Given $\mathrm{P}\left(\mathrm{x}\right)={\mathrm{x}}^4+{\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}$ such that x = 0 is the only real root of $\mathrm{P}\left(\mathrm{x}\right)=0.$ If $\mathrm{P}(-1)<\mathrm{P}\left(1\right),$ then in the interval [–1, 1]

$$\begin{gathered} P\text{'}\left(x\right)=4x^3+3a{x}^2+2bx+c \\ x=0 is only root of P\text{'}\left(x\right)=0 \\ P\text{'}\left(0\right)=0\Rightarrow c=0 \\ P\text{'}\left(x\right)=x(4x^2+3ax+2b) but P\text{'}\left(x\right) only real root x=0 \\ \Rightarrow quadratic eqn has imaginary roots \\ \Rightarrow 4x^2+3ax+2b>0 \\ P\text{'}\left(x\right)=\left\{\begin{array}{l}+ve x>0\\ -ve x<0\end{array}\right. \\ x>0 P\left(x\right) is increa\sin g \\ x>0 P\left(x\right)is decrea\sin g \\ min at x=0 \\ given P(-1)<P\left(1\right) max is P\left(1\right) \end{gathered}$$