Question 1 |
In the Taylor series expansion of e^{x} about x=2, the coefficient of (x-2)^4 is
\frac{1}{4!} | |
\frac{2^4}{4!} | |
\frac{e^2}{4!} | |
\frac{e^4}{4!} |
Question 1 Explanation:
f(x)\text{in the neighbourhood of a}
f(x)=\Sigma b_{n}(x-a)^{n} \text{where n from 0 to 0}
\begin{aligned}\quad\quad\quad\quad b_{n}&=\frac{f^{n}(a)}{n !} \\ f^{4}(x)&=e^{x} ; f^{4}(2)=e^{2}\\ \end{aligned}
\therefore \text{Coefficient of} (x-2)^{4}=b_{4}=\frac{f^{4}(2)}{4 !}=\frac{e^{2}}{4 !}
f(x)=\Sigma b_{n}(x-a)^{n} \text{where n from 0 to 0}
\begin{aligned}\quad\quad\quad\quad b_{n}&=\frac{f^{n}(a)}{n !} \\ f^{4}(x)&=e^{x} ; f^{4}(2)=e^{2}\\ \end{aligned}
\therefore \text{Coefficient of} (x-2)^{4}=b_{4}=\frac{f^{4}(2)}{4 !}=\frac{e^{2}}{4 !}
Question 2 |
Given that \ddot{x}+3x=0, and x(0) = 1, \dot{x}(0)=0, what is x(1)?
-0.99 | |
-0.16 | |
0.16 | |
0.99 |
Question 2 Explanation:
\ddot{x}+3 x=0
Auxiliary equation is
m^{2}+3=0
i.e. \qquad m=\pm \sqrt{3}
\therefore \qquad x =A \cos \sqrt{3} t+B \sin \sqrt{3} t
\dot{x} =\sqrt{3}(B \cos \sqrt{3} t-A \sin \sqrt{3} t)
\text{At} \quad t=0
\qquad 1=A \qquad ...(i)
\qquad 0=B \qquad ...(ii)
\therefore \qquad x =\cos \sqrt{3} t
x(1) =\cos (\sqrt{3} \times 57.3)=-0.16
Auxiliary equation is
m^{2}+3=0
i.e. \qquad m=\pm \sqrt{3}
\therefore \qquad x =A \cos \sqrt{3} t+B \sin \sqrt{3} t
\dot{x} =\sqrt{3}(B \cos \sqrt{3} t-A \sin \sqrt{3} t)
\text{At} \quad t=0
\qquad 1=A \qquad ...(i)
\qquad 0=B \qquad ...(ii)
\therefore \qquad x =\cos \sqrt{3} t
x(1) =\cos (\sqrt{3} \times 57.3)=-0.16
Question 3 |
The value of \lim_{x\rightarrow 8}\frac{x^{1/3}-2}{(x-8)}
1/16 | |
1/12 | |
1/8 | |
1/4 |
Question 3 Explanation:
\begin{aligned} \quad \quad(x-8) &=h(\text { say }) \\ \Rightarrow \quad x &=8+h \end{aligned}
\; \therefore \lim_{h \rightarrow 0} \frac{(8+h)^{1 / 3}-2}{h}
\text{Above form in the} \left(\frac{0}{0}\right) \text{by putting the value h=0 }
\text{Applying L hospital rule}
\lim _{h \rightarrow 0} \frac{\frac{1}{3}(8+h)\left(\frac{1}{3}-1\right)}{1}=\frac{1}{3}(8)^{-\frac{2}{3}}=\frac{1}{12}
\; \therefore \lim_{h \rightarrow 0} \frac{(8+h)^{1 / 3}-2}{h}
\text{Above form in the} \left(\frac{0}{0}\right) \text{by putting the value h=0 }
\text{Applying L hospital rule}
\lim _{h \rightarrow 0} \frac{\frac{1}{3}(8+h)\left(\frac{1}{3}-1\right)}{1}=\frac{1}{3}(8)^{-\frac{2}{3}}=\frac{1}{12}
Question 4 |
A coin is tossed 4 times. What is the probability of getting heads exactly 3 times?
1/4 | |
1/2 | |
3/8 | |
3/4 |
Question 4 Explanation:
P(H)=0.5
Probability of getting head exactly 3 times is
^{4} C_{3}(0.5)^{3}(0.5)=\frac{1}{4}
Probability of getting head exactly 3 times is
^{4} C_{3}(0.5)^{3}(0.5)=\frac{1}{4}
Question 5 |
The matrix \begin{bmatrix} 1 &2 & 4\\ 3 & 0 &6 \\ 1& 1 & p \end{bmatrix} has one Eigenvalue equal to 3. The sum of the other two Eigenvalues is
p | |
p-1 | |
p-2 | |
p-3 |
Question 5 Explanation:
Sum of the eigen values of matrix is
= sum of diagonal values present in the matrix
\therefore 1+0+p=3+\lambda_{2}+\lambda_{3} \\ \Rightarrow p+1=3+\lambda_{2}+\lambda_{3} \\ \Rightarrow \lambda_{2}+\lambda_{3}=p+1-3=p-2
= sum of diagonal values present in the matrix
\therefore 1+0+p=3+\lambda_{2}+\lambda_{3} \\ \Rightarrow p+1=3+\lambda_{2}+\lambda_{3} \\ \Rightarrow \lambda_{2}+\lambda_{3}=p+1-3=p-2
There are 5 questions to complete.