Question

Choose the correct pair(s) of equal functions.

• A. $\sqrt{x^2},x$
• B. $\sqrt{x^2-1},\sqrt{x-1}.\sqrt{x+1}$
• C. $\tan x,\frac{1}{\cot x}$
• D. $\text{sgn}(x^2+1),{\sin }^{2}x+{\cos }^{2}x$

Answer 1

The correct option is D $\text{sgn}(x^2+1),{\sin }^{2}x+{\cos }^{2}x$

Two functions $f$ and $g$ are said to be identical or equal functions, if
(i) Domain of $f$ = Domain of $g$,
(ii) Range of $f$ = Range of $g$ and
(iii) $f\left(x\right)=g\left(x\right),\forall x\in \text{Domain}$

(a) $f\left(x\right)=\tan x$ and $g\left(x\right)=\frac{1}{\cot x}$
$x=0\in$ Domain of $\tan x$ but
$\cot x$ is not defined at $x=0$
$\Rightarrow g\left(x\right)=\frac{1}{\cot x}$ is not defined at $x=0$
Domain of given functions are not equal.
$\therefore \tan x,\frac{1}{\cot x}$ are not equal functions.

(b) $f\left(x\right)=\text{sgn}(x^2+1),g\left(x\right)={\sin }^{2}x+{\cos }^{2}x$
Domain of both $f\left(x\right),g\left(x\right)$ is $R$.
Also, from the definition $f\left(x\right)=\text{sgn}(x^2+1)=1\forall x\in R,$
$g\left(x\right)={\sin }^{2}x+{\cos }^{2}x=1\forall x\in R$
Hence range sets are equal.
Also $f\left(x\right)=g\left(x\right)$ for all $x\in$ Domain
Hence these functions are equal functions.


(c) $f\left(x\right)=\sqrt{x^2-1},g\left(x\right)=\sqrt{x-1}.\sqrt{x+1}$
$f\left(x\right)=\sqrt{x^2-1}$ is defined for $x^2-1\ge 0\Rightarrow (x-1)(x+1)\ge 0$
by wavy curve method
$x\in (-\infty ,-1]\cup [1,\infty )$
but $g\left(x\right)=\sqrt{x-1}.\sqrt{x+1}$ is defined if
$x+1\ge 0$ and $x-1\ge 0$
$\Rightarrow x\in [1,\infty )$
Domain of given functions are not equal.
$\therefore \sqrt{x^2-1},\sqrt{x-1}.\sqrt{x+1}$ are not equal functions.

(d) $f\left(x\right)=\sqrt{x^2},g\left(x\right)=x$
As we know $f\left(x\right)=\sqrt{x^2}=|x|,\text{whose range is}[0,\infty )$
But range of $g\left(x\right)=x$ is $R$.
Range of given functions are not equal.
$\therefore \sqrt{x^2-1},\sqrt{x-1}.\sqrt{x+1}$ are not equal functions.