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Standalone Questions
#977
Mathematics
Relations and Functions
ASSERTION_REASON
APPLY
2025
AISSCE(Board Exam)
Competency
1 Marks
Assertion (A): Let $f(x) = e^{x}$ and $g(x) = \log x$. Then $(f + g)x = e^{x} + \log x$ where domain of $(f + g)$ is $\mathbb{R}$.
Reason (R): $\text{Dom}(f + g) = \text{Dom}(f) \cap \text{Dom}(g)$.
Key: D
Sol:
Sol:
#976
Mathematics
Vector Algebra
ASSERTION_REASON
APPLY
2025
AISSCE(Board Exam)
Competency
1 Marks
Assertion (A) : If $|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=256$ and $|\vec{b}|=8$, then $|\vec{a}|=2$.
Reason (R) : $\sin ^{2} \theta+\cos ^{2} \theta=1$ and $|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta$ and $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$.
Key: A
Sol:
Sol:
#975
Mathematics
Inverse Trigonometric Functions
ASSERTION_REASON
REMEMBER
2025
AISSCE(Board Exam)
Competency
1 Marks
Assertion (A) : Set of values of $\sec^{-1}\left(\frac{\sqrt{3}}{2}\right)$ is a null set.
Reason (R) : $\sec^{-1}$ x is defined for $x \in \mathbb{R}-(-1, 1)$.
Key: A
Sol:
Sol:
#974
Mathematics
Continuity and Differentiability
ASSERTION_REASON
REMEMBER
2025
AISSCE(Board Exam)
Competency
1 Marks
Assertion (A): $f(x) = \begin{cases} x\sin\frac{1}{x}, & x\neq 0 \\ 0, & x=0 \end{cases}$ is continuous at $x=0$.
Reason (R): When $x \to 0$, $\sin\frac{1}{x}$ is a finite value between $-1$ and $1$.
Key:
Sol:
Sol:
#971
Mathematics
Linear Programming
MCQ_SINGLE
APPLY
2025
AISSCE(Board Exam)
Competency
1 Marks
In a Linear Programming Problem (LPP), the objective function $Z = 2x + 5y$ is to be maximised under the following constraints :$x + y \leq 4$, $3x + 3y \geq 18$, $x, y \geq 0$
Study the graph and select the correct option.
(A) lies in the shaded unbounded region.
(B) lies in $\triangle AOB$.
(C) does not exist.
(D) lies in the combined region of $\triangle AOB$ and unbounded shaded region.
Key: C
Sol:
Sol:
The feasible region of a Linear Programming Problem is the set of points that satisfy all the given constraints simultaneously.Constraint 1 requires the points to be in the region where the sum of $x$ and $y$ is less than or equal to 4.Constraint 2 requires the points to be in the region where the sum of $x$ and $y$ is greater than or equal to 6.Mathematically, a number cannot be both $\le 4$ and $\ge 6$ at the same time. Visually, looking at the graph, there is a clear gap between the shaded region $\Delta AOB$ and the shaded unbounded region above $PQ$. The two regions do not overlap.3. ConclusionSince there is no common region that satisfies all constraints, the feasible region is an empty set.Without a feasible region, there are no valid values for $x$ and $y$ to substitute into the objective function $Z$. Therefore, an optimal solution (maximum value) cannot be found.Answer:The correct option is (C) does not exist.
#970
Mathematics
Linear Programming
MCQ_SINGLE
APPLY
2025
AISSCE(Board Exam)
Competency
1 Marks
For a Linear Programming Problem (LPP), the given objective function is $Z = x + 2y$. The feasible region PQRS determined by the set of constraints is shown as a shaded region in the graph.Which of the following statements is correct ?
(A) Z is minimum at $(\frac{18}{7}, \frac{2}{7})$
(B) Z is maximum at R$(\frac{7}{2}, \frac{3}{4})$
(C) (Value of Z at P) > (Value of Z at Q)
(D) (Value of Z at Q) < (Value of Z at R)
Key:
Sol:
Sol:
#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:
#932
Mathematics
Applications of Derivatives
VSA
APPLY
2023
Competency
2 Marks
If \(f(x)=a(\tan x-\cot x)\), where \(a>0\), then find whether \(f(x)\) is increasing or decreasing function in its domain.
