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UD-TRG-Maths-Second-Sem-Ans

B.E. Second Semester All Branches (C.B.S.)

Time: 3 hours
Maximum marks: 80

Notes :

  1. Question No.1 is compulsory.
  2. Answer any three question from remaining.

Q.1 a) Prove that \tan h^{-1}\left(\sin\theta\right)=\cos h^{-1}\left(sec\theta\right) [3M]

Taking R.H.S

Let \cos h^{-1}\left(sec\theta\right)=u

\begin{array}{l}\cos hu=sec\theta\\\frac{e^u+e^{-u}}2=sec\theta\\\frac{e^{2u}+1}{2e^u}=sec\theta\end{array} e^{2u}-2e^usec\theta+1=0

Taking roots we get,

\begin{array}{l}e^u=\frac{2sec\theta\pm\sqrt{4sec^2\theta-4}}2\\\;\;\;\;=\frac{2\;sec\theta\pm2\;\tan\theta}2\\\;\;\;\;=\frac1{\cos\theta}\pm\frac{\sin\theta}{\cos\theta}\end{array}

Taking +ve root

\begin{array}{l}e^u=\frac{1+\sin\theta}{\cos\theta}\\u=\log\left|\frac{1+\sin\theta}{\cos\theta}\right|-----\left(i\right)\end{array}

Now taking L.H.S

\begin{array}{l}\tan h^{-1}\left(\sin\theta\right)=v\\\tan hv=\sin\theta\\\frac{e^v-e^{-v}}{e^v+e^{-v}}=\sin\theta\end{array}

Applying C and D

\begin{array}{l}\frac{e^v-e^{-v}+e^v+e^{-v}}{e^v+e^{-v}-e^v+e^{-v}}=\frac{1+\sin\theta}{1-\sin\theta}\\\frac{2e^v}{2e^{-v}}=\frac{\left(1+\sin\theta\right)^2}{1-\sin^2\theta}\end{array}

Multiplying Nr. and Dr. by 1 + sinθ

\begin{array}{l}e^{2v}=\frac{1+\sin^2\theta+2\sin\theta}{\cos^2\theta}\\e^{2v}=\left(\frac{1+\sin\theta}{\cos\theta}\right)^2\end{array}

Taking square root

\begin{array}{l}e^v=\frac{1+\sin\theta}{\cos\theta}\\v=\log\left|\frac{1+\sin\theta}{\cos\theta}\right|-----\left(ii\right)\end{array}

form (i) and (ii) we get

u=v

Hence, L.H.S=R.H.S Proved.

b) Prove that the matrix \frac1{\sqrt3}\begin{bmatrix}1&1+i\1-i&-1\end{bmatrix} is unitary [3M]

We know that for a matrix A to be unitary

A^\theta\cdot A=I

Let, A=\frac1{\sqrt3}\begin{bmatrix}1&1+i\1-i&-1\end{bmatrix}

A^\theta=\frac1{\sqrt3}\begin{bmatrix}1&1+i\1-i&-1\end{bmatrix} \begin{array}{l}A^\theta\cdot A=\frac13\begin{bmatrix}1&1+i\1-i&-1\end{bmatrix}\begin{bmatrix}1&1+i\1-i&-1\end{bmatrix}\\\;\;\;\;\;\;\;\;\;\;=\frac13\begin{bmatrix}1+1-i^2&1+i-1-i\1-i-1+i&1-i^2+1\end{bmatrix}\\\;\;\;\;\;\;\;\;\;\;=\frac13\begin{bmatrix}3&0\0&3\end{bmatrix}\cdot\begin{bmatrix}1&0\0&1\end{bmatrix}\end{array}

Hence proved A is a unitary matrix.

c) If x=uv\;and\;y=\frac uv prove that ΤΤ^1=1 [3M]

