The approximate value of $3^{2.001}$, if $\log 3=1.0986$ is

The approximate value of $3^{2.001}$, if $\log 3=1.0986$ is
  1. 9.00898
  2. 9.0989
  3. 9.0898
  4. 9.00989

Solution

$\begin{aligned} & \text { Let } \mathrm{f}(x) \\ \therefore \quad & =3^x \\ \quad f^{\prime}(x) & =3^x \log 3 \\ \therefore \quad \text { Here, } \mathrm{a} & =2 \text { and } \mathrm{h}=0.001 \\ \therefore \quad \mathrm{f}(\mathrm{a}+\mathrm{h}) & \approx \mathrm{f}(\mathrm{a})+\mathrm{hf}^{\prime}(\mathrm{a}) \\ & \approx \mathrm{f}(2)+(0.001)+\mathrm{f}^{\prime}(2) \\ & \approx 3^2+(0.001)\left(3^2 \log 3\right) \\ & \approx 9+(0.001)(9 \times 1.0986) \\ & \approx 9+0.00989 \\ & \approx 9.00989\end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 1)

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