Question Bank [2022] — Maths (commerce) STD 12 Commerce / Arts — Question
Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Question Bank [2022]3 Marks
Question
Show that the function $f ( x )=\frac{x-2}{x+1}, x \neq-1$ is increasing
✓
Answer
$f ( x )=\frac{x-2}{x+1}$ for $x \neq-1$
For function to be increasing, $f^{\prime}(x)>0$
Then, $f ^{\prime}( x )=\frac{(x+1) \frac{ d }{ d x}(x-2)-(x-2) \frac{ d }{ d x}(x+1)}{(x+1)^2}$
$ =\frac{(x+1)-(x-2)}{(x+1)^2}$
$=\frac{x+1-x+2}{(x+1)^2}$
$=\frac{3}{(x+1)^2}>0 \quad \ldots \ldots .\left[\because(x+1) \neq 0,(x+1)^2>0\right] $
Thus, $f(x)$ is an increasing function for $x \neq-1$.
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