Let $\mathrm{P}(x)$ be a polynomial of degree 2 , with $\mathrm{P}(2)=-1, \mathrm{P}^{\prime}(2)=0,…

Let $\mathrm{P}(x)$ be a polynomial of degree 2 , with $\mathrm{P}(2)=-1, \mathrm{P}^{\prime}(2)=0, \mathrm{P}^{\prime \prime}(2)=2$, then $\mathrm{P}(1.001)$ is
  1. 0.002
  2. -0.002
  3. 0.004
  4. -0.004

Solution

$\begin{aligned} & \text { Let } \mathrm{P}(x)=\mathrm{a} x^2+\mathrm{b} x+\mathrm{c} \\ & \Rightarrow \mathrm{P}^{\prime}(x)=2 \mathrm{a} x+\mathrm{b} \\ & \Rightarrow \mathrm{P}^{\prime \prime}(x)=2 \mathrm{a} \\ & \mathrm{P}^{\prime \prime}(2)=2 \mathrm{a} \\ & \Rightarrow 2=2 \mathrm{a} \\ & \Rightarrow \mathrm{a}=1 \\ & \mathrm{P}^{\prime}(2)=2 \mathrm{a}(2)+\mathrm{b} \\ & \Rightarrow 0=4 \mathrm{a}+\mathrm{b} \\ & \Rightarrow 0=4(1)+\mathrm{b} \\ & \Rightarrow \mathrm{b}=-4 \\ & \mathrm{P}(2)=\mathrm{a}(2)^2+\mathrm{b}(2)+\mathrm{c} \\ & \Rightarrow-1=4 \mathrm{a}+2 \mathrm{~b}+\mathrm{c} \\ & \Rightarrow-1=4(1)+2(-4)+\mathrm{c} \\ & \Rightarrow \mathrm{c}=3 \\ & \mathrm{P}(x)=x^2-4 x+3 \\ & \Rightarrow \mathrm{P}^{\prime}(x)=2 x-4 \\ & x=1.001=1+0.001=\mathrm{a}+\mathrm{h} \end{aligned}$ Here, $\mathrm{a}=1, \mathrm{~h}=0.001$ $\begin{gathered} \mathrm{P}(\mathrm{a})=\mathrm{P}(1)=1-4+3=0 \\ \mathrm{P}^{\prime}(\mathrm{a})=\mathrm{P}^{\prime}(1)=2-4=-2 \end{gathered}$ $\therefore \quad \mathrm{P}(1.001)=0+(0.001)(-2)=-0.002$

Asked in: MHT CET 2023 (14 May Shift 2)

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