Show that $f(x) > 1$ for all x > 0.

Solve the equation $f(x) = 4$.

EITHER

$f(x) - 1 = \frac{{1 + {3^{ - x}}}}{{{3^x} - {3^{ - x}}}}$     M1A1

> 0 as both numerator and denominator are positive     R1

OR

${3^x} + 1 > {3^x} > {3^x} - {3^{ - x}}$     M1A1

Note: Accept a convincing valid argument the numerator is greater than the denominator.

numerator and denominator are positive     R1

hence $f(x) > 1$     AG

[3 marks]

one line equation to solve, for example, $4({3^x} - {3^{ - x}}) = {3^x} + 1$, or equivalent     A1

$\left( {3{y^2} - y - 4 = 0} \right)$

attempt to solve a three-term equation     M1

obtain $y = \frac{4}{3}$     A1

$x = {\log _3}\left( {\frac{4}{3}} \right)$ or equivalent     A1

Note: Award A0 if an extra solution for x is given.

[4 marks]

(a) This is a question where carefully organised reasoning is crucial. It is important to state that both the numerator and the denominator are positive for $x > 0$. Candidates were more successful with part (b) than with part (a).

(a) This is a question where carefully organised reasoning is crucial. It is important to state that both the numerator and the denominator are positive for $x > 0$. Candidates were more successful with part (b) than with part (a).