Income \(=500 \times 100 \times \frac{15}{100}=7,500\)

If FV 100, Investment = 120
FV = 20,000,
Investment = \(\frac{120\times 20,000}{100} \) = Rs. 24000

\(\frac{\partial u}{\partial x} =8 x+4 y+4 \)
\(\frac{\partial^2 u}{\partial y \partial x} =4 \)

\(\frac{d y}{d x}=1, \frac{d z}{d x}=\frac{-1}{x^2} \quad \frac{d y}{d z}=-x^2\)

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

\(\sin \left(-420^{\circ}\right) =-\sin 420^{\circ}=-\sin (360+60) =-\sin 60^{\circ}=\frac{-\sqrt{3}}{2} \)

Equation of x axis y = 0, equation of y axis x = 0
\(\therefore\) combined equation xy = 0

On x axis y = 0
\(3 x=1 \Rightarrow x=\frac{1}{3}\)

The number of ways = 6!/3! = 120

u(x, y) = (x²+y²)^-1/2
u(tx, ty) = (t²x²+t²y²)^-1/2 = t^-1(x²+y²)^-1/2
∴ u is a homogeneous function of degree –1
By Euler’s theorem \(x.\frac { \partial u }{ \partial x } +y.\frac { \partial u }{ \partial y } =\left( -1 \right) u=-u\)
Verification:
\(u =\left(x^2+y^2\right)^{-\frac{1}{2}} \)
\(\frac{\partial u}{\partial x} =-\frac{1}{2}\left(x^2+y^2\right)^{-\frac{3}{2}} \cdot 2 x=\frac{-x}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(x \cdot \frac{\partial u}{\partial x} =\frac{-x^2}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(\frac{\partial u}{\partial y} =-\frac{1}{2}\left(x^2+y^2\right)^{-\frac{3}{2}} \cdot 2 y=\frac{-y}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(y \cdot \frac{\partial u}{\partial y} =\frac{-y^2}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(\therefore x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y}=\frac{-\left(x^2+y^2\right)}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(=(-1) \frac{1}{\sqrt{x^2+y^2}}=(-1) u=-u \)
Hence Euler’s theorem verified

Given \(P(A)=\frac { 1 }{ 2 } \Rightarrow p(\bar { A } )=1-P(A)=1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
\(P(B)=\frac { 2 }{ 3 } \Rightarrow P(\bar { B } )=1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)
\(P(C)=\frac { 3 }{ 4 } \Rightarrow P(\bar { C } )=1-P(C)=1-\frac { 3 }{ 4 } =\frac { 1 }{ 4 } \)
P(Only one of them hits the target)
\(=P(A\cap \bar { B } \cap \bar { C } )+P(\bar { A } \cap B\cap \bar { C } )+P(\bar { A } \cap \bar { B } \cap C)\)
\(=P(A).P(\bar { B) } .P(\bar { C) } +P(\bar { A } ).P(B).P(\bar { C } )+P(\bar { A } ).P(\bar { B } ).P(C)\)
\(=\frac { 1 }{ 2 } \times \frac { 1 }{ 3 } \times \frac { 1 }{ 4 } +\frac { 1 }{ 2 } \times \frac { 2 }{ 3 } \times \frac { 1 }{ 4 } +\frac { 1 }{ 2 } \times \frac { 1 }{ 3 } \times \frac { 3 }{ 4 } \)
\(=\frac { 1 }{ 24 } +\frac { 2 }{ 24 } +\frac { 3 }{ 24 } =\frac { 6 }{ 24 } =\frac { 1 }{ 4 } \)

