f ( x ) = a x − a − x a x + a − x f(x)=\frac{a^x-a^{-x}}{a^x+a^{-x}}

Algebra Level 2

For f ( x ) = a x − a − x a x + a − x ( a ≠ 1 , a > 0 ) , f(x)=\frac{a^x-a^{-x}}{a^x+a^{-x}} ( a \neq 1, a>0), if f ( k ) = 10 , f(k)=10, what is the value of f ( 2 k ) ? f(2k)?

19 101 \frac{19}{101} 20 101 \frac{20}{101} 22 101 \frac{22}{101} 21 101 \frac{21}{101}

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Mahindra Jain by Mahindra Jain

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1 solution

Sidharth Kapur
Mar 10, 2014

We start by rearranging f ( x ) f (x) .

f ( x ) = a x − a − x a x + a − x = a x − 1 a x a x + 1 a x = a 2 x − 1 a 2 x + 1 = 1 − 2 a 2 x + 1 f (x) = \dfrac{a^x - a^{-x}} {a^x + a^{-x}} = \dfrac{a^x - \frac{1}{a^x} }{a^x + \frac{1}{a^x}} = \dfrac{a^{2 x} - 1} {a^{2 x} + 1} = 1 - \dfrac{2}{a^{2 x} + 1}

Therefore

f ( k ) = 1 − 2 a 2 k + 1 = 10 f ( k) = 1 - \dfrac{2}{a^{2 k} + 1} = 10

− 9 = 2 a 2 k + 1 - 9 = \dfrac{2}{a^{2 k} + 1}

a 2 k + 1 = − 2 9 a^{2 k} + 1 = - \frac{2}{9}

a 2 k = − 11 9 a^{2 k} = - \frac{11}{9}

We would like to find a 4 k a^{4 k} .

a 4 k = 121 81 a^{4 k} = \frac{121}{81}

f ( 2 k ) = 1 − 2 a 4 k + 1 = 1 − 2 121 81 + 1 f ( 2 k) = 1 - \dfrac{2}{a^{4 k} + 1} = 1 - \dfrac{2}{\frac{121}{81} + 1}

This simplifies to 20 101 \frac{20}{101} .

3 Helpful 0 Interesting 0 Brilliant 0 Confused

The third statement says 'We would like to find a^2k' I thought it could b more good if it was a^4k and so on for aftermath....

Vishal Ajwani - 7 years, 2 months ago

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Thanks for the correction! I have fixed it now.

Sidharth Kapur - 7 years, 2 months ago

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