2015 A-level H2 Mathematics (9740) Paper 2 Question 3 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(a)
(i)
f(x) = \frac{1}{1-x^2}, x \in \mathbb{R}, x>1

f'(c) = \frac{2x}{1-x^2} > 0 for all x \in \mathbb{R}, x>1

Since f(x) is increasing function, it has no turning points, f is 1-1, its inverse will exist.

(a)
(ii)
Let f(x) = y

x = \pm \sqrt{1- \frac{1}{y}} Reject -\sqrt{1- \frac{1}{y}} since x>1

\Rightarrow f^{-1}(x) = \sqrt{1- \frac{1}{x}}

D_{f^{-1}} = \{x \in \mathbb{R} | x ~ \textless ~ 0 \}

(b)
Let g(x) = y = \frac{2+x}{1-x^2}

We need b^2 - 4ac \ge 0 for all real values of g.

(1)^2 - 4(y)(2-y) \ge 0

1 - 8y + 4y^2 \ge 0

y = \frac{8 \pm \sqrt{48}}{8} = 1 \pm \frac{\sqrt{3}}{2}

y \ge 1 + \frac{\sqrt{3}}{2} \mathrm{~or~} y \le 1 - \frac{\sqrt{3}}{2}

R_g = (-\infty, 1- \frac{\sqrt{3}}{2} ] \cup [1 +  \frac{\sqrt{3}}{2}, \infty )

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