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STEP II 1992 question 6 solutionTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1992 question 6 solution 
When  is tangent to  at point of intersection,  , is the only solution of Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse. with kx=\ln{(2\sec(x))  (as can be seen from the graph above)
Both curves have the same gradient at Gradient of Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse. : \ln{(2\sec(x)) Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
Therefore \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}  . Substitute this into  to get Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse. as required.
To find x0 correct to 2 d.p, you can either use the Newton-Raphson method or the iterative formula x_{0}=\cot({x_{0})\ln{(2\sec{x_{0}}} Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse. . 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_{n+1}=\cot({x_{n})\ln{(2\sec{x_{n}})}  . 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 The value given by computer is .9144038679. Both x0=0.91 and x0=0.92 give k0=1.3 to 2 s.f. Solution by khaixiang. |
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