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:
Case-Based Questions
CASE ID: #118
Cl: CBSE Class 12
Mathematics
A shop selling electronic items sells smartphones of only three reputed companies A, B and C because chances of their manufacturing a defective smartphone are only 5%, 4% and 2% respectively. In his inventory he has 25% smartphones from company A, 35% smartphones from company B and 40% smartphones from company C.
SUBJECTIVE
APPLY
2025
AISSCE(Board Exam)
Competency
4 Marks
A person buys a smartphone from this shop.
(i) Find the probability that it was defective.
(ii) What is the probability that this defective smartphone was manufactured by company B ?
(i) Find the probability that it was defective.
(ii) What is the probability that this defective smartphone was manufactured by company B ?
Key:
Sol:
Sol:
CASE ID: #117
Cl: CBSE Class 12
Mathematics
Three students, Neha, Rani and Sam go to a market to purchase stationery items. Neha buys 4 pens, 3 notepads and 2 erasers and pays ₹ 60. Rani buys 2 pens, 4 notepads and 6 erasers for ₹ 90. Sam pays ₹ 70 for 6 pens, 2 notepads and 3 erasers.
SUBJECTIVE
APPLY
2025
AISSCE(Board Exam)
Competency
4 Marks
(i) Form the equations required to solve the problem of finding the price of each item, and express it in the matrix form $AX = B$.
(ii) Find $|A|$ and confirm if it is possible to find $A^{-1}$.
(iii) (a) Find $A^{-1}$, if possible, and write the formula to find $X$.
OR
(iii) (b) Find $A^2 - 8I$, where $I$ is an identity matrix.
(ii) Find $|A|$ and confirm if it is possible to find $A^{-1}$.
(iii) (a) Find $A^{-1}$, if possible, and write the formula to find $X$.
OR
(iii) (b) Find $A^2 - 8I$, where $I$ is an identity matrix.
Key:
Sol:
Sol:
CASE ID: #116
Cl: CBSE Class 12
Mathematics
Based upon the results of regular medical check-ups in a hospital, it was found that out of 1000 people, 700 were very healthy, 200 maintained average health and 100 had a poor health record.
Let $A_1$: People with good health,
$A_2$: People with average health,
and $A_3$: People with poor health.
During a pandemic, the data expressed that the chances of people contracting the disease from category $A_1$, $A_2$ and $A_3$ are 25%, 35% and 50%, respectively.
SUBJECTIVE
APPLY
2025
AISSCE(Board Exam)
Competency
4 Marks
(i) A person was tested randomly. What is the probability that he/she has contracted the disease ?
(ii) Given that the person has not contracted the disease, what is the probability that the person is from category $A_2$ ?
(ii) Given that the person has not contracted the disease, what is the probability that the person is from category $A_2$ ?
Key:
Sol:
Sol:
Based on the information provided, here are the step-by-step calculations:Let the events be defined as:$A_1$: Person has good health$A_2$: Person has average health$A_3$: Person has poor health$E$: Person contracts the diseaseFrom the data given:$P(A_1) = \frac{700}{1000} = 0.7$$P(A_2) = \frac{200}{1000} = 0.2$$P(A_3) = \frac{100}{1000} = 0.1$The conditional probabilities of contracting the disease are:$P(E|A_1) = 25\% = 0.25$$P(E|A_2) = 35\% = 0.35$$P(E|A_3) = 50\% = 0.50$(i) Probability that a randomly selected person has contracted the diseaseWe use the Law of Total Probability:$$P(E) = P(E|A_1)P(A_1) + P(E|A_2)P(A_2) + P(E|A_3)P(A_3)$$Substitute the values:$$P(E) = (0.25 \times 0.7) + (0.35 \times 0.2) + (0.50 \times 0.1)$$$$P(E) = 0.175 + 0.07 + 0.05$$$$P(E) = 0.295$$Answer: The probability that the person has contracted the disease is 0.295 (or 29.5%).(ii) Probability that the person is from category $A_2$, given that they have not contracted the diseaseLet $E'$ be the event that the person has not contracted the disease.We need to find $P(A_2|E')$.First, calculate the probability of not contracting the disease, $P(E')$:$$P(E') = 1 - P(E) = 1 - 0.295 = 0.705$$Next, calculate the probability of not contracting the disease given the person is in category $A_2$:$$P(E'|A_2) = 1 - P(E|A_2) = 1 - 0.35 = 0.65$$Now, apply Bayes' Theorem:$$P(A_2|E') = \frac{P(E'|A_2) \cdot P(A_2)}{P(E')}$$Substitute the values:$$P(A_2|E') = \frac{0.65 \times 0.2}{0.705}$$$$P(A_2|E') = \frac{0.13}{0.705}$$$$P(A_2|E') = \frac{130}{705} = \frac{26}{141}$$$$P(A_2|E') \approx 0.1844$$Answer: The probability that the person is from category $A_2$ given they have not contracted the disease is $\frac{26}{141}$ or approximately 0.184.
