Relations & Functions

Miscellaneous Exercise • Chapter 2
1. Show that f is a function and g is not.
Analysis of f(x)

Defined as $x^2$ for $0 \le x \le 3$ and $3x$ for $3 \le x \le 10$.

At the split point $x = 3$:

  • LHL ($x^2$): $3^2 = 9$
  • RHL ($3x$): $3(3) = 9$

Since both definitions give the same value (9), the image is unique.
f is a function

Analysis of g(x)

Defined as $x^2$ for $0 \le x \le 2$ and $3x$ for $2 \le x \le 10$.

At the split point $x = 2$:

  • Case 1 ($x^2$): $2^2 = 4$
  • Case 2 ($3x$): $3(2) = 6$

Since $4 \neq 6$, the element 2 has two different images.
g is not a function

2. If $f(x) = x^2$, find $\frac{f(1.1) – f(1)}{1.1 – 1}$.

Given $f(x) = x^2$.

$$f(1.1) = (1.1)^2 = 1.21$$

$$f(1) = (1)^2 = 1$$

Value = $$\frac{1.21 – 1}{1.1 – 1} = \frac{0.21}{0.1} = \frac{2.1}{1}$$

Answer: 2.1
3. Find the domain of $f(x) = \frac{x^2 + 2x + 1}{x^2 – 8x + 12}$.

For a rational function to be defined, the denominator cannot be zero.

$$x^2 – 8x + 12 \neq 0$$ $$(x – 6)(x – 2) \neq 0$$ $$x \neq 6 \text{ and } x \neq 2$$

Domain: $R – \{2, 6\}$

4. Find Domain and Range of $f(x) = \sqrt{x-1}$.

Domain: Quantity under root must be non-negative.

$$x – 1 \ge 0 \Rightarrow x \ge 1$$

Domain: $[1, \infty)$


Range: For $x \ge 1$, $\sqrt{x-1}$ produces non-negative real numbers.

Range: $[0, \infty)$

5. Find Domain and Range of $f(x) = |x-1|$.

Domain: Modulus function is defined for all real numbers.

Domain: R


Range: The absolute value is always non-negative.

$$|x-1| \ge 0$$

Range: $[0, \infty)$

6. Determine range of $f(x) = \frac{x^2}{1+x^2}$.

Let $y = \frac{x^2}{1+x^2}$.

Since $x^2 \ge 0$, then $y \ge 0$.
Also, $x^2 < 1 + x^2$ (Denominator is always greater than numerator).
Therefore, the fraction is always less than 1.

Combining these: $0 \le y < 1$.

Range: $[0, 1)$

7. Find $f+g$, $f-g$, $f/g$.

Given $f(x) = x + 1$ and $g(x) = 2x – 3$.

  • (f + g)(x): $(x+1) + (2x-3) = 3x – 2$
  • (f – g)(x): $(x+1) – (2x-3) = -x + 4$
  • (f / g)(x): $\frac{x+1}{2x-3}, \quad x \neq \frac{3}{2}$
8. Determine integers a, b for $f(x) = ax + b$.

Given pairs: $(1,1), (2,3), (0,-1)$.

Using $(0, -1)$: $f(0) = a(0) + b = -1 \Rightarrow b = -1$.
Using $(1, 1)$: $f(1) = a(1) + b = 1$.
Substitute b: $a – 1 = 1 \Rightarrow a = 2$.

Answer: $a = 2, b = -1$.

9. Relation $R = \{(a, b) : a = b^2\}$ on N. True or False?

(i) $(a, a) \in R$ for all $a \in N$.

False
For $a=2$, $2 \neq 2^2$.

(ii) $(a, b) \in R \implies (b, a) \in R$.

False
Let $a=4, b=2$. $4 = 2^2$ is true. But $2 \neq 4^2$.

(iii) Transitive Property.

False
Let $a=16, b=4, c=2$.
$16 = 4^2$ (True), $4 = 2^2$ (True).
But $16 \neq 2^2$ (False).
10. Check if f is a relation/function.

Set $A = \{1, 2, 3, 4\}$, Set $B = \{1, 5, 9, 11, 15, 16\}$.
$f = \{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)\}$.

(i) Is f a relation?

True

It is a subset of $A \times B$.

(ii) Is f a function?

False
The element 2 in the domain corresponds to two different images: 9 and 11.
11. Is $f = \{(ab, a+b)\}$ a function?

We check if one input ($ab$) yields unique output ($a+b$).

No, it is not a function
Counter-Example
Consider the product 6.
Case 1: $a=1, b=6 \Rightarrow ab=6$. Output $a+b = 7$. Pair $(6, 7)$.
Case 2: $a=2, b=3 \Rightarrow ab=6$. Output $a+b = 5$. Pair $(6, 5)$.
Since input 6 maps to 7 and 5, it is not a function.
12. Find the range of f(n) = highest prime factor of n.

Domain $A = \{9, 10, 11, 12, 13\}$.

  • $n=9 = 3 \times 3 \rightarrow$ Max Prime: 3
  • $n=10 = 2 \times 5 \rightarrow$ Max Prime: 5
  • $n=11 = 11 \rightarrow$ Max Prime: 11
  • $n=12 = 2^2 \times 3 \rightarrow$ Max Prime: 3
  • $n=13 = 13 \rightarrow$ Max Prime: 13

Range: $\{3, 5, 11, 13\}$

learncbsehub.in