\(\text {(i) Number of shares }=\frac{\text { Investment }}{M \cdot V}\)
\(=\frac{8000}{80}=100\)
\(\text {Amount of dividend }=100 \times 100 \times \frac{7}{100}= 700\)
\(\text {(ii) SP of } 100 \text { shares }=75 \times 100= 7500\)
\(\text {Investment }=\mathrm{SP}+\text { Dividend }\)
\(=7500+700= 8200\)
\(\text {Income }=\frac{\text { Investment }}{\mathrm{M} . \mathrm{V}} \times \mathrm{FV} \times \text { Rate of dividend }\)
\(=\frac{8200}{41} \times 25 \times \frac{18}{100}= 900 \text {. }\)
\(\text {(iii) Increase in his return }=900-700\)
\(\text { = } 200\)
\(\% \text { of increase in return }=\frac{\text { Increase in return }}{\text { Investment }} \times 100\)
\(=\frac{200}{8000} \times 100=2.5 \% \)

LHS = \({ cot }^{ -1 }\left[ \frac { \sqrt { 1+sinx } +\sqrt { 1-sinx } }{ \sqrt { 1+sinx } -\sqrt { 1-sinx } } \right] \)
=  \({ cot }^{ -1 }\left[ \frac { \sqrt { 1+sinx } +\sqrt { 1-sinx } }{ \sqrt { 1+sinx } -\sqrt { 1-sinx } } \times \frac { \sqrt { 1+sinx } +\sqrt { 1+sinx } }{ \sqrt { 1+sinx } +\sqrt { 1+sinx } } \right] \)    [multiplying the numerator and denominator by the conjugate of the denominator]
=  \({ cot }^{ -1 }\left[ \frac { \left( \sqrt { 1+sinx } +\sqrt { 1-sinx } \right) ^{ 2 } }{ (\sqrt { 1+sinx } )^{ 2 }-(\sqrt { 1-sinx } )^{ 2 } } \right] \)

= \({ cot }^{ -1 }\left[ \frac { 1+sinx+1-sinx+2\sqrt { (1+sin\quad x)(1-sin\quad x) } }{ (1+sin\quad x)-(1-sin\quad x) } \right] \)
=  \({ cot }^{ -1 }\left[ \frac { 2+2\sqrt { 1-{ sin }^{ 2 }x } }{ 2\quad sin\quad x } \right] ={ cot }^{ -1 }\left[ \frac { 2+2\quad cos\quad x }{ sin\quad x } \right] \left[ \because 1-{ sin }^{ 2 }x={ cos }^{ 2 }x \right] \)
= \({ cot }^{ -1 }\left[ \frac { 1+cos\quad x }{ sin\quad x } \right] ={ cot }^{ -1 }\left[ \frac { 2{ cos }^{ 2 }\frac { x }{ 2 } }{ 2sin\frac { x }{ 2 } cos\frac { x }{ 2 } } \right] \)    \(\left[ \because sin\quad A=2 \ \sin\frac { A }{ 2 } cos\frac { A }{ 2 } \quad and\quad 1+cos\quad x=2{ cos }^{ 2 }\frac { x }{ 2 } \right] \)
= \({ cot }^{ -1 }\left[ \frac { cos\frac { x }{ 2 } }{ sin\frac { x }{ 2 } } \right] ={ cot }^{ -1 }\left( cot\frac { x }{ 2 } \right) \)
= \(\frac { x }{ 2 } \) =RHS
Hence proved.

Given \(f(3)=6\)
\(\lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3} \frac{x^2-9}{x-3}\)
\(=\lim _{x \rightarrow 3} \frac{(x+3)(x-3)}{x-3}\)
\(=3+3=6=f(3)\)
\(\lim _{x \rightarrow 3} f(x)=f(3)\)
\(\therefore\) The function is continuous at x = 3.

Given \(\Delta =\left| \begin{matrix} { cosec }^{ 2 }\theta & { cot }^{ 2 }\theta & 1 \\ { cot }^{ 2 }\theta & { cosec }^{ 2 }\theta & -1 \\ 42 & 40 & 2 \end{matrix} \right| =0\)
Applying C₁\(\rightarrow\)C₁ - C₂, we get,
\(\Delta =\left| \begin{matrix} { cosec }^{ 2 }\theta -{ cot }^{ 2 }\theta & { cot }^{ 2 }\theta & 1 \\ { cot }^{ 2 }\theta -{ cosec }^{ 2 }\theta & { cosec }^{ 2 }\theta & -1 \\ 42-40 & 40 & 2 \end{matrix} \right| \)
\(=\left| \begin{matrix} 1 & { cot }^{ 2 }\theta & 1 \\ -1 & { cosec }^{ 2 }\theta & -1 \\ 2 & 40 & 2 \end{matrix} \right| \)  [\(\because\) cosec²  \(\theta\)  - cot²  \(\theta\)  =1]   
= 0
\(\Delta =0\)    [\(\because\)  C₁ \(\equiv \) C₃]  

