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Continuity of sin(pi[x]/4)/[x] with the Greatest Integer Function | JEE

JEE Maths question with a full step-by-step solution.

Question
Let [x][x] be the greatest integer function. Then
f(x)=sin⁡(14π[x])[x]f(x) = \frac{\sin\left(\dfrac14\pi[x]\right)}{[x]}
 is
Anot continuous at any point
Bcontinuous at 32\dfrac32correct
Cdiscontinuous at 22correct
Ddifferentiable at 43\dfrac43correct
Solution
Step 1: Observe that ff depends on xx only through [x][x], which is constant on each interval [n, n+1)[n,\ n+1). So ff is constant on each such interval - a step function. Step 2: Compute the value on 1≤x<21 \le x < 2, where [x]=1[x] = 1.
f(x)=sin⁡π41=12.f(x) = \frac{\sin\dfrac{\pi}{4}}{1} = \frac{1}{\sqrt2} .
 Step 3: Compute the value on 2≤x<32 \le x < 3, where [x]=2[x] = 2.
f(x)=sin⁡π22=12.f(x) = \frac{\sin\dfrac{\pi}{2}}{2} = \frac12 .
 Step 4: Test (1). Inside any interval [n,n+1)[n, n+1) with n≠0n \ne 0 the function is constant, hence continuous there. So it is certainly continuous at many points. False. Step 5: Test (2). The point 32\dfrac32 lies strictly inside [1,2)[1,2), where ff is the constant 12\dfrac{1}{\sqrt2}, so ff is continuous there. True. Step 6: Test (3). At x=2x = 2 the left-hand limit is 12≈0.707\dfrac{1}{\sqrt2} \approx 0.707, while f(2)=12f(2) = \dfrac12. These differ, so ff is discontinuous at 22. True. Step 7: Test (4). The point 43\dfrac43 also lies strictly inside [1,2)[1,2), where ff is constant, so f′(43)=0f'\left(\dfrac43\right) = 0 exists. True. Answer: (2), (3) and (4).
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