Let $\text{f}(\text{t})=\begin{vmatrix}\cos\text{t}&\text{t}&1\\2\sin\text{t}&\text{t}&2\text{t}\\\sin\text{t}&\text{t}&\text{t}\end{vmatrix} ,$ then $\lim\limits_{\text{t}\rightarrow0}\frac{\text{f(t)}}{\text{t}^2}$ is equal to:
- 0
- -1
- 2
- 3
Solution:
We have, $\text{f}(\text{t})=\begin{vmatrix}\cos\text{t}&\text{t}&1\\2\sin\text{t}&\text{t}&2\text{t}\\\sin\text{t}&\text{t}&\text{t}\end{vmatrix} $
$\Rightarrow\ \frac{\text{f(x)}}{\text{t}^2}=\frac{1}{\text{t}^2}\begin{vmatrix}\cos\text{t}&\text{t}&1\\2\sin\text{t}&\text{t}&2\text{t}\\\sin\text{t}&\text{t}&\text{t}\end{vmatrix} $
$=\begin{vmatrix}\cos\text{t}&\text{t}&1\\\frac{2\sin\text{t}}{\text{t}}&1&2\\\frac{\sin\text{t}}{\text{t}}&1&1\end{vmatrix} $ $\big[\text{Dividing R}_2\text{ and R}_3\text{ by }'\text{t}'\big]$
$\Rightarrow\ \lim\limits_{\text{t}\rightarrow0}\frac{\text{f(t)}}{\text{t}^2}=\begin{vmatrix} \lim\limits_{\text{t}\rightarrow0}\text{t}\cos\text{t}&\lim\limits_{\text{t}\rightarrow0}\text{t}&\lim\limits_{\text{t}\rightarrow0}1\\\lim\limits_{\text{t}\rightarrow0}\text{t}\frac{2\sin}{\text{t}}&\lim\limits_{\text{t}\rightarrow0}1&\lim\limits_{\text{t}\rightarrow0}2\\\lim\limits_{\text{t}\rightarrow0}\text{t}\frac{\sin\text{t}}{\text{t}}&\lim\limits_{\text{t}\rightarrow0}1&\lim\limits_{\text{t}\rightarrow0}1\end{vmatrix}$
$=\begin{vmatrix}1&0&1\\2&1&2\\1&1&1\end{vmatrix}$ $\bigg(\because\lim\limits_{\text{t}\rightarrow 0}\frac{\sin\text{t}}{\text{t}}=1\bigg)$
$=1(1-2)-0+1(2-1)$
$=0$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.