Step 1: Understanding the Concept:
A row-matrix is a matrix with only one horizontal row of elements. The order of a matrix is given as (number of rows) × (number of columns).

Step 2: Key Formula or Approach:
A matrix qualifies as a row-matrix if its order is of the form 1 × n, where n ≥ 1.

Step 3: Detailed Explanation:

(A) 2 × 1: Contains 2 rows and 1 column → more than one row → a column-matrix, not a row-matrix.
(B) 1 × 2: Contains 1 row and 2 columns → satisfies row-matrix definition.
(C) 1 × 1: Single row and single column → can be considered a row-matrix (and also a column-matrix).
(D) 1 × n: General form of a row-matrix with n columns.

Step 4: Final Answer:
The order 2 × 1 cannot represent a row-matrix. Option (A) is correct.

3. Properties of Transpose of Matrices

Question:
Which of the following properties is/are true for two matrices of suitable orders?

(A + B)′ = A′ + B′
(A − B)′ = B′ − A′
(AB)′ = A′B′
(kAB)′ = kB′A′ (k is a scalar)

(A) (i) only
(B) (i), (ii) and (iii)
(C) (i) and (iii)
(D) (i) and (iv)

Correct Answer: (D) (i) and (iv)

Solution:

Step 1: Understanding the Concept:
The transpose of a matrix, denoted A′, is obtained by interchanging rows and columns. Standard properties of transpose include addition, subtraction, multiplication, and scalar multiplication rules.

Step 2: Key Formulas:

(A + B)′ = A′ + B′
(A − B)′ = A′ − B′
(AB)′ = B′A′
(kA)′ = kA′

Step 3: Detailed Explanation:

(i) (A + B)′ = A′ + B′ → Correct, follows addition property.
(ii) (A − B)′ = B′ − A′ → Incorrect; correct formula: (A − B)′ = A′ − B′.
(iii) (AB)′ = A′B′ → Incorrect; correct formula: (AB)′ = B′A′ (order reverses).
(iv) (kAB)′ = kB′A′ → Correct; scalar k remains, and transpose of product reverses order.

Step 4: Final Answer:
Statements (i) and (iv) are true. Option (D) is correct.

4. Relation Between Two Determinants

Question:
Find the relation between ∆1 and ∆2:

(A) ∆1 = 2∆2
(B) ∆2 = −2∆1
(C) ∆1 = ∆2
(D) ∆2 = −∆1

Correct Answer: (B) ∆2 = −2∆1

Solution:

Step 1: Understanding the Concept:
The determinant of a matrix can be calculated using expansion or determinant properties. Key points:

For a diagonal matrix, the determinant equals the product of the diagonal elements.
If two rows or columns are interchanged, the determinant changes sign.

Step 2: Key Formula or Approach:

For a 3 × 3 diagonal matrix D = diag(a, b, c), |D| = abc
If rows Ri and Rj are swapped → New determinant = −(Original determinant)

Step 3: Detailed Explanation:

Evaluate ∆1 (diagonal matrix): ∆1 = 1 × 2 × 3 = 6
Evaluate ∆2 by expanding along the third row (contains two zeros):
Compute the 2 × 2 determinant part: |0 2; 1 0| = (0×0 − 2×1) = −2
Then, ∆2 = 6 × (−2) = −12
Compare: ∆2 / ∆1 = −12 / 6 = −2 → ∆2 = −2∆1

Step 4: Final Answer:
The correct relation is ∆2 = −2∆1, which is option (B).

5. Value of x for a 2×2 Determinant Equation

Question:
Find one value of x for which:

| cos x sin x |
| −cos x sin x | = 1

(A) 0
(B) π/4
(C) π/3
(D) π/2

Correct Answer: (B) π/4

Solution:

Step 1: Understanding the Concept:
For a 2×2 matrix |a b; c d|, the determinant is ad − bc. Expanding gives a trigonometric equation.

Step 2: Key Formula or Approach:

Determinant expansion: (cos x)(sin x) − (sin x)(−cos x)
Trigonometric identity: 2 sin x cos x = sin(2x)

Step 3: Detailed Explanation:

Expand determinant: (cos x)(sin x) − (sin x)(−cos x) = sin x cos x + sin x cos x = 2 sin x cos x
Set equal to 1: 2 sin x cos x = 1 → sin(2x) = 1
Sine equals 1 at angles 2x = π/2, 5π/2, …
Principal solution: 2x = π/2 → x = π/4

Step 4: Final Answer:
A value of x that satisfies the equation is x = π/4, option (B).

6. Skew-Symmetric Matrices

Question:
If A and B are skew-symmetric matrices of the same order, which of the following matrices is also skew-symmetric?

