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Class 10 Mathematics Chapter 4 Test 1 | Functions and Graphs | Punjab Board 2027
Class 10 Math Unit 4 Test 1 new Syllabus 2027
Class 10 Mathematics Chapter 4 Test 1 Exercise 4.1 to Exercise 4.2 – Functions and Graphs is designed according to the latest Punjab Board syllabus and examination pattern. This Class 10 Math Chapter 4 Test covers important concepts of Functions and Graphs, including functions, one-to-one functions, composition of functions, inverse functions, inequalities, and related questions. Students can use this Mathematics Test 1 for chapter-wise preparation, class tests, revision, and Punjab Board Class 10 Mathematics 2027 exam practice.
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Exercise 4.1 to 4.2
T-1
Unit 4: Functions and Graphs
Time: 5 minutes
(Objective Type)
Total Marks: 5
1. Tick the correct answer. (1 × 5 = 5)
(i) If f(x) = x² − 1 then f(−1) =
(ii) Which of the following is not a function?
(a)
(b)
(c)
Answer: (c). In diagram (c) one element of the domain has no arrow, so it is not a function (every element of the domain must have exactly one image).
(iii) If f(x) = x + 1, g(x) = x² then (g ○ f)(x) =
(iv) Which of the following is not a one-to-one function?
(v) If f(x) = x then its inverse function is f−1(x) =
Answer Key: (i) c (ii) c (iii) b (iv) b (v) a
(i) f(−1) = (−1)² − 1 = 0. (iii) (g○f)(x) = g(x + 1) = (x + 1)² = x² + 2x + 1. (iv) x² gives the same value for x and −x. (v) f(x) = x ⇒ f−1(x) = x.
(i) f(−1) = (−1)² − 1 = 0. (iii) (g○f)(x) = g(x + 1) = (x + 1)² = x² + 2x + 1. (iv) x² gives the same value for x and −x. (v) f(x) = x ⇒ f−1(x) = x.
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Exercise 4.1 to 4.2
T-1
Unit 4: Functions and Graphs
Time: 35 minutes
(Subjective Type)
Total Marks: 20
2. Write short answers to the following questions: (6 × 2 = 12)
(i) If f(x) = x + 2, g(x) = 2x + 4, then find
g(x)
f(x)
f(x)
g(x)f(x) = 2x + 4x + 2 = 2(x + 2)x + 2 = 2, x ≠ −2
(ii) For the function f and g, find (f○f)(x) where
f(x) =
2
x
, x ≠ 0 ; g(x) = 2x² − 1
(f○f)(x) = f(f(x)) = f(2x) = 22x = 2 × x2 = x, x ≠ 0
(iii) Given that f(x) = 3x + 2 and g(x) = 2x + 3 Find f(g(−2))
g(−2) = 2(−2) + 3 = −1
f(g(−2)) = f(−1) = 3(−1) + 2 = −1
f(g(−2)) = f(−1) = 3(−1) + 2 = −1
(iv) Find f−1(x) in each of the following:
f(x) = 4x³ − 1
Let y = 4x³ − 1 ⇒ 4x³ = y + 1 ⇒ x³ = y + 14 ⇒ x = ∛(y + 14)
f−1(x) = ∛(x + 14)
f−1(x) = ∛(x + 14)
(v) Solve the inequality |x − 1| ≥ 5 and express the solution on number line.
|x − 1| ≥ 5 ⇒ x − 1 ≥ 5 or x − 1 ≤ −5 ⇒ x ≥ 6 or x ≤ −4
Solution set = (−∞, −4] ∪ [6, ∞)
Solution set = (−∞, −4] ∪ [6, ∞)
(vi) Solve and express the solution on number line. |7 − 2x| = 1
7 − 2x = 1 ⇒ x = 3; or 7 − 2x = −1 ⇒ x = 4
Solution set = {3, 4}
Solution set = {3, 4}
Note: Answer the following Questions. (4 + 4 = 8)
3. Find the value of k, such that (f○g)(x) = (g○f)(x), where f(x) = 3x + 2, g(x) = 6x − k.
(f○g)(x) = f(6x − k) = 3(6x − k) + 2 = 18x − 3k + 2
(g○f)(x) = g(3x + 2) = 6(3x + 2) − k = 18x + 12 − k
Equating: −3k + 2 = 12 − k ⇒ −2k = 10 ⇒ k = −5
(g○f)(x) = g(3x + 2) = 6(3x + 2) − k = 18x + 12 − k
Equating: −3k + 2 = 12 − k ⇒ −2k = 10 ⇒ k = −5
4. Solve and express the solution on number line: |3x − 2| ≤ 12
|3x − 2| ≤ 12 ⇒ −12 ≤ 3x − 2 ≤ 12 ⇒ −10 ≤ 3x ≤ 14 ⇒ −103 ≤ x ≤ 143
Solution set = [−103, 143]
Solution set = [−103, 143]
