ExamSpark Class 12 Mathematics

Mock Test 09 Performance Solutions

Subject: CBSE Class 12 Mathematics (NCERT Full-Syllabus Practice)

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Detailed Question Analysis

Q1. Let \(A\) and \(B\) be symmetric matrices of the same order. Then the matrix \((AB - BA)\) is a:

Correct Answer: Option B (Skew-symmetric matrix)

Concept Explanation: \((AB - BA)' = (AB)' - (BA)' = B'A' - A'B'\). Since \(A\) and \(B\) are symmetric (\(A'=A, B'=B\)), this becomes \(BA - AB = -(AB - BA)\). A matrix whose transpose is its negative is skew-symmetric.

Q2. Let \(R\) be a relation on the set \(N\) of natural numbers defined by \(n R m\) if \(n\) divides \(m\). Then relation \(R\) is:

Correct Answer: Option D (Reflexive, transitive but not symmetric)

Concept Explanation: Every number divides itself, so \(n R n\) (Reflexive). If \(n\) divides \(m\) and \(m\) divides \(p\), then \(n\) divides \(p\) (Transitive). However, \(2\) divides \(4\), but \(4\) does not divide \(2\), so it is not symmetric.

Q3. The domain of the function \(f(x) = \sin^{-1}(\sqrt{x - 1})\) is:

Correct Answer: Option A (\([1, 2]\))

Concept Explanation: For \(\sin^{-1}(y)\), the domain is \(-1 \le y \le 1\). So, \(-1 \le \sqrt{x - 1} \le 1\). Since a principal square root is always non-negative, \(0 \le \sqrt{x - 1} \le 1\). Squaring gives \(0 \le x - 1 \le 1\), which simplifies to \(1 \le x \le 2\).

Q4. If \(A\) is a square matrix of order 3 such that \(|A| = 5\), then the value of \(|adj(adj A)|\) is:

Correct Answer: Option C (\(625\))

Concept Explanation: The standard formula is \(|adj(adj A)| = |A|^{(n-1)^2}\). Here, the order \(n=3\). So, the value is \(|A|^{(3-1)^2} = |A|^4 = 5^4 = 625\).

Q5. If \(\vec{a}\) and \(\vec{b}\) are two unit vectors such that \(\vec{a} + \vec{b}\) is also a unit vector, then the angle between \(\vec{a}\) and \(\vec{b}\) is:

Correct Answer: Option C (\(\frac{2\pi}{3}\))

Concept Explanation: Given \(|\vec{a}+\vec{b}| = 1\). Squaring both sides: \(|\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos\theta = 1\). Substituting unit vector values (\(1\)): \(1 + 1 + 2(1)(1)\cos\theta = 1 \implies \cos\theta = -\frac{1}{2} \implies \theta = \frac{2\pi}{3}\).

Q6. A water tank has the shape of an inverted right circular cone. Water is poured into it at a constant rate. How does the rate of change of the water level behave as the tank fills?

Correct Answer: Option C (It decreases continuously.)

Concept Explanation: Volume \(V = \frac{1}{3}\pi r^2 h\). As the height \(h\) increases, the cross-sectional area gets wider. To maintain a constant volume rate (\(\frac{dV}{dt}\)), the rate at which the height increases (\(\frac{dh}{dt}\)) must decrease because more water is needed to raise the level by the same height.

Q7. Three schools want to award their students for honesty, punctuality, and obedience. If we represent the prize amounts as a \(3 \times 1\) column matrix \(X\) and the number of students per category as a \(3 \times 3\) matrix \(P\), then the product \(PX\) represents:

Correct Answer: Option B (Total amount spent by each school on all awards.)

Concept Explanation: Matrix \(P\) (Schools \(\times\) Categories) multiplied by Matrix \(X\) (Categories \(\times\) 1) yields a new matrix of order (Schools \(\times\) 1). This resulting column matrix explicitly shows the total money spent by each individual school.

Q8. The population of a city grows at a rate proportional to its current population. If this scenario is modeled mathematically by a differential equation, what is the order and degree of this equation?

Correct Answer: Option A (Order 1, Degree 1)

Concept Explanation: The mathematical model is \(\frac{dP}{dt} = kP\). This involves only the first derivative with a power of one. Therefore, it is a first-order, first-degree linear ordinary differential equation.

Q9. Geometrically, the definite integral \(\int_0^2 |x - 1| dx\) represents:

Correct Answer: Option A (The algebraic sum of areas of two triangles above the x-axis.)

Concept Explanation: The modulus function \(|x - 1|\) is strictly non-negative. Its graph forms two identical \(V\)-shaped triangles resting on the x-axis from \(x=0\) to \(1\), and \(x=1\) to \(2\). The integral calculates the combined positive area of these two triangles.

Q10. A medical test for a rare disease is 99% accurate. However, only 0.1% of the population actually has the disease. If a person tests positive, the probability they actually have the disease is surprisingly low. Based on Bayes' Theorem logic, why does this happen?

Correct Answer: Option B (The very low base rate means the 1% false positives from healthy people vastly outnumber the true positives.)

Concept Explanation: This is a classic application of conditional probability and the Base Rate Fallacy. Since 99.9% of people are healthy, a 1% error rate on that massive group produces far more false positives than the 99% accurate true positives found in the tiny 0.1% infected group.

Q11. A factory makes tennis rackets (\(x\)) and cricket bats (\(y\)). A racket takes 1.5 hours of machine time, and a bat takes 3 hours. The factory has a maximum of 42 hours of machine time available. To maximize profit, this is formulated as an LPP . The correct machine time constraint is:

Correct Answer: Option A (\(1.5x + 3y \le 42\))

Concept Explanation: Total machine time consumed by \(x\) rackets and \(y\) bats is \(1.5x + 3y\). The word "maximum" implies the usage cannot exceed the available hours, forming the inequality \(1.5x + 3y \le 42\).

