Welcome to Education Hub

Education

Class 10 Mathematics Chapter 4 Full Unit Test 2027 | Functions and Graphs | Punjab Boards Paper Pattern

Class 10 Math Chapter 4 Full unit Test

Photo of admin admin Send an email September 29, 2026
0 70 13 minutes read

Class 10 Mathematics Chapter 4 Full Unit Test 2027 is designed for students preparing for the Punjab Boards examinations according to the latest paper pattern. This Functions and Graphs test covers important short-answer and long-answer questions, including functions, composition of functions, inverse functions, absolute value functions, graphs, domain and range, one-to-one and bijective functions, and related problem-solving questions. It is useful for Class 10 Mathematics exam preparation, chapter-wise practice, school tests, and Punjab Board 2027 preparation.

786Times786Times786Times786Times786Times 786Times786Times786Times786Times786Times 786Times786Times786Times786Times786Times
Full Unit Test
Unit 4: Functions and Graphs
New Board Paper Pattern 2027
Objective Type
Time Allowed: 20 Min. Max. Marks: 15
Note: Four possible answer 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 ink pen in the answer book. Cutting filling two or more circles will result in zero mark in that question.
(i) If f : X → Y is a function, then y is called:
(A) domain
(B) range
(C) co-domain
(D) none of these
Answer: (C). In f : X → Y, the set Y is the co-domain.
(ii) If f(x) = x2 − 1 then f(−1) =
(A) −2
(B) −1
(C) 0
(D) 1
Answer: (C). f(−1) = (−1)2 − 1 = 0.
(iii) Which of the following is not a function?
(A)
(B)
(C)
(D) both (B) and (C)
Answer: (A). In diagram (A) one element of the domain has two arrows, so it is not a function. Diagrams (B) and (C) give every element exactly one image.
(iv) For any function f, which of the following is true?
(A) Range ⊆ Domain
(B) Range ⊇ Domain
(C) Range ⊆ Co-domain
(D) Range ⊇ Co-domain
Answer: (C). The range is always a subset of the co-domain.
(v) If f(x) = x + 1, g(x) = x2 then (f + g)(x) =
(A) x2 + x + 1
(B) x2 + 2x + 1
(C) x2 + 1
(D) x3 + x2
Answer: (A). (f + g)(x) = (x + 1) + x2 = x2 + x + 1.
(vi) Which of the following is not a one-to-one function?
(A) f(x) = x
(B) f(x) = x2
(C) f(x) = x3
(D) f(x) = −x
Answer: (B). x2 gives the same value for x and −x (e.g. f(2) = f(−2) = 4).
(vii) If f(x) = x + 1, g(x) = x2 then g(f(1)) =
(A) 1
(B) 2
(C) 4
(D) none of these
Answer: (C). f(1) = 2, then g(2) = 22 = 4.
(viii) Graph of the function f(x) = 10 |x| is:
(A)
(B)
(C)
(D)
Answer: (B). 10|x| is never negative, so the graph is a V opening upward with its vertex at the origin, and the factor 10 makes it much steeper (narrower) than y = |x|.
(ix) A function from A to B is represented by 4
(A) f : A → B
(B) f : A ← B
(C) f : B − A
(D) f : BA
Answer: (A). A function from A to B is written f : A → B.
(x) If f(x) = x2 − 1 and g(x) = x − 1 then fg(x) =
(A) x − 1
(B) x + 1
(C) 1x + 1
(D) 1x − 1
Answer: (B). (x2 − 1)/(x − 1) = (x − 1)(x + 1)/(x − 1) = x + 1.
(xi) Shape of the graph of parabola function is:
(A) U-shaped
(B) V-shaped
(C) W-shaped
(D) none of these
Answer: (A). The graph of a quadratic (parabola) function is U-shaped.
(xii) Which of the following is graph of a one-to-one function?
(A)
(B)
(C)
(D)
Answer: (D). By the horizontal line test, only the always-rising curve (D) is cut by every horizontal line at most once. Graphs (A), (B) and (C) are cut twice.
(xiii) Which of the following function is such that f(x) = f−1(x)?
(A) f(x) = x
(B) f(x) = 1x
(C) f(x) = −x
(D) all of these
Answer: (D). The inverse of x is x, the inverse of 1/x is 1/x, and the inverse of −x is −x.
(xiv) Solution of |v − 5| = 100, v =
(A) −95
(B) 95
(C) 105
(D) both (A) & (C)
Answer: (D). v − 5 = 100 ⇒ v = 105;  v − 5 = −100 ⇒ v = −95.
(xv) If salary function of a worker manufacturing n units is f(n) = 500n + 5000, then find the salary of the worker who has manufactured 20 units is:
(A) 5000
(B) 5500
(C) 15000
(D) 55000
Answer: (C). f(20) = 500(20) + 5000 = 10000 + 5000 = 15000.
Answer Key: (i) C   (ii) C   (iii) A   (iv) C   (v) A   (vi) B   (vii) C   (viii) B   (ix) A   (x) B   (xi) A   (xii) D   (xiii) D   (xiv) D   (xv) C
Class 10 Mathematics | Unit 4: Functions and Graphs | Full Unit Test

