STEP II 1992 question 6 solution

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STEP II 1992 question 6 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1992 question 6 solution

When $k_{0}x$ is tangent to $\ln{(2\sec(x))}$ at point of intersection, $x_{0}$ , is the only solution of $kx=\ln{(2\sec(x))$ with $0\leq x< \frac{\pi}{2}$ (as can be seen from the graph above)

Both curves have the same gradient at $x_{0}$ .

Gradient of $k_{0}x$ at $x_{0}$ is $k_{0}$ Gradient of $\ln{(2\sec(x))$ : $\frac{d}{dx}(\ln{(2\sec(x)))=\frac{2\tan(x)\sec(x)}{2\sec(x)} =\tan(x)=\tan(x_{0}) \text{ at } x_{0}$

Therefore $k_{0}=\tan(x_{0})$ . Substitute this into $k_{0}x_{0}=\ln(2\sec(x_{0}))$ to get $x_{0}=\cot({x_{0})\ln{(2\sec{x_{0}}}$ as required.

To find x0 correct to 2 d.p, you can either use the Newton-Raphson method or the iterative formula $x_{n+1}=\cot({x_{n})\ln{(2\sec{x_{n}})}$ . However, the latter converges very very slowly (about 30 iterations when initial value of 0.5 is used) and is slightly inaccurate (somehow). It gives to 2 d.p. $x_{0}=0.92$ . Also, remember that your initial value must be between 0 and pi/2.

Newton-Raphson method on the other hand converges very very quickly, about 1-2 iterations for initial value of 0.5, it gives to 2 d.p $x_{0}=0.91$ . The iterative formula for Newton-Raphson method is $x_{n+1}=x_{n}+\frac{\sin{2x_{n}}}{2}-\frac{(\sin^{2}{x_{n}})x_{n}}{\ln{(2\sec{x_{n}})}}$