JEE Main 2026 — Mathematics Calculus

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

$$f(x) = \begin{cases} \cos \pi x & x \to 1^- \\ \dfrac{-\sin(x-1)}{(x-1)} & x \to 1^+ \end{cases}$$f(x)=⎩⎨⎧​cosπx(x−1)−sin(x−1)​​x→1−x→1+​

Continuity at $x = 1$x=1:

$$\text{RHL} = \lim_{x \to 1^+} \frac{-\sin(x-1)}{(x-1)} = -1$$RHL=x→1+lim​(x−1)−sin(x−1)​=−1

$\displaystyle \text{LHL} = \lim_{x \to 1^-} \cos \pi x = -1$LHL=x→1−lim​cosπx=−1, $f(1) = -1$f(1)=−1

$\Rightarrow f(x)$⇒f(x) is continuous at $x = 1$x=1. So Statement (I) is false.

Continuity at $x = -1$x=−1:

$$f(x) = \begin{cases} \dfrac{-\sin(x-1)}{-(x-1)} & x \to -1^- \\ \cos \pi x & x \to -1^+ \end{cases}$$f(x)=⎩⎨⎧​−(x−1)−sin(x−1)​cosπx​x→−1−x→−1+​

$$\text{RHL} = \lim_{x \to -1^+} \cos \pi x = -1$$RHL=x→−1+lim​cosπx=−1

$$\text{LHL} = \lim_{x \to -1^-} \frac{-\sin(x-1)}{-(x-1)} = \frac{\sin 2}{-2}$$LHL=x→−1−lim​−(x−1)−sin(x−1)​=−2sin2​

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

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