Q74 / 160 · AP EAMCET 2021 (24 Aug Shift 2) · Mathematics

If $\mathbf{a}=t^2 \hat{\mathbf{i}}+e^t \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}+t^2 \hat{\mathbf{j}}+\log t \hat{\mathbf{k}}$ and $f(t)=\mathbf{a} \cdot \mathbf{b}$, then $f^{\prime}(\mathrm{l})$ is equal to

  • A
    $3+5 e$
  • Correct answer
    B
    $5+3 e$
  • C
    $3+3 e$
  • D
    $4+3 e$

Solution

We have, $\mathbf{a}=t^2 \hat{\mathbf{i}}+e^t \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}+t^2 \hat{\mathbf{j}}+\log t \hat{\mathbf{k}}$ $\begin{aligned} & f(t)=\mathbf{a} \cdot \mathbf{b} \\ & f(t)=2 t^2+t^2 e^t+\log t \\ & f^{\prime}(t)=4 t+2 t e^t+t^2 e^t+\frac{1}{t} \\ & f^{\prime}(1)=4+2 e+e+1 \\ & f^{\prime}(1)=5+3 e\end{aligned}$

Hence, the correct answer is option: 'B'.

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