Given P(x)=x^4+a x³+b x²+c x+d such that x=0 is the only real root of P'(x)=0. If P(-1) [AIEEE 2009]

Correct Answer is : P(−1)P(-1)P(−1) is not minimum but P(1)P(1)P(1) is the maximum of PPP

P(-1) is the minimum but P(1) is not the maximum of P

Neither P(-1) is the minimum nor P(1) is the maximum of P

P(-1) is the minimum and P(1) is the maximum of P

Correct Answer is : P(−1)P(-1)P(−1) is not minimum but P(1)P(1)P(1) is the maximum of PPP

Test Index

AIEEE 2009 Solved Paper

Section: Mathematics

Question : 74 of 89

Marks: +1, -0

Given

P(x)=x^4+a x^3+b x^2+c x+d

x=0

P'(x)=0

P(-1)< P(1)

[-1,1]

Solution:

We have

P(x)=x^4+a x^3+b x^2+c x+d

\Rightarrow P'(x)=4 x^3+3 a x^2+2 b x+c

But

P'(0)=0 \Rightarrow c=0

\therefore P(x)=x^4+a x^3+b x^2+d

As given that

P(-1)< P(a)

\Rightarrow 1-a+b+d< 1+a+b+d \Rightarrow a>0

Now

P'(x)=4 x^3+3 a x^2+2 b x=x (4 x^2+3 a x+2 b)

P'(x)=0

, there is only one solution

x=0

therefore

4 x^2+3 a x+2 b=0

should not have any real roots i.e.

D< 0

\Rightarrow 9 a^2-32 b< 0

\Rightarrow b>\;\frac{9 a^2}{32}>0

Hence

a, b>0 \Rightarrow P'(x)=4 x^3+3 a x^2+2 b x>0

\forall x>0

\therefore P(x)

is an increasing function on

(0,1)

\therefore P(0)< P(a)

Similarly we can prove

P(x)

is decreasing on

(-1,0)

\therefore P(-1)>P(0)

So we can conclude that

\text{Max} P(x)=P(1) \; \text{and} \; \text{Min} P(x)=P(0)

P(-1)

is not minimum but

P(1)

is the maximum of

P

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