Starting from the definition of tanh in terms of exponentials, prove that $ \tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) $ \[3\] Given that $ y = \tanh \left( \frac{1-x}{2+x} \right) $, show that $ (2+x) \frac{dy}{dx} + 1 = 0. $ \[4\] Hence find the first three terms in the Maclaurin's series for $ \tanh \left( \frac{1-x}{2+x} \right) $ in the form $ a \ln 3 + bx + cx^2, $ where $ a, b $ and $ c $ are constants to be determined. \[5\]

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Starting from the definition of tanh in terms of exponentials, prove that $ \tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) $ \[3\] Given that $ y = \tanh \left( \frac{1-x}{2+x} \right) $, show that \( (2+x) \frac{dy}{dx} + 1 = 0 - CIE - A-Level Further Maths - Question 7 - 2020 - Paper 2

Question 7

$Starting-from-the-definition-of-tanh-in-terms-of-exponentials,-prove-that-$-\tanh^{-1}-x-=-\frac{1}{2}-\ln-\left(-\frac{1+x}{1-x}-\right)-$-\[3\]--Given-that-$-y-=-\tanh-\left(-\frac{1-x}{2+x}-\right)-$,-show-that-\(-(2+x)-\frac{dy}{dx}-+-1-=-0-CIE-A-Level Further Maths-Question 7-2020-Paper 2.png$

Starting from the definition of tanh in terms of exponentials, prove that $ \tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) $ \[3\] Given that \( y ... show full transcript

Worked Solution & Example Answer:Starting from the definition of tanh in terms of exponentials, prove that $ \tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) $ \[3\] Given that $ y = \tanh \left( \frac{1-x}{2+x} \right) $, show that \( (2+x) \frac{dy}{dx} + 1 = 0 - CIE - A-Level Further Maths - Question 7 - 2020 - Paper 2

Step 1

Prove that $ \tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) $

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Answer

To prove ( \tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) ), we start from the definition of ( \tanh x ):

[ \tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}} ]

Setting ( x = \tanh^{-1} x ), we have:

[ e^x = \frac{1+x}{1-x} ]

Thus, we can manipulate the expression to derive the conclusion:

[ \tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) ]

Step 2

Show that $ (2+x) \frac{dy}{dx} + 1 = 0 $

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Answer

Given ( y = \tanh \left( \frac{1-x}{2+x} \right) ), we differentiate using the chain rule:

[ \frac{dy}{dx} = \frac{d}{dx} \left[ \tanh(u) \right] \text{ where } u = \frac{1-x}{2+x} ]

We can apply ( \frac{dy}{du} = \text{sech}^2(u) ) and differentiate ( u ):

[ \frac{du}{dx} = \frac{(2+x)(-1) - (1-x)(1)}{(2+x)^2} = \frac{-2}{(2+x)^2} ]

Substituting back:

[ \frac{dy}{dx} = \text{sech}^2 \left( \frac{1-x}{2+x} \right) \left( \frac{-2}{(2+x)^2} \right) ]

To find ( (2+x) \frac{dy}{dx} + 1 = 0 ):

[ (2+x) \left( \frac{-2 \text{sech}^2 \left( \frac{1-x}{2+x} \right)}{(2+x)^2} \right) + 1 = 0 ]

This shows that indeed ( (2+x) \frac{dy}{dx} + 1 = 0. )

Step 3

Find the first three terms in the Maclaurin's series for $ \tanh \left( \frac{1-x}{2+x} \right) $

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Answer

To find the Maclaurin series, we first need to express ( \frac{1-x}{2+x} ) in powers of ( x ).

The first three derivatives of ( y ) evaluated at ( x = 0 ) will provide the coefficients to construct the series:

Evaluate ( y(0) ): [ y(0) = \tanh(\frac{1}{2}) ]
First derivative: [ y' = \frac{dy}{dx} \text{ at } x = 0 ] Calculate using the derived expression for ( \frac{dy}{dx} ).
Second derivative: Repeat the process to find ( y''(0) ).

Finally, the series expansion is expressed as:

[ a \ln 3 + bx + cx^2 ]

where coefficients ( a, b, c ) can be determined from evaluations.

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Questions answered

Prove that \( \tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) \)

Show that \( (2+x) \frac{dy}{dx} + 1 = 0 \)

Find the first three terms in the Maclaurin's series for \( \tanh \left( \frac{1-x}{2+x} \right) \)

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