Question

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Answer 1

Given: $f(x)=\dfrac{\log(2x-3)}{\sqrt{x-1} } +\sqrt{5-2x}$

To find: The domain of definition.

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Things to Know

1. Domain of Definition

• Domain of $\log(2x-3)$

$\log_{a}b$ is defined if $a>0,a\neq 1$ and $b>0$.

• Domain of $\sqrt{5-2x}$

The square root function is defined when the radicand is greater than or 0.

• Domain of $\dfrac{1}{\sqrt{x-1} }$

Rational functions are defined when the denominator is not 0. But since the denominator is a square root function, the radicand should be positive.

2. Interval Notation

When the interval includes the number, it is shown by the '[' or ']' bracket, or when it does not include it, it is shown by the '(' or ')' bracket.

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Solution

The domain of $f(x)$ is '$2x-3>0\ \text{and}\ 5-2x\geq 0\ \text{and}\ x-1>0$'

$\implies x>\dfrac{3}{2} \ \text{and}\ x\leq \dfrac{5}{2} \ \text{and}\ x>1$

$\implies \boxed{\dfrac{3}{2} <x\leq \dfrac{5}{2}}$

Hence, the domain of the function is $\boxed{(\dfrac{3}{2}, \dfrac{5}{2}]}$ which is choice (d).