Find comprehensive, solved answers for the NCERT Class 11 Maths Chapter 2 Miscellaneous Exercise . Master advanced concepts: proving a relation is not a function by checking overlapping intervals, finding the Domain and Range of complex functions ($f(x) = \frac{x^2}{1+x^2}$, $\sqrt{x-1}$), solving for unknown constants in a function, and checking the properties of relations (Reflexivity, Symmetry, Transitivity). Essential preparation for higher-level math.
This exercise reviews key concepts, including identifying functions, finding the domain and range of real functions, function algebra, and properties of relations.
1. Show $f$ is a function and $g$ is not a function
A relation is a function if every element in its domain is mapped to exactly one element in its range. The only potential issue arises where the defining intervals meet (i.e., at the point of overlap).
Relation $f$:
$$f(x) = \begin{cases} x^2, & 0 \le x \le 3 \\ 3x, & 3 \le x \le 10 \end{cases}$$
The domains overlap at $x=3$. We check the image of $x=3$ in both definitions:
- Using $f(x) = x^2$: $f(3) = 3^2 = 9$.
- Using $f(x) = 3x$: $f(3) = 3(3) = 9$.Since $f(3)$ is uniquely defined as 9 by both parts, the relation $f$ maps $x=3$ to only one image.Therefore, $f$ is a function.
Relation $g$:
$$g(x) = \begin{cases} x^2, & 0 \le x \le 2 \\ 3x, & 2 \le x \le 10 \end{cases}$$
The domains overlap at $x=2$. We check the image of $x=2$ in both definitions:
- Using $g(x) = x^2$: $g(2) = 2^2 = 4$.
- Using $g(x) = 3x$: $g(2) = 3(2) = 6$.Since $g(2)$ is defined as 4 by the first rule and 6 by the second rule, the element $x=2$ has two distinct images (4 and 6). This means $g$ contains the ordered pairs $(2, 4)$ and $(2, 6)$.Therefore, $g$ is not a function.
2. Evaluating a Function Expression
Given $f(x) = x^2$. We need to find the value of $\frac{f(1.1) – f(1)}{1.1 – 1}$.
- Calculate $f(1.1)$ and $f(1)$:$$f(1.1) = (1.1)^2 = 1.21$$$$f(1) = (1)^2 = 1$$
- Substitute into the expression:$$\frac{f(1.1) – f(1)}{1.1 – 1} = \frac{1.21 – 1}{0.1}$$$$= \frac{0.21}{0.1}$$$$= \mathbf{2.1}$$
3. Finding the Domain of a Rational Function
Given the function $f(x) = \frac{x^2 + 2x + 1}{x^2 – 8x + 12}$.
The domain of a rational function is the set of all real numbers ($\mathbf{R}$) except for the values of $x$ that make the denominator equal to zero.
- Set the denominator to zero:$$x^2 – 8x + 12 = 0$$
- Factor the quadratic equation:$$(x – 2)(x – 6) = 0$$
- Find the excluded values of $x$:$$x = 2 \quad \text{or} \quad x = 6$$
The domain is all real numbers excluding 2 and 6.
$$\mathbf{\text{Domain}(f) = \mathbf{R} – \{2, 6\}}$$
4. Domain and Range of $f(x) = \sqrt{x – 1}$
(i) Domain:
For the function to be real-valued, the expression under the square root must be non-negative:
$$x – 1 \ge 0$$
$$x \ge 1$$
$$\mathbf{\text{Domain}(f) = [1, \infty)}$$
(ii) Range:
Since $x \ge 1$, we have $x – 1 \ge 0$. The square root function always yields a non-negative result.
Thus, $f(x) = \sqrt{x – 1} \ge 0$.
$$\mathbf{\text{Range}(f) = [0, \infty)}$$
5. Domain and Range of $f(x) = |x – 1|$
(i) Domain:
Since $x$ can be any real number, the absolute value is defined for all $\mathbf{R}$.
$$\mathbf{\text{Domain}(f) = \mathbf{R}}$$
(ii) Range:
The absolute value function, regardless of the expression inside, always yields a non-negative result. The minimum value of $|x – 1|$ is 0 (when $x=1$).
Thus, $f(x) = |x – 1| \ge 0$.
$$\mathbf{\text{Range}(f) = [0, \infty)}$$
6. Determining the Range of $f(x) = \frac{x^2}{1 + x^2}$
Let $y = f(x)$. We want to find the possible values of $y$.
- Lower Bound: Since $x^2 \ge 0$, the denominator $1 + x^2$ is always greater than the numerator $x^2$.
- Also, since $x^2 \ge 0$ and $1 + x^2 > 0$, the fraction $y = \frac{x^2}{1 + x^2}$ must be $\ge 0$.
- When $x=0$, $y = 0/1 = 0$.
- Thus, $y \ge 0$.
- Upper Bound: We can rewrite the expression:$$y = \frac{x^2}{1 + x^2} = \frac{1 + x^2 – 1}{1 + x^2} = 1 – \frac{1}{1 + x^2}$$
- Since $x^2 \ge 0$, we have $1 + x^2 \ge 1$.
- Therefore, $0 < \frac{1}{1 + x^2} \le 1$.
- Multiplying by $-1$ reverses the inequality: $-1 \le -\frac{1}{1 + x^2} < 0$.
