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

By KS Teng

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 )

Leave a Comment

Contact Us

CONTACT US We would love to hear from you. Contact us, or simply hit our personal page for more contact information

Not readable? Change text. captcha txt

Start typing and press Enter to search

2015 A-level H2 Mathematics (9740) Paper 2 Question 2 Suggested SolutionsJC Mathematics
2015 A-level H2 Mathematics (9740) Paper 2 Question 4 Suggested SolutionsJC Mathematics