A Place for Latest Exam wise Questions, Videos, Previous Year Papers, Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
4
Let $f(t) =
\begin{vmatrix}
\cos t & 1 & 1 \\
2\sin t & 2t & 1 \\
\sin t & t & t
\end{vmatrix}$,
then $\displaystyle \lim_{t \to 0}\dfrac{f(t)}{t^2}$ is equal to:
Solution
Expand the determinant using first row:
$f(t) = \cos t
\begin{vmatrix}
2t & 1 \\ t & t
\end{vmatrix}
- 1
\begin{vmatrix}
2\sin t & 1 \\ \sin t & t
\end{vmatrix}
+ 1
\begin{vmatrix}
2\sin t & 2t \\ \sin t & t
\end{vmatrix}$
Simplify and expand around $t \to 0$ using $\sin t \approx t$ and $\cos t \approx 1 - t^2/2$.
After simplification, $\dfrac{f(t)}{t^2} \to 3$.
$\boxed{\text{Answer: (D) 3}}$
Online Test Series, Information About Examination, Syllabus, Notification and More.