The function f(x)=(x-1)/(x(x^2-1)),\ x ≠ 1,\ f(1)=1 , is discontinuou…

The correct answer is Option (3) → Exactly three points

Note: The given answer is as per the official NTA Answer Key. However, the question does not say whether the points where f is not defined (x = 0 and x = −1) are counted as points of discontinuity, and if continuity is checked only at the points where f is defined, Option (1) could also be a possible answer, as explained below.

Factorising the denominator:

    \( \displaystyle f(x) = \frac{x-1}{x(x-1)(x+1)} \)

For \(x \ne 1\) the factor \(x - 1\) cancels:

    \( \displaystyle f(x) = \frac{1}{x(x+1)},\quad x \ne 0, -1, 1 \)

At x = 1, f is defined, with f(1) = 1, and

    \( \displaystyle \lim_{x \to 1} f(x) = \frac{1}{1 \times 2} = \frac{1}{2} \)

The limit \(\frac{1}{2}\) is not equal to f(1) = 1, so f is discontinuous at x = 1.

At x = 0 and x = −1 the denominator is zero, so f is not defined there, and |f(x)| becomes very large near these points.

So the graph of f breaks at three points, x = −1, 0 and 1, and x = 1 is the only one of them where f is defined.

Option (1) Exactly one point — Should also be correct, though not the answer given by NTA. If only the points where f is defined are checked, the only failure is x = 1, where the limit is 1/2 but f(1) = 1.

Option (2) Exactly two points — Incorrect. The count is three (x = −1, 0, 1) if the points where f is not defined are included, and one (x = 1) if they are not; it is never two.

Option (3) Exactly three points — Correct. The two points where f is not defined (x = 0 and x = −1), together with x = 1, give three points.

Option (4) No point — Incorrect. In either reading f is discontinuous at x = 1, where the limit is 1/2 but f(1) = 1.