Question

If $f(t)=\left|\begin{array}{ccc}\cos t& t& 1\\ 2\sin t& t& 2t\\ \sin t& t& t\end{array}\right|$, then $\underset{t\to 0}{\lim }\frac{f(t)}{t^2}$ is equal to

• A. 0
• B. $-1$
• C. 2
• D. 3

Answer 1

Answer: (A)

Consider, $f(t)=\left|\begin{array}{ccc}\cos t& t& 1\\ 2\sin t& t& 2t\\ \sin t& t& t\end{array}\right|$
Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$,
$\Rightarrow f(t)=\left|\begin{array}{ccc}\cos t& t& 1\\ 2\sin t-\cos t& 0& 2t-1\\ \sin t-\cos t& 0& t-1\end{array}\right|$
By expanding along $C_2$, we get
$f(t)=-t\left|\begin{array}{cc}2\sin t-\cos t& 2t-1\\ \sin t-\cos t& t-1\end{array}\right|$
$=(-t)[(t-1)(2\sin t-\cos t)-(2t-1)(\sin t-\cos t)]$
$=(-t)[2\sin t\cdot t-\cos t\cdot t-2\sin t+\cos t$
$-2t\cdot \sin t+2t\cdot \cos t+\sin t-\cos t]$
$\Rightarrow f(t)=(-t)[t\cdot \cos t-\sin t]$
$\Rightarrow f(t)=t\sin t-t^2\cos t$
$\therefore \underset{t\to 0}{\lim }\frac{f(t)}{t^2}=\underset{t\to 0}{\lim }\frac{t\sin t-t^2\cos t}{t^2}$
$=\underset{t\to 0}{\lim }\frac{\sin t}{t}-\underset{t\to 0}{\lim }\cos t$
$=1-1=0$