JAMIA 2021 — Mathematics PYQ

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Question

JAMIA 2021 — Mathematics PYQ

JAMIA | Mathematics | 2021

Let $f (t) = cos t 2 sin t sin t am p; t am p; t am p; t am p; 1 am p; 2 t am p; t$ , then $lim_{t \to 0} \frac{f ( t )}{t ^{2}}$ is equal to

Choose the correct answer:

Explanation

1. Define the Function $f(t)$

Assuming $f(t)$ represents the determinant of the matrix:

f(t) = \begin{vmatrix} \cos t & t & 1 \\ 2\sin t & t & 2t \\ \sin t & t & t \end{vmatrix}

2. Calculate the Determinant

We expand the determinant along the first row:

f(t) = \cos t \begin{vmatrix} t & 2t \\ t & t \end{vmatrix} - t \begin{vmatrix} 2\sin t & 2t \\ \sin t & t \end{vmatrix} + 1 \begin{vmatrix} 2\sin t & t \\ \sin t & t \end{vmatrix}

Now, evaluate each $2 \times 2$ minor:

Minor 1: $\begin{vmatrix} t & 2t \\ t & t \end{vmatrix} = (t \cdot t) - (2t \cdot t) = t^2 - 2t^2 = -t^2$
Minor 2: $\begin{vmatrix} 2\sin t & 2t \\ \sin t & t \end{vmatrix} = (2\sin t \cdot t) - (2t \cdot \sin t) = 2t \sin t - 2t \sin t = 0$
Minor 3: $\begin{vmatrix} 2\sin t & t \\ \sin t & t \end{vmatrix} = (2\sin t \cdot t) - (t \cdot \sin t) = 2t \sin t - t \sin t = t \sin t$

Substitute these back into the expression for $f(t)$ :

f(t) = \cos t (-t^2) - t(0) + 1(t \sin t)

f(t) = t \sin t - t^2 \cos t

3. Evaluate the Limit

We need to find $\lim_{t\to 0} \dfrac{f(t)}{t^2}$ :

\lim_{t\to 0} \frac{t \sin t - t^2 \cos t}{t^2}

Split the fraction into two parts:

\lim_{t\to 0} \left( \frac{t \sin t}{t^2} - \frac{t^2 \cos t}{t^2} \right)

\lim_{t\to 0} \left( \frac{\sin t}{t} - \cos t \right)

4. Apply Standard Limits

Using the standard limit $\lim_{t\to 0} \frac{\sin t}{t} = 1$ and the fact that $\lim_{t\to 0} \cos t = 1$ :

1 - 1 = 0

Final Result

The value of the limit is:

0