Differentiation (Part-38) | S N De

In the previous article, we discussed Long Answer Type Questions (Questions 1–6) of the Relation and Mapping Chapter from the S.N. Dey Math Exercise. In this post, we complete the remaining Long Answer Type Questions (Questions 7–12) as a part of our step-by-step S.N. Dey Math solution series. Let's get started.

Question 7 (i)

Find the domain of definition of $f(x) = \sqrt{x+1} + \sqrt{4-x}$.

Solution

$f(x)$ will be defined if $x+1 \geq 0$ and $4-x \geq 0$ simultaneously.

\[ x \geq -1 \quad \text{and} \quad x \leq 4 \implies -1 \leq x \leq 4 \]

Therefore, the domain of definition of $f(x)$ is:

\[ \text{Domain} = \{x \in \mathbb{R} : -1 \leq x \leq 4\} \]

Question 7 (ii)

Find the domain of definition of $\phi(x) = \cos^{-1}\left(\frac{x-4}{3}\right) + \log(5-x)$.

Solution

The function $\phi(x)$ will be defined if $5-x > 0$ and $-1 \leq \frac{x-4}{3} \leq 1$.

\[ 5-x > 0 \implies x < 5 \] \[ -1 \leq \frac{x-4}{3} \leq 1 \implies -3 \leq x-4 \leq 3 \implies 1 \leq x \leq 7 \]

Taking the intersection of $x < 5$ and $1 \leq x \leq 7$ gives $1 \leq x < 5$.

So, the domain of definition of $\phi(x)$ is:

\[ \text{Domain} = \{x \in \mathbb{R} : 1 \leq x < 5\} \]

Question 8

The function $f(x)$ is defined in $0 \leq x \leq 1$. Find the domain of definition of:

• (i) $f(2x-1)$
• (ii) $f(x^2)$

Solution to 8(i)

The function $f(x)$ is defined for $0 \leq x \leq 1$. Thus, $f(2x-1)$ is defined when:

\[ 0 \leq 2x-1 \leq 1 \] \[ \implies 1 \leq 2x \leq 2 \] \[ \implies \frac{1}{2} \leq x \leq 1 \]

So, the domain of definition is: $\{x \in \mathbb{R} : \frac{1}{2} \leq x \leq 1\}$.

Solution to 8(ii)

The function $f(x^2)$ will be defined if:

\[ 0 \leq x^2 \leq 1 \] \[ \implies -1 \leq x \leq 1 \]

So, the domain of definition is: $\{x \in \mathbb{R} : -1 \leq x \leq 1\}$.

Question 9

If two real functions $f(x)$ and $\phi(x)$ are defined respectively by $f(x) = \sqrt{x-2}$ and $\phi(x) = x+3$, find each of the following functions along with their domain of definition:

• (i) $\frac{1}{f}$
• (ii) $\frac{1}{\phi}$
• (iii) $f + \phi$
• (iv) $f \phi$
• (v) $\frac{f}{\phi}$

Solutions

(i) $\left(\frac{1}{f}\right)(x) = \frac{1}{f(x)} = \frac{1}{\sqrt{x-2}}$

This is defined when $x-2 > 0 \implies x > 2$.

\[ \left(\frac{1}{f}\right)(x) = \frac{1}{\sqrt{x-2}}, \quad 2 < x < \infty \]

(ii) $\left(\frac{1}{\phi}\right)(x) = \frac{1}{\phi(x)} = \frac{1}{x+3}$

This is defined when $x+3 \neq 0 \implies x \neq -3$.

\[ \left(\frac{1}{\phi}\right)(x) = \frac{1}{x+3}, \quad x \in \mathbb{R} \setminus \{-3\} \]

(iii) $(f + \phi)(x) = f(x) + \phi(x) = \sqrt{x-2} + x + 3$

This is defined when $x-2 \geq 0 \implies x \geq 2$.

\[ (f + \phi)(x) = \sqrt{x-2} + x + 3, \quad 2 \leq x < \infty \]

(iv) $(f \phi)(x) = f(x) \phi(x) = \sqrt{x-2}(x+3)$

This is defined when $x-2 \geq 0 \implies x \geq 2$.

\[ (f \phi)(x) = \sqrt{x-2}(x+3), \quad 2 \leq x < \infty \]

