Question 1 |
The determinant of matrix A is 5 and the determinant of matrix B is 40. The
determinant of matrix AB is ______.
200 | |
100 | |
50 | |
45 |
Question 1 Explanation:
Determinant of A=5
Determinant of B=40
Determinant of AB=|A||B|
\begin{array}{l} \quad=5 \times 40 \\ \quad=200 \end{array}
Determinant of B=40
Determinant of AB=|A||B|
\begin{array}{l} \quad=5 \times 40 \\ \quad=200 \end{array}
Question 2 |
Let X be a random variable which is uniformly chosen from the set of positive
odd numbers less than 100. The expectation, E [X], is _____.
100 | |
50 | |
25 | |
10 |
Question 2 Explanation:
E[X]=\frac{1+2+3+\cdots 99}{50}=\frac{2500}{50}=50
Question 3 |
For 0\leq t \leq \infty
, the maximum value of the function f(t) =e^{-t}-2e^{-2t} occurs at
t=log_{e}4 | |
t=log_{e}2 | |
t=0 | |
t=log_{e}8 |
Question 3 Explanation:
\begin{aligned} f(t)&=e^{-t}-2 e^{-2 t} \\ f^{\prime}(t)&=-e^{-t}+4 e^{-2 t} \\ \text{For maximum value} &P(t)=0 \\ f^{\prime}(t)&=0=-e^{-t}+4 e^{-2 t} \\ \Rightarrow \quad 4 e^{-2 t}&=e^{t} \\ 4 e^{t}&=1 \\ \therefore \quad t&=\log _{e} 4 \end{aligned}
Question 4 |
The value of
\lim_{x\rightarrow \infty }(1+\frac{1}{x})^{x}
is
\lim_{x\rightarrow \infty }(1+\frac{1}{x})^{x}
is
ln 2 | |
1 | |
e | |
\infty |
Question 4 Explanation:
\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e^{\lim _{x \rightarrow \infty} \frac{1}{x} \cdot x}=e^{1}=e
Question 5 |
If the characteristic equation of the differential equation
\frac{d^{2}y}{dx^{2}}+2\alpha \frac{dy}{dx}+y=0
has two equal roots, then the values of a are
\frac{d^{2}y}{dx^{2}}+2\alpha \frac{dy}{dx}+y=0
has two equal roots, then the values of a are
\pm 1 | |
0,0 | |
\pm j | |
\pm 1/2 |
Question 5 Explanation:
\frac{d^{2} y}{d x^{2}}+2 \alpha\frac{d y}{d x}+y=0
The characteristic equation is given as
\begin{aligned} \left(m^{2}+2(x)+1\right) &=0 \\ m_{1}, m_{2} &=\frac{-2 x_{1} \pm \sqrt{4 x^{2}-4}}{2} \end{aligned}\\ \text{since both roots are equal i.e.} \\ \begin{aligned} m_{1}=& m_{2} \\ \frac{-2 \alpha+\sqrt{4 \alpha^{2}-4}}{2} &=\frac{-2\alpha \cdot-\sqrt{4 a^{2}-4}}{2} \\ \sqrt{4\left(1^{2}-4\right.} &=-\sqrt{4 c^{2}-4} \\ 2 \sqrt{4 c^{2}-4} &=0 \\ 4 \alpha^{2}-4 &=0 \\ \alpha^{2} &=1 \\ \alpha &=\pm 1 \end{aligned}
The characteristic equation is given as
\begin{aligned} \left(m^{2}+2(x)+1\right) &=0 \\ m_{1}, m_{2} &=\frac{-2 x_{1} \pm \sqrt{4 x^{2}-4}}{2} \end{aligned}\\ \text{since both roots are equal i.e.} \\ \begin{aligned} m_{1}=& m_{2} \\ \frac{-2 \alpha+\sqrt{4 \alpha^{2}-4}}{2} &=\frac{-2\alpha \cdot-\sqrt{4 a^{2}-4}}{2} \\ \sqrt{4\left(1^{2}-4\right.} &=-\sqrt{4 c^{2}-4} \\ 2 \sqrt{4 c^{2}-4} &=0 \\ 4 \alpha^{2}-4 &=0 \\ \alpha^{2} &=1 \\ \alpha &=\pm 1 \end{aligned}
There are 5 questions to complete.