y = A sin x + B cos x
y₁ = A cos x – B sin x
y₂ = –A sin x –B cos x
y₂ = –y
y₂ + y = 0

nP _r= 360 = 36 \(\times\) 10
= 3 \(\times\) 3 \(\times\) 4 \(\times\) 5 \(\times\) 2
= 6 \(\times\) 5 \(\times\) 4 \(\times\) 3 = 6P₄
Therefore n = 6 and r = 4

(i) Variables:
Let x₁, x₂ represent bags of brand P and brand Q. 
(ii) Objective function:
Let Z be the amount of nitrogen. Since the amount of nitrogen is to be minimized, we have minimize Z = 3x + 3.5y
(iii) Constraints:
For Phosphoric acid,\({ x }_{ 1 }+2{ x }_{ 2 }\ge 240\)
For Potash, \({ 3x }_{ 1 }+{ 1.5x }_{ 2 }\ge 270\)
For Chlorine, \({ 1.5x }_{ 1 }+2{ x }_{ 2 }\ge 310\)
(iv) Non-negative restrictions:
Since the number of bags of brand P and Q, cannot be negative,\({ x }_{ 1 },{ x }_{ 2 }\ge 0\).
Here Mathematical form of the LPP is Minimize Z = 3x₁ + 3.5x₂ 
Subject to the constraints
\({ x }_{ 1 }+2{ x }_{ 2 }\ge 240\)
\({ 3x }_{ 1 }+{ 1.5x }_{ 2 }\ge 270\)
\( { 1.5x }_{ 1 }+2{ x }_{ 2 }\ge 310\)
and x₁, x₂ ≥ 0.

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

Here a = 18,000 ; i = 0.15 and k = 12.
P = \({\frac{a}{{i}/k}} = {\frac{18,000}{0.15/12}}\) 
= \({\frac{18,000}{0.15}} \times {12}\)
\(=\frac{18,00,000\times 12}{15}=14,40,000\)
Therefore the amount to be deposited is Rs. 14,40,000.

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

Let \(A=\begin{bmatrix} 1&-1\\2&3 \end{bmatrix}\)
\(|A|=\begin{vmatrix} 1&-1\\2&3 \end{vmatrix}=3+2=5\) ≠ 0
\(\therefore A^{-1} \text { exists }\)
Now, \({A}^{-1}={{1}\over{|A|}}adj\ A={{1}\over{5}}\begin{bmatrix} 3 & 1 \\ -2 & 1\end{bmatrix}\)

\(\frac { 4x+1 }{ (x-2)(x+1) } =\frac { A }{ x-2 } +\frac { B }{ x+1 } \)
⇒ \(\frac { 4x+1 }{ (x-2)(x+1) } =\frac { A(x+1)+B(x-2) }{ (x-2)(x+1) } \)
⇒ 4x + 1 = A(x + 1) + B(x - 2)
If x = 2
9 = A(2 + 1)
\(\Rightarrow A=\frac{9}{3}=3\)
If x = -1
-4 + 1 = B(-1 - 2)
⇒ -3 = -3B  ⇒ \(\ B=1 \)
\(\frac { 4x+1 }{ (x-2)(x+1) } =\frac { 3 }{ x-2 } +\frac { 1 }{ x+1 } \)

\(P =\frac{\frac{a}{i}}{k} \)
\(=\frac{\frac{2000}{0.1}}{12}=2,40,000\)

Amount = 200 x 50 = Rs. 10,000

\(\frac{d C}{d x}=M C=\frac{5 e^{5 x}}{25}=\frac{1}{5} e^{5 x}\)

Since x₁ = 2.5, x₂ = 35 satisfies all the Constraints

(fg) (x) = 2x³ + x² = fg (0) = 0

\(\sin \left(\cos ^{-1} \frac{3}{5}\right)=\sin \left(\sin ^{-1} \frac{4}{5}\right)=4 / 5\)

\(\sec A(-\cos A)=\frac{1}{\cos A}(-\cos A)=-1\)

nP₂ = 20
n(n - 1) = 5 x 4
n = 5

\(\left|\begin{array}{lll} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{array}\right|=0 \Rightarrow-1\left[x^2-x\right]=0\)
\(\Rightarrow x(x-1)=0 \Rightarrow x=0,1\)

