Question 1 |
A pair of six-faced dice is rolled thrice. The probability that the sum of the
outcomes in each roll equals 4 in exactly two of the three attempts is ______.
(round off to three decimal places)
0.045 | |
0.078 | |
0.018 | |
0.025 |
Question 1 Explanation:
Event, E = {(1, 3)(3, 1)(2, 2)}
n(E) = 3
n(S) = 36
p=P(E)=\frac{3}{36}=\frac{1}{12}
q=P(\bar{E})=1-\frac{1}{12}=\frac{11}{12}
P(x)=3C_2(p^2)(q^1)=3 \times \left (\frac{1}{12} \right )^2\left (\frac{11}{12} \right )=0.02
n(E) = 3
n(S) = 36
p=P(E)=\frac{3}{36}=\frac{1}{12}
q=P(\bar{E})=1-\frac{1}{12}=\frac{11}{12}
P(x)=3C_2(p^2)(q^1)=3 \times \left (\frac{1}{12} \right )^2\left (\frac{11}{12} \right )=0.02
Question 2 |
Let y be a non-zero vector of size 2022 x 1. Which of the following statement(s) is/are TRUE?
yy^T is a symmetric matrix. | |
y^Ty is an eigenvalue of yy^T | |
yy^T has a rank of 2022. | |
yy^T is invertible. |
Question 2 Explanation:
Let vector
\begin{aligned} y&=\begin{bmatrix} 4\\ 4\\ 4 \end{bmatrix}_{3 \times 1}\\ y^T&=\begin{bmatrix} 4& 4& 4 \end{bmatrix}_{1 \times 3}\\ yy^T&=\begin{bmatrix} 4\\ 4\\ 4 \end{bmatrix}\begin{bmatrix} 4& 4& 4 \end{bmatrix}\\ &=\begin{bmatrix} 16 & 16 & 16\\ 16 & 16 & 16\\ 16 & 16 & 16 \end{bmatrix}\\ y^Ty&=[4^2+4^2+4^2]_{1 \times 1}\\ &=[48]_{1 \times 1}\\ \rho (y)&=\rho (y^T)=\rho (yy^T)=\rho (y^Ty)=1 \end{aligned}
From above information
yy^T is asymmetric.
y^Ty is an eigen value of yy^T .
yy^T has rank 1
det(yy^T) =0 so, yy^T is not invertible.
\begin{aligned} y&=\begin{bmatrix} 4\\ 4\\ 4 \end{bmatrix}_{3 \times 1}\\ y^T&=\begin{bmatrix} 4& 4& 4 \end{bmatrix}_{1 \times 3}\\ yy^T&=\begin{bmatrix} 4\\ 4\\ 4 \end{bmatrix}\begin{bmatrix} 4& 4& 4 \end{bmatrix}\\ &=\begin{bmatrix} 16 & 16 & 16\\ 16 & 16 & 16\\ 16 & 16 & 16 \end{bmatrix}\\ y^Ty&=[4^2+4^2+4^2]_{1 \times 1}\\ &=[48]_{1 \times 1}\\ \rho (y)&=\rho (y^T)=\rho (yy^T)=\rho (y^Ty)=1 \end{aligned}
From above information
yy^T is asymmetric.
y^Ty is an eigen value of yy^T .
yy^T has rank 1
det(yy^T) =0 so, yy^T is not invertible.
Question 3 |
Consider the polynomial f(x)=x^3-6x^2+11x-6 on the domain S given by 1 \leq x \leq 3. The first and second derivatives are f'(x) and f''(x).
Consider the following statements:
I. The given polynomial is zero at the boundary points x = 1 and x = 3.
II. There exists one local maxima of f(x) within the domain S.
III. The second derivative f''(x) \gt 0 throughout the domain S.
IV. There exists one local minima of f(x) within the domain S.
The correct option is:
Consider the following statements:
I. The given polynomial is zero at the boundary points x = 1 and x = 3.
II. There exists one local maxima of f(x) within the domain S.
III. The second derivative f''(x) \gt 0 throughout the domain S.
IV. There exists one local minima of f(x) within the domain S.
The correct option is:
Only statements I, II and III are correct. | |
Only statements I, II and IV are correct. | |
Only statements I and IV are correct. | |
Only statements II and IV are correct. |
Question 3 Explanation:
\begin{aligned}
f(x)&=x^3-6x^2+11x-6 \\
f(1)&=1-6+11-6=0 \\
f(3)&=3^3-6 \times 3^2+11 \times 3-6=0 \\
&\text{Statement (I) is correct.} \\
f'(x)&=3x^2-12x+11 \\
f'(x)&=0\Rightarrow x=2\pm \frac{1}{\sqrt{3}}
\end{aligned}
f(x) has local maxima at x=2-\frac{1}{\sqrt{3}}
Statement (II) is also true.
Now,
\begin{aligned} f''(x)&=6x-12\\ f''(x) &\gt 0\\ 6x-12 &\gt 0\\ x & \gt 0 \end{aligned}
Statement (III) is incorrect statement (IV) is also correct.
