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Class 10 Mathematics Chapter 3 & 4 Board Pattern Paper 2027 | Matrices and Determinants, Functions and Graphs
CLASS 10 Math Chapter 3 & 4 Board Pattern Paper 2027
Class 10 Mathematics Board Pattern Paper 2027 for Chapter 3: Matrices and Determinants and Chapter 4: Functions and Graphs is designed according to the latest Punjab Board-style examination pattern. This Mathematics paper includes objective and subjective-type questions covering important concepts, short questions, matrix operations, determinants, functions, inverse functions, domain and range, absolute value functions, and related problem-solving practice. Students preparing for the Class 10 Mathematics 2027 board examination can use this chapter-wise paper for revision, self-assessment, and exam preparation.
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BOARD PAPER PATTERN
MATHEMATICS
CHAPTER
WISE
WISE
Roll No. (in Figures):
………………………………….
(in Words):
………………………………….
Unit 3: Matrices and Determinants
Time : 20 Minutes
Unit 4: Functions and Graphs
Marks: 15
OBJECTIVE TYPE
1 ABCD
2 ABCD
3 ABCD
4 ABCD
5 ABCD
6 ABCD
7 ABCD
8 ABCD
9 ABCD
10 ABCD
11 ABCD
12 ABCD
13 ABCD
14 ABCD
15 ABCD
Note: Four possible answers A, B, C and D to each question are given.
The choice which you think is correct, fill that circle in front of that question with
Marker or Pen ink. Cutting or filling two or more circles will result in zero mark in
that question.
1.
A =
3 5 0
is a ______ matrix.
2.
B =
2
3 is a ______ matrix.
3 is a ______ matrix.
3.
If
A =
3 4
,
B =
7 8
,
then A + B =
4.
3 1
0 4 =
0 4 =
5. Identify the order of the matrix
7 8 9
.
6.
If
2x 1
=
10 1
then what is the value of x?
7. A square matrix A is symmetric if:
8. Commutative law of Addition of Matrices is:
9. A function f from X to Y is represented by:
10.
(fog)(x) =
11. What is the shape of the graph of an absolute value function?
12. If f(x) = x3, then
f(−2) =
13. If f(x) = x2 + 3x + 1,
what is the value of f(−2)?
14. If f(x) = x2 and
g(x) = 2x−1, then the value of
(f−g)(x) is:
15. Domain of f = __________.
Here is Subjective Part of paper.
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BOARD PAPER PATTERN
MATHEMATICS
CHAPTER
WISE
WISE
Roll No. (in Figures): ………………………………….
(in Words): ………………………………….
Unit 3: Matrices and Determinants
Time : 2:10 Hours
Unit 4: Functions and Graphs
Marks: 60
SUBJECTIVE TYPE (PART – I)
Q2. Write short answers to any SIX (6) questions:
(6×2=12)
(i) If
2x + 1 4
7 5 = 9 4
7 y , then find the values of x and y.
7 5 = 9 4
7 y , then find the values of x and y.
Solution:
Equal matrices have equal corresponding elements.
Comparing 1st row, 1st column: 2x + 1 = 9
2x = 9 − 1 = 8 ⟹ x = 4
Comparing 2nd row, 2nd column: 5 = y
x = 4, y = 5
(ii) Find transpose of the matrix:
A = 2 3
4 5
4 5
Solution:
Transpose means rows become columns.
Row 1 (2, 3) becomes Column 1; Row 2 (4, 5) becomes Column 2.
At =
(iii) Find negative of the matrix:
A = −3 0
5 6
5 6
Solution:
Negative of a matrix: multiply every element by −1, i.e. −A = (−1)A.
−A =
−A =
(iv) Show that
L = 2 3 4
3 −2 5
4 5 0 is a symmetric matrix.
3 −2 5
4 5 0 is a symmetric matrix.
Solution:
A matrix is symmetric if Lt = L.
L =
Interchange rows and columns:
Lt =
Check: L12 = L21 = 3, L13 = L31 = 4, L23 = L32 = 5.
Since Lt = L, L is a symmetric matrix. ∎
(v) If
A = 2 3
4 5 and B = 7 6
5 8 , then verify that (At)t = A.
