Question 1 |
Newton-Raphson method is to be used to find root of equation 3x-e^{x}+\sin x=0. If the initial trial value for the root is taken as 0.333, the next approximation for the root would be _________
0.33 | |
0.54 | |
0.36 | |
0.76 |
Question 1 Explanation:
According to Newton-Raphson Method:
\begin{aligned} x_{N+1}&=X_{N}-\frac{f\left ( X_{N} \right )}{{F}'\left ( X_{N} \right )} \\ f\left ( x \right )&=3x-e^{x}+\sin x \\ {f}'\left ( x \right )&=3-e^{x}+\cos x \\ \Rightarrow \;\; X_{1}&=X_{0}-\frac{f\left ( 0.333 \right )}{{f}'\left ( 0.333 \right )} \\ &=0.333-\frac{3\times 0.333-e^{0.333}+\sin 0.333}{3-e^{0.333}+\cos 0.333} \\ \therefore \;\; X_{1}&=0.36 \end{aligned}
\begin{aligned} x_{N+1}&=X_{N}-\frac{f\left ( X_{N} \right )}{{F}'\left ( X_{N} \right )} \\ f\left ( x \right )&=3x-e^{x}+\sin x \\ {f}'\left ( x \right )&=3-e^{x}+\cos x \\ \Rightarrow \;\; X_{1}&=X_{0}-\frac{f\left ( 0.333 \right )}{{f}'\left ( 0.333 \right )} \\ &=0.333-\frac{3\times 0.333-e^{0.333}+\sin 0.333}{3-e^{0.333}+\cos 0.333} \\ \therefore \;\; X_{1}&=0.36 \end{aligned}
Question 2 |
The type of partial differential equation \frac{\partial^2 P}{\partial x^2} + \frac{\partial^2 P}{\partial y^2} +3\frac{\partial^2 P}{\partial x \partial y } +2\frac{\partial P}{\partial x}-\frac{\partial P}{\partial y}=0 is
elliptic | |
parabolic | |
hyperbolic | |
none of these |
Question 2 Explanation:
Comparing the given equation with the general form of second order partial differential equation, we have A=1, B=3, C=1 \Rightarrow\; B^{2}-4AC=5\gt0
\therefore PDE is Hyperbola.
\therefore PDE is Hyperbola.
Question 3 |
If the entries in each column of a square matrix M add up to 1, then an eigen value of M is
4 | |
3 | |
2 | |
1 |
Question 3 Explanation:
Consider the '2\times 2' square matrix M=\begin{bmatrix} a & b\\ c & d \end{bmatrix}
\Rightarrow \;\; \lambda ^{2}-\left ( a+d \right )\lambda +\left ( ad-bc \right )=0 \;\;...(i)
Putting \lambda =1, we get
1-(a+d)+ad-bc=0
1-a-d+ad-(1-d)(1-a)=0
1-a-d+ad-1+a+d-ad=0
0=0 which is true.
\therefore \;\;\lambda =1 satisfied the eq.(i) but \lambda =2,3,4 does not satisfy the eq.(i). For all possible values of a, d.
\Rightarrow \;\; \lambda ^{2}-\left ( a+d \right )\lambda +\left ( ad-bc \right )=0 \;\;...(i)
Putting \lambda =1, we get
1-(a+d)+ad-bc=0
1-a-d+ad-(1-d)(1-a)=0
1-a-d+ad-1+a+d-ad=0
0=0 which is true.
\therefore \;\;\lambda =1 satisfied the eq.(i) but \lambda =2,3,4 does not satisfy the eq.(i). For all possible values of a, d.
