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Home  >>  CBSE XII  >>  Math  >>  Determinants
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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} $

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1 Answer

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Toolbox:
  • 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:$\lim_{t\to 0}\frac{sin t}{t}=1$
$\;\;\;=-1+1=0.$
Hence A is the correct answer.
answered Mar 26, 2013 by sreemathi.v
 

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