Question

When \[{x^{1999}} \div {x^2} - 1\] , Find remainder?

Answer 1

Hint: In order to solve the question given above, we have first multiplied the divisor by the dividend. After doing this, we get a simple result with the help of which we can simplify our answer and find the remainder easily. We can use one more method to solve the above question, where we make use of \[{p_n}\left( x \right) = {x^n} + {c_{n - 1}}{x^{n - 1}} + ..... + {c_0}\] .
Formula used:
We can solve the above question using two simple methods. In the second method we will use the formula \[{p_n}\left( x \right) = {x^n} + {c_{n - 1}}{x^{n - 1}} + ..... + {c_0}\] .

Complete step by step solution:
We have to find the remainder of \[{x^{1999}} \div {x^2} - 1\] .
We know that,
 \[
  {x^{1999}} \div {x^2} - 1 \\
  \left( {{x^2} - 1} \right)\left( {{x^{1997}} + {x^{1995}} + {x^{1993}} + ........ + {x^3} + x} \right) \\
   \Rightarrow {x^2}\left( {{x^{1997}} + {x^{1995}} + {x^{1993}} + .... + {x^3} + x} \right) - \left( {{x^{1997}} + {x^{1995}} + {x^{1993}} + ... + {x^3} + x} \right) \;
 \] .
Now, open the brackets and subtract the necessary terms, we get,
 \[{x^{1999}} - x\] .
So,
 \[{x^{1999}} = \left( {{x^2} - 1} \right)\left( {{x^{1997}} + {x^{1995}} + {x^{1993}} + ........ + {x^3} + x} \right) + x\] .
This can be written as:
 \[\dfrac{{{x^{1999}}}}{{\left( {{x^2} - 1} \right)}} = {x^{1997}} + {x^{1995}} + {x^{1993}} + ........ + {x^3} + x\] , with the remainder \[x\] .
So, the correct answer is “x ”.

Note: This question can be solved in yet another way.
We know that,
 \[{p_n}\left( x \right) = {x^n} + {c_{n - 1}}{x^{n - 1}} + ..... + {c_0}\]
Now,
 \[{x^n} = {p_{n - 2}}\left( {{x^2} - 1} \right) + ax + b\] where \[n \geqslant 2\] .
So, from this, we have:
 \[{\left( { - 1} \right)^n} = a\left( { - 1} \right) + b\]
 \[ \Rightarrow 1 = a + b\] .
On solving this, we have:
 \[a = \dfrac{{1 - {{\left( { - 1} \right)}^n}}}{2}\] ,
And \[b = \dfrac{{1 + {{\left( { - 1} \right)}^n}}}{2}\] .
Now, in the above question we have \[n = 1999\] , so we have:
 \[a = 1\] , \[b = 0\] and the remainder is \[x\] .