Q12. A company designs a cylindrical can to hold 1 liter (\(1000 \text{ cm}^3\)) of oil. To minimize manufacturing cost, they must minimize the total surface area. What relation between height (\(h\)) and radius (\(r\)) yields the most economical can?

Correct Answer: Option B (\(h = 2r\))

Concept Explanation: Surface area \(S = 2\pi r^2 + 2\pi rh\). Volume \(V = \pi r^2 h = 1000 \implies h = \frac{1000}{\pi r^2}\). Substituting \(h\) into \(S\) and minimizing by setting \(\frac{dS}{dr}=0\) yields \(2\pi r^3 = 1000\). Substituting this back gives \(h = 2r\) (height equals diameter).

Q13. An insurance company insured 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers. The probability of them having an accident is 0.01, 0.03, and 0.15 respectively. If an insured person meets with an accident, finding the probability that they are a scooter driver directly applies which concept?

Correct Answer: Option C (Bayes' Theorem)

Concept Explanation: We are given the final outcome (an accident has occurred) and need to calculate the "reverse probability" of a specific prior cause (it was a scooter driver). This scenario is the defining use-case of Bayes' Theorem.

Q14. Assertion (A): The function \(f(x) = |x|\) is continuous at \(x = 0\) but not differentiable at \(x = 0\). Reason (R): A function is differentiable at a point only if it is continuous at that point, but the converse is not always true.

Correct Answer: Option B (Both A and R are true but R is not the correct explanation of A)

Concept Explanation: Both statements are mathematically true. However, R is just a general theorem. The specific reason A is true is because at \(x=0\), the Left Hand Derivative is \(-1\) and the Right Hand Derivative is \(1\) (\(LHD \neq RHD\)), causing a sharp corner.

Q15. Assertion (A): The lines \(\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}\) and \(\frac{x-3}{1} = \frac{y-k}{2} = \frac{z}{1}\) intersect if \(k = \frac{9}{2}\). Reason (R): Two lines intersect if the shortest distance between them is zero.

Correct Answer: Option A (Both A and R are true and R is the correct explanation of A)

Concept Explanation: Two non-parallel lines in 3D space intersect if and only if their shortest distance is zero. Applying the coplanarity determinant \((\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0\) yields \(k = \frac{9}{2}\). R perfectly explains the method to prove A.

Q16. Case Study Passage (Questions 16 & 17) A drone is flying in a 3D coordinate space. Its position vector at any time \(t\) (in seconds) is given by \(\vec{r} = (2t)\hat{i} + (t^2)\hat{j} + (3)\hat{k}\). The control tower is situated at the origin \((0,0,0)\).

What is the mathematical path followed by the drone's shadow on the XY-plane?

Correct Answer: Option C (A parabola)

Concept Explanation: On the XY-plane, we only look at \(x\) and \(y\) components. \(x = 2t \implies t = \frac{x}{2}\). The y- coordinate is \(y = t^2\). Substituting \(t\) gives \(y = (\frac{x}{2})^2 = \frac{x^2}{4}\), which represents a parabola.

Q17. Case Study Passage (Questions 16 & 17) A drone is flying in a 3D coordinate space. Its position vector at any time \(t\) (in seconds) is given by \(\vec{r} = (2t)\hat{i} + (t^2)\hat{j} + (3)\hat{k}\). The control tower is situated at the origin \((0,0,0)\).

At \(t=1\), what is the straight-line distance of the drone from the control tower?

Correct Answer: Option A (\(\sqrt{14}\) units)

Concept Explanation: At \(t=1\), substituting \(1\) into the position vector gives \(\vec{r} = 2\hat{i} + 1\hat{j} + 3\hat{k}\). The distance from the origin \((0,0,0)\) is the magnitude \(|\vec{r}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}\).

Q18. The value of \(\int_{-1}^1 (x^3 + x\cos x + \tan^5 x + 1) dx\) is:

Correct Answer: Option B (\(2\))

Concept Explanation: Let \(f(x) = x^3 + x\cos x + \tan^5 x\). Note that \(f(-x) = -f(x)\), making it an odd function. Thus, \(\int_{- 1}^1 f(x) dx = 0\). The integral simplifies to \(\int_{-1}^1 1 dx = [x]_{-1}^1 = 1 - (-1) = 2\).

Q19. The integrating factor (I.F.) of the differential equation \(x\frac{dy}{dx} + 2y = x^2\) (\(x \neq 0\)) is:

Correct Answer: Option B (\(x^2\))

Concept Explanation: Divide the equation by \(x\) to get standard linear form: \(\frac{dy}{dx} + (\frac{2}{x})y = x\). Here \(P = \frac{2}{x}\). The Integrating Factor is \(e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2\ln x} = e^{\ln(x^2)} = x^2\).

Q20. Two independent events \(A\) and \(B\) are such that \(P(A) = 0.3\) and \(P(A \cup B) = 0.6\). The value of \(P(B)\) is:

Correct Answer: Option D (\(\frac{3}{7}\))

Concept Explanation: For independent events, \(P(A \cap B) = P(A)P(B)\). Using the addition theorem: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). Substituting values: \(0.6 = 0.3 + P(B) - 0.3P(B) \implies 0.3 = 0.7P(B) \implies P(B) = \frac{0.3}{0.7} = \frac{3}{7}\). (Option A is a trap for mutually exclusive events).

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