(1 + x)²ⁿ
2n is even
Middle term is \(t_{\frac{n}{2}+1}=t_{\frac{2 n}{2}+1}=t_{n+1}\)
\(r=n\)
\(t_{r+1}=n C_r x^{n-r} a^r\)
\(t_{n+1}=2 n C_n(1)^{2 n-n} x^n\)
\(=2 n C_n \cdot x^n\)
\(n C_r=\frac{n !}{r !(n-r) !}\)
\(=\frac{(2 n) !}{n !(2 n-n) !} x^n\)
\(=\frac{(2 n)(2 n-1)(2 n-2)(2 n-3) \ldots 5 \cdot 4 \cdot 3 \cdot 2.1}{n ! n !} x^n\)
\(=\frac{(2 n)(2 n-2) \ldots 4.2(2 n-1)(2 n-3) \ldots 5.3 .1}{n ! n !} x^n\)
\(=\frac{2^n(n(n-1) \ldots 2.1)(2 n-1)(2 n-3) \ldots 5.3 .1}{n ! n !} x^n\)
\(=\frac{2^n \cdot n !(2 n-1)(2 n-3) \ldots 5 \cdot 3 \cdot 1}{n ! n !} x^n\)
\(=\frac{1.3 .5 \ldots(2 n-3)(2 n-1) 2^n x^n}{n !}\)

\({ \left( x+\frac { 1 }{ x } \right) }^{ 2n }\)
t_{r+ 1} = nC_r x^n-r a^r
\(\therefore\) t_r+1 = 2nC_r x^2n-r \({\left( {{1}\over{x}} \right)}^{r}\) = 2nC_r x^2n-r \({\left( {{1}\over{x}} \right)}^{r}\) = 2nC_r x^2n-2r
To find the term independent of x, equate the power of x as 0
 2n = 2r \(\Rightarrow\) r = n
\(\therefore\) t_r+1 = 2nC_n.x^2n-2n = 2nC_n
\(={{2n!}\over{n!n!}}={{(2n)(2n-1)(2n-2)(2n-3)......4,3,2,1}\over{n!n!}}\)
\(={{[(2n)(2n-2)(2n-4)...4.2][(2n-1)(2n-3)......3.1]}\over{n!n!}}\)
\(={{2^n(n)(n-1)(n-2).2.1.[(2n-1)(2n-3)....3.1]}\over{n!n!}}\) 
\(=\frac{2^n . n !(2 n-1) \ldots 5.3 .1}{n ! n !}\)
\(=\frac{1.3 .5 \ldots(2 n-1) 2^n}{n !}\) 
Hence proved.

a = Rs.500, i = 0.07, n = 7
p=\(\frac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
\(= \frac { 500 }{ 0.07 } \left[ \left(1.07 \right) ^{ 7 }-1 \right] \)
= 7142.86(1.6058-1)
= 7142.86(0.6058)
= Rs.4328.14

Calculation of Geometric Mean

Class of people	Income per head in 1990 (Rs) X	No of Families f	log X	f log X
Landlords	1000	1	3.0000	3.0000
Cultivators	80	50	1.9031	95.1550
Landless labourers	40	25	1.6021	40.0525
Money- lenders	750	2	2.8751	5.7502
School teachers	100	3	2.0000	6.0000
Shop-keepers	150	4	2.1761	8.7044
Carpenters	120	3	2.0792	6.2376
Weavers	60	5	1.7782	8.8910
		N = 93		\(\sum\) = 173.7907

