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Question Number 165164 by mnjuly1970 last updated on 26/Jan/22

	Answered by aleks041103 last updated on 27/Jan/22
	$f(x)=∣(x+a)(x^2 +(a+2)x+1)∣ for this to be non−diff at only one point we need (x+a)(x^2 +(a+2)x+1) to change signs only once. Since one always exists at x=−a, we need x^2 +(a+2)x+1 to not have real roots or to have exactly one with algebraic multiplicity of 2 or one of the roots of x^2 +(a+2)x+1 to coincide with the root of x+a, i.e x=−a. ⇒D=(a+2)^2 −4≤0 ⇒a(a+4)≤0 a∈[−4,0] or (−a)^2 +(a+2)(−a)+1=0 a^2 −a^2 +1−2a=0 ⇒a=0.5 then the other roots of f(x) are −0.5,−2,−0.5, which makes for one change in sign. ⇒Ans. a∈[−4,0]∪{(1/2)}$
	$f (x) =∣ (x + a) (x^{2} + (a + 2) x + 1) ∣$ $f o r t h i s t o b e n o n - d i f f a t o n l y o n e p o i n t$ $w e n e e d (x + a) (x^{2} + (a + 2) x + 1) t o$ $c h a n g e s i g n s o n l y o n c e . S i n c e o n e a l w a y s$ $e x i s t s a t x = - a, w e n e e d x^{2} + (a + 2) x + 1$ $t o n o t h a v e r e a l r o o t s o r t o h a v e e x a c t l y o n e$ $w i t h a l g e b r a i c m u l t i p l i c i t y o f 2 o r o n e$ $o f t h e r o o t s o f x^{2} + (a + 2) x + 1 t o c o i n c i d e$ $w i t h t h e r o o t o f x + a, i . e x = - a .$ $\Rightarrow D = {(a + 2)}^{2} - 4 ⩽ 0$ $\Rightarrow a (a + 4) ⩽ 0$ $a \in [- 4, 0]$ $o r$ ${(- a)}^{2} + (a + 2) (- a) + 1 = 0$ $a^{2} - a^{2} + 1 - 2 a = 0$ $\Rightarrow a = 0.5$ $t h e n t h e o t h e r r o o t s o f f (x) a r e$ $- 0.5, - 2, - 0.5, w h i c h m a k e s f o r$ $o n e c h a n g e i n s i g n .$ $\Rightarrow A n s . a \in [- 4, 0] \cup {\frac{1}{2}}$
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