\({{x-2}\over{(x+2){(x-1)}^{2}}}={{A}\over{x+2}}+{{B}\over{x-1}}+{{C}\over{{(x-1)}^{2}}}\)
\(\frac{x-2}{(x+2)(x-1)^2}=\frac{A(x-1)^2+B(x+2)(x-1)+C(x+2)}{(x+2)(x-1)^2}\)
x - 2 = A(x - 1)²+ B(x + 2)(x - 1)+ C(x + 2)     ..(1)
x = -2 in (1) we get,
-2 - 2 =  A (-3)2 \(\Rightarrow\) - 4 = 9 A \(\Rightarrow\) A = \({{-4}\over{9}}\)
x = 1 in (1) we get,
1- 2 = C(1 + 2) \(\Rightarrow\) -1 = 3C \(\Rightarrow\) C = \({{-1}\over{3}}\)
Equate coefficient of x² on both sides of (1)
0 = A + B
\(B=-A=\frac{4}{9}\)  
\(\therefore\)   \({{x-2}\over{(x+1){(x-1)}^{2}}}-{{-{{4}\over{9}}}\over{x+2}}+{{{{4}\over{9}}}\over{x-1}}+{{-{{1}\over{3}}}\over{{(x-1)}^{2}}}+{{-4}\over{9(x+2)}}+{{4}\over{9(x-1)}}-{{1}\over{3{(x+1)}^{2}}}\)

Since \(\cfrac { 3\pi }{ 2 } <\left( A,B \right) <2\pi \), both A and B lie in the fourth quadrant,
 \(\therefore \) sinA and sinB are negative
Given \(cosA=\cfrac { 4 }{ 5 } \ cosB=\cfrac { 12 }{ 13 } \)
Therefore  \(\sin A=-\sqrt{1-\cos ^2 A}\)
\(=-\sqrt{1-\frac{16}{25}}\)
\(=-\sqrt{\frac{25-16}{25}}\)
\(=-\frac{3}{5}\)
\(\operatorname{Sin} B =-\sqrt{1-\cos ^2 B} \)
\(=-\sqrt{1-\frac{144}{169}} \)
\(=-\sqrt{\frac{169-144}{169}} \)
\(=-\frac{5}{13}\)
\(sin(A-B)=sinAcosB-cosAsinB\\ =\left( \cfrac { -3 }{ 5 } \right) \left( \cfrac { 12 }{ 13 } \right) -\left( \cfrac { 4 }{ 5 } \right) \left( \cfrac { -5 }{ 13 } \right) \\ =\cfrac { -36 }{ 65 } +\cfrac { 20 }{ 65 } =\cfrac { -6 }{ 65 } \)

Let the given statement P(n) be defined as \(P(n)={ 2 }^{ 3n }-1\) 
Step 1: put \(n=1\)
\(\therefore P(1)={ 2 }^{ 3 }-1\) 
= 7  is divisible by 7
i.e., P(1) is true.
Step 2: Let us assume that the statement is true for n = k i.e., p(k) is true 
We assume 2^3k-1 is divisible by 7 
\(\Rightarrow { 2 }^{ 3k }-1=7m\) 
Step 3: To prove that P(k + 1) is true
\(P(k+1)={ 2 }^{ 3(k+1) }-1\)
\(={ 2 }^{ (3k+3) }-1\)
\(=2^{3 k} \cdot 2^3-1\)
\(= { 2 }^{ 3k }.8-1\)
\(= { 2 }^{ 3k }.\left( 7+1 \right) -1\)
\(={ 2 }^{ 3k }.7+{ 2 }^{ 3k }-1\)
\( ={ 2 }^{ 3k }.7+7m=7\left( { 2 }^{ 3k }+m \right) \) 
which is divisible by 7
P(k +1) is true whenever p(k) is true 
\(\therefore P(n)\) ) is true for all \(n \in N\) 
Hence the proof.