x = uv, y=\frac uv

\begin{array}{l}\frac{\partial x}{\partial u}=v,\;\;\frac{\displaystyle\partial y}{\displaystyle\partial u}=\frac1v\\\frac{\displaystyle\partial x}{\displaystyle\partial v}=u,\;\;\frac{\displaystyle\partial y}{\displaystyle\partial v}=\frac{-u}{v^2}\end{array} \begin{array}{l}\Tau=\begin{vmatrix}\;\frac{\displaystyle\partial x}{\displaystyle\partial u}&\;\frac{\displaystyle\partial x}{\displaystyle\partial v}\\;\frac{\displaystyle\partial y}{\displaystyle\partial u}&\;\frac{\displaystyle\partial y}{\displaystyle\partial v}\end{vmatrix}\\\;\;\;=\begin{vmatrix}v&u\\frac1v&\frac{-u}{v^2}\end{vmatrix}\\\;\;\;=\frac{-u}v-\frac uv\\\;\;\;=\frac{-2u}v\;-----\left(i\right)\end{array} \Tau'=\begin{vmatrix}\;\frac{\displaystyle\partial u}{\displaystyle\partial x}&\;\frac{\displaystyle\partial u}{\displaystyle\partial y}\\;\frac{\displaystyle\partial v}{\displaystyle\partial x}&\;\frac{\displaystyle\partial v}{\displaystyle\partial y}\end{vmatrix}

Since, x = uv and \frac uv

we get u = vy, x = v2y ⇒ v^2\sqrt{\frac xy}

u\;=\sqrt{xy} \begin{array}{l}\frac{\displaystyle\partial u}{\displaystyle\partial x}=\frac{\sqrt y}{2\sqrt x},\;\;\frac{\partial u}{\partial y}=\frac{\sqrt x}{2\sqrt y}\\\frac{\displaystyle\partial v}{\displaystyle\partial x}=\frac1{2\sqrt{xy}},\;\;\frac{\displaystyle\partial v}{\displaystyle\partial y}=\frac{-\sqrt x}{2y\sqrt y}\end{array}

By substituting we get,

\begin{array}{l}\Tau'=\begin{vmatrix}\frac{\sqrt y}{2\sqrt x}&\frac{\sqrt x}{2\sqrt y}\\frac1{2\sqrt{xy}}&\frac{-\sqrt x}{2y\sqrt y}\end{vmatrix}\\\;\;\;\;=\frac{-1}{4y}-\frac1{4y}\\\;\;\;\;=\frac{-1}{2y}\\\Tau'=\frac{-v}{2u}-----\left(ii\right)\end{array}

Form (i) and (ii)

\begin{array}{l}\therefore\Tau\cdot\Tau'=\frac{-2u}v\times\frac{-v}{2u}\\\Tau\cdot\Tau'=1\end{array}

Hence proved.

d) If Z=\tan^{-1}\left(\frac xy\right) where x=2t,\;y=1-t^2,\;prove\;that\;\frac{dz}{dt}=\frac2{1+t^2} [3M]

Z=\tan^{-1}\left(\frac xy\right)

x = 2t, y=1 – t2

T.P.T \frac{dz}{dt}=\frac2{1+t^2}

z=\tan^{-1}\left(\frac xy\right) \begin{array}{l}\frac{\partial z}{\partial x}=\frac1{1+{\displaystyle\frac{x^2}{y^2}}}\times\frac1y\\\;\;\;\;=\frac y{y^2+x^2}\end{array} \begin{array}{l}\frac{\partial z}{\partial y}=\frac1{1+{\displaystyle\frac{x^2}{y^2}}}\times\frac{-x}{y^2}\\\;\;\;\;=\frac{-x}{y^2+x^2}\end{array} \frac{dz}{dt}=\frac{\partial z}{\partial x}\times\frac{dx}{dt}+\frac{\partial z}{\partial y}\times\frac{dy}{dt} \begin{Bmatrix}x=2t&y=1-t^2\\frac{dx}{dt}=2&\frac{dy}{dt}=-2t\end{Bmatrix} \therefore\frac{dz}{dt}=\frac{2y}{x^2+y^2}+\frac{-x\times\left(-2t\right)}{x^2+y^2}

Taking in terms of t we get,

\begin{array}{l}\frac{dz}{dt}=\frac{2\left(1-t^2\right)+2\times2t^2}{4t^2+\left(1-t^2\right)^2}\\\;\;\;\;=\frac{2+2t^2}{1+t^4+2t^2}\\\;\;\;\;=\frac{2\left(1+t^2\right)}{\left(1+t^2\right)^2}\\\frac{dz}{dt}=\frac2{1+t^2}\end{array}

Hence proved.

e) Find the nth derivative of \left(\cos5x\cdot\cos3x\cdot\cos x\right) [4M]

