
The fractional part of a real number x is $x - \left[ x \right]$, where $\left[ x \right]$ is the greatest integer less than or equal to x. Let ${F_1}$ and ${F_2}$ be the fractional parts of ${\left( {44 + \sqrt {2017} } \right)^{2017}}$ and ${\left( {44 - \sqrt {2017} } \right)^{2017}}$ respectively. Then ${F_1} + {F_2}$ lies between the numbers.
A. 0and 0.45
B. 0.45 and 0.9
C. 0.9 and 1.35
D. 1.35 and 1.8
Answer
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Hint: In the question we will use the expansion ${\left( {a - b} \right)^n}{ = ^n}{C_0}{a^n}{b^0}{ - ^n}{C_1}{a^{\left( {n - 1} \right)}}{b^1}{ + ^n}{C_2}{a^{(n - 2)}}{b^2} + ........{ + ^n}{C_n}{a^0}{b^n}$ and find the integral part and fractional part i.e. the largest that doesn’t exceeds x is an integral part and the difference $\left\{ x \right\} = x - \left[ x \right]$ is called fractional part. Use this to find between which numbers ${F_1} + {F_2}$ lies.
Complete step by step answer:
According to the given information, we have $x - \left[ x \right]$which is a fractional part of the real number x where $\left[ x \right] \leqslant x$ and we have fractional parts ${F_1}$ and ${F_2}$ which are fractional part of ${\left( {44 + \sqrt {2017} } \right)^{2017}}$ and ${\left( {44 - \sqrt {2017} } \right)^{2017}}$
For ${\left( {44 + \sqrt {2017} } \right)^{2017}}$using the formula of expansion which is given as ${\left( {a - b} \right)^n}{ = ^n}{C_0}{a^n}{b^0}{ - ^n}{C_1}{a^{\left( {n - 1} \right)}}{b^1}{ + ^n}{C_2}{a^{(n - 2)}}{b^2} + ........{ + ^n}{C_n}{a^0}{b^n}$ we get
${\left( {44 - \sqrt {2017} } \right)^{2017}} = {44^{2017}}{{\text{ }}^{2017}}{C_0} - {44^{2016}}{{\text{ }}^{2017}}{C_1}\left| {\sqrt {2017} } \right| + {44^{2015}}{{\text{ }}^{2017}}{C_2}{\left( {\sqrt {2017} } \right)^2} - {44^{2014}}{{\text{ }}^{2017}}{C_3}{\left( {\sqrt {2017} } \right)^3} + ........$
So, only odd powers of $\sqrt {2017} $ have fractional part and other terms have zero fractional part
Let the sum of odd powers is m.abc where $m = \left[ x \right]$ and abc is$\left\{ x \right\}$
Also let the sum of even powers in n which is a positive integer
Therefore, ${\left( {44 - \sqrt {2017} } \right)^{2017}} = n - m.abc$
$ \Rightarrow $$n - m - 0.abc$
$ \Rightarrow $$\underbrace {\left( {n - m - 1} \right)}_{Integralpart\left[ x \right]} + \underbrace {\left( {1 - 0.abc} \right)}_{fractionalpart\left\{ x \right\}}$
Similarly, for${\left( {44 + \sqrt {2017} } \right)^{2017}} = n + m.abc = \underbrace {\left( {n + m} \right)}_{\operatorname{int} egerpart} + \underbrace {\left( {0.abc} \right)}_{fractionalpart}$
Since fractional part of ${\left( {44 - \sqrt {2017} } \right)^{2017}}$and ${\left( {44 + \sqrt {2017} } \right)^{2017}}$are (1 – 0.abc) and (0.abc)
Therefore, the sum of ${F_1} + {F_2} = \left( {1 - 0.abc} \right) + \left( {0.abc} \right) = 1$
Thus ${F_1} + {F_2} = 1$
Therefore, ${F_1} + {F_2}$ 1 lies between 0.9 & 1.35
So, the correct answer is “Option C”.
Note:
In such types of questions it is advisable to remember some binomial expansions to identify what is an integral part and fractional part, so it saves a lot of time. In the beginning it will be difficult to learn every expansion but with time and practice everything gets easier.
Complete step by step answer:
According to the given information, we have $x - \left[ x \right]$which is a fractional part of the real number x where $\left[ x \right] \leqslant x$ and we have fractional parts ${F_1}$ and ${F_2}$ which are fractional part of ${\left( {44 + \sqrt {2017} } \right)^{2017}}$ and ${\left( {44 - \sqrt {2017} } \right)^{2017}}$
For ${\left( {44 + \sqrt {2017} } \right)^{2017}}$using the formula of expansion which is given as ${\left( {a - b} \right)^n}{ = ^n}{C_0}{a^n}{b^0}{ - ^n}{C_1}{a^{\left( {n - 1} \right)}}{b^1}{ + ^n}{C_2}{a^{(n - 2)}}{b^2} + ........{ + ^n}{C_n}{a^0}{b^n}$ we get
${\left( {44 - \sqrt {2017} } \right)^{2017}} = {44^{2017}}{{\text{ }}^{2017}}{C_0} - {44^{2016}}{{\text{ }}^{2017}}{C_1}\left| {\sqrt {2017} } \right| + {44^{2015}}{{\text{ }}^{2017}}{C_2}{\left( {\sqrt {2017} } \right)^2} - {44^{2014}}{{\text{ }}^{2017}}{C_3}{\left( {\sqrt {2017} } \right)^3} + ........$
So, only odd powers of $\sqrt {2017} $ have fractional part and other terms have zero fractional part
Let the sum of odd powers is m.abc where $m = \left[ x \right]$ and abc is$\left\{ x \right\}$
Also let the sum of even powers in n which is a positive integer
Therefore, ${\left( {44 - \sqrt {2017} } \right)^{2017}} = n - m.abc$
$ \Rightarrow $$n - m - 0.abc$
$ \Rightarrow $$\underbrace {\left( {n - m - 1} \right)}_{Integralpart\left[ x \right]} + \underbrace {\left( {1 - 0.abc} \right)}_{fractionalpart\left\{ x \right\}}$
Similarly, for${\left( {44 + \sqrt {2017} } \right)^{2017}} = n + m.abc = \underbrace {\left( {n + m} \right)}_{\operatorname{int} egerpart} + \underbrace {\left( {0.abc} \right)}_{fractionalpart}$
Since fractional part of ${\left( {44 - \sqrt {2017} } \right)^{2017}}$and ${\left( {44 + \sqrt {2017} } \right)^{2017}}$are (1 – 0.abc) and (0.abc)
Therefore, the sum of ${F_1} + {F_2} = \left( {1 - 0.abc} \right) + \left( {0.abc} \right) = 1$
Thus ${F_1} + {F_2} = 1$
Therefore, ${F_1} + {F_2}$ 1 lies between 0.9 & 1.35
So, the correct answer is “Option C”.
Note:
In such types of questions it is advisable to remember some binomial expansions to identify what is an integral part and fractional part, so it saves a lot of time. In the beginning it will be difficult to learn every expansion but with time and practice everything gets easier.
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