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,

(A) Neither (I) nor (II) is True

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Continuity and Differentiability

Let

f (x) = lim_{θ \to 0} (\frac{c o s π x - x ^{(\frac{2}{θ})} s i n ( x - 1 )}{1 + x ^{(\frac{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,

Answer: (A)

θ \to 0^{+}

\frac{2}{θ} \to \infty

, so

x^{2/ θ} \to 0

when

∣ x ∣ < 1

, equals

1

when

∣ x ∣ = 1

, and

\to \infty

when

∣ x ∣ > 1

.

At

x = 1

: from the side

∣ x ∣ < 1

f (x) = cos π x \to - 1

; from the side

∣ x ∣ > 1

, dividing numerator and denominator by

x^{2/ θ}

gives

f (x) \to \frac{- sin ( x - 1 )}{x - 1} \to - 1

; and

f (1) = \frac{cos π - sin 0}{1 + 0} = - 1

. All three agree, so

f

is continuous at

x = 1

and statement (I) is false.

At

x = - 1

: from

x \to - 1^{-}

(where

∣ x ∣ > 1

f (x) \to \frac{- sin ( x - 1 )}{x - 1} = \frac{- sin ( - 2 )}{- 2} = - \frac{sin 2}{2}

; from

x \to - 1^{+}

(where

∣ x ∣ < 1

f (x) = cos π x \to cos (- π) = - 1

. The one-sided limits differ, so

f

is discontinuous at

x = - 1

and statement (II) is false.

Neither (I) nor (II) is true.

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option