(A) AB
(B) AB + BA
(C) (A + B)²
(D) A − B

Correct Answer: (D) A − B

Solution:

Step 1: Understanding the Concept:
A matrix M is skew-symmetric if its transpose satisfies M′ = −M. Since A and B are skew-symmetric:

A′ = −A
B′ = −B

Step 2: Key Formulas:

(X ± Y)′ = X′ ± Y′
(XY)′ = Y′ X′

Step 3: Examine Each Option:

(A) (AB)′ = B′A′ = (−B)(−A) = BA ≠ −AB → Not skew-symmetric
(B) (AB + BA)′ = B′A′ + A′B′ = (−B)(−A) + (−A)(−B) = AB + BA → Symmetric, not skew-symmetric
(C) ((A + B)²)′ = ((A + B)(A + B))′ = (A + B)′(A + B)′ = (−A − B)(−A − B) = (A + B)² → Symmetric
(D) (A − B)′ = A′ − B′ = (−A) − (−B) = −A + B = −(A − B) → Skew-symmetric

Step 4: Final Answer:
The matrix A − B is skew-symmetric. Option (D) is correct.

7. Least Value of f(x) = x³ − 12x on [0, 3]

Question:
Find the least value of f(x) = x³ − 12x for x ∈ [0, 3].

(A) −16
(B) −9
(C) 0
(D) 16

Correct Answer: (A) −16

Solution:

Step 1: Understanding the Concept:
The absolute minimum of a continuous function over a closed interval [a, b] occurs at critical points inside the interval or at the endpoints.

Step 2: Key Formula / Approach:

Find f′(x) and solve f′(x) = 0 to locate critical points.
Evaluate f(x) at all critical points and at the endpoints a and b.

Step 3: Detailed Explanation:
Function: f(x) = x³ − 12x, x ∈ [0, 3]

Derivative: f′(x) = 3x² − 12
Set derivative to zero: 3x² − 12 = 0 → x² = 4 → x = ±2
Only x = 2 lies in [0, 3]; discard x = −2

Step 4: Evaluate f(x) at relevant points:

f(0) = 0³ − 12·0 = 0
f(2) = 2³ − 12·2 = 8 − 24 = −16
f(3) = 3³ − 12·3 = 27 − 36 = −9

Comparing values 0, −16, −9, the least value is −16.

Step 5: Final Answer:
The least value of f(x) on [0, 3] is −16, option (A).

8. Value of A in the Integral

Question:
Evaluate A if:

∫ 3ax / (b² + c²x²) dx = A log |b² + c²x²| + K

(A) 3a
(B) 3a²b²
(C) 3a / (b²c²)
(D) 3a / (2c²)

Correct Answer: (D) 3a / 2c²

Solution:

Step 1: Understanding the Concept:
The integral can be solved using substitution. The numerator is proportional to the derivative of the denominator, suggesting a logarithmic integration.

Step 2: Key Formula:
∫ f′(x)/f(x) dx = log|f(x)| + C

Step 3: Detailed Explanation:

Let I = ∫ 3ax / (b² + c²x²) dx
Set u = b² + c²x² → du = 2c²x dx → x dx = du / 2c²
Substitute: I = ∫ (3a / u) · (du / 2c²) = (3a / 2c²) ∫ (1/u) du
Integrate: I = (3a / 2c²) log |u| + K
Back-substitute u = b² + c²x²: I = (3a / 2c²) log |b² + c²x²| + K

Comparing with the given form A log|b² + c²x²| + K, we find:

A = 3a / 2c²

Step 4: Final Answer:
A = 3a / 2c², option (D).

9. Definite Integral of an Odd Function

Question:
Evaluate:

∫₋₁¹ x³ / (x² + 2|x| + 1) dx

(A) 0
(B) log 2
(C) 2 log 2
(D) 1/2 log 2

Correct Answer: (A) 0

Solution:

Step 1: Understanding the Concept:
For a definite integral over a symmetric interval [−a, a], check if the integrand is even or odd:

Step 2: Key Property:
∫_−a^a f(x) dx = 0 if f(x) is odd, i.e., f(−x) = −f(x)

Step 3: Detailed Explanation:

Let f(x) = x³ / (x² + 2|x| + 1)
Evaluate f(−x):

f(−x) = (−x)³ / ((−x)² + 2|−x| + 1) = −x³ / (x² + 2|x| + 1) = −f(x)

Since f(−x) = −f(x), f(x) is an odd function.
Integral over symmetric interval [−1, 1] → ∫₋₁¹ f(x) dx = 0

Step 4: Final Answer:
∫₋₁¹ x³ / (x² + 2|x| + 1) dx = 0, option (A).

10. Area Bounded by y = x|x|, x-axis, and x = −1 to x = 1

Question:
Find the area bounded by the curve y = x|x|, the x-axis, and the ordinates x = −1 and x = 1.

