Mathematics · 2019
JEE Main · 10 January 2019, Shift 1 · Q70
For each t ε R, let [t] be the greatest integer less than or equal to t. Then, lim_x → 1+ ((1-|x|+ sin |1-x|) sin ((π)/2[1-x]))/(|1-x|[1-x])
For each $\displaystyle \mathrm{t} \mathbf{\epsilon} \mathbf{R}$, let [t] be the greatest integer less than or equal to t. Then,
$$\lim _{x \rightarrow 1+} \frac{(1-|x|+\sin |1-x|) \sin \left(\dfrac{\pi}{2}[1-x]\right)}{|1-x|[1-x]}
$$
Official answer
(1)
equals $\displaystyle 0$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.