Struggling with Unit 1: Functions and Graphs in your 12th Math FBISE NBF book? For example, we solve every question from Exercise 1.1 to 1.5 plus the review exercise. Also, view the full 163-page PDF directly — no download needed. Therefore, students, teachers, lecturers, and experts quickly master domain, range, inverses, bijectivity, and graphing using these easy solutions.
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- Full Coverage: First, we solve all parts of Exercises 1.1–1.5 and the review exercise.
- Simple Steps: Next, we use clear language and active voice.
- Exam-Focused: Moreover, we match the latest FBISE NBF syllabus.
- Visual Help: Finally, we include graphs and tables.
For instance, students revise fast. In addition, teachers save prep time. Similarly, lecturers assign ready homework. Likewise, experts verify proofs quickly.
Exercise 1.1 – Domain, Range & Function Types
Key Concepts for Domain and Range
- Domain: The set of all possible input values (x-values) for which the function is defined. Because we look for values that cause problems, like dividing by zero or taking the square root of a negative number.
- Range: The set of all possible output values (y-values) that result from using the domain.
1. Find the Domain
| Function | Domain | Why? |
|---|---|---|
| (i) ( f(x) = x^2 – 6 ) | ( \mathbb{R} ) | Because it’s a polynomial — defined everywhere |
| (ii) ( g(x) = \frac{x}{x+3} ) | ( (-\infty, -3) \cup (-3, \infty) ) | Since denominator ≠ 0 → ( x \neq -3 ) |
| (iii) ( h(x) = \frac{x+4}{x^2-9} ) | ( (-\infty, -3) \cup (-3, 3) \cup (3, \infty) ) | As ( x = \pm 3 ) make denominator zero |
| (iv) ( i(x) = \frac{x}{5x+2} ) | ( (-\infty, -\frac{2}{5}) \cup (-\frac{2}{5}, \infty) ) | Due to ( 5x+2 \neq 0 ) |
| (v) ( j(x) = \frac{x}{x^2+4} ) | ( \mathbb{R} ) | Given that ( x^2+4 \geq 4 > 0 ) |
| (vi) ( k(x) = \sqrt{x+1} ) | ( [-1, \infty) ) | Provided inside square root ≥ 0 |
2. Find the Domain and Range of the Functions
| Function | Domain | Range | Key Point |
|---|---|---|---|
| (i) ( f(x) = x + 7 ) | ( \mathbb{R} ) | ( \mathbb{R} ) | It’s linear |
| (ii) ( f(x) = 2x^2 + 1 ) | ( \mathbb{R} ) | ( [1, \infty) ) | Minimum at ( x=0 ), ( f(0)=1 ), parabola opens upwards |
| (iii) ( f(x) = 2\sqrt{x-5} ) | ( [5, \infty) ) | ( [0, \infty) ) | Starts at ( x=5 ), ( f(5)=0 ) |
| (iv) ( f(x) = |x-2| – 3 ) | ( \mathbb{R} ) | ( [-3, \infty) ) | Lowest at ( x=2 ), ( f(2)=-3 ) |
| (v) ( f(x) = 1 + \sin x ) | ( \mathbb{R} ) | ( [0, 2] ) | Since ( \sin x \in [-1,1] ) |
| (vi) ( f(x) = 3 + \sqrt{x-2} ) | ( [2, \infty) ) | ( [3, \infty) ) | Minimum at ( x=2 ), ( f(2)=3 ) |
| (vii) ( f(x) = 3e^x ) | ( \mathbb{R} ) | ( (0, \infty) ) | Exponential > 0 |
| (viii) ( f(x) = \frac{x^2-16}{x+4} ) | ( \mathbb{R} \setminus {-4} ) | ( \mathbb{R} \setminus {-8} ) | Hole at ( x=-4 ), simplifies to ( x-4 ) |
| (ix) ( f(x) = (x-1)^2 + 1 ) | ( \mathbb{R} ) | ( [1, \infty) ) | Minimum at ( x=1 ), ( f(1)=1 ) |
| (x) ( f(x) = \frac{1}{x-1} ) | ( \mathbb{R} \setminus {1} ) | ( \mathbb{R} \setminus {0} ) | Never zero |
| (xi) ( f(x) = \frac{x-2}{x+3} ) | ( \mathbb{R} \setminus {-3} ) | ( \mathbb{R} \setminus {1} ) | Horizontal asymptote |
| (xii) ( f(x) = \frac{x^2-x-6}{x-3} ) | ( \mathbb{R} \setminus {3} ) | ( \mathbb{R} \setminus {5} ) | Hole at ( x=3 ), simplifies to ( x+2 ) |
3. One-to-One & Onto
A = {0, 1, 2, 3}, B = {p, q, r, s}, f = {(0, p), (1, q), (2, r), (3, s)}
One-to-One: Outputs distinct → Yes
Onto: Range equals B → Yes
Answer: One-to-one and onto.
