2023 SJI P1 Q6

2023 SJI P1 Q6

IB Year 5 | Grade 11
10 marks

It is given that \[\mathrm f(x)=\frac{x}{1-x^2},\quad\text{where}\hspace{0.5em}-1<x<1.\]

  1. Complete the following table, by stating whether \(\mathrm f(x)\) is positive, negative or zero for the given values of \(x\).[2]
Value of \(x\)\(-1<x<0\)0\(0<x<1\)
Value of \(\mathrm f(x)\)zero
  1. Show by differentiation that \(\mathrm f'(x)>0\) for \(-1<x<1\).[3]

The graph of \(y=\mathrm g(x)\) for \(-1<x<1\) with a minimum turning point at \((0,3)\) is shown below.

  1. It is given that \(\mathrm w(x)=\mathrm f(x)\cdot\mathrm g(x)\). By finding \(\mathrm w'(x)\), show that \(\mathrm w(x)\) is increasing on \(-1<x<1\).[5]

Solution:

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Answer:(a) Negative, zero, positive (b) \(\mathrm f'(x)=\dfrac{1+x^2}{(1-x^2)^2}>0\) (c) \(\mathrm w'(x)=\mathrm f'\mathrm g+\mathrm f\mathrm g'>0\)

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