Question

# Let f(x)=2x3−3(2+p)x2+12px+ln(16−p2). If f(x) has exactly one local maximum and one local minimum, then the number of possible integral values of p is

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Solution

## Given : f(x)=2x3−3(2+p)x2+12px+ln(16−p2) Differentiating w.r.t. x, we get f′(x)=6x2−6(2+p)x+12p ⇒f′(x)=6(x2−(2+p)x+2p) For maximum or minimum, f′(x)=0 ⇒x2−(2+p)x+2p=0 ⇒x=2,p Since f(x) has exactly one local maximum and one local minimum, therefore p≠2. Also, 16−p2>0 ⇒p2−16<0 ⇒p∈(−4,4) Possible integral values of p are −3,−2,−1,0,1,3 Hence, number of possible integral values of p is 6.

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