4 5 and B = 7 6
5 8 , then verify that (At)t = A.
Solution:
A =
Step 1: At =
Step 2: (At)t = transpose of At =
(At)t = = A. Hence verified. ∎
(vi) If
A = 4 2
7 6 , B = 5 5
8 8 , then find (A − B)t.
7 6 , B = 5 5
8 8 , then find (A − B)t.
Solution:
Step 1: A − B =
A − B =
Step 2: Take transpose (rows ↔ columns):
(A − B)t =
(vii) If
A = 4 2
7 6 , B = 5 5
8 8 , then find Bt − At.
7 6 , B = 5 5
8 8 , then find Bt − At.
Solution:
Step 1: Bt = and At =
Step 2: Bt − At =
Bt − At =
(viii) If
A = 7 3
2 1 and B = 5 3
2 4 , then find |B|.
2 1 and B = 5 3
2 4 , then find |B|.
Solution:
For a 2×2 matrix, |B| = (a)(d) − (b)(c).
|B| =
|B| = (5)(4) − (3)(2) = 20 − 6
|B| = 14
(ix) If
A = 7 3
2 1 , B = 5 3
2 4 , then find Adj B.
2 1 , B = 5 3
2 4 , then find Adj B.
Solution:
For B = , Adj B = (swap the diagonal elements, change signs of the other two).
Here a = 5, b = 3, c = 2, d = 4.
Adj B =
Q3. Write short answers to any SIX (6) questions:
(6×2=12)
(i) Define a row in the context of a matrix.
Answer:
A horizontal arrangement of elements (numbers) in a matrix is called a row.
Example: in , (1 2) is the 1st row and (3 4) is the 2nd row.
(ii) Define a column in context of a matrix.
Answer:
A vertical arrangement of elements (numbers) in a matrix is called a column.
Example: in , the 1st column is 1, 3 and the 2nd column is 2, 4.
(iii) Define equal matrices.
Answer:
Two matrices A and B are equal if (1) they have the same order (same number of rows and columns), and (2) their corresponding elements are equal.
We write A = B.
(iv) Find the value of x when
A = x 6
5 15 is a singular matrix.
5 15 is a singular matrix.
Solution:
A matrix is singular if |A| = 0.
|A| = = (x)(15) − (6)(5) = 15x − 30
Put |A| = 0: 15x − 30 = 0
15x = 30 ⟹ x = 30 ÷ 15
x = 2
(v) What is a “square matrix”.
Answer:
A matrix in which the number of rows is equal to the number of columns (m = n) is called a square matrix.
Example: is a square matrix of order 2×2.
(vi) Define “Scalar matrix”.
Answer:
A square matrix in which all non-diagonal elements are zero and all diagonal elements are equal (to the same non-zero scalar) is called a scalar matrix.
Example:
(vii) Find the negative of matrix
B = −1 0
4 7
4 7
Solution:
−B = (−1)B: multiply every element by −1.
−B =
−B =
(viii) If
A = 9 −8
8 4 , then show that it is a symmetric matrix.
8 4 , then show that it is a symmetric matrix.
Solution:
A matrix is symmetric if At = A, i.e. a12 = a21.
At =
Here a12 = −8 but a21 = 8, so At ≠ A.
⚠ As printed, A is NOT symmetric (it looks like a typing error).
If the matrix is A = (a12 = a21 = 8), then At = = A, so it is symmetric. ∎
(ix) Find the determinant of
A = −7 35
−1 5
−1 5
Solution:
|A| =
|A| = (−7)(5) − (35)(−1)
|A| = −35 + 35
|A| = 0 (so A is a singular matrix)
Q4. Write short answers to any SIX (6) questions:
(6×2=12)
(i) If f(x) = 2x + 5, g(x) = 3x − 2, then find f(x) + g(x).
Solution:
f(x) + g(x) = (2x + 5) + (3x − 2)
= 2x + 3x + 5 − 2
f(x) + g(x) = 5x + 3
(ii) If f(x) = 2x − 3 then find f−1(x).
Solution:
Step 1: Put y = f(x): y = 2x − 3
Step 2: Solve for x: y + 3 = 2x ⟹ x = (y + 3)/2
Step 3: Replace y by x:
f−1(x) = (x + 3)/2
(iii) A function B(t) = 5,000 + 200t represents the total balance (in rupees) after t months. What will be the balance after 6 months?
