Available Questions 255 found Page 10 of 13
Standalone Questions
#959
Mathematics
Relations and Functions
MCQ_SINGLE
APPLY
2025
JEE Main 2025
Competency
0 Marks
Let A = { (α, β) ∈ R x R : |α - 1| ≤ 4 and |β - 5| ≤ 6} and B = { (α, β) ∈ R × R: 16(α-2)²+9(β-6)² ≤ 144}. Then
(A) A ⊂ B
(B) B ⊂ A
(C) neither A ⊂ B nor B ⊂ A
(D) A ∪ B = {(x, y) : -4 ≤ x ≤ 4, -1 ≤ y ≤ 11}
Key: B
Sol:
Sol:
#940
Mathematics
Inverse Trigonometric Functions
VSA
APPLY
2024
Competency
2 Marks
Find value of k if \(\sin^{-1}[k~\tan(2~\cos^{-1}\frac{\sqrt{3}}{2})]=\frac{\pi}{3}.\)
Key:
Sol:
Sol:
#935
Mathematics
Differential Equations
SA
APPLY
2023
Competency
3 Marks
Solve the differential equation given by:$$x \, dy - y \, dx - \sqrt{x^{2} + y^{2}} \, dx = 0$$
Key:
Sol:
Sol:
#934
Mathematics
Linear Programming
SA
APPLY
2023
Competency
3 Marks
Solve graphically the following linear programming problem : Maximise \(z = 6x + 3y\), subject to the constraints\begin{align}
4x + y &\ge 80 \\
3x + 2y &\le 150 \\
x + 5y &\ge 115 \\
x, y &\ge 0
\end{align}
4x + y &\ge 80 \\
3x + 2y &\le 150 \\
x + 5y &\ge 115 \\
x, y &\ge 0
\end{align}
Key:
Sol:
Sol:
#933
Mathematics
Linear Programming
LA
APPLY
2023
Competency
5 Marks
Solve the following Linear Programming Problem graphically: Maximize: \(P = 70x + 40y\) subject to: \(3x + 2y ≤ 9, 3x + y ≤ 9, x ≥ 0, y ≥ 0\)
Key:
Sol:
Sol:
#930
Mathematics
Applications of Integrals
LA
APPLY
2023
Competency
5 Marks
The area of the region bounded by the line \(y=mx (m>0)\), the curve \(x^{2}+y^{2}=4\) and the \(x\)-axis in the first quadrant is \(\frac{\pi}{2}\) units. Using integration, find the value of m.
Key:
Sol:
Sol:
#929
Mathematics
Applications of Integrals
LA
APPLY
2023
Competency
5 Marks
33. Using integration, find the area of the region bounded by the parabola $y^{2}=4ax$ and its latus rectum.
Key:
Sol:
Sol:
#928
Mathematics
Linear Programming
SA
APPLY
2023
Competency
3 Marks
Determine graphically the minimum value of the following objective function : $z=500x+400y$ subject to constraints $x+y\le200, x\ge20, y\ge4x, y\ge0$
Key:
Sol:
Sol:
#923
Mathematics
Relations and Functions
LA
APPLY
2023
Competency
5 Marks
Show that a function $f:\mathbb{R}\rightarrow\mathbb{R}$ defined as $f(x)=\frac{5x-3}{4}$ is both one-one and onto.
Key:
Sol:
Sol:
#922
Mathematics
Relations and Functions
LA
APPLY
2023
Competency
5 Marks
Let $f : \mathbb{R} - \left\{ \frac{4}{3} \right\} \to \mathbb{R}$ be a function defined as:$$f(x) = \frac{4x}{3x+4}$$Show that $f$ is a one-one function. Also, check whether $f$ is an onto function or not.
Key:
Sol:
Sol:
#921
Mathematics
Relations and Functions
LA
APPLY
2023
Competency
5 Marks
34. (a) If N denotes the set of all natural numbers and R is the relation on $N \times N$ defined by $(a, b) R (c, d)$, if $ad(b+c)=bc(a+d)$. Show that R is an equivalence relation.