\because x=2+\frac{1}{\sqrt{3}} is point of minima.
f(x) has local maxima at x=2-\frac{1}{\sqrt{3}}
Statement (II) is also true.
Now,
\begin{aligned} f''(x)&=6x-12\\ f''(x) &\gt 0\\ 6x-12 &\gt 0\\ x & \gt 0 \end{aligned}
Statement (III) is incorrect statement (IV) is also correct.
\because x=2+\frac{1}{\sqrt{3}} is point of minima.
Question 4 |
P and Q are two square matrices of the same order. Which of the following statement(s) is/are correct?
If P and Q are invertible, then [PQ]^{-1}=Q^{-1}P^{-1} | |
If P and Q are invertible, then [QP]^{-1}=P^{-1}Q^{-1} | |
If P and Q are invertible, then [PQ]^{-1}=Q^{-1}P^{-1} | |
If P and Q are not invertible, then [PQ]^{-1}=P^{-1}Q^{-1} |
Question 4 Explanation:
If P and Q are invertible then (PQ)^{-1} = Q^{-1}P^{-1} is correct. Let,
\begin{aligned} PQ&=C \\ P^{-1}PQ&=P^{-1}C \\ Q&=P^{-1} C\\ Q^{-1}Q&=Q^{-1}P^{-1}C \\ I&=Q^{-1}P^{-1}C \\ IC^{-1}&=Q^{-1}P^{-1}CC^{-1} \\ C^{-1}&= Q^{-1}P^{-1}\\ (PQ)^{-1}&= Q^{-1}P^{-1} \end{aligned}
Hence, proved. Similarly, we can prove if P, Q are invertible then (QP)^{-1}= P^{-1}Q^{-1}
\begin{aligned} PQ&=C \\ P^{-1}PQ&=P^{-1}C \\ Q&=P^{-1} C\\ Q^{-1}Q&=Q^{-1}P^{-1}C \\ I&=Q^{-1}P^{-1}C \\ IC^{-1}&=Q^{-1}P^{-1}CC^{-1} \\ C^{-1}&= Q^{-1}P^{-1}\\ (PQ)^{-1}&= Q^{-1}P^{-1} \end{aligned}
Hence, proved. Similarly, we can prove if P, Q are invertible then (QP)^{-1}= P^{-1}Q^{-1}
Question 5 |
Match the following attributes of a city with the appropriate scale of
measurements.
\begin{array}{|l|l|}\hline \text{Attribute}&\text{Scale of measurement} \\ \hline \text{(P) Average temperature } (^{\circ}C)\text{ of a city} & \text{((I) Interval} \\ \hline \text{(Q) Name of a city} & \text{(II) Ordinal}\\ \hline \text{(R) Population density of a city}& \text{(III) Nominal}\\ \hline \text{(S) Ranking of a city based on ease of business}& \text{(IV) Ratio}\\ \hline \end{array}
Which one of the following combinations is correct?
\begin{array}{|l|l|}\hline \text{Attribute}&\text{Scale of measurement} \\ \hline \text{(P) Average temperature } (^{\circ}C)\text{ of a city} & \text{((I) Interval} \\ \hline \text{(Q) Name of a city} & \text{(II) Ordinal}\\ \hline \text{(R) Population density of a city}& \text{(III) Nominal}\\ \hline \text{(S) Ranking of a city based on ease of business}& \text{(IV) Ratio}\\ \hline \end{array}
Which one of the following combinations is correct?
(P)-(I), (Q)-(III), (R)-(IV), (S)-(II) | |
(P)-(II), (Q)-(I), (R)-(IV), (S)-(III) | |
(P)-(II), (Q)-(III), (R)-(IV), (S)-(I) | |
(P)-(I), (Q)-(II), (R)-(III), (S)-(IV) |
Question 5 Explanation:
Meaning of
Nominal -> a name or term
Ordinal -> in an ordered sequence
Ratio -> quantitative relation between two things
Interval -> indicates average of a range
Nominal -> a name or term
Ordinal -> in an ordered sequence
Ratio -> quantitative relation between two things
Interval -> indicates average of a range
Question 6 |
\int \left ( x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+... \right )dx is equil to
\frac{1}{1+x}+constant | |
\frac{1}{1+x^2}+constant | |
-\frac{1}{1-x}+constant | |
-\frac{1}{1-x^2}+constant |
Question 6 Explanation:
MTA- Marks to All
I=\int \left ( x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...\infty \right )dx
I=\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}+...
Option (A)
\frac{1}{1+x}=(1+x)^{-1}=1-x+x^2-x^3...\infty
So, its incorrect.
Option (B)
\frac{1}{1+x^2}=(1+x^2)^{-1}=1-x^2+x^4-x^6...\infty
So, its incorrect.
Similarly option (C) and (D) both are incorrect.
No-correct choice given.
I=\int \left ( x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...\infty \right )dx
I=\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}+...