Here is Subjective Part of Model Paper for Annual Examination 2027 under Punjab Boards.

786Times786Times786Times786Times786Times786Times786Times786Times786Times786Times786Times786Times786Times786Times786Times
Time Allowed: 2:10 hours
Subjective Type
(Part-I)
Max. Marks: 60
2. Write short answers to any six (06) questions: (2 × 6 = 12)

(i) If f(x) = x2 + 3x + 1, then evaluate f(x + 1).

f(x + 1) = (x + 1)2 + 3(x + 1) + 1 = x2 + 2x + 1 + 3x + 3 + 1 = x2 + 5x + 5

(ii) Let f(x) = 5x + 2, g(x) = x2 − 3. Find f(x) − g(x).

f(x) − g(x) = (5x + 2) − (x2 − 3) = −x2 + 5x + 5

(iii) Find the domain and range of f(x) = (x + 1)2.

Domain = all real numbers R. Since a square is never negative, Range = {y : y ≥ 0} = [0, ∞).

(iv) Let f(x) = 2x + 3 and g(x) = x2 + x + 1. Add f(x) and g(x).

(f + g)(x) = 2x + 3 + x2 + x + 1 = x2 + 3x + 4

(v) If f(x) = 2x2 + 5x then find f(x + 2).

f(x + 2) = 2(x + 2)2 + 5(x + 2) = 2x2 + 8x + 8 + 5x + 10 = 2x2 + 13x + 18

(vi) Let f(x) = x + 1, g(x) = 2x − 3, find f(x) · g(x).

(x + 1)(2x − 3) = 2x2 − 3x + 2x − 3 = 2x2 − x − 3

(vii) Let f(x) = x2 − 4, g(x) = x + 2. Find fg(x), where x ≠ −2.

fg(x) = x2 − 4x + 2 = (x − 2)(x + 2)x + 2 = x − 2,  x ≠ −2

(viii) Define Composition of Function.

If f : A → B and g : B → C are functions, the composition g○f : A → C is the function defined by (g○f)(x) = g(f(x)); f is applied first and then g is applied to its result.

(ix) Let f(x) = 2x + 3, g(x) = x2, then find (gof)(x).

(g○f)(x) = g(f(x)) = g(2x + 3) = (2x + 3)2 = 4x2 + 12x + 9
3. Write short answers to any six (06) questions: (2 × 6 = 12)

(i) Define Bijective Function.

A function that is both one-to-one (injective) and onto (surjective) is bijective: each element of the codomain has exactly one pre-image in the domain.

(ii) Define Horizontal Line Test.

A function is one-to-one if and only if every horizontal line intersects its graph at most once. If any horizontal line cuts the graph at two or more points, the function is not one-to-one.

(iii) Define One to One function.

A function f is one-to-one (injective) if different elements of the domain have different images, i.e. f(x1) = f(x2) ⇒ x1 = x2.