Key:
Sol:
Sol:
#920
Mathematics
Relations and Functions
LA
APPLY
2023
Competency
5 Marks
A relation $R$ is defined on a set of real numbers $\mathbb{R}$ as:$$R = \{(x, y) : x \cdot y \text{ is an irrational number}\}$$Check whether $R$ is reflexive, symmetric, and transitive or not.
Key:
Sol:
Sol:
#912
Mathematics
Probability
SA
APPLY
2023
Competency
3 Marks
There are two coins. One of them is a biased coin such that P (head): P (tail) is 1:3 and the other coin is a fair coin. A coin is selected at random and tossed once. If the coin showed head, then find the probability that it is a biased coin.
Key:
Sol:
Sol:
#911
Mathematics
Probability
SA
APPLY
2023
Competency
3 Marks
A pair of dice is thrown simultaneously. If $X$ denotes the absolute difference of numbers obtained on the pair of dice, then find the probability distribution of $X$.
Key:
Sol:
Sol:
#909
Mathematics
Probability
SA
APPLY
2023
Competency
3 Marks
The probability distribution of a random variable X is given below :
$$\begin{array}{|c|c|c|c|}
\hline
X & 1 & 2 & 3 \\
\hline
P(X) & \frac{k}{2} & \frac{k}{3} & \frac{k}{6} \\
\hline
\end{array}$$
(i) Find the value of $k$.
(ii) Find $P(1\le X<3)$.
(iii) Find $E(X)$, the mean of $X$.
OR
$A$ and $B$ are independent events such that $P(A\cap\overline{B})=\frac{1}{4}$ and $P(\overline{A}\cap B)=\frac{1}{6}$ Find $P(A)$ and $P(B)$.
$$\begin{array}{|c|c|c|c|}
\hline
X & 1 & 2 & 3 \\
\hline
P(X) & \frac{k}{2} & \frac{k}{3} & \frac{k}{6} \\
\hline
\end{array}$$
(i) Find the value of $k$.
(ii) Find $P(1\le X<3)$.
(iii) Find $E(X)$, the mean of $X$.
OR
$A$ and $B$ are independent events such that $P(A\cap\overline{B})=\frac{1}{4}$ and $P(\overline{A}\cap B)=\frac{1}{6}$ Find $P(A)$ and $P(B)$.
Key:
Sol:
Sol:
The sum of probabilities in a probability distribution must equal $1$.$$\sum P(X) = 1$$$$\frac{k}{2} + \frac{k}{3} + \frac{k}{6} = 1$$Find the common denominator (which is $6$):$$\frac{3k}{6} + \frac{2k}{6} + \frac{1k}{6} = 1$$$$\frac{6k}{6} = 1$$$$k = 1$$(ii) Find $P(1 \le X < 3)$The range $1 \le X < 3$ includes $X=1$ and $X=2$, but not $X=3$.$$P(1 \le X < 3) = P(X=1) + P(X=2)$$Substituting $k=1$:$$P(X=1) = \frac{1}{2}, \quad P(X=2) = \frac{1}{3}$$$$P(1 \le X < 3) = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$$(iii) Find $E(X)$, the mean of $X$The expectation (mean) is defined as $E(X) = \sum X \cdot P(X)$.Using $k=1$:$$P(X=1) = \frac{1}{2}, \quad P(X=2) = \frac{1}{3}, \quad P(X=3) = \frac{1}{6}$$$$E(X) = \left(1 \times \frac{1}{2}\right) + \left(2 \times \frac{1}{3}\right) + \left(3 \times \frac{1}{6}\right)$$$$E(X) = \frac{1}{2} + \frac{2}{3} + \frac{3}{6}$$$$E(X) = \frac{1}{2} + \frac{2}{3} + \frac{1}{2}$$$$E(X) = 1 + \frac{2}{3} = \frac{5}{3}$$OR Part: Independent EventsProblem:$A$ and $B$ are independent events such that:$P(A \cap \overline{B}) = \frac{1}{4}$$P(\overline{A} \cap B) = \frac{1}{6}$Solution:Let $P(A) = x$ and $P(B) = y$.Since events are independent:$$P(A \cap \overline{B}) = P(A) \cdot P(\overline{B}) = x(1-y)$$$$P(\overline{A} \cap B) = P(\overline{A}) \cdot P(B) = (1-x)y$$We