Question

Which of the following pair(s) of functions is/are not identical

• A. $f\left(x\right)=\text{sgn}\left(x\right),g\left(x\right)=\frac{|x|}{x}$
• B. $f\left(x\right)=\frac{|x|}{x},g\left(x\right)=1$
• C. $f\left(x\right)=\cos x,g\left(x\right)=\frac{1}{\sec x}$
• D. $f\left(x\right)=\sin x,g\left(x\right)=\frac{1}{\text{cosec}x}$

Answer 1

Answer: (A), (B), (C), (D)

$f\left(x\right)=\sin x,g\left(x\right)=\frac{1}{\text{cosec}x}$

For identical function:
$D_f=D_g,R_f=R_g$ and btoh functions must have same graph.

Option$\left(a\right)$:
$f\left(x\right)=\text{sgn}\left(x\right),g\left(x\right)=\frac{|x|}{x}$
$\Rightarrow D_f\in R,D_g\in R-\left\{0\right\}\Rightarrow D_f\ne D_g$
So, not identical functions.

Option$\left(b\right)$:
$f\left(x\right)=\frac{|x|}{x},g\left(x\right)=1$
$\Rightarrow D_f\in R-\left\{0\right\},D_g\in R\Rightarrow D_f\ne D_g$
So, not identical functions.

Option$\left(c\right)$:
$f\left(x\right)=\cos x,g\left(x\right)=\frac{1}{\sec x}$
$\Rightarrow D_f\in R,D_g\in R-\left\{\frac{(2n+1)\pi }{2}\right\},n\in Z$
$\Rightarrow D_f\ne D_g$
So, not identical functions.

Option$\left(d\right)$:
$f\left(x\right)=\sin x,g\left(x\right)=\frac{1}{\text{cosec}x}$
$\Rightarrow D_f\in R,D_g\in R-\left\{n\pi \right\},n\in Z\Rightarrow D_f\ne D_g$
So, not identical functions.