Q34 / 75 · JEE Main 2026 (28 January Shift 2) · Mathematics

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

  • A
    Only (II) is True
  • Correct answer
    B
    Neither (I) nor (II) is True
  • C
    Both (I) and (II) are True
  • D
    Only (I) is True

Solution

Taking the limit as $ \theta \to 0 $, we note $ x^{(2/\theta)} \to 0 $ for $ |x| < 1 $ and $ \to \infty $ for $ |x| > 1 $. This gives:

$ f(x) = \begin{cases} \cos \pi x & x \to 1^- \\ \dfrac{-\sin(x-1)}{(x-1)} & x \to 1^+ \end{cases} $

Continuity at $ x = 1 $:
$ \text{RHL} = \lim_{x \to 1^+} \frac{-\sin(x-1)}{(x-1)} = -1 $
$ \text{LHL} = \lim_{x \to 1^-} \cos \pi x = -1 $, $ f(1) = -1 $
$ \Rightarrow f(x) $ is continuous at $ x = 1 $. So Statement (I) is false.

Continuity at $ x = -1 $:
$ f(x) = \begin{cases} \dfrac{-\sin(x-1)}{-(x-1)} & x \to -1^- \\ \cos \pi x & x \to -1^+ \end{cases} $

$ \text{RHL} = \lim_{x \to -1^+} \cos \pi x = -1 $
$ \text{LHL} = \lim_{x \to -1^-} \frac{-\sin(x-1)}{-(x-1)} = \frac{\sin 2}{-2} $

Since LHL $ \neq $ RHL, $ f(x) $ is discontinuous at $ x = -1 $. So Statement (II) is also false.

Hence, neither (I) nor (II) is true.

Hence, the correct answer is option: 'B'.

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