CASE ID: #115
Cl: CBSE Class 12
Mathematics
Camphor is a waxy, colourless solid with strong aroma that evaporates through the process of sublimation, if left in the open at room temperature.
A cylindrical camphor tablet whose height is equal to its radius (r) evaporates when exposed to air such that that the rate of reduction of its volume is proportional to its total surface area. Thus, $\frac{dV}{dt} = kS$ is the differential equation, where V is the volume, S is the surface area and t is the time in hours.
SUBJECTIVE
REMEMBER
2025
AISSCE(Board Exam)
Competency
1 Marks
(i) Write the order and degree of the given differential equation.
(ii) Substituting $V = \pi r^3$ and $S = 2\pi r^2$, we get the differential equation $\frac{dr}{dt} = \frac{2}{3}k$. Solve it, given that $r(0) = 5$ mm.
(iii) (a) If it is given that $r = 3$ mm when $t = 1$ hour, find the value of k. Hence, find t for $r = 0$ mm.
OR
(iii) (b) If it is given that $r = 1$ mm when $t = 1$ hour, find the value of k. Hence, find t for $r = 0$ mm.
(ii) Substituting $V = \pi r^3$ and $S = 2\pi r^2$, we get the differential equation $\frac{dr}{dt} = \frac{2}{3}k$. Solve it, given that $r(0) = 5$ mm.
(iii) (a) If it is given that $r = 3$ mm when $t = 1$ hour, find the value of k. Hence, find t for $r = 0$ mm.
OR
(iii) (b) If it is given that $r = 1$ mm when $t = 1$ hour, find the value of k. Hence, find t for $r = 0$ mm.
Key:
Sol:
Sol:
CASE ID: #109
Cl: CBSE Class 12
Mathematics
A technical company is designing a rectangular solar panel installation on a roof using 300 metres of boundary material. The design includes a partition running parallel to one of the sides dividing the area (roof) into two sections.
Let the length of the side perpendicular to the partition be $x$ metres and with parallel to the partition be $y$ metres.,
SUBJECTIVE
APPLY
2025
AISSCE(Board Exam)
Competency
1 Marks
Write the equation for the total boundary material used in the boundary and parallel to the partition in terms of
x and y..
x and y..
Key:
Sol:
Sol:
k
SUBJECTIVE
APPLY
2025
AISSCE(Board Exam)
Competency
1 Marks
Write the area of the solar panel as a function of $x$
Key:
Sol:
Sol:
SUBJECTIVE
APPLY
2025
AISSCE(Board Exam)
Competency
1 Marks
Find the critical points of the area function. Use second derivative test to determine critical points at the maximum area. Also, find the maximum area.
Key:
Sol:
Sol:
SUBJECTIVE
APPLY
2025
AISSCE(Board Exam)
Competency
2 Marks
Using first derivative test, calculate the maximum area the company can enclose with the 300 metres of boundary material, considering the parallel partition.
Key:
Sol:
Sol:
CASE ID: #108
Cl: CBSE Class 12
Mathematics
A bank offers loan to its customers on different types of interest namely, fixed rate, floating rate and variable rate. From the past data with the bank, it is known that a customer avails loan on fixed rate, floating rate or variable rate with probabilities 10%, 20% and 70% respectively. A customer after availing loan can pay the loan or default on loan repayment. The bank data suggests that the probability that a person defaults on loan after availing it at fixed rate, floating rate and variable rate is 5%, 3% and 1% respectively.
VSA
APPLY
2025
AISSCE(Board Exam)
Competency
2 Marks
What is the probability that a customer after availing the loan will default on the loan repayment?
Key:
Sol:
Sol:
VSA
APPLY
2025
AISSCE(Board Exam)
Competency
2 Marks
A customer after availing the loan, defaults on loan repayment. What is the probability that he availed the loan at a variable rate of interest?
Key:
Sol:
Sol: