If x = 2/5 + 1.3/2! ≤ft(2/5)² + 1·3·5/3! ≤ft(2/5)³ + ... , then x + 1/x =

Correct Answer is : 55−34{5 {5} - 3}{4}455−3

Correct Answer is : 55−34{5 {5} - 3}{4}455−3

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AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 18 of 160

Marks: +1, -0

If x =

\frac{2}{5} + \frac{1.3}{2!} \left(\frac{2}{5}\right)^2

\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^3

\frac{1}{x}

Solution:

Let,

x = (1+y)^{n} - 1

So,

x = n y + \frac{n(n-1)}{2!} y^{2}

+ \frac{n(n-1)(n-1)}{3!} y^{3} + \dots

= \frac{2}{5} + \frac{1.3}{2!} \left( \frac{2}{5} \right)^{2} + \frac{1\cdot3\cdot5}{3!} \left( \frac{2}{5} \right)^{3}

+ \dots

Compare first three terms then,

n y = \frac{2}{5}, \quad \frac{n(n-1)}{2} y^{2} = \frac{1.3}{2!} \left( \frac{2}{5} \right)^{2}

, \quad \frac{n(n-1)(n-2)}{3!} y^{3} = \frac{1\cdot3\cdot5}{3!} \left( \frac{2}{5} \right)^{3}

From first two relation,

\frac{n y (n y - y)}{2!} = \frac{ \frac{2}{5} \left( \frac{2}{5} - y \right) }{2!}

= \frac{1\cdot3}{2!} \left( \frac{2}{5} \right)^{2}

y = \frac{2}{5} - \frac{6}{5}

= - \frac{4}{5}

And,

n = - \frac{1}{2}

This implies ,

\frac{ \left( - \frac{1}{2} \right) \left( - \frac{1}{2} - 1 \right) \left( - \frac{1}{2} - 2 \right) }{3!} \left( - \frac{4}{5} \right)^{3}

= \frac{1\cdot3\cdot5}{3!} \left( \frac{2}{5} \right)^{3}

= \text{RHS}

So,

x = \left(1 - \frac{4}{5}\right)^{- \frac{1}{2}} - 1

= \sqrt{5} - 1

\frac{1}{x} = \frac{ \sqrt{5} + 1 }{4}

This implies,

x + \frac{1}{x} = \sqrt{5} - 1 + \frac{ \sqrt{5} + 1 }{4}

= \frac{5 \sqrt{5} - 3}{4}

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