GM = Anti \(\log { \left( \frac { \sum { flogX } }{ N } \right) } \)
= Anti \(\log { \left( \frac { 173.7907 }{ 93 } \right) } \)
= Anti log(1.8687)
GM = 73.91

Given \(y=\sqrt { x } +\frac { 1 }{ \sqrt { x } } .\Rightarrow y=\frac { x+1 }{ \sqrt { x } } \)  ......(1)
Differentiating with respect to 'x' we get,
\(\frac { dy }{ dx } =\frac { \sqrt { x } .\frac { d }{ dx } (x+1)-(x+1).\frac { d }{ dx } (\sqrt { x } ) }{ { (\sqrt { x } })^{ 2 } } \)
\(=\frac { \sqrt { x } (1)-(x+1).\frac { 1 }{ 2\sqrt { x } } }{ x } =\frac { \frac { 2x-(x+1) }{ 2\sqrt { x } } }{ x } \)
\(\frac { dy }{ dx } =\frac { 2x-x-1 }{ 2x.\sqrt { x } } =\frac { x-1 }{ 2x\sqrt { x } } \)...(2)
LHS=\(2x\left( \frac { dy }{ dx } \right) +y\)
\(=2x\left( \frac { x-1 }{ 2x\sqrt { x } } \right) +\frac { x+1 }{ \sqrt { x } }\)
 [From (1) and (2)]
\(=\frac { x-1 }{ \sqrt { x } } +\frac { x+1 }{ \sqrt { x } } =\frac { x-1+x+1 }{ \sqrt { x } } =\frac { 2x }{ \sqrt { x } } =\frac { 2\sqrt { x } .\sqrt { x } }{ \sqrt { x } } =2\sqrt { x } \) = RHS.
Hence proved

\(y=\frac { 1 }{ 5 } { x }^{ 2 }-6x+100\)
\(y=\frac{x^2-30 x+500}{5}\)
\(\Rightarrow\)  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.

\(f\left( x \right) =\frac { 1 }{ 2x+1 }\)
\(f\left( f\left( x \right) \right) =f\left( \frac { 1 }{ 2x+1 } \right) \)
\(=\frac { 1 }{ 2\left( \frac { 1 }{ 2x+1 } \right) +1 } =\frac { 1 }{ \frac { 2+2x+ 1 }{ 2x+1 } } =\frac { 2x+1 }{ 2x+3 } \)

5x² + 5y² +4x - 8y - 16 = 0
[Divide by 5]
x² + y² +  \(\frac { 4 }{ 5 } x-\frac { 8 }{ 5 } y-\frac { 16 }{ 5 } =0\)
Here 2g  = \(\frac { 4 }{ 5 } \) \(\Rightarrow\)  \(g=+\frac { 2 }{ 5 } \)
2f = \(-\frac { 8 }{ 5 } \)     \(\Rightarrow\) \(f=-\frac { 4 }{ 5 } \)
and c = \(-\frac { 16 }{ 5 } \)
Center of the circle is (-g, -f) \(\Rightarrow \) \(\left( -\frac { 2 }{ 5 } ,\frac { 4 }{ 5 } \right) \)
Radius of the circle is \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
\(\Rightarrow\) r = \(\sqrt { \frac { 4 }{ 25 } +\frac { 16 }{ 25 } +\frac { 16 }{ 5 } } =\sqrt { \frac { 20 }{ 25 }+ { \frac { 16 }{ 5 }} } \)
\(\Rightarrow\) \(r=\sqrt { \frac { 20 }{ 5 } } \) = \(\sqrt { 4 } \) = 2 units

LHS = \(\begin{vmatrix} a & a+b&a+b+c \\2a &3a+2b &4a+3b+2c\\3a&6a+3b&10a+6b+3c \end{vmatrix}\)
Applying R2 \(\rightarrow\) R2 - 2R1 and R3 \(\rightarrow\) 3R1 we get,
\(=\begin{vmatrix} a & a+b&a+b+c \\0 &0 &2a+b\\0&3a&7a+3b\end{vmatrix}\)

Expanding along C₁ we get
\(=a\begin{vmatrix}a&2a+b\\3a&7a+b\end{vmatrix}-0+0\)
= a(7a² + 3ab - 6a² - 3ab)
= a(7a² - 6a²) = a(a²) = a³ = RHS
Hence proved.