  \(cos2A=\cfrac { 1-{ tan }^{ 2 }A }{ 1+{ tan }^{ 2 }A } =\cfrac { 1-\frac { 1 }{ 49 } }{ 1+\frac { 1 }{ 49 } } =\cfrac { 48 }{ 49 } \times \cfrac { 49 }{ 50 } \) = \(\cfrac { 24 }{ 25 } \)   ..(1)
Now, sin4B = 2sin2B cos2B
\(\\ \\ \\ =2\cfrac { 2tanB }{ 1+{ tan }^{ 2 }B } \times \cfrac { 1-{ tan }^{ 2 }B }{ 1+{ tan }^{ 2 }B } \)   
\( =\cfrac { 4\times \frac { 1 }{ 3 } }{ 1+\frac { 1 }{ 9 } } \times \cfrac { 1-\frac { 1 }{ 9 } }{ 1+\frac { 1 }{ 9 } } =\cfrac { 24 }{ 25 } \) ..(2)
From(1) and (2) we get, cos2A = sin4B.

Let A(7, –5)
Equation of circle is
\({ x }^{ 2 }+{ y }^{ 2 }-6x+4y-12=0\) 
Substitute (7, -5) for (x, y), we get
\({ x }^{ 2 }+{ y }^{ 2 }-6x+4y-12={ 7 }^{ 2 }+\left( -5 \right) ^{ 2 }-6(7)+4(-5)-12\) 
= \(49+25-42-20-12=0\) 
\(\therefore \left( 7,-5 \right) \) lies on the circle
Here g = –3 and f = 2
\(\therefore Centre=C\left( 3,-2 \right) \) 
Let the other end of the diameter be B (x, y)
Midpoint of  AB = \(\left(\frac{x+7}{2}, \frac{y-5}{2}\right)=C(3,-2)\)
\( \cfrac { x+7 }{ 2 } =3\)
\(x=-1\) 
\(\cfrac { y-5 }{ 2 } =-2\)
\(y=1\)
Other end of the diameter is (–1 , 1).

Given the lines are concurrent
Therefore \(\left| \begin{matrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{matrix} \right| =0\) 
\(\left| \begin{matrix} 3 & -5 & -11 \\ 5 & 3 & -7 \\ 1 & k & 0 \end{matrix} \right| =0\)
\(1(35+33)-k(-21+55)=0\)
\( \Rightarrow 34k=68\therefore k=2\)

\(A=\left[ \begin{matrix} 2 & 3 \\ 1 & 4 \end{matrix} \right] \)
adj A = [A_ij]^T
\(=\left[ \begin{matrix} 4 & -3 \\ -1 & 2 \end{matrix} \right] \)

Since the decision variables x₁ ,x₂ are non-negative, the solution lies in the I quadrant of the plane.
Consider the equations
\(-{ x }_{ 1 }+3{ x }_{ 2 }=10\)

\({ x }_{ 1 }+{ x }_{ 2 }=6\)

\({ x }_{ 1 }{ -x }_{ 2 }=2\)

The feasible region is OABCD and its co-ordinates are O(0, 0)A(2, 0) B(4, 2) C(2, 4) and D(0, 10/3)

Maximum of Z occurs at\(D\left( 0,\frac { 10 }{ 3 } \right) \). Hence, the solution is \({ x }_{ 1 }=0,{ x }_{ 2 }=\frac { 10 }{ 3 } \quad and\quad { Z }_{ max }=\frac { 20 }{ 3 } \)

Given A = Rs.3783,i = 0.05,n = 3
A = \(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
3783 = \(\cfrac { a }{ i } \) [(1.05)³-1]
3783 x 0.05 = a[1.1576-1]
a = \(\cfrac { 189.15 }{ 0.1576 } \) = 1200.19
\(\therefore\) a =Rs.1200
(1.05)³ = 3 log (1.05)
= 3(0.212)
= 0.636
Antilog of 0.636 is 1.1576

n=12
Rank correlation co-efficient
\(\rho =1-\frac { 6\sum { { d }^{ 2 } } }{ N({ N }^{ 2 }-1) } \)
=1-\(\frac { 6\times 416 }{ 12({ 12 }^{ 2 }-1) } =1-\frac { 16 }{ 11 } =\frac { -5 }{ 11 } \)
\(\rho\)=-0.45

\(\bar{X}\) = 6, \(\bar{Y}\) = 8, \(\sigma _x=5\)
\(\sigma_y=\frac{40}{3}, \quad r=\frac{8}{15} \)
\(b_{y x}=r \frac{\sigma_x}{\sigma_y}=\frac{8}{15}\left(\frac{40}{3 \times 5}\right)=\frac{64}{45}=1.422\)
Regression line of Y on X is
\(Y-\bar{Y}=b_{y x}(X-\bar{X}) \)
Y - 8 = 1.422(X - 6)
Y = 1.422 X - 8.532 + 8
Y = 1.422 X - 0.532 .
If  X = Rs 100,
Y = 142.2 - 0.532 = Rs 141.67