Wages (Rs) (x)	No. of labourers(f)	Cumulative Frequency (c.f.)
20-30	3	3
30-40	5	8
40-50	20	28
50-60	10	38
60-70	5	N = 43

Q₁= Size of \({ \left( \frac { N }{ 4 } \right) }^{ th }\)value = size of\({ \left( \frac { 43 }{ 4 } \right) }^{ th }\) value
= Size of 10.75^th value
Q₁ lies in the interval (40-50) and its corresponding values are L = 40, f = 20, pcf = 8, c = 10
\({ Q }_{ 1 }=L+\left( \frac { \frac { N }{ 4 } -pcf }{ f } \right) \times c=40+\frac { 10.75-8 }{ 20 } \times 10\)
\(=40+\frac { 2.75 }{ 2 } =40+1.375\)
Q₁ = 43.38
Q₃=Size of \({ \left( \frac { 3N }{ 4 } \right) }^{ th }\) value = size of\({ \left( \frac { 3\times 43 }{ 4 } \right) }^{ th }\) value
= Size of (32.25)^th value
\(\therefore\) Q₃ lies in (50-60) and its corresponding values are L = 50, f = 10, pcf = 28 and c = 10
\({ Q }_{ 3 }=L+\left( \frac { \frac { 3N }{ 4 } -pcf }{ f } \right) \times c=50+\frac { 32.25-28 }{ 10 } \times 10\)
\(\therefore\) Q₃ = 50 + 4.25 = 54.25
\(\therefore\) Quartile deviation QD = \(\frac { 1 }{ 2 } ({ Q }_{ 3 }-{ Q }_{ 1 })=\frac { 1 }{ 2 } (54.25-43.38)=\frac { 1 }{ 2 } (10.87)=5.435\)

Let the events E₁, E₂, E₃ and A be defined as
E₁ = I urn is chosen
E₂ = II urn is chosen
E₃ = III urn is chosen
A - Ball drawn is white in color
\(\therefore\) P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)
P(A/E₁) \(=\frac{7}{17},\)
P(A/E₂ ) \(=\frac{5}{17},\)
P(A/E₃ )  \(=\frac{17}{17},\)
(i)  \(P(E_{ 1 }/A)\)
\(=\frac { P({ E }_{ 1 }).P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 })+P({ E }_{ 3 }).P(A/{ E }_{ 3 }) } \)
(By Baye's theorem)
\(=\frac { \frac { 1 }{ 3 }. \frac { 7 }{ 17 } }{ \frac { 1 }{ 3 }. \frac { 7 }{ 17 } +\frac { 1 }{ 3 } . \frac { 5 }{ 17 } +\frac { 1 }{ 3 } . \frac { 17 }{ 17 } } \)
\(=\frac {\frac {7}{51}}{\frac {7+5+17}{51}}=\frac{7}{29}\)  
(ii) \(P(E_{ 2 }/A)\) \(=\frac {\frac {1}{3}.\frac{5}{17}}{\frac {29}{51}}=\frac{5}{29}\)
(iii)  \(P(E_{ 3 }/A)\) \(=\frac {\frac {1}{3}.\frac{17}{17}}{\frac {29}{51}}=\frac{17}{29}\)

Let investment in 6% stock be x
\(\therefore\) Investment in 5% stock = 13,500 - x.
At 6% stock
If Investment = 140, Income = 6
If Investment = x, Income \(=\frac { 6x }{ 140 }=\frac { 3x }{ 70 } \)
At 5% stock
If Investment = 125, Income = 5
If Investment = 13,500 - x,
Income = \(\frac { 5(13,500-x )}{ 125 } \)  
\(=\frac { 13,500 -x}{ 25 } =540-\frac { x }{ 25 } \)
Total income = 560
\( \frac { 3x }{ 70 } +540-\frac { x }{ 25 } =560\)
\(\frac { 3x }{ 70 } -\frac { x }{ 25 } =20\)
\(\frac { 15x-14x }{ 350 } =20\) 
x = 350 x 20 = Rs. 7000
\(\therefore\) Amount invested in 6% stock = Rs. 7000
Amount invested in 5% stock = 13500 - 7000
= Rs. 6500

X	Y	X2	Y2	XY
23	18	529	324	414
27	22	729	484	594
28	23	784	529	644
29	24	841	576	696
30	25	900	625	750
31	26	961	676	806
33	28	1089	784	924
35	29	1225	841	1015
36	30	1296	900	1080
39	32	1521	1024	1248
\(\Sigma X\) = 311	\(\Sigma Y\) = 257	\(\Sigma X^2\) = 9875	\(\Sigma Y^2\) = 6763	\(\Sigma XY\) = 8171