have the system of equations:$x(1-y) = \frac{1}{4} \implies x - xy = \frac{1}{4}$$(1-x)y = \frac{1}{6} \implies y - xy = \frac{1}{6}$Subtracting equation (2) from equation (1):$$(x - xy) - (y - xy) = \frac{1}{4} - \frac{1}{6}$$$$x - y = \frac{3}{12} - \frac{2}{12} = \frac{1}{12}$$$$x = y + \frac{1}{12}$$Substitute $x$ back into equation (2):$$\left(1 - \left(y + \frac{1}{12}\right)\right)y = \frac{1}{6}$$$$\left(\frac{11}{12} - y\right)y = \frac{1}{6}$$$$\frac{11y}{12} - y^2 = \frac{1}{6}$$Multiply by 12 to clear the denominator:$$11y - 12y^2 = 2$$$$12y^2 - 11y + 2 = 0$$Solving the quadratic equation for $y$:$$y = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(12)(2)}}{2(12)}$$$$y = \frac{11 \pm \sqrt{121 - 96}}{24} = \frac{11 \pm \sqrt{25}}{24} = \frac{11 \pm 5}{24}$$Two possible values for $y$:$y_1 = \frac{16}{24} = \frac{2}{3}$$y_2 = \frac{6}{24} = \frac{1}{4}$Finding corresponding $x$ values using $x = y + \frac{1}{12}$:If $y = \frac{2}{3}$: $x = \frac{2}{3} + \frac{1}{12} = \frac{8+1}{12} = \frac{9}{12} = \frac{3}{4}$If $y = \frac{1}{4}$: $x = \frac{1}{4} + \frac{1}{12} = \frac{3+1}{12} = \frac{4}{12} = \frac{1}{3}$Answer:There are two possible sets of values:$P(A) = \frac{3}{4}$ and $P(B) = \frac{2}{3}$$P(A) = \frac{1}{3}$ and $P(B) = \frac{1}{4}$
#908
Mathematics
Probability
LA
APPLY
2023
Competency
5 Marks
(a) In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/5 be the probability that he knows the answer and 2/5 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/3. What is the probability that the student knows the answer, given that he answered it correctly? OR (b) A box contains 10 tickets, 2 of which carry a prize of ₹8 each, 5 of which carry a prize of ₹4 each, and remaining 3 carry a prize of ₹2 each. If one ticket is drawn at random, find the mean value of the prize.
Key:
Sol:
Sol:
#906
Mathematics
Applications of Derivatives
LA
UNDERSTAND
2023
Competency
5 Marks
(a) The median of an equilateral triangle is increasing at the rate of 2√3 cm/s. Find the rate at which its side is increasing. OR (b) Sum of two numbers is 5. If the sum of the cubes of these numbers is least, then find the sum of the squares of these numbers.
Key:
Sol:
Sol:
#891
Mathematics
Matrices and Determinants
LA
APPLY
2023
Competency
5 Marks
If $A=\begin{bmatrix}3 & 2\\ 5 & -7\end{bmatrix}$, then find $A^{-1}$ and use it to solve the following system of equations : $3x+5y=11, 2x-7y=-3$.
Key:
Sol:
Sol:
#889
Mathematics
Matrices and Determinants
LA
APPLY
2023
Competency
5 Marks
32. Find the inverse of the matrix $A=\begin{bmatrix}1 & -1 & 2\\ 0 & 2 & -3\\ 3 & -2 & 4\end{bmatrix}$. Using the inverse, solve the system of linear equations $x-y+2z=1; 2y-3z=1; 3x-2y+4z=3$.
Key:
Sol:
Sol:
#887
Mathematics
Three Dimensional Geometry
LA
APPLY
2023
Competency
5 Marks
Find the equations of all the sides of the parallelogram ABCD whose vertices are $A(4,7,8), B(2,3,4), C(-1,-2,1)$ and $D(1,2,5)$. Also, find the coordinates of the foot of the perpendicular from A to CD.
Key:
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