } \) + 20\(\sqrt 2\) + 5.\(\sqrt 2\)] = 2[29.\( \sqrt { 2 } \)] = 58\( \sqrt 2\)

Let x, y and z be the investments at the rate of interest 6%, 7% and 8% per annum respectively.
Then x + y + z = 5000  ...(1)
Also, \(\frac{6x}{100}+\frac{7y}{100}+\frac{8z}{100}=358\)
\(\Rightarrow\)   6x + 7y + 8z = 35800  ...(2)
And \(\frac{6x}{100}+\frac{7y}{100}=70+\frac{8z}{100}\) (Given)
\(\Rightarrow\)   6x + 7y - 8z = 7000  ...(3)
From (1), (2) and (3),
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 5000 \\ 35800 \\ 7000 \end{matrix} \right] \)
AX = B where A = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8 \end{matrix} \right] ,\quad X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,\quad Z=\left[ \begin{matrix} 5000 \\ 35800 \\ 7000 \end{matrix} \right] \) 
 \(\therefore |A|\) \(=\begin{bmatrix} 1&1&1\\6&7&8\\6&7&-8 \end{bmatrix}=1(-56-56)-1(-48-48)1(42-42)\)
\(=-16\neq0\Rightarrow{A}^{-1}\) exists.
A₁₁  = -112, A₁₂ = 96, A₁₃= 0
A₂₁ = 15, A₂₂ = -14  A₂₃ = -1
A₃₁ = 1, A₃₂ = -2, A₃₃ = -1
\(\therefore \ adj\ A=\begin{bmatrix} -112&96&0\\15&-14&-1\\1&-2&1 \end{bmatrix}^{T}=\begin{bmatrix} -112&15&1\\96&-14&-2\\0&-1&1\end{bmatrix}\)
\(\therefore \quad A^{ -1 }=\quad \frac { 1 }{ |A| } adjA=-\frac { 1 }{ 16 } \left[ \begin{matrix} -112 & 15 & 1 \\ 96 & -14 & -2 \\ 0 & -1 & 1 \end{matrix} \right] \)
Hence, the solution is given by
\(X={A}^{-1}B=-\frac{1}{16}\begin{bmatrix} -112&15&1 \\96 &-4&-2\\0&-1&1 \end{bmatrix}\begin{bmatrix} 5000\\35800\\7000 \end{bmatrix}=-{{1}\over{16}}\begin{bmatrix} -560000+537000+7000\\480000-501200-14000\\0-35800+7000 \end{bmatrix}\)
\(X=\begin{bmatrix} 1000\\2200\\1800 \end{bmatrix}\)  [ \(\because\) x = 1000, y = 2200, z = 1800]
Hence, the three investment are of Rs. 1000,Rs. 2200 and Rs. 1800.