Class 11 Maths Ch 2: Relations HOTS Worksheet

Q1. Find the domain and range of the function $f(x) = \frac{x}{1+x^2}$.

Worked Solution:
Domain: Denominator $1+x^2 \ne 0$, which is true for all real $x$. So, domain is $R$.
Range: Let $y = \frac{x}{1+x^2}$.
$\Rightarrow y + x^2y = x \Rightarrow yx^2 - x + y = 0$.
For x to be real, $D \ge 0$.
$(-1)^2 - 4(y)(y) \ge 0 \Rightarrow 1 - 4y^2 \ge 0 \Rightarrow 4y^2 \le 1 \Rightarrow y^2 \le \frac{1}{4}$.
$\Rightarrow -\frac{1}{2} \le y \le \frac{1}{2}$.
Range = $[-\frac{1}{2}, \frac{1}{2}]$.
Q2. Find the domain of the function $f(x) = \sqrt{x-1} + \sqrt{5-x}$.

Worked Solution:
For definition, $x-1 \ge 0 \Rightarrow x \ge 1$.
And $5-x \ge 0 \Rightarrow x \le 5$.
Combining both, $1 \le x \le 5$.
Domain = $[1, 5]$.
Q3. If $f(x) = \log\left(\frac{1+x}{1-x}\right)$, show that $f(x) + f(y) = f\left(\frac{x+y}{1+xy}\right)$.

Worked Solution:
$f(x) + f(y) = \log\left(\frac{1+x}{1-x}\right) + \log\left(\frac{1+y}{1-y}\right) = \log\left(\frac{(1+x)(1+y)}{(1-x)(1-y)} ight)$
$= \log\left(\frac{1+x+y+xy}{1-x-y+xy}\right)$.
Now, $f\left(\frac{x+y}{1+xy}\right) = \log\left(\frac{1 + \frac{x+y}{1+xy}}{1 - \frac{x+y}{1+xy}}\right) = \log\left(\frac{1+xy+x+y}{1+xy-x-y}\right)$
$= \log\left(\frac{1+x+y+xy}{1-x-y+xy}\right)$.
Hence proved.
Q4. Find the range of $f(x) = 1 - |x-3|$.

Worked Solution:
We know that $|x-3| \ge 0$ for all $x \in R$.
Multiply by -1: $-|x-3| \le 0$.
Add 1: $1 - |x-3| \le 1$.
So, $f(x) \le 1$.
Range = $(-\infty, 1]$.
Q5. Let $f(x) = \sqrt{x}$ and $g(x) = x$. Find the domain of $(f+g)(x)$.

Worked Solution:
$(f+g)(x) = f(x) + g(x) = \sqrt{x} + x$.
Domain of $f$ is $[0, \infty)$. Domain of $g$ is $R$.
Domain of sum is intersection of domains.
Domain = $[0, \infty) \cap R = [0, \infty)$.
Q6. Find the domain of $f(x) = \frac{1}{\sqrt{9-x^2}}$.

Worked Solution:
For $f(x)$ to be defined, $9-x^2 > 0$ (strictly greater because it's in denominator).
$x^2 < 9 \Rightarrow -3 < x < 3$.
Domain = $(-3, 3)$.
Q7. If $f(x) = \frac{ax+b}{cx+d}$, then find $(fof)(x)$ provided $a^2 + bc = 0$ is false (check condition for identity). Actually, let's find $f(f(x))$.

Worked Solution:
$f(f(x)) = f\left(\frac{ax+b}{cx+d}\right) = \frac{a(\frac{ax+b}{cx+d})+b}{c(\frac{ax+b}{cx+d})+d}$
$= \frac{a^2x + ab + bcx + bd}{acx + bc + cdx + d^2}$
$= \frac{x(a^2+bc) + b(a+d)}{cx(a+d) + (bc+d^2)}$.
(Usually specific values are given to simplify to x).
Q8. Relation $R$ is defined on the set of integers $Z$ as $(x, y) \in R \Leftrightarrow x^2 + y^2 = 25$. Express R as a set of ordered pairs and find limit domain.

Worked Solution:
We need integer solutions to $x^2 + y^2 = 25$.
Possible squares $\le 25$: 0, 1, 4, 9, 16, 25.
Pairs summing to 25: (0, 25) and (9, 16).
So possible $(x^2, y^2)$ pairs: $(0, 25), (25, 0), (9, 16), (16, 9)$.
Usually means $x = 0, y = \pm 5$; $x = \pm 5, y = 0$; $x = \pm 3, y = \pm 4$; $x = \pm 4, y = \pm 3$.
$R = \{(0, 5), (0, -5), (5, 0), (-5, 0), (3, 4), (3, -4), (-3, 4), (-3, -4), (4, 3), (4, -3), (-4, 3), (-4, -3)\}$.
Domain = $\{0, \pm 3, \pm 4, \pm 5\}$.
Q9. Find the range of $f(x) = [x] - x$, where $[x]$ is the greatest integer function.

Worked Solution:
$f(x) = - (x - [x]) = - \{x\}$, where $\{x\}$ is fractional part.
We know $0 \le \{x\} < 1$.
So $0 \ge -\{x\} > -1$.
Range = $(-1, 0]$.
Q10. Let $f(x) = x^2$ and $g(x) = |x|$. Find the values of $x$ for which $(f+g)(x) = 0$.

Worked Solution:
$(f+g)(x) = f(x) + g(x) = x^2 + |x| = 0$.
Since $x^2 \ge 0$ and $|x| \ge 0$, their sum can be zero only if each term is zero.
$x^2 = 0 \Rightarrow x = 0$.
$|x| = 0 \Rightarrow x = 0$.
So, $x = 0$.

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Chapter 2: Relations and Functions