Let y=\cos5x\cdot\cos3x\cdot\cos x\;\times\frac22 Multiply and divide by 2

\begin{array}{l}y=\frac{\left(2\cos5x\cdot\cos3x\right)\cos x}2\\\;\;\;=\frac{\left(\cos8x+\cos2x\right)\cos x}2\times\frac22\\\;\;\;=\frac{2\left(\cos8x\cos x+\cos x\cos2x\right)}4\\y=\frac{\cos9x+\cos7x+\cos3x+\cos x}4\end{array}

differentiate the above equation n times we get,

y_n=\frac14\left[9^n\cos\left(9x+\frac{\pi_n}2\right)+7^n\cos\left(7x+\frac{n\pi}2\right)+3^n\cos\left(3x+\frac{n\pi}2\right)+\cos\left(x+\frac\pi2\right)\right]

f) Evaluate \lim_{x\rightarrow0}\left(X\right)\frac1{1-x} [4M]

Q.2 a) Find all values of \left(1+i\right)^\frac13 and show that their continued product is \left(1+i\right) [6M]

Let 1+i=Q

\begin{array}{l}=\sqrt2\left(\frac1{\sqrt2}+\frac{i1}{\sqrt2}\right)\\=\sqrt2\left(\cos\frac\pi4+i\sin\frac\pi4\right)\\=\sqrt2\left[\cos\left(2n\pi+\frac\pi4\right)+i\sin\left(2n\pi+\frac\pi4\right)\right]\\=\sqrt2\left[\cos\left(\frac{8n\pi+\pi}4\right)+i\sin\left(\frac{8n\pi+\pi}4\right)\right]\end{array} \begin{array}{l}\left(1+i\right)^\frac13=2^\frac16\left[\cos\left(\frac{8n+1}4\right)\pi+i\sin\left(\frac{8n+1}4\right)\pi\right]^\frac13\\\=2^\frac16\left[\cos\left(\frac{8n+1}{12}\right)\pi+i\sin\left(\frac{8n+1}{12}\right)\pi\right]\end{array}

By De Moirre’s Theorem

n=0, 1, 2, …

\begin{array}{l}a_1=2^\frac16\left[\cos\frac\pi{12}+i\sin\frac\pi{12}\right]\\a_2=2^\frac16\left[\cos\frac{9\pi}{12}+i\sin\frac{9\pi}{12}\right]\\a_3=2^\frac16\left[\cos\frac{17\pi}{12}+i\sin\frac{17\pi}{12}\right]\end{array}

Taking product of a1, a2, and a3 we get,

\begin{array}{l}a_1\cdot a_2\cdot a_3=2^\frac16\left(\cos\frac\pi{12}+i\sin\frac\pi{12}\right)\\=2^\frac16\left(\cos\frac{9\pi}{12}+i\sin\frac{9\pi}{12}\right)\\=2^\frac16\left(\cos\frac{17\pi}{12}+i\sin\frac{17\pi}{12}\right)\\=2^\frac36\left[\cos\left(\frac\pi{12}+\frac{9\pi}{12}+\frac{17\pi}{12}\right)+i\sin\left(\frac\pi{12}+\frac{9\pi}{12}+\frac{17x}{12}\right)\right]\\=\sqrt2\left[\cos\frac{9\pi}4+i\sin\frac{9\pi}4\right]\\=\sqrt2\left[\frac{\sqrt2}2+\frac{i\sqrt2}2\right]\\=1+i\end{array}

Hence proved

a_1\cdot a_2\cdot a_3=1+i

This is the continued product

b) Find non-singular matrices P and Q is in normal form where

A=\begin{bmatrix}2&-2&3\3&-1&2\1&2&-1\end{bmatrix} [6M]

c) Find the maximum and minimum values of f\left(x,\;y\right)=x^3+3xy^2-15x^2-15y^2+72x [8M]

Let  f\left(x,\;y\right)=x^3+3xy^2-15x^2-15y^2+72x

\begin{array}{l}\frac{\partial f}{\partial x}=3x^2+3y^2-30x+72\\\frac{\displaystyle\partial f}{\displaystyle\partial y}=6xy-30y\\\frac{\displaystyle\partial f}{\displaystyle\partial x}=\frac{\displaystyle\partial f}{\displaystyle\partial y}=0\\3x^2=3y^2-30x+72=0\;-----\left(i\right)\\6xy-30y=0\\6y\left(x-5\right)=0\\x=5,\;\;y=0\end{array}

