Question 1 |
Consider the matrix
P=\begin{bmatrix} 1 & 1 &0 \\ 0&1 &1 \\ 0& 0 & 1 \end{bmatrix}
The number of distinct eigenvalues of P is
P=\begin{bmatrix} 1 & 1 &0 \\ 0&1 &1 \\ 0& 0 & 1 \end{bmatrix}
The number of distinct eigenvalues of P is
0 | |
1 | |
2 | |
3 |
Question 1 Explanation:
\text { Given: } A=\left[\begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]
It is an upper triangular matrix. It's diagonal elements are eigen values.
The eigen values of the matrix are 1, 1, 1.
\therefore Number of distinct eigen values = 1
Hence, option (B) is correct.
It is an upper triangular matrix. It's diagonal elements are eigen values.
The eigen values of the matrix are 1, 1, 1.
\therefore Number of distinct eigen values = 1
Hence, option (B) is correct.
Question 2 |
A parabola x=y^2 \; with \; 0\leq x\leq 1
is shown in the figure. The volume of the solid of rotation obtained by rotating the shaded area by 360^{\circ} around the x-axis is
\frac{\pi}{4} | |
\frac{\pi}{2} | |
\pi | |
2 \pi |
Question 2 Explanation:
\text { Given: } y^{2}=x, 0 \leq x \leq 1
The value of solid obtained by rotating the area bounded by the curve
\begin{aligned} \mathrm{y}^{2}&=\mathrm{x}, 0 \leq \mathrm{x} \leq 1 \text{ about x-axis is}\\ V &=\int_{a}^{b} \pi y^{2} d x \\ V &=\int_{0}^{1} \pi x d x \\ &=\left(\frac{\pi \mathrm{x}^{2}}{2}\right)_{0}^{1} \\ &=\frac{\pi}{2} \end{aligned}
The value of solid obtained by rotating the area bounded by the curve
\begin{aligned} \mathrm{y}^{2}&=\mathrm{x}, 0 \leq \mathrm{x} \leq 1 \text{ about x-axis is}\\ V &=\int_{a}^{b} \pi y^{2} d x \\ V &=\int_{0}^{1} \pi x d x \\ &=\left(\frac{\pi \mathrm{x}^{2}}{2}\right)_{0}^{1} \\ &=\frac{\pi}{2} \end{aligned}
Question 3 |
For the equation \frac{dy}{dx} + 7x^2 y=0, if y(0)=3/7, then the value of y(1)is
\frac{7}{3}e^{-7/3} | |
\frac{7}{3}e^{-3/7} | |
\frac{3}{7}e^{-7/3}
| |
\frac{3}{7}e^{-3/7}
|
Question 3 Explanation:
Given \frac{d y}{d x}+7 x^{2} y=0 \ldots(1)
With y(0)=\frac{3}{7} \ldots(2)
Now,(1) is written as
\Rightarrow \int \frac{1}{y} d y+\int 7 x^{2} d x=C
\Rightarrow \log y+\frac{7 x^{3}}{3}=C
\Rightarrow y=e^{\frac{7 x^{3}}{3}+c} \; \; ...(3)
Using (2),(3) becomes \frac{3}{7}=\mathrm{e}^{0} \cdot \mathrm{e}^{\mathrm{C}}(\mathrm{or}) \mathrm{e}^{\mathrm{C}}=\frac{3}{7} \; \; ...(4)
\therefore The solution of (1) with (3) &(4) is given by
y=y(x)=e^{\frac{-7 x^{3}}{3}+c}=e^{\frac{-7 x^{3}}{3}} \cdot e^{c}=\frac{3}{7} \cdot e^{\frac{-7 x^{3}}{3}}
Hence, y(1)=y=\frac{3}{7} \cdot e^{-\frac{7}{3}}
With y(0)=\frac{3}{7} \ldots(2)
Now,(1) is written as
\Rightarrow \int \frac{1}{y} d y+\int 7 x^{2} d x=C
\Rightarrow \log y+\frac{7 x^{3}}{3}=C
\Rightarrow y=e^{\frac{7 x^{3}}{3}+c} \; \; ...(3)
Using (2),(3) becomes \frac{3}{7}=\mathrm{e}^{0} \cdot \mathrm{e}^{\mathrm{C}}(\mathrm{or}) \mathrm{e}^{\mathrm{C}}=\frac{3}{7} \; \; ...(4)
\therefore The solution of (1) with (3) &(4) is given by
y=y(x)=e^{\frac{-7 x^{3}}{3}+c}=e^{\frac{-7 x^{3}}{3}} \cdot e^{c}=\frac{3}{7} \cdot e^{\frac{-7 x^{3}}{3}}
Hence, y(1)=y=\frac{3}{7} \cdot e^{-\frac{7}{3}}
Question 4 |
The lengths of a large stock of titanium rods follow a normal distribution with a mean (\mu) of 440 mm and a standard deviation (\sigma) of 1mm. What is the percentage of rods whose lengths lie between 438 mm and 441 mm?
81.85% | |
68.40% | |
99.75% | |
86.64% |
Question 4 Explanation:
\begin{array}{l} \mathrm{Z}=\frac{x-\mu}{\sigma} \\ \mathrm{Z}(\mathrm{x}=438)=\frac{438-440}{1}=-2 \\ \mathrm{P}(\mathrm{Z}=-2)=2.28 \% \\ \mathrm{Z}(\mathrm{x}=441)=\frac{441-440}{1}=1 \\ \mathrm{P}(\mathrm{Z}=1)=84.13 \% \\ \end{array}
The percentage of rods whose lengths lie between 438 mm and 441 mm =
\begin{array}{c} =\mathrm{P}(\mathrm{Z}=1)-\mathrm{P}(\mathrm{Z}=-2) \\ =84.13 \%-2.28 \%=81.85 \% \end{array}
Question 5 |
A flat-faced follower is driven using a circular eccentric cam rotating at a constant angular velocity \omega. At time t=0, the vertical position of the followeris y(0)=0, and the system is in the configuration shown below.
The vertical position of the follower face, y(t) is given by
The vertical position of the follower face, y(t) is given by
e \sin \omega t | |
e(1+ \cos 2\omega t) | |
e(1- \cos \omega t) | |
e \sin 2 \omega t |
Question 5 Explanation:
\begin{aligned} \mathrm{AA}_{1}=\mathrm{y} &=\mathrm{e}(1-\cos \theta) \\ &=\mathrm{e}(1-\cos \omega \mathrm{t}) \end{aligned}
There are 5 questions to complete.