HM = \(\frac { N }{ \sum { \left( \frac { f }{ x } \right) } } =\frac { 37 }{ 0.9771 } \) = 37.86 km/hr

C = \(\frac { 2 }{ 3 } x+\frac { 35 }{ 2 } \)
(i) At x = 4 units,
\(c=\frac{8}{3}+\frac{35}{2}=\frac{16+105}{6} = Rs.\frac { 121 }{ 6 } \)
(ii) A.C. = Average cost \(=\frac{C}{x}=\frac{2}{3}+\frac{35}{2 x}\)
At x = 10
A.C \(=\frac{2}{3}+\frac{35}{20}=\frac{40+105}{60}=\frac{29}{12}\)
(iii) Marginal cost = M.C \(=\frac{d C}{d x}=\frac{2}{3}\)
At x = 3,
M.C \(= \frac{2}{3}\)

(i) Variables: Let the mixture contains x₁ kg of food F₁ and x₂ kg of food F₂
(ii) Objective function: Cost of x₁ kg of food F₁ = 50 x₁
Cost of x₂ kg of food F₂ = 70x₂
The cost is to be minimized
Therefore minimize Z = 50 x₁ + 70x₂
(iii) Constraints:
We make the following table from the given data

4x₁ + 5x₂ ≥ 6 (since the mixture contains ‘atleast 6’ units of vitamin A, we have the inequality of the type ≥ )
6x₁ + 3x₂ ≥ 9 (since the mixture contains ‘atleast 9’ units of vitamin B, we have the inequality of the type ≥ )
(iv) Non-negative restrictions:
Since the number of kgs of vitamin A and vitamin B are non-negative, we have x₁, x₂ ≥ 0
Thus, we have the following linear programming model
Minimize Z = 50x₁ + 70x₂ subject to 4x₁+ 5x₂ ≥ 6
6x₁ + 3x₂ ≥ 9 and x₁, x₂ ≥ 0

Given equation is
3x² + 7xy + 2y² + 5x + 5y + 2 = 0
Factorizing 3x² + 7xy + 2y² = (x + 2y) (3x + y)
\(\therefore \) 3x² + 7xy + 2y² + 5x + 5y + 2 = (x + 2y + l)(3x + y + m)
Equating the x and y Co-ordinates both sides,

We get 5 = m + 3l  ...(1)
5 = 2m + l   ....(2)

Substituting l = 1 in (2) we get,
5 = 2m + 1 \(\Rightarrow \) 2m = 4 \(\Rightarrow \) m = 2.
Hence the separate equations are
x + 2y + 1 = 0 and 3x + y + 2 = 0.

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

f(x) = |x - 2|
y = x² if x ≥ 0
= 2 - x if  x ≥ 2

Here the technology matrix is given under

The technology matrix is B = \(\left[ \begin{matrix} 0.4 & 0.1 \\ 0.7 & 0.6 \end{matrix} \right] \)
I - B = \(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)-  \(\left[ \begin{matrix} 0.4 & 0.1 \\ 0.7 & 0.6 \end{matrix} \right] \) = \(\left[ \begin{matrix} 0.6 & -0.1 \\ -0.7 & 0.4 \end{matrix} \right] \)
|I - B| = \(\left| \begin{matrix} 0.6 & -0.1 \\ -0.7 & 0.4 \end{matrix} \right|\)
= (0.6)(0.4) – (–0.7)(–0.1)
= 0.24 – 0.07 = 0.17
Since the diagonal elements of I – B are positive and value of |I-B| is positive, the system is viable.
adj(I - B) = \(\left[ \begin{matrix} 0.4 & 0.1 \\ 0.7 & 0.6 \end{matrix} \right] \)
(I - B)^-1 = \(\frac{1}{|I -B |} \)adj(I - B)
= \(\frac{1}{0.17} \) \(\left[ \begin{matrix} 0.4 & 0.1 \\ 0.7 & 0.6 \end{matrix} \right] \)
X = (I – B)^–1D, where D =\(\left[ \begin{matrix} 50 \\ 100 \end{matrix} \right] \)
= \(\frac{1}{0.17} \)\(\left[ \begin{matrix} 0.4 & 0.1 \\ 0.7 & 0.6 \end{matrix} \right] \)\(\left[ \begin{matrix} 50 \\ 100 \end{matrix} \right] \)
= \(\frac{1}{0.17} \)\(\left[ \begin{matrix} 30 \\ 95 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 176.5 \\ 558.8 \end{matrix} \right] \)
Steel output = 176.5 tonnes
Coal output = 558.8 tonnes
Total labour days required
= 5(steel output) + 2( coal output)
= 5(176.5) + 2(558.8)
= 882.5 + 1117.6 = 2000.1
\(\simeq\) 2000 labour days.

