Question

Solve 10th and 4th ......​

Answer 1

4.  Given,

$\longrightarrow\sf{f(x)=\sqrt{\log_{10}x^2}}$

Actually we have to find the domain of $\sf{f(x),}$ the set of values of $\sf{x}$ for which $\sf{f(x)}$ is real.

We know the function $\sf{f(x)=\sqrt x}$ is defined for $\sf{x\geq0.}$ Then we have,

$\longrightarrow\sf{\log_{10}x^2\geq0}$

Taking antilogs,

$\longrightarrow\sf{x^2\geq10^0}$

$\longrightarrow\sf{x^2\geq1}$

$\longrightarrow\sf{x^2-1\geq0}$

$\longrightarrow\sf{(x-1)(x+1)\geq0}$

$\Longrightarrow\sf{\underline{\underline{x\leq-1\quad OR\quad x\geq1}}}$

Hence,

$\longrightarrow\sf{\underline{\underline{x\in(-\infty,\ -1]\cup[1,\ \infty)}}}$

10.  Given,

$\longrightarrow\sf{f(x)=\sqrt{\log_{\frac{1}{2}}\left(\dfrac{x-1}{x+5}\right)}}$

Here we also have to find the domain of $\sf{f(x).}$ Then,

$\longrightarrow\sf{\log_{\frac{1}{2}}\left(\dfrac{x-1}{x+5}\right)\geq0}$

Because $\sf{f(x)=\sqrt x}$ is defined for $\sf{x\geq0.}$

Since $\sf{\log_ba=\dfrac{\log_{10}a}{\log_{10}b},}$

$\longrightarrow\sf{\dfrac{\log_{10}\left(\dfrac{x-1}{x+5}\right)}{\log_{10}\left(\dfrac{1}{2}\right)}\geq0}$

$\longrightarrow\sf{\dfrac{\log_{10}\left(\dfrac{x-1}{x+5}\right)}{\log_{10}\left(2^{-1}\right)}\geq0}$

$\longrightarrow\sf{\dfrac{\log_{10}\left(\dfrac{x-1}{x+5}\right)}{-\log_{10}\left(2\right)}\geq0}$

$\longrightarrow\sf{\dfrac{\log_{10}\left(\dfrac{x-1}{x+5}\right)}{\log_{10}\left(2\right)}\leq0}$

Well denominator is a positive real number here.

$\longrightarrow\sf{\log_{10}\left(\dfrac{x-1}{x+5}\right)}\leq0}$

Taking antilogs,

$\longrightarrow\sf{\dfrac{x-1}{x+5}\leq10^0}$

$\longrightarrow\sf{\dfrac{x-1}{x+5}\leq1}$

$\longrightarrow\sf{\dfrac{x-1}{x+5}-1\leq0}$

$\longrightarrow\sf{\dfrac{-6}{x+5}\leq0}$

$\longrightarrow\sf{\dfrac{6}{x+5}\geq0}$

This implies the denominator should also be positive, since the numerator is positive. Hence,

$\longrightarrow\sf{x+5\ \textgreater\ 0}$

$\longrightarrow\sf{x\ \textgreater\ -5}$

$\longrightarrow\sf{x\in(-5,\ \infty)\quad\quad\dots(1)}$

But the function $\sf{f(x)=\log_{10}x}$ is defined for $\sf{x\ \textgreater\ 0.}$ Hence from $\sf{f(x)=\log_{10}\left(\dfrac{x-1}{x+5}\right),}$

$\longrightarrow\sf{\dfrac{x-1}{x+5}\ \textgreater\ 0}$

Case 1:- Both the numerator and denominator should be positive at one time.

$\longrightarrow\sf{x-1\ \textgreater\ 0\quad AND\quad x+5\ \textgreater\ 0}$

$\longrightarrow\sf{x\ \textgreater\ 1\quad AND\quad x\ \textgreater\ -5}$

$\Longrightarrow\sf{x\ \textgreater\ 1}$

Case 2:- Both the numerator and denominator should be negative at one time.

$\longrightarrow\sf{x-1\ \textless\ 0\quad AND\quad x+5\ \textless\ 0}$

$\longrightarrow\sf{x\ \textless\ 1\quad AND\quad x\ \textless\ -5}$

$\Longrightarrow\sf{x\ \textless\ -5}$

Thus,

$\longrightarrow\sf{x\ \textless\ -5\quad OR\quad x\ \textgreater\ 1}$

$\longrightarrow\sf{x\in(-\infty,\ -5)\cup(1,\ \infty)\quad\quad\dots(2)}$

Set of values of $\sf{x}$ is given by intersection of (1) and (2).

$\longrightarrow\sf{x\in(-5,\ \infty)\cap[(-\infty,\ -5)\cup(1,\ \infty)]}$

$\longrightarrow\sf{x\in[(-5,\ \infty)\cap(-\infty,\ -5)]\cup[(-5,\ \infty)\cap (1,\ \infty)]}$

$\longrightarrow\sf{\underline{\underline{x\in(1,\ \infty)}}}$