Given such that x = 0 is the only real root of P'(x) = 0. If P (-1) < P

JEE MainMathematicsToughMonotonocity

Given $P(x) = x^{4} + ax^{3} + bx^{2} + cx + d$ such that x = 0
is the only real root of P'(x) = 0. If P (-1) < P

Show answer and solution

P^{'}(x) = 4x^{3} + 3ax^{2} + 2bx + c

\ As\ P^{'}(0) = 0\ so\ c = 0

\therefore

P^{'}(x) = x\left( 4x^{2} + 3ax + 2b \right)

As x = 0 is the only real root of

P^{'}(x)

0, roots of 4x^3 +
3ax + 2b must have imaginary roots,
Therefore

4x^{2} + 3ax + 2b > 0\forall x \in R

Thus

P^{'}(x) < 0\ for\ x < 0

P' (x) > 0 for x > 0

Source: Grass-G / M-12 Mathematics Paper 2 Nov.docx

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