Question 1 |
A quadratic function of two variables is given as
f\left(x_{1}, x_{2}\right)=x_{1}^{2}+2 x_{2}^{2}+3 x_{1}+3 x_{2}+x_{1} x_{2}+1
The magnitude of the maximum rate of change of the function at the point (1,1) is ____(Round off to the nearest integer).
f\left(x_{1}, x_{2}\right)=x_{1}^{2}+2 x_{2}^{2}+3 x_{1}+3 x_{2}+x_{1} x_{2}+1
The magnitude of the maximum rate of change of the function at the point (1,1) is ____(Round off to the nearest integer).
10 | |
12 | |
8 | |
16 |
Question 1 Explanation:
Given :
\mathrm{F}\left(\mathrm{x}, \mathrm{x}_{2}\right)=\mathrm{x}_{1}^{2}+2 \mathrm{x}_{2}^{2}+\mathrm{x}_{1}+\mathrm{x}_{1} \mathrm{x}_{2}+1
\nabla \cdot F=\left(2 x_{1}+3+x_{2}\right) \hat{a}_{x_{1}}+\left(4 x_{2}+3+x_{1}\right) \hat{a}_{x_{z}}
At (1,1)
\begin{aligned} \nabla \cdot F & =6 \hat{a}_{x_{1}}+7 \hat{a}_{x_{2}} \\ |\nabla \cdot F| & =\sqrt{6^{2}+8^{2}}=10 \end{aligned}
\mathrm{F}\left(\mathrm{x}, \mathrm{x}_{2}\right)=\mathrm{x}_{1}^{2}+2 \mathrm{x}_{2}^{2}+\mathrm{x}_{1}+\mathrm{x}_{1} \mathrm{x}_{2}+1
\nabla \cdot F=\left(2 x_{1}+3+x_{2}\right) \hat{a}_{x_{1}}+\left(4 x_{2}+3+x_{1}\right) \hat{a}_{x_{z}}
At (1,1)
\begin{aligned} \nabla \cdot F & =6 \hat{a}_{x_{1}}+7 \hat{a}_{x_{2}} \\ |\nabla \cdot F| & =\sqrt{6^{2}+8^{2}}=10 \end{aligned}
Question 2 |
The closed curve shown in the figure is described by
r=1+\cos \theta, where r=\sqrt{x^{2}+y^{2}} x=r \cos \theta, y=r \sin \theta
The magnitude of the line integral of the vector field F=-y \hat{i}+x \hat{j} around the closed curve is ___(Round off to 2 decimal places).
r=1+\cos \theta, where r=\sqrt{x^{2}+y^{2}} x=r \cos \theta, y=r \sin \theta
The magnitude of the line integral of the vector field F=-y \hat{i}+x \hat{j} around the closed curve is ___(Round off to 2 decimal places).