(A) 0
(B) 1/3
(C) 2/3
(D) 4/3

Correct Answer: (C) 2/3

Solution:

Step 1: Understanding the Concept:
The area under a curve between x = a and x = b is given by:

Area = ∫_a^b |y| dx

Since area is non-negative, take absolute value of y.

Step 2: Split the Function Based on x:
y = x|x| →

For x ≥ 0: |x| = x → y = x·x = x²
For x < 0: |x| = −x → y = x(−x) = −x²

Step 3: Set Up the Integral:

Area = ∫₋₁⁰ |−x²| dx + ∫₀¹ |x²| dx = ∫₋₁⁰ x² dx + ∫₀¹ x² dx

Step 4: Evaluate the Integrals:

∫₋₁⁰ x² dx = [x³ / 3]₋₁⁰ = 0 − (−1/3) = 1/3
∫₀¹ x² dx = [x³ / 3]₀¹ = 1/3 − 0 = 1/3

Total Area = 1/3 + 1/3 = 2/3

Step 5: Final Answer:
The area bounded by the curve and the x-axis is 2/3 square units, option (C).

11. The integrating factor of differential equation R dx dy + P x = Q where P, Q, R are functions of y is (A) e ´ P Q dy (B) e ´ P dy (C) e ´ P R dy (D) e ´ P R dx Correct Answer: (C) e ´ P R dy Solution: Step 1: Understanding the Concept: A first-order linear differential equation in x (with x as the dependent variable) is written in the standard form dx dy + h(y)x = g(y). Step 2: Key Formula or Approach: For a differential equation in this form, the integrating factor (IF) is IF = e ´ h(y) dy . Step 3: Detailed Explanation: The given differential equation is R dx dy + P x = Q. To convert it into the standard form, divide the entire equation by R: dx dy + P R x = Q R . Comparing this with dx dy + h(y)x = g(y), we identify h(y) = P R . 14 Hence the integrating factor becomes IF = e ´ h(y) dy = e ´ P R dy . Thus the integrating factor is e ´ P R dy . Step 4: Final Answer: Integrating Factor = e ´ P R dy .

13. The value of p for which vectors ˆi + 2ˆj + 3ˆk and 2ˆi − pˆj + ˆk are perpendicular to each other is

(A) 0
(B) 1
(C) 5/2
(D) −5/2

Answer is as follow: (C) 5/2

Solution:

Two non-zero vectors are perpendicular (orthogonal) if and only if their scalar (dot) product is zero. Let ⃗a = 1ˆi + 2ˆj + 3ˆk and ⃗b = 2ˆi − pˆj + 1ˆk. Then,

⃗a · ⃗b = (1)(2) + (2)(−p) + (3)(1) = 0

2 − 2p + 3 = 0 → 5 − 2p = 0 → 2p = 5 → p = 5/2

14. The value of m for which the points with position vectors −ˆi−ˆj +2ˆk, 2ˆi+mˆj +5ˆk and 3ˆi + 11ˆj + 6ˆk are collinear is

(A) 8
(B) −8
(C) 2
(D) 5/2

Answer is as follow: (A) 8

Solution:

Three points A, B, C are collinear if the vectors AB⃗ and BC⃗ are parallel. Parallel vectors have proportional direction ratios.

Let A(−1, −1, 2), B(2, m, 5), and C(3, 11, 6).

AB⃗ = B − A = (2 − (−1))ˆi + (m − (−1))ˆj + (5 − 2)ˆk = 3ˆi + (m + 1)ˆj + 3ˆk

BC⃗ = C − B = (3 − 2)ˆi + (11 − m)ˆj + (6 − 5)ˆk = 1ˆi + (11 − m)ˆj + 1ˆk

For collinearity, component ratios must be equal:

3/1 = (m + 1)/(11 − m) = 3/1

3(11 − m) = m + 1 → 33 − 3m = m + 1 → 32 = 4m → m = 8

Hence, the points are collinear for m = 8.

15. If |⃗a| = 8, |⃗b| = 3 and |⃗a ×⃗b| = 12, then the value of |⃗a · ⃗b| is

(A) 6√3
(B) 8√3
(C) 12√3
(D) 3√12

Answer is as follow: (C) 12√3

Solution:

The magnitudes of the dot product and cross product are related through the angle between the vectors by Lagrange’s identity:

|⃗a · ⃗b|² + |⃗a × ⃗b|² = |⃗a|² |⃗b|²

Let D = |⃗a · ⃗b| and C = |⃗a × ⃗b|. We are given |⃗a| = 8, |⃗b| = 3, and C = 12.

