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    I don't know what to do for question 5
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    Try to visualise these functions: hopefully you know that a root of f(x)-g(x)=0 means finding a point of intersection of f(x) and g(x). So sin(x-6) is very close to sin(x-2pi) which is very close to sin(x). Equally, ln(x^2+1) translates the curve such that its root is at the origin, and the x^2 means it looks like a ln(x) graph forward and back (even function). Draw them or use a graphing software to see that they will cross near the start.

    At this stage, try the value k=1 since it is the lowest positive integer. Calculate f(1) and f(2) and the change of sign proves that a root exists in the boundary (1,2).

    For part (b), write sin(x-6)=ln(x^2+1), then remove the log by taking e^LHS = e^RHS. Rearrange and you're done.

    For part a what they are looking for is two integers between which sin(x-6)-ln(x^2+1) changes sign. Hence there is a root in between. This gives k.
    For b, all you have to do is use the iteration formula
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