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Given:
To find: The domain of definition.
____________________
Things to Know
1. Domain of Definition
is defined if and .
The square root function is defined when the radicand is greater than or 0.
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.
____________________
Solution
The domain of is ''
Hence, the domain of the function is which is choice (d).
Step-by-step explanation:
Given: f(x)=\dfrac{\log(2x-3)}{\sqrt{x-1} } +\sqrt{5-2x}f(x)=
x−1
log(2x−3)
+
5−2x
To find: The domain of definition.
____________________
Things to Know
1. Domain of Definition
Domain of \log(2x-3)log(2x−3)
\log_{a}blog
a
b is defined if a > 0,a\neq 1a>0,a
=1 and b > 0b>0 .
Domain of \sqrt{5-2x}
5−2x
The square root function is defined when the radicand is greater than or 0.
Domain of \dfrac{1}{\sqrt{x-1} }
x−1
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.
____________________
Solution
The domain of f(x)f(x) is '2x-3 > 0\ \text{and}\ 5-2x\geq 0\ \text{and}\ x-1 > 02x−3>0 and 5−2x≥0 and x−1>0 '
\implies x > \dfrac{3}{2} \ \text{and}\ x\leq \dfrac{5}{2} \ \text{and}\ x > 1⟹x>
2
3
and x≤
2
5
and x>1
\implies \boxed{\dfrac{3}{2} < x\leq \dfrac{5}{2}}⟹
2
3
<x≤
2
5
Hence, the domain of the function is \boxed{(\dfrac{3}{2}, \dfrac{5}{2}]}
(
2
3
,
2
5
]
which is choice (d).