UD-TRG-Maths2-Mumbai-Q4-Ans

Q. 4 b) If y=e^{\tan^{-1}x}
T.P.T
latexy_{n+2}+\left[2(n+1)x-1\right]y_{n+2}+n(n+1)y_n=0[/latex]

y=e^{\tan^{-1}x}

diff.both sides w.r.t x

y_1=\frac{e^{\tan^{-1}x}}{1+x^2}=\frac y{1+x^2} y_1=\frac{e^{\tan^{-1}x}}{1+x^2}=\frac y{1+x^2}

diff. again w.r.t.x

latexy_2+2xy_1=y_{1…….(1)}[/latex]

diff.(3) by Leibnitz’s theorem,
latexy_{n+2}+{}^nC_1(2x)y_{n+1}+{}^nC_2(2)y_n+2xy_{n+1}+{}^nC_1(2)y_n=y_{n+1}[/latex]

latexy_{n+2}+2_nxy_{n+1}+\frac{(n-1)n}2y_n+2xy_{n+1}+2_ny_n=y_{n+1}[/latex]

latexy_{n+2}+2_{(n+1)}xy_{n+1}-y_{n+1}+\left[n(n-1)+2_n\right]\;y_n=0[/latex]

latexy_{n+2}+\left[2(n+1)x-1\right]y_{n+1}+n(n+1)y_n=0[/latex]
Hence proved

Q.4 c)
(1) f(x)=2x^3+3x^2-8x+7

By Taylors theorem, f(x)-f(a)\;+(x-a)f'(a)+\frac{(x-a)^2}{2\;!}f"(a)\frac{f(x-a)^3}{3\;!}f'"(a)+…….

Put a = 2

f(x)=f(2)\;+(x-2)f'(2)+\frac{(x-2)^2}{2\;!}f"(2)+\frac{(x-2)^3}{3\;!}f'"(2)…….(1) f(x)=2x^3+3x^2-8x+7\;;\;f(2)=16+12+7-16=19 f'(x)=6x^2+6x-8;\;f'(2)=24+12-8=28 f"(x)=12x+6\;;\;f"(2)=24+6=30 f'''(x)\;=12\;\;\;\;\;\;\;;\;\;\;\;\;\;f'''(2)=12

Substracting the values in eq.(1) we get,

f(x)=19+(x-2)28+\frac{(x-2)^2\times30}{2\;!}+\frac{(x-2)^3\times12}{3\;!} f(x)=19+28(x-2)+15(x-2)^2+2(x-2)^3

Q.4
(c)(2) T.P.T.

\tan^{-1}\left(x\right)=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+…..

Let y\;=\;\tan^{-1}\left(x\right)

(diff. w.r.t. x)

\frac{dy}{dx}=\frac1{1+x^2}=(1+x^2)-1 =1-x^2+x^4-x^6+x^8……

( Binominal Theorem )
Intefrating both sides l\w 0 and x

y={\left[\tan^{-1}x\right]^x}_0={\left[x-\frac{x^3}3+\frac{x^5}5- \frac{x^7}7+…..\right]^x}_0 \begin{array}{l}\left[\tan^{-1}x\right]=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+…….\\end{array}

Hence proved

Q.5
(a) \begin{array}{l}z=x^2\tan^{-1}\left(\frac yx\right)-y^2\tan^{-1}\left(\frac xy\right)\\end{array}

T.P.T \begin{array}{l}\frac{2^2z}{2y2x}=\frac{x^2-y^2}{x^2+y^2}\\end{array}
Solution