9.42 | |
6.36 | |
2.45 | |
7.54 |
Question 2 Explanation:
\begin{aligned}
I & =\int_{0}^{2 \pi} \vec{F} \cdot \overrightarrow{d l} \\
& =\int_{0}^{2 \pi}(-y \hat{i}+x)(d x \hat{i}+d y) \\
& =\int_{0}^{2 \pi}(-y d x+x d y)
\end{aligned}
Given : \quad x=r \cos \theta and y=r \sin \theta
\begin{aligned} \therefore I&=\int_{0}^{2 \pi}[(-r \sin \theta)(-r \sin \theta) d \theta+(r \cos \theta)(r \cos \theta) d \theta] \\ &=\int_{0}^{2 \pi} r^{2} d \theta \\ &=\int_{0}^{2 \pi}(1+\cos \theta)^{2} d \theta \\ &=\int_{0}^{2 \pi}\left(1+\cos ^{2} \theta+2 \cos \theta\right) d \theta \\ &=\int_{0}^{2 \pi}\left(1+\frac{1+\cos 2 \theta}{2}+2 \cos \theta\right) \mathrm{d} \theta \\ &=3 \pi=9.425 \end{aligned}
Given : \quad x=r \cos \theta and y=r \sin \theta
\begin{aligned} \therefore I&=\int_{0}^{2 \pi}[(-r \sin \theta)(-r \sin \theta) d \theta+(r \cos \theta)(r \cos \theta) d \theta] \\ &=\int_{0}^{2 \pi} r^{2} d \theta \\ &=\int_{0}^{2 \pi}(1+\cos \theta)^{2} d \theta \\ &=\int_{0}^{2 \pi}\left(1+\cos ^{2} \theta+2 \cos \theta\right) d \theta \\ &=\int_{0}^{2 \pi}\left(1+\frac{1+\cos 2 \theta}{2}+2 \cos \theta\right) \mathrm{d} \theta \\ &=3 \pi=9.425 \end{aligned}
Question 3 |
The expected number of trials for first occurrence of a "head" in a biased coin is known to be 4. The probability of first occurrence of a "head" in the second trial is ___ (Round off to 3 decimal places).
0.125 | |
0.188 | |
0.254 | |
0.564 |
Question 3 Explanation:
Let probability of head =\mathrm{P}
Let probability of Tail =q=P-1
\begin{array}{|c|c|c|c|c|} \hline Trial & 1 & 2 & 3 & ... \\ \hline Probability & [\mathrm{P} & \mathrm{qP} & [latex]\mathrm{q}^{2} \mathrm{P} & .... \\ \hline \end{array}
\therefore Expected No. of trail
\begin{aligned} & =\sum_{n=1}^{\infty} p+2 P q+3 q^{2} P+\ldots \\ & =P\left(1+2 q+3 q^{2}+\ldots\right) \\ & =P(1-q)^{-2} \\ & =\frac{P}{P^{2}}=\frac{1}{P} \end{aligned}
Given : Trial =4
\frac{1}{P}=4 \Rightarrow P=\frac{1}{4}
\therefore \quad \mathrm{q}=\frac{3}{4}
Now, probability of head for second trail. \begin{aligned} & =\mathrm{qP} \\ & =\frac{3}{4} \times \frac{1}{4}=\frac{3}{16}=0.1875 \end{aligned}
Let probability of Tail =q=P-1
\begin{array}{|c|c|c|c|c|} \hline Trial & 1 & 2 & 3 & ... \\ \hline Probability & [\mathrm{P} & \mathrm{qP} & [latex]\mathrm{q}^{2} \mathrm{P} & .... \\ \hline \end{array}
\therefore Expected No. of trail
\begin{aligned} & =\sum_{n=1}^{\infty} p+2 P q+3 q^{2} P+\ldots \\ & =P\left(1+2 q+3 q^{2}+\ldots\right) \\ & =P(1-q)^{-2} \\ & =\frac{P}{P^{2}}=\frac{1}{P} \end{aligned}
Given : Trial =4
\frac{1}{P}=4 \Rightarrow P=\frac{1}{4}
\therefore \quad \mathrm{q}=\frac{3}{4}
Now, probability of head for second trail. \begin{aligned} & =\mathrm{qP} \\ & =\frac{3}{4} \times \frac{1}{4}=\frac{3}{16}=0.1875 \end{aligned}
Question 4 |
Consider the following equation in a 2-D realspace.
\left|x_{1}\right|^{p}+\left|x_{2}\right|^{p}=1 for p \gt 0
Which of the following statement(s) is/are true.
\left|x_{1}\right|^{p}+\left|x_{2}\right|^{p}=1 for p \gt 0
Which of the following statement(s) is/are true.
When \mathrm{p}=2, the area enclosed by the curve is \pi. | |
When p tends to \infty, the area enclosed by the curve tends to 4. | |
When p tends to 0 , the area enclosed by the curve is 1. | |
When p=1, the area enclosed by the curve is 2. |
Question 4 Explanation:
Check option (A),
put P=2
x_{1}^{2}+x_{2}^{2}=1
Which is equation of circle whose radius is 1 .
