Question

The domain of the function $f\left(x\right)=\text{exp}\left(\sqrt{5x-3-2x^2}\right)$ is

• A. $\left[\frac{3}{2},\infty \right]$
• B. $[1,3,2\}$
• C. $(-\infty ,1]$
• D. $(1,\frac{3}{2}\}$

Answer 1

The correct option is D

$(1,\frac{3}{2}\}$

Explanation for correct answer Option(s)

Find the domain of the function $f\left(x\right)$

Consider the given function as,

$f\left(x\right)=\text{exp}\left(\sqrt{5x-3-2x^2}\right)$

Defined the above function as,

$$\begin{gathered} \Rightarrow 5x-3-2x^2\ge 0 \\ \Rightarrow -2x^2+5x-3\ge 0 \end{gathered}$$

We know that,

$x=\frac{-b\pm \sqrt{b^2-4ab}}{2a}$

Substitute the value of $a=-2,b=5,\text{and}c=-3$, then

$$\begin{gathered} x=\frac{-5\pm \sqrt{5^2-4\times -2\times -3}}{2\times -2} \\ =\frac{-5\pm \sqrt{25-24}}{-4} \\ =\frac{-5\pm \sqrt{1}}{-4} \\ =\frac{-5\pm 1}{-4} \end{gathered}$$

Since,

$$\begin{gathered} \Rightarrow x=\frac{-5-1}{-4} \\ =\frac{-6}{-4} \\ =\frac{3}{2} \\ \Rightarrow x=\frac{-5+1}{-4} \\ =\frac{-4}{-4} \\ =1 \end{gathered}$$

Hence, the domain of the function is $(1,\frac{3}{2}\}$

Therefore, the correct answer is Option D.