Winter 2015 – Q.5
Q.5 a) If $latex \cos^6\theta+\sin^6\theta=\alpha\cos4\theta+\beta$ then prove that $latex \alpha+\beta=1$ [6M]
$latex \begin{array}{l}\cos^6\theta+\sin^6\theta=\propto\cos4\theta+\beta\\\\\end{array}$
$latex \begin{array}{l}T.P.T-\propto+\beta=1\\\\\end{array}$
taking L.H.S
$latex \begin{array}{l}\cos^6\theta+\sin^6\theta\\\\\end{array}$
$latex \begin{array}{l}=\left(\sin^2\theta\right)^3+\left(\cos^2\theta\right)^3\\\\\end{array}$
$latex \begin{array}{l}=\left(\sin^2\theta+\cos^2\theta\right)-3\sin^2\theta\cos^2\theta\left(\sin^2\theta+\cos^2\theta\right)\\\\\end{array}$
$latex \begin{array}{l}=1^3-3\sin^2\theta\cos^2\theta\left(1\right)\\\\=1-3\sin^2\theta\cos^2\theta\\\\=1-3\left(1-\cos^2\theta\right)\cos^2\theta\\\\\end{array}$
$latex \begin{array}{l}=1-3\cos^2\theta+3\cos^4\theta\\\\=1-3\cos\theta.\cos\theta\frac{x2}2+\frac34\left[\left(2\cos^2\theta\right)\left(2\cos^2\theta\right)\right]\\\\\end{array}$
$latex \begin{array}{l}=1-\frac32\cos2\theta-\frac32+\frac34\left[\left(\cos2\theta+1\right)\right]\\\\\end{array}$
$latex \begin{array}{l}=\frac{-1}2-\frac32\cos2\theta+\frac34\left[\cos^22\theta+1+2\cos2\theta\right]\\\\\end{array}$
$latex \begin{array}{l}=\frac{-1}2-\frac32\cos2\theta+\frac34\left[\frac{\cos4\theta+1}2+1+2\cos2\theta\right]\\\\\end{array}$
$latex \begin{array}{l}=\frac{-1}2-\frac32\cos2\theta+\frac{3\left(\cos4\theta+1\right)}8+\frac34+\frac{b\cos2\theta}4\\\\\end{array}$
$latex \begin{array}{l}=\frac58+\frac38\cos4\theta\\\\=\propto\cos4\theta+\beta=R.H.S.\\\\\end{array}$
from this we get,
$latex \begin{array}{l}\propto=\frac38,\beta=\frac58\\\\\end{array}$
$latex \begin{array}{l}2+\beta=\frac38+\frac58\\\\\;\;\;\;\;\;\;\;\;=1\\\\\propto+\beta=1\\\\\end{array}$
Hence proved
b) Find the values of a, b and such that $latex \lim_{x\rightarrow0}\frac{ae^x-be^{-x}+cx}{x-\sin x}=4$ [6M]
As the denominator being 0 for x=0,the fraction will tend to a finite limit if and if.the neumerator is also zero for x=0 this requires, a-b+0=0
i.e. a=b
$latex =\underset{x\rightarrow0}{lim}\frac{ae^x+be^{-x}+c}{1-\cos x}\left[L’Hospital\;rule\right]$
similarly,we get a+b+c=0
$latex =\underset{x\rightarrow0}{lim}\frac{ae^x-be^{-x}}{\sin x}\left[L’Hospital\;rule\right]$
Again, denominator being 0 for x=0, requires a-b=0
i.e. a=b
$latex \begin{array}{l}=\underset{x\rightarrow0}{lim}\;\frac{ae^x+be^{-x}}{\cos x}=4\\\\a+b=4\\\\a=b\left[from\;1\right]\end{array}$
so,we get 2a=4
a=2 and b=2
a+b+c=0 (from 2)
2+2+c=0 So c=-4
c) If $latex x=\cos\left[\log\left(y^\frac1m\right)\right]$ then prove that $latex \left(1-x^2\right)y_{n+2}-\left(2n+1\right)xy_{n+1}-\left(m^2+n^2\right)y_n=0$ [8M]
$latex \begin{array}{l}x=\cos\left[log\left(y^{vm}\right)\right]\\\\\\\\\\\end{array}$
$latex \begin{array}{l}\cos^{-1}x=\frac1mlog\;y\\\\\\\\\\\end{array}$
$latex \begin{array}{l}m\cos^{-1}x=10g\;y\\\\\\\\\\\end{array}$
$latex \begin{array}{l}y=e^{m\cos^{-1}x}\\\\y=e……1\\\\\\\\\\\end{array}$
differentiating 1 we get,
$latex \begin{array}{l}y_1=\frac{-me}{\sqrt{1-x^2}}=\frac{-my}{\sqrt{1-x^2}}\\\\\end{array}$
$latex \begin{array}{l}y_1^2\left(/-x^2\right)-m^2y^2=0\\\\\end{array}$