Substitute into the identity:

D² + 12² = 8² × 3² → D² + 144 = 64 × 9 → D² + 144 = 576 → D² = 432

Take the square root:

D = √432 = √(144 × 3) = 12√3

Hence, |⃗a · ⃗b| = 12√3.

16. The length of perpendicular drawn from point (2, 5, 7) on line x/1 = y/0 = z/0 is

(A) 2
(B) 5
(C) √74
(D) √78

Answer is as follow: (C) √74

Solution:

The given line has direction ratios (1, 0, 0), so it is parallel to the x-axis and passes through (0, 0, 0).

The perpendicular distance of a point (x1, y1, z1) from the x-axis is:

d = √(y1² + z1²)

Point P = (2, 5, 7). The foot of the perpendicular on the x-axis is M = (2, 0, 0).

Distance d = √((2 − 2)² + (5 − 0)² + (7 − 0)²) = √(0 + 25 + 49) = √74

Hence, the length of the perpendicular is √74.

17. The feasible region of a linear programming problem with objective function Z = 5x + 7y is shown below: [Triangle with vertices (0,2), (3,4), (7,0)]. The maximum value of Z − minimum value of Z is

(A) 8
(B) 29
(C) 35
(D) 43

Answer is as follow: (D) 43

Solution:

According to the Corner Point Theorem, the optimal values of a linear objective function occur at the vertices of the feasible region.

Calculate Z = 5x + 7y at each corner point:

At (0, 0): Z = 5(0) + 7(0) = 0
At (7, 0): Z = 5(7) + 7(0) = 35
At (3, 4): Z = 5(3) + 7(4) = 15 + 28 = 43
At (0, 2): Z = 5(0) + 7(2) = 14

Maximum Z = 43 (at (3, 4)), Minimum Z = 0 (at (0, 0))

Difference = Max Z − Min Z = 43 − 0 = 43

Hence, the required difference is 43.

18. The degree of an objective function of a linear programming problem is

(A) 0
(B) 1
(C) 2
(D) Any natural number

Answer is as follow: (B) 1

Solution:

Linear programming problems (LPP) are called “linear” because all relationships—constraints and the objective function—are linear equations or inequalities.

The objective function is typically written as Z = ax + by, where variables x and y are to the first power.

The highest power of variables in a linear expression is always 1. Therefore, the degree of any linear objective function is 1.

21(a). Check whether the function f(x) defined as

f(x) = { ( |x−3| / 2(x−3) , x < 3 )
( (x−6) / 6 , x ≥ 3 ) }

is continuous at x = 3 or not?

Answer is as follow: Continuous at x = 3

Solution:

A function is continuous at x = a if lim_x→a− f(x) = lim_x→a+ f(x) = f(a).

For x < 3, |x − 3| = −(x − 3).

Left Hand Limit (LHL):

lim_x→3− f(x) = lim_x→3− (−(x−3) / 2(x−3)) = −1/2 = −0.5

Right Hand Limit (RHL):

lim_x→3+ f(x) = lim_x→3+ ((x−6)/6) = (3−6)/6 = −0.5

Value of the function: f(3) = (3−6)/6 = −0.5

Since LHL = RHL = f(3) = −0.5, the function is continuous at x = 3.

21(b). If √3(x² + y²) = 4xy, then find dy/dx at (1/2, √3/2)

Answer is as follow: dy/dx = √3

Solution:

Since y is not isolated, use implicit differentiation.

Differentiating both sides with respect to x (using the product rule for xy):

√3 (2x + 2y y′) = 4 (x y′ + y)

Divide both sides by 2:

√3 x + √3 y y′ = 2 x y′ + 2 y

Substitute x = 1/2, y = √3/2:

√3(1/2) + √3(√3/2) y′ = 2(1/2) y′ + 2(√3/2)

√3/2 + (3/2) y′ = y′ + √3

Rearranging: (3/2)y′ − y′ = √3 − √3/2 → (1/2)y′ = √3/2 → y′ = √3

Hence, dy/dx = √3 at the given point.

22. A room freshener bottle in the shape of an inverted cone sprays the perfume at a rate such that the volume decreases at 1 mm³/min. Find the rate of level drop when the level is 10 mm and the semi-vertical angle is π/6.

Answer: 3 / 100π mm/min

Solution:

This is a related rates problem. The volume of a cone is related to the height:

V = (1/3) π r² h, and r = h tan α

Given α = π/6 → r = h tan(π/6) = h / √3

Then, V = (1/3) π (h/√3)² h = π h³ / 9

Differentiating both sides with respect to time t:

dV/dt = (π h² / 3) dh/dt

Given dV/dt = −1 and h = 10:

−1 = (π * 100 / 3) dh/dt → dh/dt = −3 / 100π

Hence, the rate of level drop is 3 / 100π mm/min.

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Last Updated on : March 26, 2026