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STEP I 1997 question 2 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1997 question 2 solution

i

\displaystyle f(x) = \arctan x + \arctan \left(\frac{1-x}{1+x}\right)

\displaystyle f'(x) = \frac{1}{1+x^2} + \frac{1}{1+\left(\frac{1-x}{1+x}\right)^2} \times \frac{-(1+x)-(1-x)}{(1+x)^2}

\displaystyle f'(x) = \frac{1}{1+x^2} - \frac{2}{(1+x)^2 + (1-x)^2}

\displaystyle f'(x) = \frac{1}{1+x^2} - \frac{2}{2 + 2x^2} = 0

\therefore f(x) = C

f(1) = \arctan 1 + \arctan 0 = \frac{\pi}{4} = C

\therefore f(x) = \frac{\pi}{4}

Hmm on second thought should this be (8n+1)pi/4?

ii'

x = y \sin y^2

\displaystyle\frac{dx}{dy} = \sin y^2 + y \times 2y \cos y^2 = \sin y^2 + 2y^2 \cos y^2

\displaystyle \sin y^2 = \frac{x}{y}

y^2 - x^2 = y^2 - y^2 \sin ^2y^2 = y^2(1- \sin ^2y^2) = y^2 \cos ^2y^2

\sqrt{y^2 - x^2} = y \cos y^2

\displaystyle\frac{dx}{dy} = \frac{x}{y} + 2y\sqrt{y^2 - x^2} = \frac{x + 2y^2\sqrt{y^2-x^2}}{y}

\displaystyle\frac{dy}{dx} = \frac{y}{x + 2y^2\sqrt{y^2-x^2}}

Solution by datr

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