Given [P( x ) = x^4 + ax^3 + bx^2 + cx + d] such that x=0 is the only real root of [P',( x…

Question

Given $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$$ such that $x=0$ is the only

real root of $$P'\,\left( x \right) = 0.$$ If $$P\left( { - 1} \right) < P\left( 1 \right),$$ then in the interval $$\left[ { - 1,1} \right]:$$

Accepted Answer

We have $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$$ $$ \Rightarrow P'\left( x \right) = 4\,{x^3} + 3a{x^2} + 2bx + c$$ But $$P'\left( 0 \right) = 0 \Rightarrow c = 0$$ $$\therefore$$ $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + d$$ As given that $$P\left( { - 1} \right) < P\left( a \right)$$ $$ \Rightarrow 1 - a + b + d\,\, < \,\,1 + a + b + d \Rightarrow a > 0$$ Now $$P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx = x\left( {4{x^2} + 3ax + 2b} \right)$$ As $$P'\left( x \right) = 0,$$ there is only one solution $x = 0,$ therefore $4{x^2} + 3ax + 2b = 0$ should not have any real roots i.e. $D < 0$ $$ \Rightarrow 9{a^2} - 32b < 0$$ $$ \Rightarrow b > {{9{a^2}} \over {32}} > 0$$ Hence $$a,b > 0 \Rightarrow P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx > 0$$ $$\forall x > 0$$ $$\therefore$$ $P(x)$ is an increasing function on $$\left( {0,1} \right)$$ $$\therefore$$ $$P\left( 0 \right) < P\left( a \right)$$ Similarly we can prove $$P\left( x \right)$$ is decreasing on $$\left( { - 1,0} \right)$$ $$\therefore$$ $$P\left( { - 1} \right) > P\left( 0 \right)$$ So we can conclude that Max $$P\left( x \right) = P\left( 1 \right)$$ and Min $$P\left( x \right) = P\left( 0 \right)$$ $$ \Rightarrow P\left( { - 1} \right)$$ is not minimum but $$P\left( 1 \right)$$ is the maximum of $P.$

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