Solution:
Put t = 6 in B(t):
B(6) = 5,000 + 200(6) = 5,000 + 1,200
B(6) = Rs. 6,200
(iv) Describe the range of a function.
Answer:
The range of a function is the set of all possible output values (y-values or f(x) values), i.e. the set of all second elements of the ordered pairs of the function.
Example: for f(x) = x², range = {y : y ≥ 0}.
(v) If g(x) = (x−1)3, then find g(−4).
Solution:
g(−4) = (−4 − 1)3
= (−5)3 = (−5)(−5)(−5)
g(−4) = −125
(vi) Find the domain and range for function: f(x) = 3√x
Solution:
The cube root exists for every real number (positive, negative or zero), e.g. ³√8 = 2, ³√(−8) = −2, ³√0 = 0.
Also, the cube root can give any real value as output.
Domain = ℝ, Range = ℝ
(vii) If f(x) = {(3,a),(7,b),(1,c)} then find f−1(x).
Solution:
To find the inverse, interchange the components of each ordered pair.
(3, a) → (a, 3); (7, b) → (b, 7); (1, c) → (c, 1)
f−1(x) = {(a,3), (b,7), (c,1)}
(viii) Define an absolute value function.
Answer:
A function of the form f(x) = |x| is called the absolute value function, defined as:
f(x) = x if x ≥ 0 and f(x) = −x if x < 0.
It always gives a non-negative output.
(ix) What is the characteristic shape of the graph of f(x) = |x|?
Answer:
The graph is V-shaped.
Its vertex (turning point) is at the origin (0, 0), it opens upward, and it is symmetric about the y-axis. It is made of two straight rays: y = x for x ≥ 0 and y = −x for x < 0.
(PART – II)
Note: Attempt any two (02) questions.
(2×8=16)
Q5. (a) If
p + q 5
11 p − 2q = 6 5
11 0 , then find the values of p and q. 4
11 p − 2q = 6 5
11 0 , then find the values of p and q. 4
Solution:
Equal matrices have equal corresponding elements, so:
p + q = 6 ……(1)
p − 2q = 0 ……(2)
From (2): p = 2q ……(3)
Put p = 2q in (1): 2q + q = 6 ⟹ 3q = 6 ⟹ q = 2
Put q = 2 in (3): p = 2(2) = 4
Check: p + q = 4 + 2 = 6 ✓ p − 2q = 4 − 4 = 0 ✓
p = 4, q = 2
(b) A shop employs, 5 men and 3 women, pays total daily wages Rs. 3500. If the number of men is reduced to 2 and 3 extra women are taken on, the daily wages amount to Rs. 5000. Find daily wage of a man and a woman by using matrices.
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Solution:
Let daily wage of a man = Rs. x and of a woman = Rs. y.
Case 1: 5 men, 3 women → 5x + 3y = 3500
Case 2: 2 men, (3 + 3) = 6 women → 2x + 6y = 5000
Matrix form AX = B:
=
|A| = (5)(6) − (3)(2) = 30 − 6 = 24 ≠ 0, so A−1 exists.
Adj A =
A−1 = (1/24)
X = A−1B = (1/24)
= (1/24)
= (1/24) = (1/24)
=
Check: 5(250) + 3(750) = 1250 + 2250 = 3500 ✓ 2(250) + 6(750) = 500 + 4500 = 5000 ✓
Daily wage of a man = Rs. 250, of a woman = Rs. 750
Q6. (a) If
A = x 8
5 10 is a singular matrix, then find the value of x. 4
5 10 is a singular matrix, then find the value of x. 4
Solution:
A is singular, so |A| = 0.
|A| = = (x)(10) − (8)(5) = 10x − 40
10x − 40 = 0 ⟹ 10x = 40 ⟹ x = 40 ÷ 10
Check: x = 4 → |A| = 40 − 40 = 0 ✓
x = 4
(b) Find the value of x when
A = x 6
5 15 is a singular matrix. 4
5 15 is a singular matrix. 4
Solution:
A is singular, so |A| = 0.
|A| = = (x)(15) − (6)(5) = 15x − 30
15x − 30 = 0 ⟹ 15x = 30 ⟹ x = 30 ÷ 15
Check: x = 2 → |A| = 30 − 30 = 0 ✓
x = 2
Q7. (a) Find the adjoint of the following matrices.