differentiating the above eqn we get,
$latex \begin{array}{l}2y,y_2\left(1-x^2\right)-2xy^2,-2m^2yy,=0\\\\\end{array}$
Dividing by 2y,we get
$latex \begin{array}{l}\left(1-x^2\right)y_2-xy,-m^2y=0…..2\\\\\end{array}$
diff 2,n times by let beings theorom,we get
$latex \begin{array}{l}\left(1-x^2\right)y_{n+2}-2nxy_{n+1}-n\left(n-1\right)y_0-xy_{n+1}-ny_n-m^2yn=0\\\\\end{array}$
$latex \begin{array}{l}\left(1-x^2\right)y_{n+2}-\left(2n+1\right)xy_{n+1}-\left(n^2+m^2\right)y_n=0\\\\\end{array}$
Hence proved
Q.6 a) Define linear dependence and independence of vectors, Examine for linear dependence of following set of vectors and find the relation between them if dependent
$latex X_1=\begin{bmatrix}1\\-1\\1\end{bmatrix},\;X_2=\begin{bmatrix}2\\1\\1\end{bmatrix},\;X_3=\begin{bmatrix}3\\0\\2\end{bmatrix}$ [6M]
Given vectors are
$latex \begin{array}{l}x_1=\begin{bmatrix}1\\-1\\1\end{bmatrix},x_2=\begin{bmatrix}2\\1\\1\end{bmatrix},x_3=\begin{bmatrix}3\\0\\2\end{bmatrix}\\\\\end{array}$
$latex \begin{array}{l}a=\begin{bmatrix}1\\-1\\1\end{bmatrix}+b\begin{bmatrix}2\\1\\1\end{bmatrix}+c\begin{bmatrix}3\\0\\2\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\\\\\end{array}$
Let us write the vectors in matrix form.
$latex \begin{array}{l}\begin{bmatrix}1&2&3&.&0\\-1&1&0&.&0\\1&1&2&.&0\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}\\\\R_2=R_2+R_1\;\;,\;R_3=R_3-R_1\end{array}$
$latex \begin{array}{l}\begin{bmatrix}1&2&3&.&0\\0&3&3&.&0\\0&1&-1&.&0\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}\\\\R_2=R_2/3\;\;,\;R_3=R_3+R_2\end{array}$
$latex \begin{array}{l}\begin{bmatrix}1&2&3&.&0\\0&1&1&.&0\\0&0&0&.&0\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}\\\\\end{array}$
we get,
$latex \begin{array}{l}a+2b+3c=0\\\\b+c=0\\\\\end{array}$
c=k
b=k
a=-k
Let c=k
b=-k
$latex \begin{array}{l}a=-\left(2b+3c\right)\\\\\;\;\;=-\left(-2k+3k\right)\\\\\;\;=-k\\\\\end{array}$
Hence, the vectors are finaly dependent.
b) If $latex z=f\left(u,\;v\right)u=x^2-y^2,\;v=2xy$ then prove that $latex \frac{\partial^2z}{\partial x^2}+\frac{\partial^2z}{\partial y^2}=4\sqrt{u^2+v^2}\left(\frac{\displaystyle\partial^2z}{\displaystyle\partial u^2}+\frac{\displaystyle\partial^2z}{\displaystyle\partial v^2}\right)$ [6M]
Given,$latex \begin{array}{l}u=x^2-y^2,v=2xy\\\\\end{array}$
T.P.T. $latex \begin{array}{l}\frac{\partial^2z}{\partial x^2}+\frac{\partial^2z}{\partial y^2}=4\sqrt{4^2+v^2}\left(\frac{\partial^2z}{\partial v^2}\right)\\\\\end{array}$
$latex \begin{array}{l}u=x^2-y^2\;\;,\;v=2xy\\\\\frac{\partial u}{\partial x}2x\;\;,\;\frac{\partial v}{\partial x}2y\\\\\frac{\partial u}{\partial y}-2y\;\;,\;\frac{\partial v}{\partial y}2x\\\\\end{array}$
$latex \begin{array}{l}\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}.\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}.\frac{\partial v}{\partial x}\\\\\;\;\;\;\;\;\;=2x\left(\frac{\partial z}{\partial u}\right)+2y\left(\frac{\partial z}{\partial v}\right)\\\\\end{array}$
$latex \begin{array}{l}\frac\partial{\partial x}=2x\left(\frac\partial{\partial u}\right)+2y\left(\frac\partial{\partial v}\right)\\\\\end{array}$
