Let f(x)=limθ→0(cos(πx)-x(2θ)sin(x-1)1+x(2θ)(x-1)), x∈ℝ.
Consider the following two statements:
I. f(x) is discontinuous at x = 1
II. f(x) is continuous at x = -1
Then, [2026]
(3)
f(x)={cos(πx)x→1--sin(x-1)x-1x→1+
RHL=limx→1-sin(x-1)x-1=-1
LHL=limx→1cos(πx)=-1, f(1)=-1
f(x) is continuous at x=1
f(x)={-sin(x-1)-(x-1)x→-1-cos(πx)x→-1+
RHL=limx→-1cos(πx)=-1
LHL=limx→-1-sin(x-1)x-1=sin2-2
f(x) is discontinuous at x=-1