Differentiation (Part-38) | S N De

In the previous article, we discussed several Short Answer Type Questions (Questions 1–12) of the Relation and Mapping chapter from the S.N. Dey Math Exercise. In this post, we continue with the remaining problems (Questions 13–24) as part of our comprehensive S.N. Dey Math solution series. Let's get started.

Question 13

If $f(x) = \tan^{-1} x$, find the relation connecting $f(x)$, $f(y)$, and $f(x+y)$.

Solution

Given $f(x) = \tan^{-1} x \implies x = \tan[f(x)]$  --- (1)

Similarly, $y = \tan[f(y)]$  --- (2)

Replacing $x$ with $x+y$ in (1):

\[ x+y = \tan[f(x+y)] \]

Substituting values from (1) and (2):

\[ \tan[f(x)] + \tan[f(y)] = \tan[f(x+y)] \]

Question 14

Let the function $\phi(x)$ be defined as: \[ \phi(x) = \begin{cases} 4 & : x \text{ is rational} \\[1ex] -4 & : x \text{ is irrational} \end{cases} \] Find the values of $\phi(0)$, $\phi(2.4)$, $\phi(\sqrt{3})$, $\phi(3.6)$, $\phi(\pi)$, $\phi(e)$, and $\phi\left(\sin \frac{\pi}{4}\right)$.

Solution

Since $0$, $2.4$, and $3.6$ are rational numbers:

\[ \phi(0) = \phi(2.4) = \phi(3.6) = 4 \]

Since $\sqrt{3}$, $\pi$, $e$, and $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ are irrational numbers:

\[ \phi(\sqrt{3}) = \phi(\pi) = \phi(e) = \phi\left(\sin \frac{\pi}{4}\right) = -4 \]

Question 15

Let: \[ f(x) = \begin{cases} 2x^2 - 3 & : x \leq 2 \\[1.5ex] 2x + 1 & : 2 < x \leq 3 \\[1.5ex] \frac{1}{2x - 1} & : x > 3 \end{cases} \] Find $f(0)$, $f(3)$, $f(-2)$, $f(4)$, $f(\sqrt{2})$, and $f(e)$.

Solution

(i) $f(0) = 2(0)^2 - 3 = -3$  [Since $x \leq 2$]

(ii) $f(3) = 2(3) + 1 = 7$  [Since $2 < x \leq 3$]

(iii) $f(-2) = 2(-2)^2 - 3 = 8 - 3 = 5$  [Since $x \leq 2$]

(iv) $f(4) = \frac{1}{2(4) - 1} = \frac{1}{7}$  [Since $x > 3$]

(v) $f(\sqrt{2}) = 2(\sqrt{2})^2 - 3 = 4 - 3 = 1$  [Since $\sqrt{2} \approx 1.414 \leq 2$]

(vi) $f(e) = 2e + 1$  [Since $e \approx 2.718 \implies 2 < e < 3$]

Question 16

Find the values of $x$ for which the following functions are undefined:

• (i) $\frac{x}{x+2}$
• (ii) $\sqrt{4x - 4x^2 - 1}$
• (iii) $\frac{x}{\sin x}$
• (iv) $\frac{x^2+x-6}{2x^2-x-6}$
• (v) $\sqrt{x^2 - 4x + 3}$

Solutions

(i) $\frac{x}{x+2}$ is undefined when denominator is zero:

\[ x + 2 = 0 \implies x = -2 \]

(ii) $f(x) = \sqrt{4x - 4x^2 - 1}$ is undefined when radicand is negative:

\[ 4x - 4x^2 - 1 < 0 \implies -(4x^2 - 4x + 1) < 0 \implies -(2x - 1)^2 < 0 \]

Since $(2x-1)^2 > 0$ for all $x \neq \frac{1}{2}$, the expression $-(2x-1)^2 < 0$ holds for all $x \in \mathbb{R} \setminus \{\frac{1}{2}\}$.

Thus, $f(x)$ is undefined for all $x \in \mathbb{R}$ except $x = \frac{1}{2}$.

(iii) $\frac{x}{\sin x}$ is undefined when denominator is zero:

\[ \sin x = 0 \implies x = n\pi, \quad n \in \mathbb{Z} \]

(iv) $\frac{x^2+x-6}{2x^2-x-6}$ is undefined when denominator is zero:

\[ 2x^2 - x - 6 = 0 \implies (2x+3)(x-2) = 0 \implies x = 2, -\frac{3}{2} \]

(v) $\sqrt{x^2 - 4x + 3}$ is undefined when radicand is negative:

\[ x^2 - 4x + 3 < 0 \implies (x-3)(x-1) < 0 \implies 1 < x < 3 \]

Question 17

If the maximum and minimum values of $f(x) = a + b \sin x$ are $7$ and $1$ respectively, find the value of $f\left(\frac{\pi}{6}\right)$.

Solution

Case 1: $b > 0$

\[ -1 \leq \sin x \leq 1 \implies a - b \leq a + b \sin x \leq a + b \]

Max value $= a + b = 7$, Min value $= a - b = 1$. Solving yields $a = 4, b = 3$.

\[ f(x) = 4 + 3\sin x \implies f\left(\frac{\pi}{6}\right) = 4 + 3\sin\left(\frac{\pi}{6}\right) = 4 + \frac{3}{2} = \frac{11}{2} \]

Case 2: $b < 0$

Max value $= a - b = 7$, Min value $= a + b = 1$. Solving yields $a = 4, b = -3$.

\[ f(x) = 4 - 3\sin x \implies f\left(\frac{\pi}{6}\right) = 4 - 3\sin\left(\frac{\pi}{6}\right) = 4 - \frac{3}{2} = \frac{5}{2} \]

Question 18

If $f(x) = ax^2 + bx + c$ and $f(x+1) = f(x) + x + 1$ is an identity, find the values of $a$ and $b$.