\begin{array}{l}z=x^2\tan^{-1}\left(\frac yx\right)-y^2\tan^{-1}\left(\frac xy\right)\\end{array} \begin{array}{l}\frac{2z}{2x}=2x\tan^{-1}\left(\frac yx\right)+x^2\frac1{1+{\displaystyle\frac{y^2}{x^2}}}\left(\frac{-y}{x^2}\right)-y^2\frac1{1+{\displaystyle\frac{x^2}{y^2}}}\left(\frac1y\right)\\end{array} \begin{array}{l}=2x\tan^{-1}\frac yx-\frac{x^2y}{x^2+y^2}-\frac{y^3}{x^2+y^2}\\end{array} \begin{array}{l}=2x\tan^{-1}\frac yx-\frac{y(x^2+y^2)}{x^2+y^2}\\end{array} \begin{array}{l}=2x\tan^{-1}\frac yx-y\\end{array} \begin{array}{l}\frac{2^2z}{2y\;2z}=2x\frac1{1+{\displaystyle\frac{y^2}{x^2}}}\times\frac1x-1=\frac{2x^2}{x^2+y^2}-1\\end{array} \begin{array}{l}\frac{2^2z}{2y\;2z}=\frac{x^2-y^2}{x^2+y^2}\\end{array}

Q.5
b) 2x + 3x + 5z = 9
7x + 3y – 2z = 8

\begin{array}{l}2x+3y+\lambda z=\mu\\end{array}

Let us write the above eqation in matrix form
\begin{array}{l}\begin{bmatrix}2&3&5\7&3&-2\2&3&\lambda\end{bmatrix}\;\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}9\8\\mu\end{bmatrix}\\end{array}
AX = B
\begin{array}{l}C\left[A:B\right]=\begin{bmatrix}2&3&5&:&9\7&3&-2&:&8\2&3&\lambda&:&\mu\end{bmatrix}\\end{array}

\begin{array}{l}C\left[A:B\right]=\begin{bmatrix}2&3&5&:&9\0&\frac{-15}2&\frac{-39}2&:&\frac{-47}2\0&0&\lambda-5&:&\mu-9\end{bmatrix}\\end{array} \begin{array}{l}\R_2\rightarrow R_2-\frac72R_1\end{array} \begin{array}{l}\R_3\rightarrow R_3-R_1\end{array}

1) No solution
Rank (A) not eqal to Rank (C)

\lambda-5=0\;\;\;\;\;or\;\;\lambda=5 \mu-9\neq0\;\;\;or\;\;\mu\neq9

2) Unique solution
Rank (A)= Rank (c)
= No. of unknowns

\lambda-5\neq0\;\;\;\;\;,\;\;\;\;\lambda\neq5

3) An infinite no. of solutions
Rank (A) = Rank (C) = 2
λ – 5 = 0, μ – 9 = 0
λ = 5 , μ = 9

Q.5 C) Using NR method, find approx.
root of x^3-2x-5\;=\;0

f(x)=x^3-2x-5
f(2)\;=\;8-4-5\;=-1
f(2.5)=(2.5)^3-2(2.5)-5
= 5.625
We get to know that f(2) and f(2.5)are of app.signs
So, the root of the equation lies b/w 2 and 2.5
f(2) is nearer to zero hence it is more approximate root than f(2.5)
f'(x)\;=\;3x^2-2

\begin{array}{l}f'(2)=12-2=10\\end{array}

Let 2 be an approximate root of (1) by N.R.method

\begin{array}{l}a_1=a\frac{-f(a)}{f'(a)}=\frac{2-f(2)}{f'(2)}=\frac{2-(-1)}{10}=2.1\\end{array} \begin{array}{l}f(2.1)=2.1^3-2\times2.1-5=9.261-4.2-5=0.061\\end{array} \begin{array}{l}f'(2.1)=3(2.1)^2-2=11.23\\end{array} \begin{array}{l}a_2=\frac{2.1-f(2.1)}{f'(2.1)}\\end{array} \begin{array}{l}=\frac{2.1-0.061}{11.23}\\end{array} \begin{array}{l}=2.1-0.00543=2.09457\\end{array} \begin{array}{l}f(2.09457)=(2.09457)^3-2(2.09457)-5\\end{array} \begin{array}{l}=8.80558-4.18914-5=-0.38356\\end{array} \begin{array}{l}\f'(2.09457)=3(2.09457)^2-2\end{array} \begin{array}{l}\13.16167-2\end{array} \begin{array}{l}\=11.16167\end{array} a_3=2.09457-\frac{f(2.09457)}{f'(2.09457)} 2.09457-\frac{-0.38356}{11.16167} =2.09457+0.034364 = 2.128934 f(2.128934)=9.649095-4.257868-5=0.391227 f'(2.128934)=3(2.128934)^2 =13.59708-2=11.59708 a_4=2.128934-\frac{f(2.128934)}{f'(2.128934)} _4=2.128934-\frac{0.391227}{11.59708}</p> <p>[latex] =2.128934-0.03373 =2.09461