\therefore \quad Area =\pi r^{2}=\pi
Check option (B),
If x_{2} \lt 1 Then, P \rightarrow \infty,\left|x_{2}\right|^{P} \rightarrow 0
\therefore For equation \left|x_{1}\right|^{\mathrm{P}}+\left|\mathrm{x}_{2}\right|^{\mathrm{P}}=1 satisfaction,
\mathrm{x}_{1} \rightarrow 1 \text { or } \dot{Y}
Curve :
\therefore Area tends to 4 .
Check option (D),
Put P=1,
\left|x_{1}\right|+\left|\mathrm{x}_{2}\right|=1
Curve:
\therefore Area of curve (square) =(\sqrt{2})^{2}=2
put P=2
x_{1}^{2}+x_{2}^{2}=1
Which is equation of circle whose radius is 1 .
\therefore \quad Area =\pi r^{2}=\pi
Check option (B),
If x_{2} \lt 1 Then, P \rightarrow \infty,\left|x_{2}\right|^{P} \rightarrow 0
\therefore For equation \left|x_{1}\right|^{\mathrm{P}}+\left|\mathrm{x}_{2}\right|^{\mathrm{P}}=1 satisfaction,
\mathrm{x}_{1} \rightarrow 1 \text { or } \dot{Y}
Curve :
\therefore Area tends to 4 .
Check option (D),
Put P=1,
\left|x_{1}\right|+\left|\mathrm{x}_{2}\right|=1
Curve:
\therefore Area of curve (square) =(\sqrt{2})^{2}=2
Question 5 |
Three points in the x-y plane are (-1,0.8) (0,2.2) and (1,2.8). The value of the slope of the best fit straight line in the least square sense is ___ (Round off to 2 decimal places).
0.25 | |
0.5 | |
0.75 | |
1 |
Question 5 Explanation:
Straight line equation, y=a x+b [Let]
where, a= slope
By lest approximation,
\sum \mathrm{y}_{i}=a \sum \mathrm{x}_{\mathrm{i}}+\mathrm{bn}
and \quad \sum x_{i} y_{i}=a \sum x_{1}^{2}+b \sum x_{i}
\begin{array}{|c|c|c|c|} \hline x & y & x^{2} & xy \\ \hline -1 & 0.8 & 1 & -0.8 \\ 0 & 2.2 & 0 & 0 \\ 1 & 2.8 & 1 & 2.8 \\ \hline \sum x=0 & \sum y=5.8 &\sum x^{2}=2 & \sum x y=2 \\ \hline \end{array}
From eqn. (1), we get
\begin{aligned} 5.8 & =a(0)+3 b \Rightarrow 5.8=3 b \\ 2 & =a(2)+b(0) \\ \Rightarrow \quad a & =1 \end{aligned}
where, a= slope
By lest approximation,
\sum \mathrm{y}_{i}=a \sum \mathrm{x}_{\mathrm{i}}+\mathrm{bn}
and \quad \sum x_{i} y_{i}=a \sum x_{1}^{2}+b \sum x_{i}
\begin{array}{|c|c|c|c|} \hline x & y & x^{2} & xy \\ \hline -1 & 0.8 & 1 & -0.8 \\ 0 & 2.2 & 0 & 0 \\ 1 & 2.8 & 1 & 2.8 \\ \hline \sum x=0 & \sum y=5.8 &\sum x^{2}=2 & \sum x y=2 \\ \hline \end{array}
From eqn. (1), we get
\begin{aligned} 5.8 & =a(0)+3 b \Rightarrow 5.8=3 b \\ 2 & =a(2)+b(0) \\ \Rightarrow \quad a & =1 \end{aligned}
There are 5 questions to complete.