Question

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Answer 1

Answer:

$\mathrm{Domain\:of\:}\:\dfrac{\log _{10}\left(2x-3\right)}{\sqrt{x-1}}+\sqrt{5-2x}\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:\dfrac{3}{2}<x\le \dfrac{5}{2}\:\\ \:\mathrm{Interval\:Notation:}&\:\left(\dfrac{3}{2},\:\dfrac{5}{2}\right]\end{bmatrix}$

Step-by-step explanation:

$\green{\mathrm{Domain\:De finition}}$

• The domain of a function is the set of input or argument values for which the function is real and defined

$\blue{\mathrm{Find\:non-negative\:values\:for\:radicals}:}$

$\gray{\sqrt{f\left(x\right)}\quad \Rightarrow \quad \:f\left(x\right)\ge 0\:}$

$\black{\mathrm{Solve\:}\:x-1\ge \:0:\quad x\ge \:1}$

$\displaystyle\black{\mathrm{Solve\:}\:5-2x\ge \:0:\quad x\le \frac{5}{2}}$

$\gray{\mathrm{Combine\:the\:intervals}}$

$\displaystyle1\le \:x\le \frac{5}{2}$

$\blue{\mathrm{Find\:positive\:values\:for\:logs}:}$

$\gray{\log _af\left(x\right)\quad \Rightarrow \quad \:f\left(x\right)>0}$

$\displaystyle\black{\mathrm{Solve\:}\:2x-3>0:\quad x>\frac{3}{2}}$

$\blue{\mathrm{Find\:unde fined\:\left(singularity\right)\:points}:}$

$\displaystyle\gray{\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{\log _{10}\left(2x-3\right)}{\sqrt{x-1}}+\sqrt{5-2x}\mathrm{\:and\:compare\:to\:zero}}$

$\black{\mathrm{Solve\:}\:\sqrt{x-1}=0:\quad x=1}$

$\gray{\mathrm{The\:following\:points\:are\:unde fined}}$

$x=1$

$\gray{\mathrm{Combine\:real\:regions\:and\:unde fined\:points\:for\:final\:function\:domain}}$

$\displaystyle\frac{3}{2}<x\le \frac{5}{2}$

Answer 2

We're given the function,

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

First consider the denominator of the fraction,

$\sqrt{x-1}$

Since it's a denominator, it can't be 0.

$\sqrt{x-1}\neq 0\ \implies\ x-1\neq 0$

We know the square of a real number is always non-negative. But in this case,

$x-1>0\ \implies\ x>1$

But consider the numerator of the fraction,

$\log(2x-3)$

We know the number of which logarithm is taken and the base number of the logarithm should be positive.

$\log_b(a)\in\mathbb{R}\ \iff\ a,\ b>0$

Here base is 10, so it is not considered. So,

$2x-3>0\ \implies\ x>\dfrac{3}{2}>1$

So we ignore $x>1.$

Now, consider  $\sqrt{5-2x}.$

As we said earlier, the square of this thing is non-negative.

$5-2x\geq 0\ \implies\ 2x-5\leq 0\ \implies\ x\leq \dfrac{5}{2}$

So, finally we get,

$\dfrac{3}{2}<x\leq\dfrac{5}{2}\ \implies\ \boxed{x\in\left(\dfrac{3}{2},\ \dfrac{5}{2}\right]}$

Hence option (d) is the answer.