The approximate value of $x^3-2 x^2+3 x+2$ at $x=2.01$ is
- 8.07
- 8.27
- 8.007
- 8.17
Solution
Here, $\mathrm{a}=2, \mathrm{~h}=0.01$ $\begin{aligned} \mathrm{f}(\mathrm{a}) & =(2)^3-2(2)^2+3(2)+2 \\ & =8-8+6+2 \\ & =8 \end{aligned}$ $\begin{aligned} \mathrm{f}^{\prime}(\mathrm{a}) & =3(2)^2-4(2)+3 \\ & =12-8+3 \\ & =7 \end{aligned}$ $\begin{aligned} \therefore \quad \mathrm{f}(\mathrm{a}+\mathrm{h}) & =\mathrm{f}(\mathrm{a})+\mathrm{hf}^{\prime}(\mathrm{a}) \\ & =8+(0.01) 7 \\ & =8+0.07 \\ & =8.07 \end{aligned}$
Asked in: MHT CET 2024 (10 May Shift 1)
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