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Question 106 of 128

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

(a) 1
(b) 0
(c) 2
(d) 3
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2020MCQ· 1mImportance★★★★★
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Splitting the absolute value at x=1x=1 and solving each case gives exactly two valid roots, x=0x=0 and x=1x=1.

Case x≥1x\ge1 (so ∣x−1∣=x−1|x-1|=x-1): x2+(x−1)=1⇒x2+x−2=0⇒(x+2)(x−1)=0⇒x=−2x^2+(x-1)=1 \Rightarrow x^2+x-2=0 \Rightarrow (x+2)(x-1)=0 \Rightarrow x=-2 or x=1x=1. Only x=1x=1 satisfies x≥1x\ge1.

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