if f(x)=2x^2 −12x+10. (i) sketch the graph of y=∣f(x)∣ for −1≤x≤7. (ii) find the set of values of k for which the equation ∣f(x)∣=k has 4 distinct roots.


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Question Number 113854 by ayenisamuel last updated on 15/Sep/20

$i f f (x) = 2 x^{2} - 12 x + 10 .$ $(i) s k e t c h t h e g r a p h o f y =∣ f (x) ∣ f o r$ $- 1 ⩽ x ⩽ 7 .$ $(i i) f i n d t h e s e t o f v a l u e s o f k f o r$ $w h i c h t h e e q u a t i o n ∣ f (x) ∣= k h a s 4$ $d i s t i n c t r o o t s .$
	Answered by MJS_new last updated on 15/Sep/20

	$∣ 2 x^{2} - 12 x + 10 ∣= k \Rightarrow k ⩾ 0$ ${(2 x^{2} - 12 x + 10)}^{2} = k^{2}$ $x^{4} - 12 x^{3} + 46 x^{2} - 60 x - \frac{k^{2}}{4} + 25 = 0$ $let x = t + 3$ $t^{4} - 8 t^{2} - \frac{k^{2}}{4} + 16 = 0$ $\Rightarrow t^{2} = \frac{8 \pm k}{2} \Rightarrow t = \pm \sqrt{\frac{8 \pm k}{2}} [4 solutions]$ $\Rightarrow x_{1, 2} = 3 \pm \sqrt{\frac{8 - k}{2}}; x_{3, 4} = 3 \pm \sqrt{\frac{8 + k}{2}}$ $\Rightarrow 4 solutions only if 8 - k > 0 \land 8 + k > 0$ $\Rightarrow - 8 < k < 8 but k ⩾ 0$ $\Rightarrow 0 ⩽ k < 8$
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