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Exercise 2.13 · Q13

Q.The number of solutions of x2+∣x−1∣=1x^2+|x-1|=1 is

(1) 11
(2) 00
(3) 22
(4) 33
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Step 1. Case x≥1x\ge1: ∣x−1∣=x−1|x-1|=x-1, equation becomes x2+x−2=0⇒(x+2)(x−1)=0⇒x=−2,1x^2+x-2=0\Rightarrow(x+2)(x-1)=0\Rightarrow x=-2,1; only x=1x=1 satisfies x≥1x\ge1. …

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