(v) $\left(\frac{f}{\phi}\right)(x) = \frac{f(x)}{\phi(x)} = \frac{\sqrt{x-2}}{x+3}$

This is defined when $x-2 \geq 0 \implies x \geq 2$ (since $x+3 \neq 0$ is automatically satisfied for $x \geq 2$).

\[ \left(\frac{f}{\phi}\right)(x) = \frac{\sqrt{x-2}}{x+3}, \quad 2 \leq x < \infty \]

Question 10

Find the domain of definition of the function defined by $f(x) = \frac{1}{\log_e(2-x)} + \sqrt{x+3}$.

Solution

The given function will be defined if:

1. Argument of logarithm is positive: $2-x > 0 \implies x < 2$
2. Denominator of logarithm is non-zero: $\log_e(2-x) \neq 0 \implies 2-x \neq 1 \implies x \neq 1$
3. Radicand is non-negative: $x+3 \geq 0 \implies x \geq -3$

Combining these conditions:

\[ -3 \leq x < 2 \quad \text{and} \quad x \neq 1 \]

So, the required domain of definition is:

\[ \text{Domain} = \{x \in \mathbb{R} : -3 \leq x < 1 \;\cup\; 1 < x < 2\} \]

Question 11

Find the domain of definition of each of the following functions:

• (i) $f(x) = \frac{1}{\sqrt{x + |x|}}$
• (ii) $g(x) = \frac{1}{\sqrt{x - |x|}}$
• (iii) $h(x) = \sqrt{x + [x]}$
• (iv) $\phi(x) = \sqrt{x - [x]}$

Solution to (i)

$f(x)$ is defined if $x + |x| > 0$.

For $x > 0$, $x + |x| = 2x > 0$. For $x \leq 0$, $x + |x| = 0$.

So, the required domain of definition is: $\{x \in \mathbb{R} : 0 < x < \infty\}$.

Solution to (ii)

$g(x)$ is defined if $x - |x| > 0$.

Since $|x| \geq x$ for all $x \in \mathbb{R}$, $x - |x| \leq 0$ always. Thus, $x - |x| > 0$ is impossible.

So, the required domain of definition is: $\emptyset$ (Null Set).

Solution to (iii)

$h(x)$ is defined if $x + [x] \geq 0$.

For $x \geq 0$, $[x] \geq 0 \implies x + [x] \geq 0$. For $x < 0$, $x + [x] < 0$.

So, the required domain of definition is: $\{x \in \mathbb{R} : x \geq 0\}$.

Solution to (iv)

$\phi(x)$ is defined if $x - [x] \geq 0$.

Since $x - [x] = \{x\}$ (fractional part function) and $\{x\} \geq 0$ for all $x \in \mathbb{R}$.

So, the required domain of definition is: $\mathbb{R}$ (all real numbers).

Question 12

Find the range of each of the following functions:

• (i) $f(x) = 2 - |x-2|$
• (ii) $f(x) = \frac{1}{\sqrt{x + [x]}}$
• (iii) $h(x) = \frac{|x-3|}{x-3}$

Solution to (i)

We know $0 \leq |x-2| < \infty$ for all real $x$.

\[ \implies -\infty < -|x-2| \leq 0 \] \[ \implies -\infty < 2 - |x-2| \leq 2 \]

So, the range of $f(x)$ is: $(-\infty, 2]$.

Solution to (ii)

The function is defined when $x + [x] > 0 \implies x > 0$.

For $x > 0$, $0 < x + [x] < \infty \implies 0 < \sqrt{x + [x]} < \infty$.

Taking the reciprocal gives values in $(0, \infty)$.

So, the range of the function is: $(0, \infty)$.

Solution to (iii)

Expressing $h(x)$ piecewise for $x \neq 3$:

\[ |x-3| = \begin{cases} x-3 & \text{if } x > 3 \\ -(x-3) & \text{if } x < 3 \end{cases} \] \[ \implies h(x) = \frac{|x-3|}{x-3} = \begin{cases} 1 & \text{if } x > 3 \\ -1 & \text{if } x < 3 \end{cases} \]

So, the range of $h(x)$ consists of only two elements: $\{-1, 1\}$.

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