Section-A: Very Short Answer Questions
Each Question carries 2 marks (Total: 20 marks)
1. If A = {0, 1, 2, 3, 4} and f: A → B is a surjection defined by f(x) = cos(x), then find B.
Answer: The function f(x) = cos(x) is defined on the set A = {0, 1, 2, 3, 4}. Evaluating f(x) for each element of A:
- f(0) = cos(0) = 1
- f(1) = cos(1) ≈ 0.5403
- f(2) = cos(2) ≈ -0.4161
- f(3) = cos(3) ≈ -0.9899
- f(4) = cos(4) ≈ -0.6536
2. Find the domain of f(x) = √(x – 4).
Answer:
The function is defined for all x where the expression inside the square root is non-negative:
x – 4 ≥ 0, so x ≥ 4.
Thus, the domain of f(x) is [4, ∞).
3. A certain book shop has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics books. Their selling prices are Rs. 80, Rs. 60, and Rs. 40 each respectively. Find the total amount that the book shop will receive by selling all the books using matrix algebra.
Answer:
We represent the number of books and prices as vectors:
Number of books vector: N = [10, 8, 10] (in dozens)
Price vector: P = [80, 60, 40] (in Rs.)
The total amount is calculated as the dot product of these two vectors:
Total amount = 10×80 + 8×60 + 10×40 = 800 + 480 + 400 = 1680 Rs.
4. If A = [0 1 4; 1 0 7; x 7 0] is a skew-symmetric matrix, then find x.
Answer:
A matrix is skew-symmetric if A^T = -A.
By comparing the elements of A and its transpose, we find that x must be -7 to satisfy the skew-symmetry condition.
Therefore, x = -7.
5. Find the angle between the vectors i + j + 2k and 3i + 2j - k.
Answer:
The angle θ between two vectors is given by:
cos(θ) = (A • B) / (|A| |B|)
After calculating the dot product of the vectors and their magnitudes:
- Dot product:
3 + 2 - 2 = 3 - Magnitude of A:
|A| = √6 - Magnitude of B:
|B| = √14
cos(θ) = 3 / (√6 × √14) = 3 / (2√21). Therefore, θ = cos-1(3 / 2√21).
6. If p = (2, 3) and q = (6, π), then find p × q.
Answer:
The cross product of two 2D vectors p = (p1, p2) and q = (q1, q2) is given by:
p × q = p1 * q2 - p2 * q1
Substituting the values:
p × q = (2)(π) - (3)(6) = 2π - 18
7. If a = i + j - 2k and b = i + j + 3k, then show that a + b and a - b are perpendicular to each other.
Answer:
The vectors a + b = 2i + 2j + k and a - b = -5k are perpendicular if their dot product is zero:
(2i + 2j + k) • (-5k) = 0 + 0 - 5 = -5
Since the dot product is not zero, the vectors a + b and a - b are not perpendicular.
8. If 0 0; 0 0; cos9 sin9, tan θ; cos9 sin9 and ‘θ’ is in Q3, find ‘θ’.
Answer: To solve this, we must use trigonometric identities and know the properties of angles in the third quadrant. The details for solving this specific problem would require more information on how the matrix relates to trigonometric functions. However, since θ is in the third quadrant, we know that sin(θ) < 0 and cos(θ) < 0 in this quadrant. The exact value of θ can be calculated using inverse trigonometric functions, depending on the given equation details.
9. Find the period of the function f(x) = sin(x + 3) + sin(x + 2) + sin(x + 1) + sin(x).
Answer: The period of a sine function is determined by the coefficient inside the sine term. Since all terms have the same period of 2π, the overall period of the function is 2π.
10. If sinh(x) = 5, show that x = log(5 + √26).
Answer:
Using the inverse hyperbolic sine function:
x = sinh-1(5)
By the formula for the inverse of the hyperbolic sine function:
x = log(5 + √(52 + 1)) = log(5 + √26)
Section-B: Short Answer Questions (Each question carries 4 marks)
12. If
A =
[ cosθ, sinθ ]
[ -sinθ, cosθ ]
Show that for all the positive integers n:
An =
[ cos(nθ), sin(nθ) ]
[ -sin(nθ), cos(nθ) ]
Solution:
Using induction:
- Base case: For n = 1, A = [ cosθ, sinθ ]
[ -sinθ, cosθ ] which is true. - Inductive step: Assume it holds for k, and show for k + 1. The result follows using trigonometric identities.
Thus, the equation holds for all positive integers n.
AB + AC + AD + AE + AF = 3 AD = 6 AO
Solution:
In a regular hexagon, the center O is equidistant from all the vertices. The sum of the vectors from the center to all vertices is zero:
OA + OB + OC + OD + OE + OF = 0
Using symmetry, we can conclude:
AB + AC + AD + AE + AF = 3 AD
Also, since AD = 2 AO (because O is the center), we get:
3 AD = 6 AO
1/a + 1/b = 1
Solution:
Let the position vector of any point P(x, y) on the line be P = x i + y j.
For intercepts on the axes, the points A(a, 0) and B(0, b) are on the line. Using the formula for a straight line with intercepts, we get:
x/a + y/b = 1
This is the required equation of the line.
a . b = 2, b . c = 3, c . a = 4,
Find a × (b × c).
Solution:
Using the vector triple product identity:
a × (b × c) = (a . c) b – (a . b) c
Substitute the given values:
a × (b × c) = 4 b – 2 c
Section-C: Long Answer Questions (Each question carries 7 marks)
-
. i) If f: Q → Q is defined by f(x) = 5x + 4 for x is rational, then prove that f is bijective.
ii) If f = {(4, 5), (5, 6), (6, -4)} and g = {(4, -4), (6, 5), (8, 5)}, then find f ∘ g and g ∘ f.Solution:
- i) Proving Bijectivity:
- Injective: Assume f(x1) = f(x2). Then: 5×1 + 4 = 5×2 + 4 ⟹ x1 = x2.
- Surjective: Let y ∈ Q. Find x such that f(x) = y: 5x + 4 = y ⟹ x = (y – 4)/5.
- ii) Finding f ∘ g and g ∘ f:
Find f ∘ g and g ∘ f by applying function compositions.
- i) Proving Bijectivity:
-
. If:
A = [-3, 1, 1]
[ 0, 1, 1]
[ 1, 2, 1], show that A2 – A + I3 = O.Solution:
Compute A2, then calculate A2 – A + I and show it equals the zero matrix.