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JEE Advanced 2017
Paper-2 2017
Multiple correct answers - Select all that apply
Question
Let $f(x)=\frac{1-x(1+|1-x|)}{|1-x|} \cos \left(\frac{1}{1-x}\right)$ for $x \neq 1$. Then
Select ALL correct options:
A
$\lim _{x \rightarrow 1^{-}} f(x)$ does not exist
B
$\lim _{x \rightarrow 1^{-}} f(x)=0$
C
$\lim _{x \rightarrow 1^{+}} f(x)=0$
D
$\lim _{x \rightarrow 1^{+}} f(x)$ does not exist
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
⚠ Partially correct. Some answers are missing.
Solution
$\begin{aligned} f(x)= & \frac{1-x(1+|1-x|)}{|1-x|} \cos \left(\frac{1}{1-x}\right) \\ x \rightarrow 1^{+} \quad & \lim _{h \rightarrow 0} \frac{1-(1+h)[1+h]}{(h)} \cos \left(\frac{1}{h}\right) \\ & \Rightarrow \quad \lim _{h \rightarrow 0} \frac{-\left(h^2+2 h\right)}{h} \cos \left(\frac{1}{h}\right) \\ & \Rightarrow \quad(2) \times(-1,1) \text { does not exist }\end{aligned}$ $\begin{aligned} x \rightarrow 1^{-} \quad & \lim _{h \rightarrow 0} \frac{1-(1-h)[1+h]}{(h)} \cos \left(\frac{1}{h}\right) \\ & \lim _{h \rightarrow 0}\left(\frac{1-\left(1-h^2\right)}{(h)}\right) \cos \left(\frac{1}{h}\right) \\ \Rightarrow \quad & 0\end{aligned}$
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