(iv) Plot graph for the following absolute valued functions: f(x) = |x − 2|

2 x y
Table: x = 0, 1, 2, 3, 4 ⇒ f(x) = 2, 1, 0, 1, 2. The graph is a V-shape opening upward with vertex (2, 0), i.e. the graph of |x| shifted 2 units to the right.xy-2-1123456(2, 0)

(v) Solve and express the solution on number line: |6x + 18| ≤ 24

−24 ≤ 6x + 18 ≤ 24 ⇒ −42 ≤ 6x ≤ 6 ⇒ −7 ≤ x ≤ 1, solution set = [−7, 1]-10-9-8-7-6-5-4-3-2-101234-71

(vi) Find f−1(x), if f(x) = 5x − 1, x ≠ 1.

y = 5x − 1 ⇒ x − 1 = 5y ⇒ x = 5y + 1.  f−1(x) = 5x + 1 = x + 5x,  x ≠ 0

(vii) Find the value of k, such that (fog)(x) = (gof)(x), where f(x) = 3x + 2, g(x) = 6x − k.

(f○g)(x) = 3(6x − k) + 2 = 18x − 3k + 2;  (g○f)(x) = 6(3x + 2) − k = 18x + 12 − k
−3k + 2 = 12 − k ⇒ −2k = 10 ⇒ k = −5

(viii) Find f−1(x) in each of the following: f(x) = 2x − 3.

y = 2x − 3 ⇒ x = y + 32.  f−1(x) = x + 32

(ix) Without finding f−1(x), find domain and range of f−1(x): f(x) = 12x − 3.

Domain and range of f are both R. Domain of f−1 = range of f = R; Range of f−1 = domain of f = R.
4. Write short answers to any six (06) questions: (2 × 6 = 12)

(i) Plot the graph of the following functions: g(x) = |x − 3|

3 x y
Table: x = 1, 2, 3, 4, 5 ⇒ g(x) = 2, 1, 0, 1, 2. V-shaped graph with vertex (3, 0), opening upward.xy-11234567(3, 0)

(ii) Solve |2x − 3| = 16 and express the solution on number line.

2x − 3 = 16 ⇒ x = 192;  2x − 3 = −16 ⇒ x = −132. Solution set = {−132, 192}-8-6-4-2024681012-13/219/2

(iii) Solve and express the solution on number line: |5x + 8| ≤ 3.

−3 ≤ 5x + 8 ≤ 3 ⇒ −11 ≤ 5x ≤ −5 ⇒ −115 ≤ x ≤ −1-4-3-2-1012-11/5-1

(iv) Let E(n) = 1,500n is the total earning (in Rs.) function, where n is the number of days. If labourer works 10 days, find his earning.

E(10) = 1,500 × 10 = Rs. 15,000

(v) Solve and express the solution on number line: |x − 2| = 6.

x − 2 = 6 ⇒ x = 8;  x − 2 = −6 ⇒ x = −4. Solution set = {−4, 8}-6-4-20246810-48

(vi) The cost of manufacturing fancy sofa set would be fixed charges Rs. 5500 which is modeled as f(n) = 5500n, where n is the number of sofa sets. Find the cost of 50 sofa sets.

f(50) = 5500 × 50 = Rs. 2,75,000 (275,000)

(vii) A function f(k) = 150 + 20k represents the total fare (in rupees) for after k kilometres. How much will the fare be for a 12 kilometres ride?

f(12) = 150 + 20(12) = 150 + 240 = Rs. 390

(viii) If f(x) = x3 and g(x) = 14 + 2x, then find: (fog)(x).

(f○g)(x) = f(g(x)) = (14 + 2x)3 = 8x3 + 168x2 + 1176x + 2744

(ix) Plot graph for the following absolute valued functions: f(x) = 7|x|.

x y
Table: x = −2, −1, 0, 1, 2 ⇒ f(x) = 14, 7, 0, 7, 14. V-shaped graph with vertex (0, 0), narrower (steeper) than y = |x|.xy-3-2-1123(0, 0)
Part-II
Note: Attempt any two (02) questions. (2 × 8 = 16)

5:(a) For the function f and g, find (a) (fog)(x), (b) (gof)(x), (c) (fof)(x) and (d) (gog)(x)

Where: f(x) = 2x + 3 ;   g(x) = x3

(a) (f○g)(x) = f(x3) = 2x3 + 3
(b) (g○f)(x) = g(2x + 3) = (2x + 3)3 = 8x3 + 36x2 + 54x + 27
(c) (f○f)(x) = f(2x + 3) = 2(2x + 3) + 3 = 4x + 9
(d) (g○g)(x) = g(x3) = (x3)3 = x9

(b) If f(x) = x3, find f−1(x). Also verify that (f−1(x)) = (f−1(x)) = x.

