Question

Which of the following is/are NOT a pair of identical functions:

$\square$ $f(x) = \log x^2; g(x) = 2 \log x$

$\square$ $f(x) = \sin (\sin^{-1} x); g(x) = \cos (\cos^{-1} x)$

$\square$ $f(x) = \frac{1+2 \cos x}{\sin x(2+\sec x)}, g(x) = \tan x$

$\square$ $f(x) = \cot^2 x . \cos^2 x, g(x) = \cot^2 x - \cos^2 x$

• A. $f(x) = \log x^2; g(x) = 2 \log x$
• B. $f(x) = \sin (\sin^{-1} x); g(x) = \cos (\cos^{-1} x)$
• C. $f(x) = \frac{1+2 \cos x}{\sin x(2+\sec x)}, g(x) = \tan x$
• D. $f(x) = \cot^2 x . \cos^2 x, g(x) = \cot^2 x - \cos^2 x$

Answer 1

Answer: (B), (D)

Pair 1 and Pair 3 are not identical functions.

Answer 2

To determine which pairs of functions are not identical, we need to compare their domains and their functional values over the common domain. Two functions $f(x)$ and $g(x)$ are identical if and only if:

1. Domain of $f$ = Domain of $g$.
2. $f(x) = g(x)$ for all $x$ in the common domain.

Let's examine each pair:

Pair 1: $f(x) = \log x^2; g(x) = 2 \log x$

• Domain of $f(x)$: The argument of the logarithm must be positive, so $x^2 > 0$. This is true for all $x \in \mathbb{R}$ except $x=0$. Domain of $f$ is $(-\infty, 0) \cup (0, \infty)$.
• Domain of $g(x)$: The argument of the logarithm must be positive, so $x > 0$. Domain of $g$ is $(0, \infty)$.

Since the domains are different, the functions are not identical.

Pair 2: $f(x) = \sin (\sin^{-1} x); g(x) = \cos (\cos^{-1} x)$

• Domain of $f(x)$: The function $\sin^{-1} x$ is defined for $x \in [-1, 1]$. The function $\sin y$ is defined for all real $y$. So, the domain of $f(x)$ is $[-1, 1]$. For $x \in [-1, 1]$, $\sin (\sin^{-1} x) = x$. Thus, $f(x) = x$ for $x \in [-1, 1]$.
• Domain of $g(x)$: The function $\cos^{-1} x$ is defined for $x \in [-1, 1]$. The function $\cos y$ is defined for all real $y$. So, the domain of $g(x)$ is $[-1, 1]$. For $x \in [-1, 1]$, $\cos (\cos^{-1} x) = x$. Thus, $g(x) = x$ for $x \in [-1, 1]$.

The domains are the same, $[-1, 1]$. For all $x$ in this domain, $f(x) = x$ and $g(x) = x$, so $f(x) = g(x)$. The functions are identical.

Pair 3: $f(x) = \frac{1+2 \cos x}{\sin x(2+\sec x)}, g(x) = \tan x$

• Domain of $f(x)$: Restrictions arise from denominators and the definition of $\sec x$. $\sin x \neq 0 \implies x \neq n\pi, n \in \mathbb{Z}$. $\cos x \neq 0 \implies x \neq (n + \frac{1}{2})\pi, n \in \mathbb{Z}$ (due to $\sec x$). $2 + \sec x \neq 0 \implies 2 + \frac{1}{\cos x} \neq 0 \implies \frac{2 \cos x + 1}{\cos x} \neq 0 \implies 2 \cos x + 1 \neq 0 \implies \cos x \neq -\frac{1}{2}$. $\cos x = -\frac{1}{2}$ for $x = 2n\pi \pm \frac{2\pi}{3}, n \in \mathbb{Z}$. Domain of $f$ is $\mathbb{R} \setminus \{n\pi\} \setminus \{(n + \frac{1}{2})\pi\} \setminus \{2n\pi \pm \frac{2\pi}{3}\}$.
• Domain of $g(x) = \tan x = \frac{\sin x}{\cos x}$: $\cos x \neq 0 \implies x \neq (n + \frac{1}{2})\pi, n \in \mathbb{Z}$.

The domains are different. For example, $x=\pi$ is in the domain of $g(x)$ but not in the domain of $f(x)$. Thus, the functions are not identical.

Pair 4: $f(x) = \cot^2 x . \cos^2 x, g(x) = \cot^2 x - \cos^2 x$

• Domain of $f(x)$: $\cot x = \frac{\cos x}{\sin x}$ is defined when $\sin x \neq 0$, so $x \neq n\pi, n \in \mathbb{Z}$. $\cos x$ is defined everywhere. Domain of $f$ is $\mathbb{R} \setminus \{n\pi\}$.
• Domain of $g(x)$: $\cot x = \frac{\cos x}{\sin x}$ is defined when $\sin x \neq 0$, so $x \neq n\pi, n \in \mathbb{Z}$. $\cos x$ is defined everywhere. Domain of $g$ is $\mathbb{R} \setminus \{n\pi\}$.

The domains are the same.

Now let's check if $f(x) = g(x)$ for all $x$ in the domain:

$g(x) = \cot^2 x - \cos^2 x = \frac{\cos^2 x}{\sin^2 x} - \cos^2 x = \cos^2 x \left( \frac{1}{\sin^2 x} - 1 \right) = \cos^2 x \left( \frac{1 - \sin^2 x}{\sin^2 x} \right)$

$g(x) = \cos^2 x \left( \frac{\cos^2 x}{\sin^2 x} \right) = \cos^2 x \cot^2 x = f(x)$.

Since the domains are the same and $f(x) = g(x)$ for all $x$ in the domain, the functions are identical.

The question asks which of the following is/are NOT a pair of identical functions. Based on our analysis, Pair 1 and Pair 3 are not identical functions.