First slide

Functions (XII)

Question

If $g (x) = x^{2} + x - 2$ and $\frac{1}{2} g o f (x) = 2 x^{2} - 5 x + 2$ then which is not a possible $f (x)$ ?

Moderate

A

$2 x - 3$
B

$- 2 x + 2$
C

$x - 3$
D

None of these

Solution

$\begin{array}{l} \frac{1}{2} (g o f) (x) = 2 x^{2} - 5 x + 2 \\ or \frac{1}{2} g [f (x)] = 2 x^{2} - 5 x + 2 \\ ∴ [{\{f (x)\}}^{2} + \{f (x)\} - 2] = 2 [2 x^{2} - 5 x + 2] \\ or f {(x)}^{2} + f (x) - (4 x^{2} - 10 x + 6) = 0 \\ ∴ f (x) = \frac{- 1 \pm \sqrt{1 + 4 (4 x^{2} - 10 x + 6)}}{2} \\ = \frac{- 1 \pm \sqrt{(16 x^{2} - 40 x + 25)}}{2} = \frac{- 1 \pm (4 x - 5)}{2} \\ = 2 x - 3 or - 2 x + 2 \end{array}$

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