CUET Preparation Today
If $f(x) = \frac{1}{1-x}$, then for x > 1, f(x) is : |
decreasing constant increasing neither decreasing nor increasing |
increasing |
$f(x) = \frac{1}{1-x}$ so $f'(x) = \frac{-1 ~(-1)}{(1-x)^2} = \frac{1}{(1-x)^2}$ as (1 - x)2 > 0 always ⇒ f(x) > 0 always so for x > 1 → f(x) is increasing |
