Question

The domain of the function \(f(x) = e^{\sqrt{5x-3-2x^2}}\) is

• A. \((1,2)\)
• B. \((-1,\frac{3}{2})\)
• C. \([1,\frac{3}{2}]\)
• D. \((-1,0)\)

Hint:

The exponential is defined everywhere, so only the expression under the square root must be non-negative.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
\(e^t\) is defined for every real \(t\). The only restriction comes from \(\sqrt{\ }\), which needs a non-negative quantity inside.

Step 2: Set up the inequality.
\[ 5x - 3 - 2x^2 \geq 0 \Rightarrow 2x^2 - 5x + 3 \leq 0 \]

Step 3: Solve.
Factorise: \(2x^2 - 5x + 3 = (2x - 3)(x - 1)\). The product is \(\leq 0\) between the roots \(x = 1\) and \(x = 3/2\).
\[ 1 \leq x \leq \frac{3}{2} \]

Step 4: Check the options.
The end points are included because the square root of zero is defined. So the domain is the closed interval \([1, 3/2]\). Options (A), (B), (D) are open intervals or do not match.

Final Answer:
The domain is \(\left[1, \dfrac{3}{2}\right]\), option (C). \[ \boxed{\left[1, \frac{3}{2}\right]} \]