Question:
Let [t] denote the greatest integer \(\le t\) and \(\lim_{x \to 0} x\left[\frac{4}{x}\right] = A\).
Then the function, \(f(x) = [x^2]\sin(\pi x)\) is discontinuous, when x is equal to :
Let [t] denote the greatest integer \(\le t\) and \(\lim_{x \to 0} x\left[\frac{4}{x}\right] = A\).
Then the function, \(f(x) = [x^2]\sin(\pi x)\) is discontinuous, when x is equal to :
Then the function, \(f(x) = [x^2]\sin(\pi x)\) is discontinuous, when x is equal to :
Show Hint
Find A = 4, then see where \([x^2]\) jumps (at \(x = \sqrt n\)) and whether \(\sin(\pi x)\) vanishes there; only non-integer \(\sqrt n\) gives a discontinuity.
Updated On: Nov 5, 2026
- \(\sqrt{A + 1}\)
- \(\sqrt{A + 5}\)
- \(\sqrt{A + 21}\)
- \(\sqrt{A}\)
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The Correct Option is A
Solution and Explanation
\(\left[\frac4x\right] = \frac4x - \left\{\frac4x\right\}\), so \(x\left[\frac4x\right] = 4 - x\left\{\frac4x\right\} \to 4\) as \(x \to 0\) (since \(0 \le \{\cdot\} \lt 1\)). Thus A = 4.
\([x^2]\) jumps at points where \(x^2\) is a positive integer, i.e. \(x = \sqrt n\). The function \(f = [x^2]\sin(\pi x)\) can be discontinuous only there, and only if \(\sin(\pi x) \ne 0\), i.e. x is not an integer.
(1) \(\sqrt{A + 1} = \sqrt5\): not an integer, \(\sin(\pi\sqrt5) \ne 0\), so f jumps: discontinuous.
(2) \(\sqrt{A + 5} = 3\): integer, \(\sin 3\pi = 0\), both one-sided limits are 0, continuous.
(3) \(\sqrt{A + 21} = 5\): integer, continuous.
(4) \(\sqrt{A} = 2\): integer, continuous.
Answer: (1) \(\sqrt{A + 1}\)
\([x^2]\) jumps at points where \(x^2\) is a positive integer, i.e. \(x = \sqrt n\). The function \(f = [x^2]\sin(\pi x)\) can be discontinuous only there, and only if \(\sin(\pi x) \ne 0\), i.e. x is not an integer.
(1) \(\sqrt{A + 1} = \sqrt5\): not an integer, \(\sin(\pi\sqrt5) \ne 0\), so f jumps: discontinuous.
(2) \(\sqrt{A + 5} = 3\): integer, \(\sin 3\pi = 0\), both one-sided limits are 0, continuous.
(3) \(\sqrt{A + 21} = 5\): integer, continuous.
(4) \(\sqrt{A} = 2\): integer, continuous.
Answer: (1) \(\sqrt{A + 1}\)
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