FunctionMedium
Question
If $g(x) = 3x^{2} + 2x - 3,f(0) = - 3$ and $4g(f(x)) = 3x^{2} - 32x + 72$, then $f(\text{ }g(2))$ is equal to:
Options
A.$\frac{25}{6}$
B.$- \frac{25}{6}$
C.$\frac{7}{2}$
D.$- \frac{7}{2}$
Solution
$g(2) = 13$
$$f(g(2)) = f(13) $$Now $4\text{ }g(f(x)) = 3x^{2} - 32x + 72$
$$4\left\lbrack 3f^{2}(x) + 2f(x) - 3 \right\rbrack = 3x^{2} - 32x + 72 $$Let $f(x) = t$
$${12t^{2} + 8t - \left( 3x^{2} - 32x + 84 \right) = 0 }{f(x) = \frac{- 8 \pm \sqrt{64 + 48\left( 3x^{2} - 32x + 84 \right)}}{24} }{f(x) = \frac{- 8 \pm 4(3x - 16)}{24} }$$$\because f(0) = - 3\ \therefore$ we take $+ ve$ sign
$${\therefore f(x) = \frac{- 8 + 4(3x - 16)}{24} }{\therefore f(13) = \frac{- 8 + 4 \times 23}{24} = \frac{84}{24} = \frac{7}{2}}$$
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