Question

The domain of definition of the function

$y = \frac{1}{\log_{10}(1 - x)} + \sqrt{x + 2}$

• A. (−3, −2) excluding −2.5
• B. [0, 1] excluding 0.5
• C. [-2, 1] excluding 0
• D. None of these

Answer 1

Answer: (C)

$-2, 1$ excluding 0

Answer 2

$y = \frac{1}{\log_{10}(1 - x)} + \sqrt{x - 2}$

Y = f(x) + g(x)

Then domain of given function is D_f $\cap$D_g

Now, for domain of f(x) = $\frac{1}{\log_{10}(1 - x)}$

We know it is defined only when 1 – x > 0 and 1 - x ≠ 1

⇒ x < 1 and x ≠ 0

∴ D_f = (−∞, 1) − {0}

For domain of g(x) = $\sqrt{x + 2}$

x + 2 ≥ 0 ⇒ x ≥ - 2

D_g = (−2, ∞)

∴ Common domain is [−2, 1] − {0}