JEE Main 2009 — Mathematics Calculus

$$\begin{aligned} & P(x)=x^4+a x^3+b x^2+c x+d \\ & P^{\prime}(x)=4 x^3+3 a x^2+2 b x+c \\ & \because x=0 \text { is a solution for } P^{\prime}(x)=0, \Rightarrow c=0 \\ & \therefore P(x)=x^4+a x^3+b x^2+d \end{aligned}$$​P(x)=x4+ax3+bx2+cx+dP′(x)=4x3+3ax2+2bx+c∵x=0 is a solution for P′(x)=0,⇒c=0∴P(x)=x4+ax3+bx2+d​ Also, we have $P(-1) < P(1)$P(−1)<P(1) $$\Rightarrow 1-a+b+d < 1+a+b+d \Rightarrow a>0$$⇒1−a+b+d<1+a+b+d⇒a>0 $\because P^{\prime}(x)=0$∵P′(x)=0, only when $x=0$x=0 and $P(x)$P(x) is differentiable in $(-1,1)$(−1,1), we should have the maximum and minimum at the points $x=-1,0$x=−1,0 and 1 only Also, we have $P(-1) < P(1)$P(−1)<P(1) $\therefore$∴ Max. of $P(x)=\operatorname{Max} .\{P(0), P(1)\}$P(x)=Max.{P(0),P(1)} & Min. of $P(x)=\operatorname{Min} .\{P(-1), P(0)\}$P(x)=Min.{P(−1),P(0)} In the interval $[0,1]$[0,1], $$P^{\prime}(x)=4 x^3+3 a x^2+2 b x=x\left(4 x^2+3 a x+2 b\right)$$P′(x)=4x3+3ax2+2bx=x(4x2+3ax+2b) $\because P^{\prime}(x)$∵P′(x) has only one root $x=0,4 x^2+3 a x+2 b=0$x=0,4x2+3ax+2b=0 has no real roots. $$\begin{aligned} & \therefore(3 a)^2-32 b < 0 \Rightarrow \frac{3 a^2}{32} < b \\ & \therefore b>0 \end{aligned}$$​∴(3a)2−32b<0⇒323a2​<b∴b>0​ Thus, we have $a>0$a>0 and $b>0$b>0 $$\therefore \mathrm{P}^{\prime}(\mathrm{x})=4 \mathrm{x}^3+3 a \mathrm{x}^2+2 \mathrm{bx}>0, \forall \mathrm{x} \in(0,1)$$∴P′(x)=4x3+3ax2+2bx>0,∀x∈(0,1) Hence $P(x)$P(x) is increasing in $[0,1]$[0,1] $\therefore$∴ Max. of $P(x)=P(1)$P(x)=P(1) Similarly, $P(x)$P(x) is decreasing in $[-1,0]$[−1,0] Therefore Min. $P(x)$P(x) does not occur at $x=-1$x=−1