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JEE Main Mathematics Continuity and Differentiability 2026 JEE Main 2026 (28 January Shift 2)

JEE Main Mathematics Question (2026) — Solution

Question

Let f(x)= _ 0 ( x-x^ ( 2 ) (x-1) 1+x^ ( 2 ) (x-1) ), x R . Consider the following two statements : (I) f(x) is discontinous at x=1. (II) f(x) is continous at x=-1. Then,

Options

  1. A. Only (II) is True
  2. B. Neither (I) nor (II) is True
  3. C. Both (I) and (II) are True
  4. D. Only (I) is True

Answer

B. Neither (I) nor (II) is True

Step-by-step solution

Taking the limit as 0 , we note x^ (2/ ) 0 for |x| 1 . This gives: f(x) = cases x & x 1^- \\ - (x-1) (x-1) & x 1^+ cases Continuity at x = 1 : RHL = _ x 1^+ - (x-1) (x-1) = -1 LHL = _ x 1^- x = -1 , f(1) = -1 f(x) is continuous at x = 1 . So Statement (I) is false. Continuity at x = -1 : f(x) = cases - (x-1) -(x-1) & x -1^- \\ x & x -1^+ cases RHL = _ x -1^+ x = -1 LHL = _ x -1^- - (x-1) -(x-1) = 2 -2 Since LHL RHL, f(x) is discontinuous at x = -1 . So Statement (II) is also false. Hence, neither (I) nor (II) is true.

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