LHS \(=\tan^{-1}\left(\frac{1}{2}\right)+\tan^{-1}\left(\frac{2}{11}\right)\)
\(=\tan^{-1}\left(\frac{\frac{1}{2}+\frac{2}{11}}{1-\frac{1}{2}.\frac{2}{11}}\right)\)
\(=\tan ^{-1}\left(\frac{\frac{11+4}{22}}{\frac{22-2}{22}}\right)\)
\(=\tan^{-1}\left(\frac{15}{20}\right) =\tan^{-1}\left(\frac34\right)\)
= RHS
Hence proved.

LHS = \(\frac { sin(-\theta )tan({ 90 }^{ o }-\theta )sec\left( { 180 }^{ o }-\theta \right) }{ sin(180+\theta )cot(360-\theta )cosec({ 90 }^{ o }-\theta ) } =1\)
= \(\cfrac { \left( -sin\theta \right) cot\theta \left( -sec\theta \right) }{ \left( -sin\theta \right) \left( -cot\theta \right) sec\theta } =1\)

Let A(2, –3) and B (3, –4) be the given points
Let P(x₁,y₁) be any point on the locus.
Given that PA = PB.
PA² = PB²
\(\left( { x }_{ 1 }-2 \right) ^{ 2 }+\left( { y }_{ 1 }+3 \right) ^{ 2 }=\left( { x }_{ 1 }-3 \right) ^{ 2 }+\left( { y }_{ 1 }+4 \right) ^{ 2 }\)
\({ x }_{ 1 }^{ 2 }-4{ x }_{ 1 }+4+y_{ 1 }^{ 2 }+6{ y }_{ 1 }+9={ x }_{ 1 }^{ 2 }-6{ x }_{ 1 }+9+{ y }_{ 1 }^{ 2 }+8{ y }_{ 1 }+16\)
\(i.e,{ 2x }_{ 1 }-{ 2y }_{ 1 }-12=0\)
\( i.e.,{ x }_{ 1 }-{ y }_{ 1 }-6=0\) 
The locus of \(P({ x }_{ 1 },{ y }_{ 1 })\ is \ x-y-6=0\)

Given f(x, y) =3x²+4y³+6xy-x²y³+6
Differentiating partially w.r.t. 'x' we get,
fx(x, y)=6x + 0 + 6y(1) -y³(2x) + 0
=6x + 6y- 2xy³
\(\therefore f_x(1,-1)=\) 6(1) +6(-1)-2(1)(-1)³
= 6-6+2
= 2

To find the average prices for shares we find harmonic mean:
\(\therefore\) Harmonic mean = \(\frac { n }{ \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } +\frac { 1 }{ d } } =\frac { 4 }{ \frac { 1 }{ 10 } +\frac { 1 }{ 15 } +\frac { 1 }{ 20 } +\frac { 1 }{ 30 } } \)
\(=\frac { 4 }{ \frac { 6+4+3+2 }{ 60 } } =\frac { 4 \times 60}{ 15 } \)
= 16

In  \({ \left( x+\frac { 1 }{ x } \right) }^{ 6 }\)n = 6,x = x ,a =\(\frac { 1 }{ x } \)
General term is t_{r+ 1} =nC_rx^n-ra^r
   t_r+1 = 6C_rx^6-r\({ \left( \frac { 1 }{ x } \right) }^{ r }\)
To find t₃, put r =2
∴ t₃=6C₂x^6-2\({ \left( \frac { 1 }{ x } \right) }^{ 2 }\)=\(\frac { 6\times 5 }{ 2\times 1 } \).x⁴.\(\frac { 1 }{ { x }^{ 2 } } \)=15x²