This is the required root of the following equation correct of
to 3 decimal places

Q.6 a) Find tan hx

Given 5\sin\;hx-\cos\;hx=5
divide By cos hx
5 tan hx - 1 = 5 sec x

Squaring both sides

25\tan\;h^2x+1-10\tan hx=25sec\;h^2x=25(1-\tan\;h^2x 50\tan h^2x-24-10\tan\;hx=0 50\tan h^2x-40\tan hx+30\tan hx-24=0 10\tan\;hx(5\tan\;hx-4)+6(5\tan\;hx-4=0) \tan\;hx=\frac{-6}{10}=\frac{-3}5 \tan\;hx=\frac45

Q.6
b) u=\sin^{-1}\left(\frac{x+y}{\sqrt x+\sqrt y}\right)

1) xu_x+yu_y=\frac12\tan u

2) x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=\frac{-\sin\;u\;\cos\;2u}{4\;\cos^3u}

We have,

u=\sin^{-1}\frac{x+y}{\sqrt x+\sqrt y} z=\sin u=\frac{x+y}{\sqrt x+\sqrt y}=\frac{x\left[1+{\displaystyle\frac yx}\right]}{\sqrt x\;\left[1+\sqrt{\displaystyle\frac yx}\right]}

z=f(u)=\sin u
Z is a homogeneous function of degree 1/2

By Eulex deduction I

x\frac{\delta_u}{\delta_x}+y\frac{\delta_u}{\delta_y}=n\frac{f(u)}{f'(u)}

x\frac{\delta_u}{\delta_x}+y\frac{\delta_u}{\delta_y}=\frac12\frac{\sin\;u}{\cos\;u\;} ……(problem)

x\frac{\delta_u}{\delta_x}+y\frac{\delta_u}{\delta_y}=\frac12\tan\;u

xu_x+yu_y=\frac12\tan\;u

Let, g(u) =1\2 tan u
By Euler deduction 2

x^2\frac{\delta^2u}{\delta x^2}+2xy\frac{\delta^2u}{\delta x\delta y}+y^2\frac{\delta^2u}{\delta y^2}=g(u)\left[g'(u)-1\right] \frac12\tan\;u\left(\frac12sec^2u-1\right)

=\frac14\frac{\sin\;u}{\cos\;u}\left(\frac1{\cos^2u}-2\right)

=\frac14\frac{\sin\;u}{\cos^3\;u}\left(1-2\cos^2u\right) =\frac{-\sin\;u\;\cos\;2u}{4\cos^3u}

We get, x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=\frac{-\sin4\;\cos2u}{4\cos^3u}
Hence Proved

Q.6 c) By Gauss Seidel method

\begin{array}{l}20x\;+y-2z=17\3x+20y-z=-18\2x-3y+20z=25\end{array}
Here,

x=\frac1{20}(17-y+2z)….(1) y=\frac1{20}(-18-3x+z)……(2)

z=\frac1{20}(25-2x+3y)…….(3)
Let us approximate x_0=y_0=z_0=0
put y_0=z_0=0 in (1)
We get,

x_1=\frac{17}{20}=0.85

Put x=x_1\;\;and\;\;\;z=0 in (2)

y_1=\frac1{20}(-18-3\times0.85)=-1.0275 x = x_1\;\;\;\;\;and\;\;\;\;\;y = y_1 z_1=\frac1{20}(25-2\times0.85+3\times(-1.0275)=1.011

By similar method . we get
x_2=1.002\;\;;\;\;y_2=-0.9998\;\;;z_2=0.9998
x_3=1.0000\;\;;\;\;y_3=-1.0000\;\;;\;\;z_3=1.0000
x_4=1.0000\;\;;\;\;\;y_4=-1.0000\;\;;\;\;z_4=1.0000
The last 2 sets of roots are almost the same.
So,we get the roots of the given equation by Gaues Siedel method as

x = 1
y = -1
z = 1