\(r =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{\sqrt{N \Sigma X^2-(\Sigma X)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{10(8171)-(311)(257)}{\sqrt{10(9875)-(311)^2} \sqrt{10(6763)-(257)^2}} \)
\(=\frac{81710-79927}{\sqrt{98750-96721} \times \sqrt{67630-66049}} \)
\(=\frac{1783}{\sqrt{2029 \times 1581}} \)
\(r =\frac{1783}{1791.05}=0.9955\)

(A) R = 800 units, C₃ = 5 C₁ \(=\frac{10}{100} \times 0.02=0.002\)
(i) \(E O Q=\sqrt{\frac{2 C_3 R}{C_1}}=\sqrt{\frac{2 \times 5 \times 800}{0.02}}=2000\) units
(ii) Minimum inventory Cost \(=\sqrt{2 R C_3 C_1}=\sqrt{2 \times 800 \times 5 \times 0.002}=₹ 4\)
(iii) E O Q in rupees \(=2000 \times 0.02= 40\)
(iv) E O Q in years of supply = \(\frac{q_0}{R}=\frac{2000}{800}=2.5=0.4\)
(v) Number of orders per year \(=\frac{R}{q_0}=\frac{800}{2000}=0.4\)
(B) R = 400 units, C₃ = 5
\(C_1=\frac{10}{100} \times 1=0.1\)
(i) E O Q \(=\sqrt{\frac{2 C_3 R}{C_1}}=\sqrt{\frac{2 \times 400 \times 5}{0.1}}=200\)
(ii) Minimum average cost \(=\sqrt{2 R C_3 C_1}\)
\(=\sqrt{2 \times 400 \times 5 \times 0.1}= 20\)
(iii) E O Q in rupees \(=200 \times 1= 200\)
(iv) E O Q in years of supply \(=\frac{q_0}{R}=\frac{200}{400}=0.5\)
(v) Number of orders per year \(=\frac{R}{q_0}=\frac{400}{200}=2\)
(C) R = 13800, C₃ = 2
\(C_1=\frac{10}{100} \times 0.2=0.02\)
(i) EOQ \(=\sqrt{\frac{2 C_3 R}{C_1}}=\sqrt{\frac{2 \times 5 \times 13800}{0.02}}\)
= 2627 (units)
(ii) Minimum average cost \(=\sqrt{2 C_1 C_3 R}\)
\(=\sqrt{2 \times 5 \times 0.02 \times 13800} = 52.54\)
(iii) EOQ in rupees \(=2627 \times 0.20= 525.40\)
(iv) EOQ in years of supply \(=\frac{q_0}{R}=\frac{2627}{13800} =0.19\)
(v) Number of orders per year \(=\frac{R}{q_0}=\frac{13,800}{2627} =5.25\)

First we have to find the feasible region using the given conditions.
Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant.
Write all the inequalities of the constraints in the form of equations.
Therefore we have the lines x₁ + 4x₂ = 24 ; 3x₁ + x₂ = 21; x₁ + x₂= 9
x₁ + 4x₂ = 24 is a line passing through the points (0, 6) and (24, 0)
[(0, 6) is obtained by taking x₁ = 0 in x₁ + 4x₂ = 24 , (24, 0) is obtained by taking x₂ = 0 in x₁+ 4x₂ = 24].
Any point lying on or below the line x₁ + 4x₂ = 24 satisfies the constraint x₁ + 4x₂ ≤ 24 .
3 x₁ + x₂ = 21 is a line passing through the points (0, 21) and (7, 0). Any point lying on or below the line 3 x₁ + x₂ = 21 satisfies the constraint 3 x₁ + x₂ ≤ 21.
x₁ + x₂ = 9 is a line passing through the points (0, 9) and (9, 0). Any point lying on or below the line x₁ + x₂ = 9 satisfies the constraint x₁ + x₂ ≤ 9.
Now we draw the graph

