9709 P1 - Jun 2008 - Q8
712
Functions f and g are defined by
\(f : x \mapsto 4x - 2k\) for \(x \in \mathbb{R}\), where \(k\) is a constant,
\(g : x \mapsto \frac{9}{2-x}\) for \(x \in \mathbb{R}, x \neq 2\).
(i) Find the values of \(k\) for which the equation \(fg(x) = x\) has two equal roots. [4]
(ii) Determine the roots of the equation \(fg(x) = x\) for the values of \(k\) found in part (i). [3]
Solution
(i) To find \(fg(x)\), substitute \(g(x)\) into \(f(x)\):
\(fg(x) = 4\left(\frac{9}{2-x}\right) - 2k = \frac{36}{2-x} - 2k\).
Set \(fg(x) = x\):
\(\frac{36}{2-x} - 2k = x\).
Multiply through by \(2-x\) to clear the fraction:
\(36 - 2kx = x(2-x)\).
Rearrange to form a quadratic equation:
\(x^2 + 2kx - 2x + 36 - 4k = 0\).
For the equation to have two equal roots, the discriminant must be zero:
\((2k-2)^2 = 4(36-4k)\).
Solve for \(k\):
\(4k^2 - 8k + 4 = 144 - 16k\).
\(4k^2 + 8k - 140 = 0\).
\(k^2 + 2k - 35 = 0\).
Factorize:
\((k-5)(k+7) = 0\).
Thus, \(k = 5\) or \(k = -7\).
(ii) Substitute \(k = 5\) and \(k = -7\) into the quadratic equation:
For \(k = 5\):
\(x^2 + 8x + 16 = 0\).
Roots are \(x = -4\) (double root).
For \(k = -7\):
\(x^2 - 16x + 64 = 0\).
Roots are \(x = 8\) (double root).
