(a) The functions \(f : x \to x^{2} + 1\) and \(g : x \to 5 - 3x\) are defined on the set of the real numbers, R. (i) State the domain of \(f^{-1}\), the in...

Assessment: WAEC SSCE - Further Mathematics - 2015 (Essay) Subject: Further Mathematics

Question 1 Report

(a) The functions \(f : x \to x^{2} + 1\) and \(g : x \to 5 - 3x\) are defined on the set of the real numbers, R.

(i) State the domain of \(f^{-1}\), the inverse of f ; (ii) find \(g^{-1} (2)\).

(b) Evaluate : \(\int \frac{(x + 3)}{x^{2} + 6x + 9} \mathrm {d} x\)

Answer Details

(a) \(f:x\to x^{2}+1\) and \(g:x\to 5-3x\) on \(\mathbb{R}\).

(i) Domain of \(f^{-1}\). The domain of the inverse equals the range of \(f\). Since \(x^{2}\ge 0\), \(f(x)=x^{2}+1\ge 1\). Hence

\[\text{Domain of }f^{-1}=\{x:x\ge 1\}=[1,\infty)\]

(ii) \(g^{-1}(2)\). Let \(y=5-3x\Rightarrow x=\dfrac{5-y}{3}\), so \(g^{-1}(x)=\dfrac{5-x}{3}\).

\[g^{-1}(2)=\frac{5-2}{3}=1\]

(b) Evaluate \(\displaystyle\int\frac{x+3}{x^{2}+6x+9}\,dx\).

The denominator is a perfect square: \(x^{2}+6x+9=(x+3)^{2}\).

\[\int\frac{x+3}{(x+3)^{2}}\,dx=\int\frac{1}{x+3}\,dx=\ln|x+3|+C\]

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