4. Into or Onto
A = {2, 3, 4, 5}, B = {b, c, d, e}, f = {(2, b), (3, c), (4, e), (5, e)}
One-to-One: e repeated → No
Onto: Missing d → Into
Answer: Not one-to-one, into.
5. Check One-to-One
(i) ( f(x) = 4x – 7 ) → One-to-one (assume ( f(a)=f(b) ) → ( a=b ))
(ii) ( f(x) = 6x^2 + 2 ) → Not one-to-one (( f(1)=f(-1)=8 ))
(iii) ( f(x) = \frac{x^3-1}{2} ) → One-to-one (( a^3=b^3 ) → ( a=b ))
6. Into Function
g(x) = 2x² + 3x + 1, Dom = {0,1,2,3} → Range = {1,6,15,28}
Codomain = {1,6,15,28,35} → Missing 35 → Into
7. Type of Function
h(x) = 2x + 1 → Bijective (linear, non-zero slope)
8. Show Bijective
( f(x) = \frac{x-2}{x-3} ), A = ℝ \ {3}, B = ℝ \ {1}
One-to-One: Cross-multiply → ( a=b )
Onto: Solve → ( x = \frac{3y+2}{1-y} ), valid for y ≠ 1 → Bijective
9. Find Domain and Range of ( f^{-1} ) and Verify
| Original | Inverse | Domain of ( f^{-1} ) | Range of ( f^{-1} ) | Verification |
|---|---|---|---|---|
| (i) ( f(x)=4x-3 ) | ( f^{-1}(x)=\frac{x+3}{4} ) | ( \mathbb{R} ) | ( \mathbb{R} ) | ( f(f^{-1}(x))=x ) |
| (ii) ( f(x)=\frac{x}{x-5} ) | ( f^{-1}(x)=\frac{5x}{x-1} ) | ( \mathbb{R} \setminus {1} ) | ( \mathbb{R} \setminus {5} ) | Done |
| (iii) ( f(x)=\frac{x+2}{x-1} ) | ( f^{-1}(x)=\frac{x+2}{x-1} ) | ( \mathbb{R} \setminus {1} ) | ( \mathbb{R} \setminus {1} ) | Done |
| (iv) ( f(x)=\sqrt{x+2} ) | ( f^{-1}(x)=x^2-2 ) (( x \geq 0 )) | ( [0,\infty) ) | ( [-2,\infty) ) | Done |
| (v) ( f(x)=x^2+6 ) (( x \geq 0 )) | ( f^{-1}(x)=\sqrt{x-6} ) | ( [6,\infty) ) | ( [0,\infty) ) | Done |
| (vi) ( f(x)=\frac{2x-1}{x+4} ) | ( f^{-1}(x)=\frac{-1-4x}{x-2} ) | ( \mathbb{R} \setminus {2} ) | ( \mathbb{R} \setminus {-4} ) | Done |
Exercise 1.2 – Graphing Made Easy
| Function | y-intercept | x-intercept | Slope |
|---|---|---|---|
| (i) ( f(x)=3x-2 ) | (0, -2) | (2/3, 0) | 3 |
| (ii) ( f(x)=3x ) | (0, 0) | (0, 0) | 3 |
| (iii) ( f(x)=1-2x ) | (0, 1) | (1/2, 0) | -2 |
Fig 1: Rising line (PDF Page 6)
Fig 2: Line through origin (PDF Page 8)
Fig 3: Falling line (PDF Page 10)
Exercises 1.3 to 1.5 + Review Exercise
First, we cover function composition. Then, piecewise functions. After that, limits & continuity. Finally, graph shifts.
In addition, the review exercise includes 20+ questions. Therefore, we provide step-by-step solutions aligned with FBISE marking.
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|---|---|---|
| Revise Unit 1 in 2 hours | Ready lesson plans | Fast proof checks |
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