# The function $f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}$

A. is a constant function

B. has a domain$\left( 0,1 \right)\bigcup \left( e,\infty \right)$

C. is such that $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$

D. is aperiodic

Last updated date: 20th Mar 2023

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Answer

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Hint: First try to simplify the given function $f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}$.Then proceeds to the options provided to verify one by one. Here, aperiodic means irregularity.

Complete step-by-step answer:

Here, we have a function given as$f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}\ldots \ldots (1)$.

We need to find correct answer/answers by observing all the options given.

Now, Option A. is given as ${{x}^{\dfrac{1}{\ln x}}}$is a constant function.

We can verify this option by just simplifying the function $f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}$given in the problem.

We have $f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}$

Taking log to both sides, we get

\[{{\log }_{e}}\left( f\left( x \right) \right)={{\log }_{e}}{{\left( x \right)}^{\dfrac{1}{\ln x}}}\ldots \ldots (2)\]

We know property of log as

$\log {{m}^{n}}=n\log m$

So, we can write equation (2) as

\[{{\log }_{e}}\left( f\left( x \right) \right)=\dfrac{1}{\ln x}{{\log }_{e}}\left( x \right)\]

Here ${{\log }_{e}}x$ has base ‘e’ and we know that $\ln x={{\log }_{e}}x$(same value with different representations).

${{\log }_{e}}\left( f\left( x \right) \right)=1$

Now, we know that if

${{\log }_{a}}N=x$, then

\[N={{a}^{x}}\]

Hence,${{\log }_{e}}\left( f\left( x \right) \right)=1$, can be written as

$\begin{align}

& f\left( x \right)=e'=e \\

& f\left( x \right)=e \\

\end{align}$

Hence, given function f(x) is ‘e’ i.e. Constant function.

So, Option A is correct.

Now, Coming to option B. i.e. f(x) has a domain$\left( 0,1 \right)\bigcup \left( e,\infty \right)$.

To verify option ‘B’, we need to calculate the domain of$f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}$.

The given function is in volument of ‘$\ln x$’. So, we cannot put negative values to a given function f(x) because the domain of $\ln x$ is ${{R}^{+}}$or$\left( 0,\infty \right)$.

We know the value of ln1 is 0. So, $\dfrac{1}{\ln x}$cannot take x=1 as well because $\dfrac{1}{0}$is not defined.

Hence, domain of ${{\left( x \right)}^{\dfrac{1}{\ln x}}}$ is$\left( 0,1 \right)\bigcup \left( e,\infty \right)$.

So, option B. is not the correct answer.

Now, to verify option ‘C’ i.e. function $f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}$ is such that $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$ exist,

We need to calculate LHL and RHL for x=1 i.e. $x\to {{1}^{-}}$and$x\to {{1}^{+}}$.

Now, we have already calculated that $f\left( x \right)=e$ for the domain$\left( 0,1 \right)\bigcup \left( e,\infty \right)$.

So, $\ln x$ LHL ($x\to {{1}^{-}}$) and RHL ($x\to {{1}^{+}}$).

We have constant function ‘e’. So the limit at $x\to 1$will exist.

So, option C. is the correct answer.

As ${{\left( x \right)}^{\dfrac{1}{\ln x}}}$ is simplified to ‘e’.

Hence graph of given function i.e.

$f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}$Is given as

Hence, a given function is aperiodic as well where aperiodic means irregularity or not periodic function.

Hence option D. is also correct.

Note: One can go wrong if he/she will not simplify the given relation. As one can think that a given function is written in terms of ‘x’ so how it could be a constant function. So, be careful with these kinds of problems.

One can go wrong with the domain of${{\left( x \right)}^{\dfrac{1}{\ln x}}}$. As domain of ${{x}^{a}}$is ‘R’ i.e. All real numbers. So, one can get confused with it. So, checking domain for $\dfrac{1}{\ln x}$is the key point for calculation of domain of$f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}$.

Complete step-by-step answer:

Here, we have a function given as$f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}\ldots \ldots (1)$.

We need to find correct answer/answers by observing all the options given.

Now, Option A. is given as ${{x}^{\dfrac{1}{\ln x}}}$is a constant function.

We can verify this option by just simplifying the function $f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}$given in the problem.

We have $f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}$

Taking log to both sides, we get

\[{{\log }_{e}}\left( f\left( x \right) \right)={{\log }_{e}}{{\left( x \right)}^{\dfrac{1}{\ln x}}}\ldots \ldots (2)\]

We know property of log as

$\log {{m}^{n}}=n\log m$

So, we can write equation (2) as

\[{{\log }_{e}}\left( f\left( x \right) \right)=\dfrac{1}{\ln x}{{\log }_{e}}\left( x \right)\]

Here ${{\log }_{e}}x$ has base ‘e’ and we know that $\ln x={{\log }_{e}}x$(same value with different representations).

${{\log }_{e}}\left( f\left( x \right) \right)=1$

Now, we know that if

${{\log }_{a}}N=x$, then

\[N={{a}^{x}}\]

Hence,${{\log }_{e}}\left( f\left( x \right) \right)=1$, can be written as

$\begin{align}

& f\left( x \right)=e'=e \\

& f\left( x \right)=e \\

\end{align}$

Hence, given function f(x) is ‘e’ i.e. Constant function.

So, Option A is correct.

Now, Coming to option B. i.e. f(x) has a domain$\left( 0,1 \right)\bigcup \left( e,\infty \right)$.

To verify option ‘B’, we need to calculate the domain of$f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}$.

The given function is in volument of ‘$\ln x$’. So, we cannot put negative values to a given function f(x) because the domain of $\ln x$ is ${{R}^{+}}$or$\left( 0,\infty \right)$.

We know the value of ln1 is 0. So, $\dfrac{1}{\ln x}$cannot take x=1 as well because $\dfrac{1}{0}$is not defined.

Hence, domain of ${{\left( x \right)}^{\dfrac{1}{\ln x}}}$ is$\left( 0,1 \right)\bigcup \left( e,\infty \right)$.

So, option B. is not the correct answer.

Now, to verify option ‘C’ i.e. function $f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}$ is such that $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$ exist,

We need to calculate LHL and RHL for x=1 i.e. $x\to {{1}^{-}}$and$x\to {{1}^{+}}$.

Now, we have already calculated that $f\left( x \right)=e$ for the domain$\left( 0,1 \right)\bigcup \left( e,\infty \right)$.

So, $\ln x$ LHL ($x\to {{1}^{-}}$) and RHL ($x\to {{1}^{+}}$).

We have constant function ‘e’. So the limit at $x\to 1$will exist.

So, option C. is the correct answer.

As ${{\left( x \right)}^{\dfrac{1}{\ln x}}}$ is simplified to ‘e’.

Hence graph of given function i.e.

$f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}$Is given as

Hence, a given function is aperiodic as well where aperiodic means irregularity or not periodic function.

Hence option D. is also correct.

Note: One can go wrong if he/she will not simplify the given relation. As one can think that a given function is written in terms of ‘x’ so how it could be a constant function. So, be careful with these kinds of problems.

One can go wrong with the domain of${{\left( x \right)}^{\dfrac{1}{\ln x}}}$. As domain of ${{x}^{a}}$is ‘R’ i.e. All real numbers. So, one can get confused with it. So, checking domain for $\dfrac{1}{\ln x}$is the key point for calculation of domain of$f\left( x \right)={{x}^{\dfrac{1}{\ln x}}}$.

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