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If  \large P=(7+4 \sqrt{3})^{9} and P = I + f, where I is the integer just less than P, then the value of \large P(1-f) is equal to

Option: 1

1


Option: 2

2^n


Option: 3

2


Option: 4

2^{2n}


Answers (1)

best_answer

Finding nature of an integral part of the expression.

If the given expansion is in the form of \mathrm{N}=(\mathrm{a}+\sqrt{\mathrm{b}})^{\mathrm{n}} \quad(\mathrm{n} \in \mathrm{N})

Working rule:

Step 1: \mathrm{Choose\;\;N}^{\prime}=(\mathrm{a}-\sqrt{\mathrm{b}})^{\mathrm{n}} \text { or }(\sqrt{\mathrm{b}}-\mathrm{a})^{\mathrm{n}} \text { according as a }>\sqrt{\mathrm{b}} \text { or } \sqrt{\mathrm{b}}>\mathrm{a}

Step 2: Use N + N’ or N - N’ such that result is an integer

i.e. \mathrm{(a+\sqrt{b})^n+(a-\sqrt{b})^n\;\;or\;\;(a+\sqrt{b})^n-(a-\sqrt{b})^n\;\;is\;\;an\;integer}

Step 3: Now use N = I + f, where I is an integral part of N and f is a fractional part of N (0 < f < 1)

 

Now,

\begin{array}{l}{ P=(7+4 \sqrt{3})^{9}} \\ {\text { Let } P'=(7-4 \sqrt{3})^{9} \quad \& \quad 0<7-4 \sqrt{3}<1}\end{array}

Using Binomial addition for P+P'

P+P'=2 \cdot\left[^{9} C_{0} 7^{9} + \;^{9} C_{2} 7^{7} \cdot(4 \sqrt{3})^{2}+\;^{9} C_{4} 7^{5} \cdot(4 \sqrt{3})^{4}\cdots \cdots \right]

and it is given that P + I + f , where  0 < f < 1 

Hence,

\large I+f+P'=2 \cdot\left[^{9} C_{0} 7^{9} + \;^{9} C_{2} 7^{7} \cdot(4 \sqrt{3})^{2}+\;^{9} C_{4} 7^{5} \cdot(4 \sqrt{3})^{4}\cdots \cdots \right]

\large So, \,\,I +f+P'=\text{ even integer}

Now it is clear that I is an integer so \large f+P' should be an integer

f lies between \large (0,1) and P' also lie between \large (0,1)

hence, \large f+P' \in(0,2)

But their sum is an integer and 1 is only integer that lies between this interval

hence, \large f+P'=1

\large P'=1-f

we need to calculate \large P(1-f)

or \large P(1-f)=PP'=\left[(7+4 \sqrt{3})^{9}(7-4 \sqrt{3})^{9}\right]

\large \begin{aligned} P(1-f) &=\left[(7+4 \sqrt{3})^{9}(7-4 \sqrt{3})^{9}\right] \\ &=[(7+4 \sqrt{3})(7-4 \sqrt{3})]^{9} \\ &=[49-48]^{9} \\ &=1^{9}=1 \end{aligned}

hence, option A is correct

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Rishi

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