$latex \begin{array}{l}\frac{\partial^2z}{\partial x^2}=\frac\partial{\partial x}\left(\frac{\partial z}{\partial x}\right)=\left[2x\left(\frac\partial{\partial u}\right)+2y\left(\frac\partial{\partial v}\right)\right]\times\left[2x\left(\frac{\partial z}{\partial u}\right)+2y\left(\frac{\partial z}{\partial v}\right)\right]\\\\\end{array}$
$latex \begin{array}{l}=4x^2\frac{\partial^2z}{\partial u^2}+4xy\frac{\partial^2z}{\partial u\partial v}+4xy\frac{\partial^2z}{\partial u\partial v}+4y^2\frac{\partial^2z}{\partial v^2}\\\\\end{array}$
$latex \begin{array}{l}=4x^2\frac{\partial^2z}{\partial u^2}+8xy\frac{\partial^2z}{\partial u\partial v}+4y^2\frac{\partial^2z}{\partial v^2}…….1\\\\\end{array}$
similarly,
$latex \begin{array}{l}\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}.\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}.\frac{\partial v}{\partial y}\\\\\;\;\;\;\;\;\;\;=\frac{\partial z}{\partial u}\left(-2y\right)+\frac{\partial z}{\partial v}\left(2x\right)\\\\\;\;\;\;\;\;\;\;=-2y\frac{\partial z}{\partial u}+2x\frac{\partial z}{\partial v}\\\\\\\\\end{array}$
$latex \begin{array}{l}\frac\partial{\partial y}=-2y\left(\frac\partial{\partial u}\right)+2x\left(\frac\partial{\partial v}\right)\\\\\frac{\partial z}{\partial y^2}=\frac\partial{\partial y}\left(\frac{\partial z}{\partial y}\right)\\\\\\\\\end{array}$
$latex \begin{array}{l}=\left(-2y\frac{\partial z}{\partial u}+\partial x\frac{\partial z}{\partial v}\right)\times\left(-2y\frac\partial{\partial u}+2z\frac\partial{\partial v}\right)\\\\\\\\\end{array}$
$latex \begin{array}{l}=4y^2\frac{\partial^2z}{\partial u^2}-4xy\frac{\partial^2z}{\partial u\partial v}+4x^2\frac{\partial^2z}{\partial v^2}-4xy\frac{\partial^2z}{\partial u\partial v}\\\\\\\\\end{array}$
$latex \frac{\partial^2z}{\partial y^2}=4x^2\frac{\partial^2z}{\partial v^2}+4y^2\frac{\partial^2z}{\partial u^2}-8xy\frac{\partial^2z}{\partial u\partial v}…..2$
Adding 1 and 2 we get,
$latex \frac{\partial^2z}{\partial x^2}+\frac{\partial^2z}{\partial y^2}=4x^2\frac{\partial^2z}{\partial4^2}+8xy\frac{\partial z}{\partial u\partial v}+4y=\frac{\partial^2z}{\partial v^2}+4x^2\frac{\partial^2z}{\partial v^2}+4y^2\frac{\partial^2z}{\partial u^2}-8xy\frac{\partial z}{\partial u\partial v}$
$latex =\frac{\partial^2z}{\partial u^2}\left[4x^2+4y^2\right]+\frac{\partial^2z}{\partial v^2}\left[4x^2+4y^2\right]$
$latex =\left(\frac{\partial^2z}{\partial u^2}+\frac{\partial^2z}{\partial v^2}\right)\left(4x^2+4y^2\right)…..3$
$latex \begin{array}{l}u=n^2-y^2\\\\v=2xy\end{array}$
$latex \begin{array}{l}u^2+v^2=\left(x^2-y^2\right)^2+4x^2y^2\\\\\;\;\;\;\;\;\;\;\;\;\;=x^4+y^4-2x^2y^2+4x^2y^2\\\\\;\;\;\;\;\;\;\;\;\;\;=\left(x^2+y^2\right)^2\end{array}$
$latex x^2+y^2=\sqrt{4^2+v^2}$
substitute in 3
$latex \frac{\partial^2z}{\partial x^2}+\frac{\partial^2z}{\partial y^2}=4\sqrt{4^2+v^2}\left(\frac{\partial^2z}{\partial u^2}+\frac{\partial^2z}{\partial v^2}\right)$
Hence proved.
c) Fit a straight line passing through points (0, 1), (1, 2), (2, 3), (3, 4, 5), (4, 6), (5, 7, 5) [8M]