Option (A)
\frac{1}{1+x}=(1+x)^{-1}=1-x+x^2-x^3...\infty
So, its incorrect.
Option (B)
\frac{1}{1+x^2}=(1+x^2)^{-1}=1-x^2+x^4-x^6...\infty
So, its incorrect.
Similarly option (C) and (D) both are incorrect.
No-correct choice given.
Question 7 |
The function f(x, y) satisfies the Laplace equation
\triangledown ^2f(x,y)=0
on a circular domain of radius r = 1 with its center at point P with coordinates x = 0, y = 0. The value of this function on the circular boundary of this domain is equal to 3.
The numerical value of f(0, 0) is:
\triangledown ^2f(x,y)=0
on a circular domain of radius r = 1 with its center at point P with coordinates x = 0, y = 0. The value of this function on the circular boundary of this domain is equal to 3.
The numerical value of f(0, 0) is:
0 | |
2 | |
3 | |
1 |
Question 7 Explanation:
According to given condition given function f(x,y) is nothing but constant function i.e. f(x,y)=3 because this is the only function whose value is 3 at any point on the boundary of unit circle and it is also satisfying Laplace equation, so
f(0,0)=3
f(0,0)=3
Question 8 |
A set of observations of independent variable (x) and the corresponding
dependent variable (y) is given below.
\begin{array}{|c|c|c|c|c|} \hline x&5&2&4&3 \\ \hline y&16&10&13&12\\ \hline \end{array}
Based on the data, the coefficient a of the linear regression model
y = a + bx
is estimated as 6.1.
The coefficient b is ______________ . (round off to one decimal place)
\begin{array}{|c|c|c|c|c|} \hline x&5&2&4&3 \\ \hline y&16&10&13&12\\ \hline \end{array}
Based on the data, the coefficient a of the linear regression model
y = a + bx
is estimated as 6.1.
The coefficient b is ______________ . (round off to one decimal place)
6.1 | |
1.9 | |
2.2 | |
3.6 |
Question 8 Explanation:
We know that, normal equation for fitting of straight lines are
\begin{aligned} \Sigma y&=na+b\Sigma x\\ \Sigma xy&=a\Sigma x+b\Sigma x^2\\ n&=4\\ 51&=4a+b(14)\\ 188&=a(14)+b(54) \end{aligned}
After solving, a=6.1 and b=1.9
\begin{aligned} \Sigma y&=na+b\Sigma x\\ \Sigma xy&=a\Sigma x+b\Sigma x^2\\ n&=4\\ 51&=4a+b(14)\\ 188&=a(14)+b(54) \end{aligned}
After solving, a=6.1 and b=1.9
Question 9 |
Consider the differential equation
\frac{dy}{dx}=4(x+2)-y
For the initial condition y = 3 at x = 1, the value of y at x = 1.4 obtained using Euler's method with a step-size of 0.2 is _________. (round off to one decimal place)
\frac{dy}{dx}=4(x+2)-y
For the initial condition y = 3 at x = 1, the value of y at x = 1.4 obtained using Euler's method with a step-size of 0.2 is _________. (round off to one decimal place)
5.4 | |
6.4 | |
2.8 | |
4.2 |
Question 9 Explanation:
\begin{aligned}
x_0&=1\rightarrow y_0=3\\
x_1&=1.2\rightarrow y_1=?\\
x_2&=1.4\rightarrow y_2=?\\
\end{aligned}
Using Euler's forward methods
\begin{aligned} y_n&=y_n+hf(x_n,y_n)\\ y_1&=y_0+hf(x_0,y_0)\\ &=3+0.2 \times (4(1+2)-3)\\ &=4.8\\ y_2&=y_1+hf(x_1,y_1)\\ &=4.8+0.2 \times (4(1.2+2)-4.8)\\ &=6.4\\ \end{aligned}
Using Euler's forward methods
\begin{aligned} y_n&=y_n+hf(x_n,y_n)\\ y_1&=y_0+hf(x_0,y_0)\\ &=3+0.2 \times (4(1+2)-3)\\ &=4.8\\ y_2&=y_1+hf(x_1,y_1)\\ &=4.8+0.2 \times (4(1.2+2)-4.8)\\ &=6.4\\ \end{aligned}
Question 10 |
Let max \{a, b\} denote the maximum of two real numbers a and b.
Which of the following statement(s) is/are TRUE about the function
f(x) = max\{3 -x, x - 1\} ?
It is continuous on its domain. | |
It has a local minimum at x = 2. | |
It has a local maximum at x = 2. | |
It is differentiable on its domain. |
Question 10 Explanation:
f(x)=max{3-x,x-1}
both intersecting at
\begin{aligned} 3-x&=x-1 \\ 2x&= 4\\ x&=2 \\ y &= max\{3-x,x-1\} \end{aligned}
both intersecting at
\begin{aligned} 3-x&=x-1 \\ 2x&= 4\\ x&=2 \\ y &= max\{3-x,x-1\} \end{aligned}
There are 10 questions to complete.
Good question answer very nice