Question

If $\dfrac{{{a}^{n}}+{{b}^{n}}}{{{a}^{n-1}}+{{b}^{n-1}}}=x$, where $a,b>0\And a\ne b$, then

P. If $n=1$, then x is	1. $\left( a+b \right)-2ab{{\left( a+b \right)}^{-1}}$
Q. If $n=\dfrac{1}{2}$, then x is	2. AM of a and b
R. If $n=0$, then x is	3. HM of a and b
S. If $n=2$, then x is	4. GM of a and b.

A. P-2, Q-4, R-3, S-1.
B. P-1, Q-2, R-3, S-4
C. P-4, Q-2, R-3, S-1
D. P-3, Q-2, R-4, S-1.

Answer 1

Hint: To solve the above question we will substitute the different values of n in the given equation one by one and compare the obtained values from the given options. According to the values obtained we will match the columns and choose the correct answer.

Complete step-by-step solution:
We have been given an equation $\dfrac{{{a}^{n}}+{{b}^{n}}}{{{a}^{n-1}}+{{b}^{n-1}}}=x$, where $a,b>0\And a\ne b$
We have to match the column and choose the correct option.
Let us consider the first statement from the left column. Then we get
If $n=1$, then x is
Now, substituting the value of n in the given equation we will get
$\Rightarrow \dfrac{{{a}^{1}}+{{b}^{1}}}{{{a}^{1-1}}+{{b}^{1-1}}}=x$
Now, simplifying the above obtained equation we will get
$\Rightarrow \dfrac{a+b}{{{a}^{0}}+{{b}^{0}}}=x$
Now, we know that ${{a}^{0}}=1$
So substituting the value in the above obtained equation we will get
\[\begin{align}
  & \Rightarrow \dfrac{a+b}{1+1}=x \\
 & \Rightarrow \dfrac{a+b}{2}=x \\
\end{align}\]
Now, we know that if we have two numbers a and b in arithmetic sequence then the arithmetic mean of two numbers is given as $\dfrac{a+b}{2}$
Hence if $n=1$, then x is AM of a and b. Correct match is P-2.
Now, let us consider the Q part. Then we will get
If $n=\dfrac{1}{2}$, then x is
Now, substituting the value of n in the given equation we will get
$\Rightarrow \dfrac{{{a}^{\dfrac{1}{2}}}+{{b}^{\dfrac{1}{2}}}}{{{a}^{\dfrac{1}{2}-1}}+{{b}^{\dfrac{1}{2}-1}}}=x$
Now, simplifying the above obtained equation we will get
$\begin{align}
  & \Rightarrow \dfrac{{{a}^{\dfrac{1}{2}}}+{{b}^{\dfrac{1}{2}}}}{{{a}^{\dfrac{1-2}{2}}}+{{b}^{\dfrac{1-2}{2}}}}=x \\
 & \Rightarrow \dfrac{{{a}^{\dfrac{1}{2}}}+{{b}^{\dfrac{1}{2}}}}{{{a}^{\dfrac{-1}{2}}}+{{b}^{\dfrac{-1}{2}}}}=x \\
\end{align}$
Now, we know that ${{a}^{\dfrac{1}{2}}}=\sqrt{a}$ and ${{a}^{\dfrac{-1}{2}}}=\dfrac{1}{\sqrt{a}}$
So substituting the value in the above obtained equation we will get
\[\Rightarrow \dfrac{\sqrt{a}+\sqrt{b}}{\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}}=x\]
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow \dfrac{\sqrt{a}+\sqrt{b}}{\dfrac{\sqrt{b}+\sqrt{a}}{\sqrt{ab}}}=x \\
 & \Rightarrow \dfrac{\sqrt{ab}\left( \sqrt{a}+\sqrt{b} \right)}{\sqrt{b}+\sqrt{a}}=x \\
 & \Rightarrow \sqrt{ab}=x \\
\end{align}\]
Now, we know that if we have two numbers a and b in a geometric sequence then the geometric mean of two numbers is given as \[\sqrt{ab}\].
Hence if $n=\dfrac{1}{2}$, then x is GM of a and b. Correct match is Q-4.
Now, let us consider the R part. Then we will get
If $n=0$, then x is
Now, substituting the value of n in the given equation we will get
$\Rightarrow \dfrac{{{a}^{0}}+{{b}^{0}}}{{{a}^{0-1}}+{{b}^{0-1}}}=x$
Now, simplifying the above obtained equation we will get
$\Rightarrow \dfrac{{{a}^{0}}+{{b}^{0}}}{{{a}^{-1}}+{{b}^{-1}}}=x$
Now, we know that ${{a}^{0}}=1$ and ${{a}^{-1}}=\dfrac{1}{a}$
So substituting the value in the above obtained equation we will get
\[\begin{align}
  & \Rightarrow \dfrac{1+1}{\dfrac{1}{a}+\dfrac{1}{b}}=x \\
 & \Rightarrow \dfrac{2}{\dfrac{b+a}{ab}}=x \\
 & \Rightarrow \dfrac{2ab}{a+b}=x \\
\end{align}\]
Now, we know that if we have two numbers a and b in arithmetic sequence then the arithmetic mean of two numbers is given as $\dfrac{2ab}{a+b}$
Hence if $n=0$, then x is HM of a and b. Correct match is R-3.
Now, let us consider part S. then we will get
If $n=2$, then x is
Now, substituting the value of n in the given equation we will get
$\Rightarrow \dfrac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2-1}}+{{b}^{2-1}}}=x$
Now, simplifying the above obtained equation we will get
$\Rightarrow \dfrac{{{a}^{2}}+{{b}^{2}}}{{{a}^{1}}+{{b}^{1}}}=x$
Now, we know that ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$
So substituting the value in the above obtained equation we will get
\[\begin{align}
  & \Rightarrow \dfrac{{{\left( a+b \right)}^{2}}-2ab}{a+b}=x \\
 & \Rightarrow {{\left( a+b \right)}^{-1}}\left( {{\left( a+b \right)}^{2}}-2ab \right)=x \\
 & \Rightarrow \left( a+b \right)-2ab{{\left( a+b \right)}^{-1}}=x \\
\end{align}\]
Hence if $n=2$, then x is $\left( a+b \right)-2ab{{\left( a+b \right)}^{-1}}$. Correct match is S-1.
Hence from the above solution we get that the correct match is P-2, Q-4, R-3, S-1.
Option A is the correct answer.

Note: To solve this particular question we can check all the options after solving the first part and eliminate the incorrect options. After the solution of part P we get the correct match as P-2 so we can eliminate all three incorrect options. But to make sure that the answer is correct we solve all the parts.