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Question

If f(x)=1|x−1|−x2 and Df,Rf denote the domain and range of the function respectively, then

A
Df=R{1±52}
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B
Df=R{1±32}
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C
Rf=(,0)[45,)
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D
Rf=(,43)[45,)
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Solution

The correct options are A Df=R−{−1±√52} C Rf=(−∞,0)∪[45,∞) f(x)=1|x−1|−x2 The function is defined when |x−1|−x2≠0 Now, |x−1|=x2 Squaring on the both sides, ⇒(x−1)2=x4 ⇒(x2)2−(x−1)2=0 ⇒(x2+x−1)(x2−x+1)=0 x2−x+1>0 as Δ<0 ∴x2+x−1=0 ⇒x=−1±√52 ∴Df=R−{−1±√52} For range: Let y=1|x−1|−x2 ⇒x2−|x−1|+1y=0, y≠0 Case :1 When x≥1 ⇒x2−x+(1y+1)=0 As x∈R ⇒Δ≥0 ⇒1−4(1y+1)≥0 ⇒−4y−3≥0 ⇒4+3yy≤0, y≠0 ⇒y∈[−43,0) ...(1) Case :2 When x<1 ⇒x2+x+(1y−1)=0 ⇒Δ≥0 ⇒1−4(1y−1)≥0 ⇒−4y+5≥0 ⇒5y−4y≥0, y≠0 ⇒y∈(−∞,0)∪[45,∞) ...(2) From (1) and (2) ⇒y∈(−∞,0)∪[45,∞) ∴Rf=(−∞,0)∪[45,∞)

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