Question

The domain of the function $f(x) = \dfrac{{\sqrt { - {{\log }_{0.3}}(x - 1)} }}{{\sqrt {{x^2} + 2x + 8} }}$ is
(a) (1, 4) (b) (− 2, 4) (c) (2, 4) (d) [2, ∞)

Answer 1

Hint: Here, find the values of x for which both numerator and denominator are defined. Apply the standard condition of logarithm function to be defined and square root function to be defined. Also the denominator must not be equal to 0 for any defined function. By applying all these facts we can get the values of x.

Complete step-by-step answer: We have numerator of f(x) is $\sqrt { - {{\log }_{0.3}}(x - 1)} $ and denominator is $\sqrt {{x^2} + 2x + 8} $.
For numerator:
As we know, the square root of a negative number is not real, so $ - {\log _{0.3}}(x - 1)$must be greater than or equal to 0.
$ - {\log _{0.3}}(x - 1) \geqslant 0$
$ \Rightarrow {\log _{0.3}}{(x - 1)^{ - 1}} \geqslant 0$ [− log a = log a−1]
$ \Rightarrow {\log _{0.3}}\left( {\dfrac{1}{{x - 1}}} \right) \geqslant 0$
$ \Rightarrow \dfrac{{{{\log }_e}\left( {\dfrac{1}{{x - 1}}} \right)}}{{{{\log }_e}0.3}} \geqslant 0$ $\left[ {\because {{\log }_x}y = \dfrac{{{{\log }_e}y}}{{{{\log }_e}x}}} \right]$
$ \Rightarrow {\log _e}\left( {\dfrac{1}{{x - 1}}} \right) \geqslant 0$
$
   \Rightarrow {e^0} \geqslant \dfrac{1}{{x - 1}} \\
   \Rightarrow 1 \geqslant \dfrac{1}{{x - 1}} \\
   \Rightarrow x - 1 \geqslant 1 \\
   \Rightarrow x \geqslant 1 + 1 \\
   \Rightarrow x \geqslant 2 \\
 $
So, the value of x should be greater than or equal to 0.
Now, denominator of f(x) is $\sqrt {{x^2} + 2x + 8} $
Here, also ${x^2} + 2x + 8$ must be greater than 0 for the denominator to be defined, because the denominator can’t be 0, and the square root of a negative number is never given real value.
So, ${x^2} + 2x + 8 > 0$
Solve the quadratic equation by middle term splitting method.
${x^2} + 4x - 2x + 8 > 0$ [2x = 4x – 2x]
$ \Rightarrow x(x + 4) - 2(x + 4) > 0$
Taking (x + 4) common, we have
⇒ (x + 4)(x – 2) > 0
⇒ x + 4 ≥ 0 or x −2 > 0
⇒ x > − 4 or x > 2
Combining all values of x in both numerator and denominator, we get a common value as x ≥ 2.
Therefore, x belongs to [2, ∞).

So, the correct answer is “Option D”.

Note: In these types of questions always try to satisfy the given function. All functions like square root, logarithm function, trigonometric function etc have many limitations and they must satisfy some condition to be defined.