Put, x = 5 in equation (i) we get

\begin{array}{l}3\times25+3y^2-150+72=0\\3y^2=3\\y=\pm1\end{array}

Put, y = 0 in equation (i) we get

\begin{array}{l}3\times x^2+30x+72=0\\x^2-10x+24=0\\\left(x-6\right)\left(x-4\right)=0\\x=6,\;\;x=4\end{array}

when x = 5, y = ±1, y = 0, x = 6, 4

\begin{array}{l}r=\frac{\partial^2f}{\partial x^2}=6x-30\\t=\frac{\partial^2f}{\partial y^2}=6x-30\\s=\frac{\partial^2f}{\partial x\partial y}=6y\end{array}

when x = 5, y = ±1

\begin{array}{l}rt-s^2=\left(6\times5-30\right)\left(6\times5-30\right)-36\times\left(\pm1\right)^2\\\;\;\;\;\;\;\;\;\;\;\;=-36<0\rightarrow-ve\end{array}[/latex] <p>when x = 6, y = 0 [latex] \begin{array}{l}rt-s^2=\left(36-30\right)\left(36-30\right)\\\;\;\;\;\;\;\;\;\;\;\;=36>0\rightarrow+ve\end{array}

when x = 4, y = 0

\begin{array}{l}rt-s^2=\left(24-30\right)\left(24-30\right)\\\;\;\;\;\;\;\;\;\;\;\;=36>0\rightarrow+ve\end{array} \begin{array}{l}r=6x-30;\;x=6\\=6>0\rightarrow minimum\;value\\r=6x-30;\;x=4\\=24-30\\=-6<0\rightarrow maximum\;value\end{array}

Hence, f(x, y) is minimum when x = 6 and y = 0 and

Maximum when x = 4 and y = 0

Q.3 a) If u=f\left(\frac{y-x}{xy},\;\frac{z-x}{xz}\right), show that x^2\frac{\partial u}{\partial z}+y^2\frac{\partial u}{\partial y}+z^2\frac{\partial u}{\partial z}=0 [6M]

u=f\left(\frac{y-x}{xy},\;\frac{z-x}{xz}\right)

T.P.T. x^2\frac{\partial u}{\partial x}+y^2\frac{\partial u}{\partial y}+z^2\frac{\partial u}{\partial z}=0

Let u = u(r, s)

where,

\begin{array}{l}r=\frac{y-x}{xy},\;s=\frac{z-x}{zx}\\r=\frac1x-\frac1y,\;s=\frac1x-\frac1z\\\begin{array}{cc}\frac{\partial r}{\partial x}=\frac{-1}{x^2},&\frac{\partial s}{\partial x}=\frac{-1}{x^2}\end{array}\\\begin{array}{cc}\frac{\partial r}{\partial y}=\frac1{y^2},\;&\frac{\partial s}{\partial z}=\frac1{z^2}\end{array}\\\begin{array}{cc}\frac{\partial r}{\partial z}=0,\;&\frac{\partial s}{\partial y}=0\end{array}\end{array}

We know that,

\begin{array}{l}\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\cdot\frac{\partial r}{\partial x}+\frac{\partial u}{\partial s}\cdot\frac{\partial s}{\partial x}\\\;\;\;\;=\frac{\partial u}{\partial r}\left(\frac{-1}{x^2}\right)+\frac{\partial u}{\partial s}\left(\frac{-1}{x^2}\right)\\\;\;\;\;=\frac{-1}{x^2}\frac{\partial u}{\partial r}-\frac1{x^2}\frac{\partial u}{\partial s}\end{array} \begin{array}{l}x^2\frac{\partial u}{\partial x}=\frac{-\partial u}{\partial r}-\frac{\partial u}{\partial s}\\\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\cdot\frac{\partial r}{\partial y}+\frac{\partial u}{\partial s}\cdot\frac{\partial s}{\partial y}\\\;\;\;\;=\frac{\partial u}{\partial r}\times\frac1{y^2}+\frac{\partial u}{\partial s}\times0\\\;\;\;\;=\frac1{y^2}\frac{\partial u}{\partial r}-----\left(i\right)\end{array} \begin{array}{l}y^2\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\\\frac{\partial u}{\partial z}=\frac{\partial u}{\partial r}\cdot\frac{\partial r}{\partial z}+\frac{\partial u}{\partial s}\cdot\frac{\partial s}{\partial z}\\\;\;\;\;=\frac{\partial u}{\partial r}\times0+\frac{\partial u}{\partial s}\times\frac1{z^2}\\\;\;\;\;=\frac1{z^2}\frac{\partial u}{\partial s}-----\left(ii\right)\end{array} z^2\frac{\partial u}{\partial z}=\frac{\partial u}{\partial s}-----\left(iii\right)

