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Question Number 24641 by math solver last updated on 23/Nov/17

Commented by math solver last updated on 23/Nov/17

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	Answered by ajfour last updated on 23/Nov/17

	$x^{2} + a x + b = {(x + \frac{a}{2})}^{2} + b - \frac{a^{2}}{4}$ $C a s e : I$ $i f - \frac{a}{2} < 0$ $f (x) i s i n c r e a s i n g i n [0, 2]$ $s o f (0) = b = 2$ $a n d f (2) = 4 + 2 a + b = 3$ $\Rightarrow a = - \frac{3}{2}; b = 2$ $C a s e I I :$ $i f 2 < - \frac{a}{2}$ $f (x) i s d e c r e a s i n g i n [0, 2]$ $S o f (0) = b = 3 a n d$ $f (2) = 4 + 2 a + b = 2$ $\Rightarrow a = - \frac{5}{2}; b = 3$ $C a s e I I I :$ $i f 0 < - \frac{a}{2} < 2$ $T h e n b - \frac{a^{2}}{4} = 2$ $a n d e i t h e r f (0) = b = 3 o r$ $f (2) = 4 + 2 a + b = 3$ $\Rightarrow c a s e I I I (a) :$ $b = 3 t h e n \frac{a^{2}}{4} = 1 o r a = \pm 2$ $t h a t i s (a, b) \equiv (\pm 2, 3)$ $c a s e I I I (b) :$ $4 + 2 a + b = 3 \Rightarrow b = - 1 - 2 a$ $s o \frac{a^{2}}{4} = b - 2 = - 1 - 2 a - 2$ $a^{2} + 8 a + 12 = 0$ $\Rightarrow {(a + 4)}^{2} = 4$ $a = - 2, - 6$ $\Rightarrow b = - 1 - 2 a = 3 o r 11$ $t h a t i s (a, b) \equiv (- 2, 3) o r$ $\equiv (- 6, 11) .$
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