EFT and LFT are same in the activities.
1 - 2, 2 - 3, 3 - 4 and 4 - 5
Hence, the critical path is 1 - 2 - 3 - 4 - 5 and the duration of project completion is 48 days .

Given q = 16 -3P₁ - 2P₂²
Differentiating partially w.r.t, 'P₁' and 'P₂' we get,
\({\partial q\over \partial p_1}=-3\ and\ {\partial q\over \partial p_2}=-4p_2\)
Partial elasticities
\({Eq\over Ep_1}={-p_1\over q}{\partial q\over \partial p_2}\)
\(={-p_1\over 16-3p_1-2p_2^2}(-3)\)
\(={3p_1\over 16-3p_1-2p_2^2}\)
and \({Eq\over Ep_2}=-{p_2\over q}{\partial q\over \partial p_2}={-p_2\over 16-3p_1-3p_2^2}(-4p_2)\)
\(={4p_2^2\over 16-{ 3p }_{ 1 }-2p_2^2}\)

Stock = Rs.9000
FV =Rs.100
Dividend Rate = 7%
Income on 7% stock =\(\frac { \text {stock} }{ 100 } \)x FV x Rate percentage
=\(\frac { 9000 }{ 100 } \times 100\times \frac { 7 }{ 100 } \) = Rs.630    ...(1)
Cost of price of one share = 80
Sale proceeds = \(\frac { \text {stock} }{ 100 } \) x Cost Price
=\(\cfrac { 9000 }{ 100 } \)x 80 =7200
Investment = Rs.7200
Market Price = Rs.120
Dividend Rate =15%
Income =\(\frac {\text { Investment} }{ \text {Market Price } }\)x Dividend Rate
=\(\frac { 7200 }{ 120 } \)x15=Rs.900 .....(2)
Change in income = 930 - 630 = Rs.270

Already the values are arranged in ascending order then Median is obtained by the following.

Median = size of \(\left( \frac { (n+1) }{ 2 } \right) ^{ th }\)value
= size of \(\left( \frac { (120+1) }{ 2 } \right) ^{ th }\) value
= size of 60.5^th item = 55
MD about Median = \(\frac { \Sigma f|X-Median| }{ N } =\frac { \Sigma f|D| }{ N } \)
Mean deviation about Median

MD about Median = \(\frac { 2220 }{ 120 } \) = 18.5
Coefficient of mean deviation about median = \(\frac { MD \ about \ Median }{ Median } \)
= \(\frac { 18.5}{ 55 } \) = 0.34

\(\overline { X } =\frac { \sum { fx } }{ \sum f } =\frac { 390 }{ 30 } =13\)
Mean deviation about mean = \(\frac { \sum { f|D| } }{ \sum f } =\frac { 130 }{ 30 } =4.34\) 

