Q90 / 150 · MHT CET 2024 (10 May Shift 1) · Mathematics
The approximate value of $x^3-2 x^2+3 x+2$ at $x=2.01$ is
- Correct answerA8.07
- B8.27
- C8.007
- D8.17
Solution
$\begin{aligned}
& \mathrm{f}(x)=x^3-2 x^2+3 x+2 \\
& \mathrm{f}^{\prime}(x)=3 x^2-4 x+3
\end{aligned}$
Here, $\mathrm{a}=2, \mathrm{~h}=0.01$ $\begin{aligned} \mathrm{f}(\mathrm{a}) & =(2)^3-2(2)^2+3(2)+2 \\ & =8-8+6+2 \\ & =8 \end{aligned}$ $\begin{aligned} \mathrm{f}^{\prime}(\mathrm{a}) & =3(2)^2-4(2)+3 \\ & =12-8+3 \\ & =7 \end{aligned}$ $\begin{aligned} \therefore \quad \mathrm{f}(\mathrm{a}+\mathrm{h}) & =\mathrm{f}(\mathrm{a})+\mathrm{hf}^{\prime}(\mathrm{a}) \\ & =8+(0.01) 7 \\ & =8+0.07 \\ & =8.07 \end{aligned}$
Here, $\mathrm{a}=2, \mathrm{~h}=0.01$ $\begin{aligned} \mathrm{f}(\mathrm{a}) & =(2)^3-2(2)^2+3(2)+2 \\ & =8-8+6+2 \\ & =8 \end{aligned}$ $\begin{aligned} \mathrm{f}^{\prime}(\mathrm{a}) & =3(2)^2-4(2)+3 \\ & =12-8+3 \\ & =7 \end{aligned}$ $\begin{aligned} \therefore \quad \mathrm{f}(\mathrm{a}+\mathrm{h}) & =\mathrm{f}(\mathrm{a})+\mathrm{hf}^{\prime}(\mathrm{a}) \\ & =8+(0.01) 7 \\ & =8+0.07 \\ & =8.07 \end{aligned}$
Hence, the correct answer is option: 'A'.