Question

The domain of ${\cos }^{-1}\left(\frac{x-3}{2}\right)-{\log }_{10}(4-x)$ is

• A. $(1,4)$
• B. $[1,4)$
• C. $(1,4]$
• D. $[1,4]$

Answer 1

The correct option is B

$[1,4)$

Explanation for the correct answer:

Let, ${\cos }^{-1}\left(\frac{x-3}{2}\right)=\theta$

$\therefore \cos \theta =\frac{x-3}{2}$

We know, that the range of $\cos \theta$ is given by $[-1,1]$, therefore the we can write it in terms of inequality as:

$$\begin{gathered} \Rightarrow -1\le \cos \theta \le 1 \\ \Rightarrow -1\le \frac{x-3}{2}\le 1 \\ \Rightarrow -2\le x-3\le 2 \\ \Rightarrow -2+3\le x\le 2+3 \\ \Rightarrow 1\le x\le 5...\left(1\right) \end{gathered}$$

Similarly for the $\log$, we know that only positive numbers are defined, hence we can write the inequality as:

$$\begin{gathered} \Rightarrow 4-x>0 \\ \Rightarrow 4>x...\left(2\right) \end{gathered}$$

On, Taking intersection of $\left(1\right)\mathrm{and}\left(2\right)$, we get:

$\Rightarrow 1\le x<4\mathrm{or}\mathrm{x}\in [1,4)$

Hence, the correct answer is option (B)