Question

If x is nearly equal to 1, $\dfrac{{{\text{m}}{{\text{x}}^{\text{m}}}{\text{ - n}}{{\text{x}}^{\text{n}}}}}{{{\text{m - n}}}}$ is equal to:
$
  {\text{A}}{\text{. }}{{\text{x}}^{{\text{m + n}}}} \\
  {\text{B}}{\text{. }}{{\text{x}}^{{\text{m - n}}}} \\
  {\text{C}}{\text{. }}{{\text{x}}^{\text{m}}} \\
  {\text{D}}{\text{. }}{{\text{x}}^{\text{n}}} \\
$

Answer 1

Hint: Here we have to consider the value of x very close to 1 but not 1. Then doing binomial expansion this problem will be solved.

Complete step-by-step answer:
It is given that x is nearly equal to 1 so we can say x = 1 + h where h is a very small number tending towards 0 but not 0.
The given equation is $\dfrac{{{\text{m}}{{\text{x}}^{\text{m}}}{\text{ - n}}{{\text{x}}^{\text{n}}}}}{{{\text{m - n}}}}$.
On putting x = 1 + h in the given equation we get,
$ \Rightarrow \dfrac{{m{{(1 + h)}^m} - n{{(1 + h)}^n}}}{{m - n}}$ ……….(1)
Then on doing the expansion by binomial theorem we get,
$
   \Rightarrow {(1 + h)^m} = {}^m{C_0}{1^m}{h^0} + {}^m{C_1}{1^{m - 1}}{h^1} + {}^m{C_2}{1^{m - 2}}{h^2}............ \\
   \Rightarrow {(1 + h)^m} = \dfrac{{m!}}{{m!}} + h.\dfrac{{m!}}{{(m - 1)!1!}} + 0.......\, = \,1 + mh\,\,\,\,\, \\
  ({\text{Higher power of h has been neglected since it is tending towards zero}}) \\
  {\text{ }} \\
   \Rightarrow {(1 + h)^n} = {}^n{C_0}{1^n}{h^0} + {}^n{C_1}{1^{n - 1}}{h^1} + {}^n{C_2}{1^{n - 2}}{h^2}............ \\
   \Rightarrow {(1 + h)^n} = \dfrac{{n!}}{{n!}} + h.\dfrac{{n!}}{{(n - 1)!1!}} + 0.......\, = \,1 + nh \\
  ({\text{Higher power of h has been neglected since it is tending towards zero}}) \\
$
Now the equation (1) becomes:
$ \Rightarrow \dfrac{{m(1 + mh) - n(1 + nh)}}{{m - n}}$
On solving it further we get,
$
   \Rightarrow \dfrac{{m(1 + mh) - n(1 + nh)}}{{m - n}} = \dfrac{{m - n + h({m^2} - {n^2})}}{{m - n}} \\
   \Rightarrow 1 + h(m + n) \\
$
The term obtained can also be obtained as $ \Rightarrow {(1 + h)^{m + n}} = 1 + h(m + n)$
Since $1 + h(m + n)$ is the expansion of ${(1 + h)^{m + n}}$ when we neglect higher powers of h.
And we can write ${(1 + h)^{m + n}} = {x^{m + n}}$ as we have assumed x = 1 + h.
Hence the answer is ${x^{m + n}}$.
So, the correct option is A.

Note: To solve this problem we have added a term which is tending to zero but not exactly zero. Then we have solved using the binomial expansion neglecting the higher power of the term tending to zero. Here generally students put the value of x = 1 and try to find out but one should think that it is written x is nearly equal to 1 it may have a purpose. Here if we put the value of x = 1 then none of the options matches our answer, that’s the other trick to check whether we are proceeding in the right or wrong way. Therefore after observing all these we come to binomial expansion and solve the problem.