Question

Match the following by appropriately matching the lists based on the information given in Column I and Column II.

$\begin{array}{cc}\text{Column I}& \\ \text{([.] denotes the greatest integer function)}& \text{Column II}\\ a.\underset{x\to 0}{lim}(\left[100\frac{\sin x}{x}\right]+\left[100\frac{\tan x}{x}\right])& p.198\\ b.\underset{x\to 0}{lim}(\left[100\frac{x}{\sin x}\right]+\left[100\frac{\tan x}{x}\right])& q.199\\ c.\underset{x\to 0}{lim}(\left[100\frac{{\sin }^{-1}x}{x}\right]+\left[100\frac{{\tan }^{-1}x}{x}\right])& r.200\\ d.\underset{x\to 0}{lim}(\left[100\frac{x}{{\sin }^{-1}x}\right]+\left[100\frac{{\tan }^{-1}x}{x}\right])& s.201\end{array}$

• A. $(a-q;b-r;c-q;d-p)$
• B. $(a-r;b-q;c-r;d-p)$
• C. $(a-q;b-r;c-q;d-s)$
• D. $(a-r;b-q;c-r;d-s)$

Answer 1

The correct option is A $(a-q;b-r;c-q;d-p)$

We know that $\underset{x\to 0}{lim}\frac{\sin x}{x}=1$ (but a value which is smaller than $1$)
$\Rightarrow \left[\underset{x\to 0}{lim}100\frac{\sin x}{x}\right]=99$
$\text{and}\left[\underset{x\to 0}{lim}100\frac{x}{\sin x}\right]=100$
Also, $\underset{x\to 0}{lim}\frac{{\sin }^{-1}x}{x}=1$ (but a value which is more than $1$)
$\text{or}\left[\underset{x\to 0}{lim}100\frac{{\sin }^{-1}x}{x}\right]=100$
$\text{and}\left[\underset{x\to 0}{lim}100\frac{x}{{\sin }^{-1}x}\right]=99$
$\underset{x\to 0}{lim}\frac{\tan x}{x}=1$ (but a value which is bigger than $1$)
$\text{or}\left[\underset{x\to 0}{lim}100\frac{\tan x}{x}\right]=100$
$\text{and}\left[\underset{x\to 0}{lim}100\frac{{\tan }^{-1}x}{x}\right]=99$
Hence,
$a.\underset{x\to 0}{lim}(\left[100\frac{\sin x}{x}\right]+\left[100\frac{\tan x}{x}\right])=199$
$b.\underset{x\to 0}{lim}(\left[100\frac{x}{\sin x}\right]+\left[100\frac{\tan x}{x}\right])=200$
$c.\underset{x\to 0}{lim}(\left[100\frac{{\sin }^{-1}x}{x}\right]+\left[100\frac{{\tan }^{-1}x}{x}\right])=199$
$d.\underset{x\to 0}{lim}(\left[100\frac{x}{{\sin }^{-1}x}\right]+\left[100\frac{{\tan }^{-1}x}{x}\right])=198$