- Adding 1: $1 – 1 \le 1 – \frac{1}{1 + x^2} < 1 + 0$.$$0 \le y < 1$$
Combining the bounds, the range is $[0, 1)$.
$$\mathbf{\text{Range}(f) = [0, 1)}$$
7. Function Algebra: $f(x) = x + 1$ and $g(x) = 2x – 3$
(i) $f + g$
$$(f + g)(x) = f(x) + g(x)$$
$$(f + g)(x) = (x + 1) + (2x – 3) = \mathbf{3x – 2}$$
(ii) $f – g$
$$(f – g)(x) = f(x) – g(x)$$
$$(f – g)(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = \mathbf{-x + 4}$$
(iii) $f/g$
$$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x + 1}{2x – 3}$$
The domain of $f/g$ is $\mathbf{R}$ excluding values where $g(x) = 0$.
$$2x – 3 = 0 \implies x = 3/2$$
$$\mathbf{\left(\frac{f}{g}\right)(x) = \frac{x + 1}{2x – 3}, \text{ Domain: } \mathbf{R} – \{3/2\}}$$
8. Determining $a$ and $b$ for $f(x) = ax + b$
Given the function $f = \{(1, 1), (2, 3), (0, -1), (-1, -3)\}$ and $f(x) = ax + b$.
We use two ordered pairs to form a system of linear equations.
- Using $(0, -1)$:$$f(0) = a(0) + b = -1 \implies \mathbf{b = -1}$$
- Using $(1, 1)$ and $b = -1$:$$f(1) = a(1) + b = 1$$$$a + (-1) = 1$$$$a = 1 + 1 \implies \mathbf{a = 2}$$
The function is $f(x) = 2x – 1$.
(Check with $(2, 3)$: $f(2) = 2(2) – 1 = 3$. Correct. Check with $(-1, -3)$: $f(-1) = 2(-1) – 1 = -3$. Correct.)
9. Properties of Relation $R = \{(a, b) : a, b \in N \text{ and } a = b^2\}$
(i) $(a, a) \in R$, for all $a \in N$ (Reflexive)?
False. For $(a, a) \in R$, we need $a = a^2$. This is only true if $a=1$.
Counterexample: Let $a=2$. Then $2 = 2^2$ is false. So $(2, 2) \notin R$.
(ii) $(a, b) \in R$ implies $(b, a) \in R$ (Symmetric)?
False. If $(a, b) \in R$, then $a = b^2$. For $(b, a) \in R$, we need $b = a^2$.
Counterexample: Let $b=2$. Then $a = 2^2 = 4$. So $(4, 2) \in R$.
However, for $(2, 4)$ to be in $R$, we need $2 = 4^2$, which is false. So $(2, 4) \notin R$.
(iii) $(a, b) \in R$, $(b, c) \in R$ implies $(a, c) \in R$ (Transitive)?
False.
- $(a, b) \in R \implies a = b^2$
- $(b, c) \in R \implies b = c^2$
- For $(a, c) \in R$, we need $a = c^2$.
Counterexample: Let $c=2$.
- $b = c^2 = 2^2 = 4$. $(4, 2) \in R$.
- $a = b^2 = 4^2 = 16$. $(16, 4) \in R$.If $R$ were transitive, $(16, 2)$ must be in $R$. But $(16, 2) \in R$ requires $16 = 2^2$, which is false.
10. Relation $f = \{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)\}$
Given $A = \{1, 2, 3, 4\}$ and $B = \{1, 5, 9, 11, 15, 16\}$.
(i) $f$ is a relation from $A$ to $B$.
True. A relation from $A$ to $B$ is any subset of $A \times B$. We check if all pairs $(x, y) \in f$ satisfy $x \in A$ and $y \in B$.
- (1, 5): $1 \in A, 5 \in B$ ($\checkmark$)
- (2, 9): $2 \in A, 9 \in B$ ($\checkmark$)
- (3, 1): $3 \in A, 1 \in B$ ($\checkmark$)
- (4, 5): $4 \in A, 5 \in B$ ($\checkmark$)
- (2, 11): $2 \in A, 11 \in B$ ($\checkmark$)Since all pairs are in $A \times B$, $f$ is a relation from $A$ to $B$.
(ii) $f$ is a function from $A$ to $B$.
False. For $f$ to be a function, every element in the domain must map to a unique element.
The element $x = 2$ is associated with two different images:
- $(2, 9)$
- $(2, 11)$Since 2 maps to both 9 and 11, $f$ is not a function.
11. Is $f = \{(ab, a + b) : a, b \in Z\}$ a function from $Z$ to $Z$?
False. We need to check if a single input (the product $ab$) can result in two different outputs (the sum $a+b$).
Let the input be $x = 6$. We can achieve $x=6$ with different integer pairs $(a, b)$:
- Case 1: $a=2, b=3$. Input $ab = 6$. Output $a+b = 2+3 = 5$. Pair is $(6, 5)$.
- Case 2: $a=1, b=6$. Input $ab = 6$. Output $a+b = 1+6 = 7$. Pair is $(6, 7)$.
Since the input $x=6$ is associated with two different outputs (5 and 7), $f$ is not a function.
$f$ is not a function.
12. Range of $f(n) = \text{highest prime factor of } n$
Given $A = \{9, 10, 11, 12, 13\}$. $f: A \to N$.
We find the highest prime factor for each element $n \in A$:
- $f(9)$: Prime factors of $9$ are $\{3\}$. Highest is $\mathbf{3}$.
- $f(10)$: Prime factors of $10$ are $\{2, 5\}$. Highest is $\mathbf{5}$.
- $f(11)$: Prime factors of $11$ are $\{11\}$. Highest is $\mathbf{11}$.
- $f(12)$: Prime factors of $12$ are $\{2, 3\}$. Highest is $\mathbf{3}$.
- $f(13)$: Prime factors of $13$ are $\{13\}$. Highest is $\mathbf{13}$.
The range is the set of the outputs $\{3, 5, 11, 3, 13\}$.
$$\mathbf{\text{Range}(f) = \{3, 5, 11, 13\}}$$