Solution

Substitute into the identity:

\[ a(x+1)^2 + b(x+1) + c = ax^2 + bx + c + x + 1 \] \[ ax^2 + 2ax + a + bx + b + c = ax^2 + (b+1)x + c + 1 \] \[ 2ax + a + b = x + 1 \]

Equating coefficients on both sides:

\[ 2a = 1 \implies a = \frac{1}{2} \] \[ a + b = 1 \implies \frac{1}{2} + b = 1 \implies b = \frac{1}{2} \]

Question 19

If $f(x) = x^2 + ax + b$, $f(1) = 1$, and $f(2) = 2$, find the value of $f(3)$.

Solution

From $f(1) = 1$:

\[ 1^2 + a(1) + b = 1 \implies a + b = 0 \quad \text{--- (1)} \]

From $f(2) = 2$:

\[ 2^2 + a(2) + b = 2 \implies 2a + b = -2 \quad \text{--- (2)} \]

Subtracting (1) from (2): $a = -2$. Substituting in (1): $b = 2$.

\[ f(x) = x^2 - 2x + 2 \] \[ f(3) = 3^2 - 2(3) + 2 = 9 - 6 + 2 = 5 \]

Question 20

If the function $f: \mathbb{R} \to \mathbb{R}$ is given by $f(x) = x, \forall x \in \mathbb{R}$ and $g: \mathbb{R}\setminus\{0\} \to \mathbb{R}$ is given by $g(x) = \frac{1}{x}, \forall x \in \mathbb{R}\setminus\{0\}$, find the functions $f+g$ and $f-g$.

Solution

The domain of definition for both $f+g$ and $f-g$ is the intersection of the domains of $f$ and $g$:

\[ D_f \cap D_g = \mathbb{R} \cap (\mathbb{R} \setminus \{0\}) = \mathbb{R} \setminus \{0\} \]

For $(f+g)$:

\[ (f+g): \mathbb{R} \setminus \{0\} \to \mathbb{R} \quad \text{defined by} \quad (f+g)(x) = x + \frac{1}{x} \]

For $(f-g)$:

\[ (f-g): \mathbb{R} \setminus \{0\} \to \mathbb{R} \quad \text{defined by} \quad (f-g)(x) = x - \frac{1}{x} \]

Question 21

If two real functions $f$ and $g$ are defined respectively by $f(x) = \sqrt{x+1}$ and $g(x) = \sqrt{x-1}$, find the values of $\frac{f}{g}(1)$ and $\frac{g}{f}(1)$, and state their domains of definition.

Solution

Evaluating at $x = 1$:

\[ \frac{f}{g}(1) = \frac{f(1)}{g(1)} = \frac{\sqrt{1+1}}{\sqrt{1-1}} = \frac{\sqrt{2}}{0} \quad \text{(Undefined)} \] \[ \frac{g}{f}(1) = \frac{g(1)}{f(1)} = \frac{\sqrt{1-1}}{\sqrt{1+1}} = \frac{0}{\sqrt{2}} = 0 \]

Domains of individual functions:

\[ D_f = [-1, \infty), \quad D_g = [1, \infty) \]

Domain of $\frac{f}{g}$: $D_f \cap D_g \setminus \{x : g(x) = 0\} = [1, \infty) \setminus \{1\} = (1, \infty)$

Domain of $\frac{g}{f}$: $D_f \cap D_g \setminus \{x : f(x) = 0\} = [1, \infty) \setminus \{-1\} = [1, \infty)$

Question 22

Find the range of the function $f(x) = 3 - |x-3|$.

Solution

For all real $x$:

\[ 0 \leq |x-3| < \infty \] \[ \implies -\infty < -|x-3| \leq 0 \] \[ \implies -\infty < 3 - |x-3| \leq 3 \]

Hence, the range of $f(x)$ is $(-\infty, 3]$.

Question 23

Find $(f \circ g \circ h)(x)$, if $f(x) = \frac{x}{x+1}$, $g(x) = x^{10}$, and $h(x) = x+3$.

Solution \[ (f \circ g \circ h)(x) = f\Big(g\big(h(x)\big)\Big) \] \[ = f\Big(g(x+3)\Big) = f\Big((x+3)^{10}\Big) = \frac{(x+3)^{10}}{(x+3)^{10} + 1} \]

Question 24

If $f(x) = \frac{4^x}{4^x + 2}$, show that $f(x) + f(1-x) = 1$. Hence, prove that: \[ f\left(\frac{1}{1997}\right) + f\left(\frac{2}{1997}\right) + \dots + f\left(\frac{1996}{1997}\right) = 998 \]

Solution

Part 1: Proof of identity

\[ f(1-x) = \frac{4^{1-x}}{4^{1-x} + 2} = \frac{\frac{4}{4^x}}{\frac{4}{4^x} + 2} = \frac{4}{4 + 2 \cdot 4^x} = \frac{2}{2 + 4^x} \] \[ f(x) + f(1-x) = \frac{4^x}{4^x + 2} + \frac{2}{4^x + 2} = \frac{4^x + 2}{4^x + 2} = 1 \]

Part 2: Series evaluation

Pairing terms of the form $f(k) + f(1-k) = 1$:

\[ \left[f\left(\frac{1}{1997}\right) + f\left(\frac{1996}{1997}\right)\right] + \left[f\left(\frac{2}{1997}\right) + f\left(\frac{1995}{1997}\right)\right] + \dots \]

There are $\frac{1996}{2} = 998$ such pairs, each summing to $1$.

\[ \text{Total sum} = 1 + 1 + \dots \text{ (998 times)} = 998 \]

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