The feasible region satisfying all the conditions is OABCD.The co-ordinates of the points are O(0,0) A(7,0);B(6,3) [the point B is the intersection of two lines x₁ + x₂ = 9 and 3 x₁ + x₂ = 21]; C(4,5) [the point C is the intersection of two lines x₁ + x₂ = 9 and x₁ + 4x₂ = 24] and D(0,6).
Corner points Z = 2x₁ + 5x₂ O(0,0) 0 A(7,0) 14 B(6,3) 27 C(4,5) 33 D(0,6) 30 Maximum value of Z occurs at C. Therefore the solution is x₁ = 4, x₂ = 5, Z_max = 33

p = 50 – 3x
\({dp\over dx}=-3\) ⇒ \({dx\over dp}=-{1\over 3}\)
Elasticity of demand: \(η_d=-{p\over x}.{dx\over dp}\)
\(=-{50-3x\over x}\left(-{1\over3}\right)\)\(={50-3x\over 3x}\ \ ...(1)\)
Now, Revenue: R = px
= (50 - 3x)x = 50x - 3x²
Average revenue: AR = p = 50 - 3x
Marginal revenue: \(MR={dR\over dx} = 50 - 6x\)
\({AR\over AR-MR}={50-3x\over (50-3x)-(50-6x)}\)
\(={50-3x\over 3x}\ \ ...(2)\)
From (1) and (2), we get
\(η_d={AR\over AR-MR}\), Hence verified.

LHS = cos 4x = cos 2(2x)
= 2 cos²2x-1 [∴ cos 2A = 2cos²A - 1]
= 2 [2 cos²x - 1]² - 1 = 2 [4 cos⁴x - 4cos2x + 1] - 1
= 8 cos⁴x - 8 cos²x + 2 - 1 = 1 - 8 cos²x + 8 cos⁴x
= 1 - 8 cos²x (1 - cos²x) = 1 - 8 cos²x . sin²x = RHS.

Since the axis of the parabola is vertical, its equation will be x² = 4ay.
Arch is 10 m high and 5 m wide at base.
​\(\therefore\)​ Point \((\frac{5}{2},10)\) lies on the parabola
​\(\therefore\)​  \(\frac{25}{4}\) = 4a(10)⇒ 4a = \(\frac{25}{40}\) = \(\frac{5}{8}\)(y)
\(\therefore\) Equation of the parabola becomes x² = \(\frac{5}{8}\)(y)
Let the width of the arch 2 m from the vertex is 2b, then point (b, 2) lies on the parabola
\(\therefore\) b² = \(\frac { 5(2) }{ 8 } =\frac { 10 }{ 8 } =\frac { 5 }{ 4 } \Rightarrow b=\frac { \sqrt { 5 } }{ 2 } \)

\(\therefore\) width of arch is  2b = 2.\(\frac{\sqrt{5}}{2}\) = √5m = 2.23m(app)

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\( \sqrt 2\)

\(\frac { \sin { \left( { 180 }^{ o }+A \right) \cos { \left( { 90 }^{ o }-A \right) \tan { \left( { 270 }^{ o }-A \right) } } } \quad \quad }{ \sec { \left( { 540 }^{ o }-A \right) \cos { \left( { 360 }^{ o }+A \right) \csc { \left( { 270 }^{ o }+A \right) } } } } \)
\(=\frac{(-\sin A)(\sin A)(\cot A)}{\sec (360+180-A) \cos A(-\sec A)} \)
\(=\frac{(-\sin A)(\sin A) \frac{\cos A}{\sin A}}{(-\sec A) \cos A(-\sec A)} \)
\(=-\frac{\sin A \cos A}{\frac{1}{\cos A} \cos A \frac{1}{\cos A}} \)
\(=-\sin A \cos ^2 A\)
= R.H.S
Hence proved.

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.

Given f(2) = 0
= 2 + 2
\(=4 \neq 0\)
\(\therefore \lim _{x \rightarrow 2} f(x) \neq f(2)\)
The function is not continuous at x = 2