\(\frac { \partial p }{ \partial { P }_{ 1 } } =2-{ p }_{ 1 }{ p }_{ 2 }\)
\(\frac { \partial p }{ \partial { P }_{ 2 } } =1-{ p }_{ 1 }^{ 2 }\)
(i) \(\frac { Eq }{ { EP }_{ 1 } } =\frac { { p }_{ 1 } }{ q } \frac { \partial p }{ \partial { P }_{ 1 } } =\frac { -p }{ 5-2{ p }_{ 1 }+{ p }_{ 2 }-{ p }_{ 1 }^{ 2 }{ p }_{ 2 } } (-2-{ p }_{ 1 }{ p }_{ 2 })\)
= \(\frac { { 2p }_{ 1 }+2{ p }_{ 1 }^{ 2 }{ p }_{ 2 } }{ 5-2{ p }_{ 1 }+{ p }_{ 2 }-{ p }_{ 1 }^{ 2 }{ p }_{ 2 } } \)
when p₁ = 3 and p₂ = 7
\(\frac { { E }q }{ { Ep_{ 1 } } } =\frac { 2(3)+2(9)(7) }{ 5-6+7-(9)(7) } =\frac { 132 }{ -57 } =\frac { -132 }{ 57 } \)
\(\frac { Eq }{ { EP }_{ 1 } } =\frac { { p }_{ 2 } }{ q } \frac { \partial p }{ \partial { P }_{ 2 } } =\frac { -p\left( 1-{ p }_{ 1 }^{ 2 } \right) }{ 5-2{ p }_{ 1 }+{ p }_{ 2 }-{ p }_{ 1 }^{ 2 }{ p }_{ 2 } } \)
\(\frac { { -p }_{ 2 }+{ p }_{ 2 }{ p }_{ 1 }^{ 2 } }{ 5-2{ p }_{ 1 }+{ p }_{ 2 }-{ p }_{ 1 }^{ 2 }{ p }_{ 2 } } \)
when p₁ - 3 and p₂ = 7
\(\frac { { E }q }{ { Ep_{ 2 } } } =\frac { -7+7(9) }{ 5-6+7-(9)(7) } =\frac { 56 }{ -57 } =\frac { -56 }{ 57 } \)

\(r(x, y)=\frac{N \Sigma X Y-\left(\sum X\right)\left(\sum Y\right)}{\sqrt{N \Sigma X^2-(\Sigma Y)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{10(10718)-(310)(340)}{\sqrt{10(10480)-(310)^2} \sqrt{10(12280)-(340)^2}} \)
\(=\frac{107180-105400}{\sqrt{104800-96100} \sqrt{122800-115600}} \)
\(=\frac{1780}{\sqrt{8700 \times 7200}}=\frac{1780}{7914.54}=0.2249\)

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.

A parallelogram is formed by choosing two straight lines from the set of m parallel lines and two straight lines from the set of n parallel lines. Two straight lines from the set of m parallel lines can be chosen in mC₂ ways and Two straight lines from the set of n parallel lines can be chosen in nC₂ ways
Hence, the number of parallelograms formed\(\Rightarrow\)mC₂\(\times \)nC₂\(=\frac { m(m-1) }{ 2\times 1 } \times \frac { n(n-1) }{ 2\times 1 } =\frac { mn(m-1)(n-1) }{ 4 } \)

Let P(n) denote the statement 5²ⁿ - 1 is divisible by 24.
Step-1: Put n = 1
5²(1) - 1 = 25 - 1 = 24 is divisible by 24
∴ P(1) is true.
Step-2:
Let us assume that P(k) is true
∴ 5^2k - 1 is divisible by 24.
⇒ 5^2k - 1 = 24m
⇒ 5^2k = 24m + 1    ...(1)
To prove that P(k + 1) is true 
P(k + 1) = 5^2(k+1) - 1
= 5^2k+2 - 1 = 5^2k.5² - 1
= (24m + 1)25 - 1
= 24m. 25 + 25 - 1
= 24m + 25 - 24
= 24(25m + 1)
Which is divisible by 24
∴ P (k + 1) is true whenever P(k) is true.
∴ p(n) is true for all \(n\in N\).

Given \(\sin { \theta = \frac { 3 }{ 5 } } \) and \(\frac { \pi }{ 2 } <\theta <\pi \)
\(\therefore\) \(\theta\) is in II quadrant, only sin and its reciprocal is positive
\(\cos { \theta } =\frac { adj }{ hyp } =-\frac { 4 }{ 5 } ,\tan { \theta } =\frac { -3 }{ 4 } \quad .....(1)\)

Also \(\tan { \varphi } =\frac { 1 }{ 2 } \) and \(\pi <\varphi <\frac { 3\pi }{ 2 } \)
\(\therefore\) \(\phi \) is in III quadrants, tan \(\varphi \) and its reciprocal alone are positive.
\(\therefore \quad \sec { \phi } =\frac { hyp }{ adj } =-\frac { \sqrt { 5 } }{ 2 } \)...(2)

\(8\tan { \theta } -\sqrt { 5 } \sec { \phi } \)
\(=8\left( \frac { -3 }{ 4 } \right) -\sqrt { 5 } \left( -\frac { \sqrt { 5 } }{ 2 } \right) \)   [using (1) and (2)]
\(=+2(-3)+\frac { 5 }{ 2 } =-6+\frac { 5 }{ 2 } \)
\(=\frac { -12+5 }{ 2 } =\frac { -7 }{ 2 } \) 
Hence Proved.