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B
f(x)=2x3+x−7 when 2x3+x−7≤0
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C
f(x)=7−2x3−x when 2x3+x−7<0
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D
f(x)=7−2x3−x when 2x3+x−7>0
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Solution
The correct options are Af(x)=2x3+x−7 when 2x3+x−7≥0 Cf(x)=7−2x3−x when 2x3+x−7<0 f(x)={2x3+x−7,2x3+x−7≥0−(2x3+x−7),2x3+x−7<0 ⇒f(x)={2x3+x−7,2x3+x−7≥07−2x3−x,2x3+x−7<0