Let y = x3 ⇒ x = ∛y. So f−1(x) = ∛x.
Verification: (f○f−1)(x) = f(∛x) = (∛x)3 = x and (f−1○f)(x) = f−1(x3) = ∛(x3) = x.  Hence (f○f−1)(x) = (f−1○f)(x) = x. ✔

6:(a) Find f−1(x) in each of the following: (iv) f(x) = x + 13x − 2, x ≠ 23

y = x + 13x − 2 ⇒ y(3x − 2) = x + 1 ⇒ 3xy − 2y = x + 1 ⇒ 3xy − x = 2y + 1 ⇒ x(3y − 1) = 2y + 1 ⇒ x = 2y + 13y − 1
f−1(x) = 2x + 13x − 1,  x ≠ 13

(b) Plot graph for the following absolute valued functions: f(x) = 2|x + 4| − 3

−4, −3 x y
Vertex = (−4, −3); opens upward; slope 2 on each side. Table: x = −6, −5, −4, −3, −2 ⇒ f(x) = 1, −1, −3, −1, 1.xy-9-8-7-6-5-4-3-2-11(-4, -3)

7:(a) Solve and express the solution on number line: |1 − 2 32x − 5| > −3

An absolute value is never negative, so |…| ≥ 0 > −3 is true for every real x.
Solution set = R = (−∞, ∞): the whole number line is shaded.-5-4-3-2-1012345

(b) A company charges Rs. 100 for an encoding work. In addition, the company charges Rs. 5 per page of printed output. The model of function f(x) = 100 + 5x, where x represents the number of pages printed out. How much will company charge for 55 page encoding and printing work?

f(55) = 100 + 5(55) = 100 + 275 = Rs. 375
Part-III
Note: Attempt any one (01) question. (1 × 8 = 8)

8:(a) Solve and express the solution on number line: |6x + 1| = 9

6x + 1 = 9 ⇒ x = 43;  6x + 1 = −9 ⇒ x = −53. Solution set = {−53, 43}-3-2-10123-5/34/3

(b) A GPS system is considered accurate if the actual position differs from the reported position by no more than 6 metres. If the actual location is at point x = 100, the allowed range is defined as: |x − r| ≤ 6, where r is the reported location. What is the range of acceptable reported locations?

|100 − r| ≤ 6 ⇒ −6 ≤ 100 − r ≤ 6 ⇒ −106 ≤ −r ≤ −94 ⇒ 94 ≤ r ≤ 106. Acceptable reported locations: from 94 m to 106 m.

9:(a) Plot the graph of the following functions: g(x) = |x − 3|

3 x y
Table: x = 1, 2, 3, 4, 5 ⇒ g(x) = 2, 1, 0, 1, 2. V-shaped graph with vertex (3, 0).xy-11234567(3, 0)

(b) Express the solution of 40|2x − 5| + 1 > 201 on number line.

40|2x − 5| > 200 ⇒ |2x − 5| > 5 ⇒ 2x − 5 > 5  or  2x − 5 < −5 ⇒ x > 5  or  x < 0
Solution set = (−∞, 0) ∪ (5, ∞) (open circles at 0 and 5)-4-3-2-101234567805
Class 10 Mathematics | Unit 4: Functions and Graphs | Subjective Type
🏠 Home 📚 Model Papers 2027 📝 Pairing Schemes 2027 📝 Study Schemes 2027 ⬅ Back to Previous Page

Photo of admin admin Send an email September 29, 2026
0 70 13 minutes read

Related Articles

Class 10 Physics Chapter 10 Test 3 – Thermal Physics

1 week ago

Class 10 Tarjuma Tul Quran Pairing Scheme 2027 – Free PDF Download

1 week ago

Class 10 Pakistan Studies Pairing Scheme 2027 PDF Download Punjab Board

1 week ago

Class 10 Math Pairing Scheme 2027 PDF Download – Punjab Board

1 week ago

Leave a Reply

Your email address will not be published. Required fields are marked *

Back to top button