Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Determinants
0 votes

Let $f(t)=\begin{vmatrix}\cos t & t & 1\\2\sin t & t & 2t\\\sin t & t & t\end{vmatrix}$, then $lim_{t \to 0}\Large \frac{f(t)}{t^2}$ is equal to

$\begin{array}{1 1} 0 \\ -1 \\ 2 \\ 3 \end{array} $

Can you answer this question?

1 Answer

0 votes
  • If $A=\begin{vmatrix}a_{11} &a_{21} & a_{31}\\a_{12} &a_{22} &a_{32}\\a_{13} &a_{23} &a_{33}\end{vmatrix}$
  • $|A|=a_{11}(a_{22}\times a_{33}-a_{32}\times a_{23})-a_{21}(a_{12}\times a_{33}-a_{32}\times a_{13})+a_{31}(a_{12}\times a_{23}-a_{22}\times a_{13})$
Let $f(t)=\begin{vmatrix}cos t & t & 1\\2sin t & t & 2t\\sin t & t & t\end{vmatrix}$
Apply $R_2\rightarrow R_2-2R_3$
$f(t)=\begin{vmatrix}cos t & t & 1\\0 & -t &0\\sin t & t & t\end{vmatrix}$
Let us take t as a common factor from $C_2$
$f(t)=t\begin{vmatrix}cos t & 1 & 1\\0 & -1 &0\\sin t & 1 & t\end{vmatrix}$
Now expand along $R_1$ we get,
f(t)=t(cos t(-t-0)-1(0)+1(sin t)
$\;\;=t(-tcost+sin t)$
$\;\;=-t^2cost+tsin t$
Now applying the limits we get
$\lim_{t \to 0}\Large \frac{f(t)}{t^2}=\normalsize \lim_{t \to 0}\Large \frac{-t^2cos t+tsin t}{t^2}$
On splitting we get,
$\;\;\;=\lim_{t \to 0}\bigg(\Large \frac{-t^2cos t}{t^2}+\frac{tsin t}{t^2}\bigg)$
$\;\;\;=\lim_{t \to 0}\bigg(-cos t+\frac{tsin t}{t}\bigg)$
On applying limits we get,
we know cos 0=1
$\;\;\;=-1+\lim_{t \to 0} \frac{sin t}{t}$
hint:$\lim_{t\to 0}\frac{sin t}{t}=1$
Hence A is the correct answer.
answered Mar 26, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App