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Class 10 Math Chapter 4 Test 3 | Functions and Graphs | Punjab Board 2027
Class 10 Math Chapter 4 Test 3 New Syllabus 2027
Class 10 Mathematics Chapter 4 Test 3 is a useful practice paper for students preparing Functions and Graphs according to the Punjab Board examination pattern. This Class 10 Math Test 3 covers important concepts from Chapter 4, including functions, graphs, domain and range, inverse functions, one-to-one and bijective functions, and absolute value functions. Students can use this Class 10 Mathematics model test for chapter-wise preparation, revision, self-assessment, and practice for the Punjab Board 2027 Mathematics examination.
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Unit 04:
Total Time: 35 min
Total Time: 35 min
CHAPTER WISE CLASS TEST
(Test # 3)
Mathematics
Total Marks: 25
Total Marks: 25
1ABCD
2ABCD
3ABCD
4ABCD
5ABCD
6ABCD
7ABCD
8ABCD
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.
Q.1:(8)
1. If f(x) = 5x + 2, h(x) = 2x − 2, then what is f(x) − h(x) =
2. If f(x) = x2 − 4, g(x) = x + 2, x ≠ −2 then f(x) / g(x) =
3. If a graph represents a function, then every vertical line must intersect it at:
4. If f(x) = 10x3 and g(x) = 5x, what is f(x) / g(x)?
5. One-to-one function is also called:
6. Domain of f−1 ________.
7. Which is the vertex form of an absolute value function?
8. The solution of equation |x| = 7 is:
Answer Key: 1-(C) 2-(D) 3-(D) 4-(A) 5-(A) 6-(B) 7-(C) 8-(C)
1) (5x + 2) − (2x − 2) = 3x + 4. 2) (x2 − 4)/(x + 2) = (x − 2)(x + 2)/(x + 2) = x − 2. 4) 10x3 / 5x = 2x2. 6) The domain of f−1 is the range of f.
1) (5x + 2) − (2x − 2) = 3x + 4. 2) (x2 − 4)/(x + 2) = (x − 2)(x + 2)/(x + 2) = x − 2. 4) 10x3 / 5x = 2x2. 6) The domain of f−1 is the range of f.
(PART – I)
Q.2: Write short answers to following questions.
(6 × 2 = 12)
(i) If f(x) = 7x − 3, g(x) = 2x2 + 5, then find (f.g)(x).
(f.g)(x) = (7x − 3)(2x2 + 5) = 14x3 + 35x − 6x2 − 15
= 14x3 − 6x2 + 35x − 15
= 14x3 − 6x2 + 35x − 15
(ii) If f(x) = 3x − 4, then find (f○f)(x).
(f○f)(x) = f(f(x)) = f(3x − 4) = 3(3x − 4) − 4 = 9x − 12 − 4 = 9x − 16
(iii) Define one-to-one function:
A function f : A → B is one-to-one (injective) if different elements of the domain have different images, i.e. f(x1) = f(x2) ⇒ x1 = x2. Every element of the range has exactly one pre-image.
(iv) Define bijective function:
A function that is both one-to-one (injective) and onto (surjective) is called a bijective function: every element of the codomain has exactly one pre-image in the domain.
(v) Find the inverse of f(x) = 5x2 + 1.
Let y = 5x2 + 1 ⇒ x2 = y − 15 ⇒ x = ±√y − 15
f−1(x) = ±√x − 15, x ≥ 1
(f is not one-to-one on all real numbers, so the inverse is a function only if the domain is restricted, e.g. x ≥ 0, which gives the positive root.)
f−1(x) = ±√x − 15, x ≥ 1
(f is not one-to-one on all real numbers, so the inverse is a function only if the domain is restricted, e.g. x ≥ 0, which gives the positive root.)
(vi) What is the range of the basic absolute value function f(x) = |x|?
|x| is never negative, so the range is all real numbers y ≥ 0, i.e. [0, ∞).
(PART – II)
Note: Attempt the following question. (5)
Q.3: 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, ∞)
⇒ x ≥ 6 or x ≤ −4
Solution set = (−∞, −4] ∪ [6, ∞)
Number Line
