Question

How many of the following pairs of function are identical?

(a) f (x) = $\frac{x^2}{x},g\left(x\right)=x$

(b) f (x) = $\frac{x^2-1}{x-1}$ g(x) = x + 1

(c) f (x) = $se{c}^{2}x-ta{n}^{2}x$ , g(x) = 1

(d) $\frac{1}{x}f\left(x\right)=\frac{x}{x^2},g\left(x\right)=\frac{1}{x}$

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Answer 1

Two functions are equal and identical if the domain and range of the 2 functions are equal.

[ f(x) = g(x) for every x in their domain]

1. Domain of f (x) is real numbers without zero or R - {0} because $\frac{x^2}{x}$ is not defined at zero.
f (x) and g(x) are not equal because their domains are not equal.
2. For this pair also the domains are not the same because $\frac{x^2-1}{x-1}$ is not defined at x = 1.
3. Domain of g(x) is R (real numbers), f(x) is not defined when x = (2n + 1)$\frac{\pi }{2}$.

$\Rightarrow$ they are not equal
4. $\frac{x}{x^2}$ is not defined at x = 0.
If x $\ne$ 0, $\frac{x}{x^2}=\frac{1}{x}$
$\frac{1}{x}$ is not defined at x = 0
$\Rightarrow$ f (x) = g(x) = $\frac{1}{x}$ when x $\ne$ 0. Since zero is not in their domain, the functions are identical.