5. Let f(x)=lim θ rightarrow 0 (frac cos π x-x (2/θ ) sin (x-1) 1+x (2/θ ) (x-1) ), x in R . Consider the following two statements : (I) f ( x ) is discontinous at x =1 . (II) f ( x ) is continous at x =-1 . Then :

Mathematics · Limits, Continuity and Differentiability

JEE Main 2026 — 28 January, Evening Shift — Question 5

Let $f(x)=\lim _{\theta \rightarrow 0}\left(\frac{\cos \pi x-x^{\left(\frac{2}{\theta}\right)} \sin (x-1)}{1+x^{\left(\frac{2}{\theta}\right)}(x-1)}\right), x \in R$ . Consider the following two statements :

(I) $\mathrm{f}(\mathrm{x})$ is discontinous at $\mathrm{x}=1$ .

(II) $\mathrm{f}(\mathrm{x})$ is continous at $\mathrm{x}=-1$ . Then :

$f(x)=\left\{\begin{array}{cc}\cos \pi x & x \rightarrow 1^{-} \\ \frac{-\sin (x-1)}{(x-1)} & x \rightarrow 1^{+}\end{array}\right.$

RHL $=\lim _{x \rightarrow 1} \frac{-\sin (x-1)}{(x-1)}=-1$

$\mathrm{LHL}=\lim _{\mathrm{x} \rightarrow 1} \cos \pi \mathrm{x}=-1, \mathrm{f}(1)=-1$

$\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=1$

$f(x)=\left\{\begin{array}{cc}\frac{-\sin (x-1)}{-(x-1)} & x \rightarrow-1^{-} \\ \cos \pi x & x \rightarrow-1^{+}\end{array}\right.$

RHL $=\lim _{\mathrm{x} \rightarrow-1} \cos \pi \mathrm{x}=-1$

$\mathrm{LHL}=\lim _{\mathrm{x} \rightarrow-1} \frac{-\sin (\mathrm{x}-1)}{(\mathrm{x}-1)}=\frac{\sin 2}{-2}$

$\mathrm{f}(\mathrm{x})$ is discontinuous at $\mathrm{x}=-1$ .

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