First slide

Functions (XII)

Question

The domain of definition of the function $y = \frac{1}{\log_{10} (1 - x)} + \sqrt{x + 2} is$

Difficult

A

$(- 3, - 2) excluding - 2.5$
B

$[0, 1] excluding 0.5$
C

$[- 2, 1) excluding 0$
D

None of these

Solution

$\begin{aligned} y = \frac{1}{\log_{10} (1 - x)} + \sqrt{x - 2} \\ y = f (x) + g (x) \end{aligned}$
$Then, the domain of given function is D_{f} \cap D_{g} .$
$Now, for the domain of f (x) = \frac{1}{\log_{10} (1 - x)},$
$we know it is defined only when 1 - x > 0 and 1 - x \neq 1$
$or x < 1 and x \neq 0 . Therefore, D_{f} = (- \infty, 1) - {0} .$
$For the domain of g (x) = \sqrt{x + 2},$
$\begin{aligned} x + 2 \geq 0 or x \geq - 2 \\ ∴ D_{g} = [- 2, \infty) \end{aligned}$
$Therefore, common domain is [- 2, 1) - {0} .$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App