By adding (i), (ii) and (iii), we get

z^2\frac{\partial u}{\partial x}+y^2\frac{\partial u}{\partial y}+z^2\frac{\partial u}{\partial x}=0

Hence Proved.

b) Using encoding matrix \begin{bmatrix}1&1\0&1\end{bmatrix}, encode and decode the message 'MUMBAI'. [6M]

Encoding matrix \begin{bmatrix}1&1\0&1\end{bmatrix}=A

Message - \begin{array}{c}M\;\;U\;\;M\;\:B\;\;A\;\;I\13\;\;21\;\;13\;\;2\;\;1\;\;9\end{array}

Encoded Matrix

\begin{array}{l}\begin{bmatrix}13&21\13&2\1&9\end{bmatrix}\;\begin{bmatrix}1&1\0&1\end{bmatrix}=\begin{bmatrix}13&34\13&15\1&10\end{bmatrix}\\A^{-1}=\begin{bmatrix}1&-1\0&1\end{bmatrix}\end{array}

Decoded Matrix

\begin{bmatrix}13&34\13&15\1&10\end{bmatrix}\;\begin{bmatrix}1&1\0&1\end{bmatrix}=\begin{bmatrix}13&21\13&2\1&9\end{bmatrix}

c) prove that \log\;\left[\tan\left(\frac\pi4+\frac{ix}2\right)\right]=i\tan^{-1}\left(\sin h\;x\right) [8M]

Q.4 a) Obtain tan5θ in terms of tanθ and show that 1-10\;\tan^2\frac\pi{10}+5\;\tan^4\frac\pi{10}=0 [6M]

Get the value of tan5θ in terms of tanθ

T.P.T = 1-10\tan^2\frac\pi{10}+5\tan^4\frac\pi{10}=0

We know

\tan n\theta=\frac{{}^nc_1\tan\theta-{}^nc_3\tan^3\theta+{}^nc_5\tan^5\theta…..}{1-{}^nc_2\tan^2\theta+{}^nc_4+\tan^4\theta-{}^nc_6\tan^6\theta…….} \tan5\theta=\frac{{}^5c_1\tan\theta-{}^5c_3\tan^3\theta+{}^5c_5\tan^5\theta}{1-{}^5c_2\tan^2\theta+{}^5c_4\tan^4\theta.} {}^nc_r=\frac{\left.n\right|}{r\left|\left(n-r\right)\right|} \tan5\theta=\frac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\;\tan^2\theta+5\tan^4\theta}

Put \theta=\frac\pi{10}, we get

\infty=\frac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\;\tan^2{\displaystyle\frac\pi{10}}+5\tan^4\frac\pi{10}}

Since,

\begin{array}{l}\tan\;5\theta=\tan\;90^\circ\\\;\;\;\;\;\;\;\;\;\;\;=\infty\end{array}

We can say the dinominator well be 0 so, we get

1-10\tan^2\frac\pi{10}+5\tan\frac\pi{10}=0

Hence Proved.

b) If y=e^{\tan^{-1}x}, prove that [6M]

c) i. Express \left(2x^3+3x^2-8x+7\right) in terms of (x - 2) using Taylor's theorem. [4M]

ii. Prove that \tan^{-1}x=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+-----\; [4M]

Q.5 a) If z=x^2\tan^{-1}\left(\frac yx\right)+y^2\tan^{-1}\left(\frac xy\right) Prove that \frac{\partial^2z}{\partial y}=\frac{x^2-y^2}{x^2+y^2} [6M]

b) Investigate for what values of λ and μ the equations,

\begin{array}{l}2x+3y+5z=9\\7x+3y-2z=8\\2x+3y+\lambda z=\mu\end{array}

Have 1) no solution
2) a unique solution
3) an infinite no. of solutions [6M]

c) Using Newton Raphson method, find approximate root of x^3-2x-5=0 (correct to three places of decimals.) [8M]

Q.6 a) Find tanhx if 5 sinhx - coshx=5 [6M]

b) If u=\sin^{-1}\left(\frac{x+y}{\sqrt x+\sqrt y}\right) prove that [6M]

i. xu_x+yu_y=\frac12\tan u

ii. x^2u_{xx}+2xy\;u_{xy}+y^2u_{yy}=\frac{-\sin u\;\cos2u}{4\cos^3u}

c) Solve the following systems of equations by Gauss-seidel method.

\begin{array}{l}20x+y-2z=17\\3x+20y-z=-18\\2x-3y+20z=25\end{array} [8M]

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