Question 4 |
Type II error in hypothesis testing is
acceptance of the null hypothesis when it is false and should be rejected | |
rejection of the null hypothesis when it is true and should be accepted | |
rejection of the null hypothesis when it is false and should be rejected | |
acceptance of the null hypothesis when it is true and should be accepted |
Question 5 |
The solution of the partial differential equation \frac{\partial u}{\partial t}=\alpha \frac{\partial^2 u}{\partial x^2} is of the form
C\cos \left ( kt \right )\left \lfloor C_{1}e^{(\sqrt{k/\alpha })x} + C_{2}e^{-(\sqrt{k/\alpha })x} \right \rfloor | |
Ce^{kt}\left \lfloor C_{1}e^{(\sqrt{k/\alpha })x} + C_{2}e^{-(\sqrt{k/\alpha })x} \right \rfloor | |
Ce^{kt}\left \lfloor C_{1}\cos (\sqrt{k/\alpha } )x+ C_{2}\sin {(-\sqrt{k/\alpha })x} \right \rfloor | |
C\sin (kt)\left \lfloor C_{1}\cos (\sqrt{k/\alpha } )x+ C_{2}\sin {(-\sqrt{k/\alpha })x} \right \rfloor |
Question 5 Explanation:
The PDE \frac{\partial u}{\partial t}=\alpha \frac{\partial^2 u}{\partial x^2} \;\;...(i)
Solution of (i) is,
u\left ( x,t \right )=\left ( A\cos px+B\sin px \right )Ce^{-p^{2}\alpha t}
Put -p^{2}\alpha =k
\Rightarrow \;\; p=\sqrt{-\frac{k}{\alpha }}=\sqrt{\frac{k}{\alpha }}i
Putting value of p in eq.(i),
\begin{aligned}u\left ( x,t \right )&=\left ( A\cos \sqrt{\frac{k}{\alpha }}x+b\sin h\sqrt{\frac{k}{\alpha }}x \right )Ce^{kt} \\ &=Ce^{kt}\left [ A\left \{ \frac{e^{\sqrt{\frac{k}{\alpha }}x}+e^{-\sqrt{\frac{k}{\alpha }}x}}{2} \right \}+B\left \{ \frac{e^{\sqrt{\frac{k}{\alpha }}x}-e^{-\sqrt{\frac{k}{\alpha }}x}}{2} \right \} \right ] \\ &=Ce^{kt}\left [ e^{\sqrt{\frac{k}{\alpha }}x}\left \{ \frac{A+B}{2} \right \}+e^{-\sqrt{\frac{k}{\alpha }}x}\left \{ \frac{A-B}{2} \right \} \right ] \\ &=Ce^{kt}\left [ c_{1}e^{\sqrt{\frac{k}{\alpha }}x}+c_{2}e^{-\sqrt{\frac{k}{\alpha }}x} \right ] \end{aligned}
Solution of (i) is,
u\left ( x,t \right )=\left ( A\cos px+B\sin px \right )Ce^{-p^{2}\alpha t}
Put -p^{2}\alpha =k
\Rightarrow \;\; p=\sqrt{-\frac{k}{\alpha }}=\sqrt{\frac{k}{\alpha }}i
Putting value of p in eq.(i),
\begin{aligned}u\left ( x,t \right )&=\left ( A\cos \sqrt{\frac{k}{\alpha }}x+b\sin h\sqrt{\frac{k}{\alpha }}x \right )Ce^{kt} \\ &=Ce^{kt}\left [ A\left \{ \frac{e^{\sqrt{\frac{k}{\alpha }}x}+e^{-\sqrt{\frac{k}{\alpha }}x}}{2} \right \}+B\left \{ \frac{e^{\sqrt{\frac{k}{\alpha }}x}-e^{-\sqrt{\frac{k}{\alpha }}x}}{2} \right \} \right ] \\ &=Ce^{kt}\left [ e^{\sqrt{\frac{k}{\alpha }}x}\left \{ \frac{A+B}{2} \right \}+e^{-\sqrt{\frac{k}{\alpha }}x}\left \{ \frac{A-B}{2} \right \} \right ] \\ &=Ce^{kt}\left [ c_{1}e^{\sqrt{\frac{k}{\alpha }}x}+c_{2}e^{-\sqrt{\frac{k}{\alpha }}x} \right ] \end{aligned}
There are 5 questions to complete.