Vidya-Sum-17-M1-Q1

Time: 3 hours
Maximum marks: 80

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

Taking R.H.S

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

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

$latex e^{2u}-2e^usec\theta+1=0$

Taking roots we get,

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

Taking +ve root

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

Now taking L.H.S

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

Applying C and D

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

Multiplying Nr. and Dr. by 1 + sinθ

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

Taking square root

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

form (i) and (ii) we get

u=v

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

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

We know that for a matrix A to be unitary

$latex A^\theta\cdot A=I$

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

$latex A^\theta=\frac1{\sqrt3}\begin{bmatrix}1&1+i\\1-i&-1\end{bmatrix}$

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

Hence proved A is a unitary matrix.

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

x = uv, $latex y=\frac uv$

$latex \begin{array}{l}\frac{\partial x}{\partial u}=v,\;\;\frac{\displaystyle\partial y}{\displaystyle\partial u}=\frac1v\\\\\frac{\displaystyle\partial x}{\displaystyle\partial v}=u,\;\;\frac{\displaystyle\partial y}{\displaystyle\partial v}=\frac{-u}{v^2}\end{array}$

$latex \begin{array}{l}J=\begin{vmatrix}\;\frac{\displaystyle\partial x}{\displaystyle\partial u}&\;\frac{\displaystyle\partial x}{\displaystyle\partial v}\\\;\frac{\displaystyle\partial y}{\displaystyle\partial u}&\;\frac{\displaystyle\partial y}{\displaystyle\partial v}\end{vmatrix}\\\\\;\;\;=\begin{vmatrix}v&u\\\frac1v&\frac{-u}{v^2}\end{vmatrix}\\\\\;\;\;=\frac{-u}v-\frac uv\\\\\;\;\;=\frac{-2u}v\;—–\left(i\right)\end{array}$

$latex J’=\begin{vmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{vmatrix}$

Since, x = uv and $latex \frac uv$

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

$latex u\;=\sqrt{xy}$

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

By substituting we get,

$latex \begin{array}{l}J’=\begin{vmatrix}\frac{\sqrt y}{2\sqrt x}&\frac{\sqrt x}{2\sqrt y}\\\frac1{2\sqrt{xy}}&\frac{-\sqrt x}{2y\sqrt y}\end{vmatrix}\\\\\;\;\;\;=\frac{-1}{4y}-\frac1{4y}\\\\\;\;\;\;=\frac{-1}{2y}\\\\J’=\frac{-v}{2u}—–\left(ii\right)\end{array}$

Form (i) and (ii)

$latex \therefore J\cdot J’=\frac{-2u}v\times\frac{-v}{2u}$

$latex \therefore J\cdot J’=1$

Hence proved.

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

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

x = 2t, y=1 – t²

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

$latex z=\tan^{-1}\left(\frac xy\right)$

$latex \begin{array}{l}\frac{\partial z}{\partial x}=\frac1{1+{\displaystyle\frac{x^2}{y^2}}}\times\frac1y\\\\\;\;\;\;=\frac y{y^2+x^2}\end{array}$

$latex \begin{array}{l}\frac{\partial z}{\partial y}=\frac1{1+{\displaystyle\frac{x^2}{y^2}}}\times\frac{-x}{y^2}\\\\\;\;\;\;=\frac{-x}{y^2+x^2}\end{array}$

$latex \frac{dz}{dt}=\frac{\partial z}{\partial x}\times\frac{dx}{dt}+\frac{\partial z}{\partial y}\times\frac{dy}{dt}$

$latex \begin{Bmatrix}x=2t&y=1-t^2\\\frac{dx}{dt}=2&\frac{dy}{dt}=-2t\end{Bmatrix}$

$latex \therefore\frac{dz}{dt}=\frac{2y}{x^2+y^2}+\frac{-x\times\left(-2t\right)}{x^2+y^2}$

Taking in terms of t we get,

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

Hence proved.

e) Find the n^th derivative of $latex \left(\cos5x\cdot\cos3x\cdot\cos x\right)$ [4M]

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

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

differentiate the above equation n times we get,

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

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