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Question

# If f(x)=(x2−1)n+1(x2+x+1),nϵN and f(x) has a local extremum at x=1, then n=

A
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B
3
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C
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D
5
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Solution

## The correct options are B 3 D 5f(x)=(x2−1)n+1(x2+x+1),nϵNdydx=(x2−1)n+1(2x+1)+(n+1)(x2−1)n2x(x2+x+1)=(x2−1)n[(x2−1)(2x+1)+2x(n+1)(x2+x+1)] Since f(x) has a local extremum dydx must change sign i.e. f′(1+)andf′(1−) should be of opposite signs.Now we know that x2+x+1 is always +ive (b2−4ac=−ive) and the first term tends to zero as x→1.Hence the sign of dydx will depend upon (x2−1)n. But x2−1=(x+1)(x−1) is +ive for x>1 and -ive for x<1. Hence if n is even then f′(1+) and f′(1−) will have the same sign, but they must be of opposite sign for the existence of extremum. Thus n must be odd. Hence (b) and (d) are the correct choices.Ans: B,D

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