Let S be the set of values of x for which the tangent to the curve y=f(x)=x3−x2−2x at (x,y) is parallel to the line segment joining the points (1,f(1)) and (−1,f(−1)), then S is equal to:
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Solution
y=f(x)=x3−x2−2x⇒f′(x)=3x2−2x−2⋯(i)
Also f(1)=1−1−2=−2 and f(−1)=−1−1+2=0⇒m=f(1)−f(−1)1+1=−2−02⇒m=−1⋯(ii)
From (i) and (ii) 3x2−2x−2=−1⇒3x2−2x−1=0⇒(3x+1)(x−1)=0⇒x=−13,1