If f(x)=ax+b(x−1)(x−4) has local maximum point at (2,−1), then (a+b) is equal to
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Solution
f(x)=ax+b(x−1)(x−4) ⇒f′(x)=(x2−5x+4)a−(ax+b)(2x−5)(x2−5x+4)2
As x=2 is a maxima , ⇒f′(2)=0 ⇒(22−5×2+4)a−(2a+b)(2×2−5)=0 ⇒−2a+2a+b=0 ⇒b=0
And f(2)=−1 f(2)=2a+b(2−1)(2−4) ⇒−1=2a+b−2 ⇒2a+b=2 ⇒a=1