a₁₁ = 30, a₁₂ = 40, x₁ = 120
a₂₁ = 20, a₂₂ = 10, x₂ = 60
\({b}_{11}={{{a}_{11}}\over{x_1}}={{30}\over{120}}={{1}\over{4}}\)
\({b}_{12}={{{a}_{12}}\over{{x}_{2}}}={{40}\over{60}}={{2}\over{3}}\)
\({b}_{21}={{{a}_{21}}\over{x_1}}={{20}\over{120}}={{1}\over{6}}\)
\({b}_{22}={{{a}_{22}}\over{{x}_{1}}}={{10}\over{60}}={{1}\over{6}}\)
The technology matrix is   B =  \(\begin{bmatrix}{{1}\over{4}}&{{2}\over{3}}\\ {{1}\over{6}}&{{1}\over{6}} \end{bmatrix}\)
I - B = \(\begin{bmatrix} 0&0\\0&1 \end{bmatrix}-\begin{bmatrix} {{1}\over{4}} &{{2}\over{3}}\\{{1}\over{6}}&{{1}\over{6}} \end{bmatrix}=\begin{bmatrix} {{3}\over{4}}&{{-2}\over{3}}\\ {{-1}\over{6}}&{{5}\over{6}} \end{bmatrix}\)
\(|I-B|=\frac{3}{4} \times \frac{5}{6}-\frac{2}{3} \times \frac{1}{6}=\frac{5}{8}-\frac{1}{9}=\frac{37}{72}=0\)
Since diagonals of I - B are positive and | I - B | is positive, the system is viable
\({(I-B)}^{-1}={{1}\over{|I-B|}}adj\ (I-B)={{72}\over{37}}\begin{bmatrix}{{5}\over{6}}&{{2}\over{3}}\\{{1}\over{6}}&{{3}\over{4}} \end{bmatrix}\)
X = (I - B)^-1 D where D = \(\begin{bmatrix} 80\\40 \end{bmatrix}\)
\(={{72}\over{37}}\begin{bmatrix}{{5}\over{6}}&{{2}\over{3}}\\{{1}\over{6}}&{{3}\over{4}} \end{bmatrix}\begin{bmatrix} 80 \\ 40 \end{bmatrix}\)
\(=\frac{72}{37}\left(\begin{array}{cc} 66.67 & +26.67 \\ 13.33 & +30 \end{array}\right)=\frac{72}{37}\left(\begin{array}{l} 93.34 \\ 43.33 \end{array}\right)\)
\(=\left(\begin{array}{c} 181.63 \\ 84.32 \end{array}\right)\)
The output for production sector X and Y are 181.63 and 84.32 respectively.

(i) Let y = x^x
Taking logarithm on both sides logy = x log x
Differentiating with respect to x,
\(\cfrac { 1 }{ y } .\cfrac { dy }{ dx } =x.\cfrac { 1 }{ x } +logx.1\)
\(\cfrac { dy }{ dx } =y\left[ 1+logx \right] \)
\( \therefore \cfrac { dy }{ dx } ={ x }^{ x }\left[ 1+logx \right] \)
 (ii) Let y = (logx)^cosx
Taking logarithm on both sides log y = cos x log(logx)
Differentiating with respect to x,
\(\cfrac { 1 }{ y } .\cfrac { dy }{ dx } =cosx\cfrac { 1 }{ logx } .\cfrac { 1 }{ x } +\left[ log\left( logx \right) \right] \left( -sinx \right) \)
\(\cfrac { dy }{ dx } =y\left[ \cfrac { cosx }{ xlogx } -sinxlog\left( logx \right) \right] \)
\(=\left( logx \right) ^{ cosx }\left[ \cfrac { cosx }{ xlogx } -sinxlog\left( logx \right) \right] \)

\(\bar{X}=36, \bar{Y}=85 \)
\(\sigma_x=11, \sigma_y=8, r=0.66 \)
\(b_{y x}=r \frac{\sigma_y}{\sigma_x}=0.66\left(\frac{8}{11}\right)=0.48 \)
\(b_{x y}=r\left(\frac{\sigma_x}{\sigma_y}\right)=0.66\left(\frac{11}{8}\right)=0.91\)
(i) Regression equation of Y on X is
\(Y-\bar{Y} =b_{y x}(X-\bar{X}) \)
Y - 85 = 0.48 X - 17.28
Y = 0.48 X + 67.72 .
Regression equation of X on Y is
\(X-\bar{X} =b_{x y}(Y-\bar{Y}) \)
X - 36 = 0.91Y - 77.35
X = 0.91 Y - 41.35
(ii) If X = 10
Y = 4.8 + 67.72 = 72.52

Y = -x² + 10 x - 15
\(\Rightarrow\) x² - 10x = - y - 15
\(\Rightarrow\) (x- 5)² = -y - 15 + 25
\(\Rightarrow\) (x- 5)² = -y + 10 
\(\Rightarrow\) (x - 5)² = (-y - 10)
The best time to end the project is when x = 5 months.