Question:
Number of points where \(f(x)=[\sin x + \cos x]\) is not continuous in \((0,2\pi)\) is:
Number of points where \(f(x)=[\sin x + \cos x]\) is not continuous in \((0,2\pi)\) is:
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Check all integer levels in range AND include maxima/minima carefully—they can also contribute.
Updated On: Apr 16, 2026
- 3
- 4
- 5
- 6
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The Correct Option is C
Solution and Explanation
Concept:
Greatest integer function \([f(x)]\) is discontinuous at points where \(f(x)\) is an integer and crosses that integer value.
Also, at extrema (max/min), even touching can create discontinuity.
Step 1: Range of function. \[ \sin x + \cos x = \sqrt{2}\sin\left(x+\frac{\pi}{4}\right) \] \[ \Rightarrow -\sqrt{2} \le \sin x + \cos x \le \sqrt{2} \] Possible integer values: \[ -1,\ 0,\ 1 \]
Step 2: Solve each case. (i) \(\sin x + \cos x = 1\) \[ x = \frac{\pi}{2},\ \frac{3\pi}{2} \] At both points, function crosses/touches → discontinuities = 2 (ii) \(\sin x + \cos x = -1\) \[ x = \pi \] Crossing occurs → discontinuities = 1 (iii) \(\sin x + \cos x = 0\) \[ x = \frac{3\pi}{4},\ \frac{7\pi}{4} \] Crossing occurs → discontinuities = 2
Step 3: Total. \[ 2 + 1 + 2 = 5 \] Conclusion: \[ {(C)\ 5} \]
Step 1: Range of function. \[ \sin x + \cos x = \sqrt{2}\sin\left(x+\frac{\pi}{4}\right) \] \[ \Rightarrow -\sqrt{2} \le \sin x + \cos x \le \sqrt{2} \] Possible integer values: \[ -1,\ 0,\ 1 \]
Step 2: Solve each case. (i) \(\sin x + \cos x = 1\) \[ x = \frac{\pi}{2},\ \frac{3\pi}{2} \] At both points, function crosses/touches → discontinuities = 2 (ii) \(\sin x + \cos x = -1\) \[ x = \pi \] Crossing occurs → discontinuities = 1 (iii) \(\sin x + \cos x = 0\) \[ x = \frac{3\pi}{4},\ \frac{7\pi}{4} \] Crossing occurs → discontinuities = 2
Step 3: Total. \[ 2 + 1 + 2 = 5 \] Conclusion: \[ {(C)\ 5} \]
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