JEE Main 2009 — Application of Derivatives Question with Solution

$$\begin{array}{rl}& P(x)=x^4+a{x}^3+b{x}^2+cx+d\\ & P^{\prime }(x)=4x^3+3a{x}^2+2bx+c\\ & \because x=0\text{ is a solution for }{P}^{\prime }(x)=0,\Rightarrow c=0\\ & \therefore P(x)=x^4+a{x}^3+b{x}^2+d\end{array}$$ Also, we have $P(-1)<P(1)$ $$\Rightarrow 1-a+b+d<1+a+b+d\Rightarrow a>0$$ $\because P^{\prime }(x)=0$, only when $x=0$ and $P(x)$ is differentiable in $(-1,1)$, we should have the maximum and minimum at the points $x=-1,0$ and 1 only Also, we have $P(-1)<P(1)$ $\therefore$ Max. of $P(x)=\mathrm{Max}.\{P(0),P(1)\}$ \& Min. of $P(x)=\mathrm{Min}.\{P(-1),P(0)\}$ In the interval $[0,1]$, $$P^{\prime }(x)=4x^3+3a{x}^2+2bx=x\left(4x^2+3ax+2b\right)$$ $\because P^{\prime }(x)$ has only one root $x=0,4x^2+3ax+2b=0$ has no real roots. $$\begin{array}{rl}& \therefore (3a{)}^2-32b<0\Rightarrow \frac{3a^2}{32}<b\\ & \therefore b>0\end{array}$$ Thus, we have $a>0$ and $b>0$ $$\therefore {\mathrm{P}}^{\prime }(\mathrm{x})=4{\mathrm{x}}^3+3a{\mathrm{x}}^2+2\mathrm{bx}>0,\forall \mathrm{x}\in (0,1)$$ Hence $P(x)$ is increasing in $[0,1]$ $\therefore$ Max. of $P(x)=P(1)$ Similarly, $P(x)$ is decreasing in $[-1,0]$ Therefore Min. $P(x)$ does not occur at $x=-1$