A = a b
c d , B = 5 −2
3 7 , C = −3 5
−3 2 4
c d , B = 5 −2
3 7 , C = −3 5
−3 2 4
Solution:
Rule: for , interchange a and d, and change the signs of b and c.
(1) A = ⟹ Adj A =
(2) B = (a = 5, b = −2, c = 3, d = 7)
Adj B = =
(3) C = (a = −3, b = 5, c = −3, d = 2)
Adj C = =
Adj A = , Adj B = , Adj C =
(b) Find the inverse of f(x) = 2x + 3. Also find domain and range of f−1(x).
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Solution:
Step 1: Put y = f(x): y = 2x + 3
Step 2: Solve for x: y − 3 = 2x ⟹ x = (y − 3)/2
Step 3: Replace y by x: f−1(x) = (x − 3)/2
Check: f(f−1(x)) = 2·(x − 3)/2 + 3 = x − 3 + 3 = x ✓
f−1(x) is a linear function: it is defined for every real x, and takes every real value.
f−1(x) = (x − 3)/2; Domain of f−1 = ℝ, Range of f−1 = ℝ
(PART – III)
Note: Attempt any one (01) question.
(1×8=8)
Q8. (a) If f(x) = x3, find f−1(x). Also verify that f(f−1(x)) = f−1(f(x)) = x.
4
Solution:
Step 1: Put y = f(x): y = x³
Step 2: Take cube root: x = ³√y
Step 3: Replace y by x: f−1(x) = ³√x
Verification:
f(f−1(x)) = f(³√x) = (³√x)³ = x ……(1)
f−1(f(x)) = f−1(x³) = ³√(x³) = x ……(2)
From (1) and (2): f(f−1(x)) = f−1(f(x)) = x. Hence verified. ∎
(b) Given that f(x) = x2 + 9 and g(x) = x + 21. Find the values of a such that: f(a) = g(a).
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Solution:
f(a) = a² + 9 and g(a) = a + 21
Put f(a) = g(a): a² + 9 = a + 21
a² − a + 9 − 21 = 0 ⟹ a² − a − 12 = 0
Factorise (find two numbers with product −12 and sum −1: −4 and +3):
a² − 4a + 3a − 12 = 0
a(a − 4) + 3(a − 4) = 0 ⟹ (a − 4)(a + 3) = 0
a − 4 = 0 or a + 3 = 0
Check: a = 4 → f = 16 + 9 = 25, g = 25 ✓ a = −3 → f = 9 + 9 = 18, g = 18 ✓
a = 4 or a = −3
Q9. (a) Solve 2|x − 3| = 16 and express the solution on number line.
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Solution:
2|x − 3| = 16
Divide both sides by 2: |x − 3| = 8
So x − 3 = 8 or x − 3 = −8
Case 1: x − 3 = 8 ⟹ x = 8 + 3 = 11
Case 2: x − 3 = −8 ⟹ x = −8 + 3 = −5
Check: |11 − 3| = 8 ✓ |−5 − 3| = 8 ✓
Solution set = {−5, 11}
(b) Express the solution of 40|2x − 5| + 1 > 201 on number line.
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Solution:
40|2x − 5| + 1 > 201
Subtract 1: 40|2x − 5| > 200
Divide by 40: |2x − 5| > 5
So 2x − 5 > 5 or 2x − 5 < −5
Case 1: 2x − 5 > 5 ⟹ 2x > 10 ⟹ x > 5
Case 2: 2x − 5 < −5 ⟹ 2x < 0 ⟹ x < 0
Check: x = 6 → 40|7| + 1 = 281 > 201 ✓ x = −1 → 40|−7| + 1 = 281 > 201 ✓ x = 2 → 40|−1| + 1 = 41, not > 201 ✓
Solution: x < 0 or x > 5, i.e. (−∞, 0) ∪ (5, ∞)
On the number line, 0 and 5 are open circles (not included); shade to the left of 0 and to the right of 5.
