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If f (x) = [x sin p x] { where [x] denotes greatest integer function}, then f (x) is
- a)Continuous at x = 0
- b)Continuous in (-1, 0)
- c)Differentiable at x = 1
- d)Differentiable in (-1, 1)
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
If f (x) = [x sinpx] { where [x] denotes greatest integer function}, t...
If −1≤x≤1, then 0≤xsinπx≤1/2
∴ f(x)=[xsinπx]=0, for −1≤x≤1
If 1<x<1+h, where h is a small positive real number, then
π<πx<π+πh
⇒−1<sinπx<0
⇒−1<xsinπx<0
∴ f(x)=[xsinπx]=−1 in the right neighbourhood of x=1.
Thus, f(x) is constant and equal to zero in [−1,1] and so f(x) is differentiable and hence continuous on (−1,1).
∴ f(x)=[xsinπx]=0, for −1≤x≤1
If 1<x<1+h, where h is a small positive real number, then
π<πx<π+πh
⇒−1<sinπx<0
⇒−1<xsinπx<0
∴ f(x)=[xsinπx]=−1 in the right neighbourhood of x=1.
Thus, f(x) is constant and equal to zero in [−1,1] and so f(x) is differentiable and hence continuous on (−1,1).
Most Upvoted Answer
If f (x) = [x sinpx] { where [x] denotes greatest integer function}, t...
Solution:
To determine the continuity and differentiability of f(x), we need to evaluate left-hand limit, right-hand limit and the derivative of f(x) at every point in the domain of f(x).
Left-hand limit of f(x):
As x approaches 0 from the left, [x] approaches -1 and sin(px) oscillates between -1 and 1. Therefore, f(x) oscillates between -x and x. Thus, the left-hand limit of f(x) as x approaches 0 is:
lim x→0- f(x) = lim x→0- [x sin(px)] = lim x→0- (-x) = 0.
Right-hand limit of f(x):
As x approaches 0 from the right, [x] approaches 0 and sin(px) oscillates between -1 and 1. Therefore, f(x) oscillates between 0 and x. Thus, the right-hand limit of f(x) as x approaches 0 is:
lim x→0+ f(x) = lim x→0+ [x sin(px)] = lim x→0+ (0) = 0.
Since the left-hand limit and right-hand limit are equal, f(x) is continuous at x=0.
Derivative of f(x):
For x≠0, f(x) is piecewise linear with slope sin(px) and jump discontinuity at every integer multiple of 1/p. Therefore, f(x) is not differentiable at these points.
Since f(x) is continuous at x=0 and not differentiable at any other point, the correct answer is option 'A'.
To determine the continuity and differentiability of f(x), we need to evaluate left-hand limit, right-hand limit and the derivative of f(x) at every point in the domain of f(x).
Left-hand limit of f(x):
As x approaches 0 from the left, [x] approaches -1 and sin(px) oscillates between -1 and 1. Therefore, f(x) oscillates between -x and x. Thus, the left-hand limit of f(x) as x approaches 0 is:
lim x→0- f(x) = lim x→0- [x sin(px)] = lim x→0- (-x) = 0.
Right-hand limit of f(x):
As x approaches 0 from the right, [x] approaches 0 and sin(px) oscillates between -1 and 1. Therefore, f(x) oscillates between 0 and x. Thus, the right-hand limit of f(x) as x approaches 0 is:
lim x→0+ f(x) = lim x→0+ [x sin(px)] = lim x→0+ (0) = 0.
Since the left-hand limit and right-hand limit are equal, f(x) is continuous at x=0.
Derivative of f(x):
For x≠0, f(x) is piecewise linear with slope sin(px) and jump discontinuity at every integer multiple of 1/p. Therefore, f(x) is not differentiable at these points.
Since f(x) is continuous at x=0 and not differentiable at any other point, the correct answer is option 'A'.
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Community Answer
If f (x) = [x sinpx] { where [x] denotes greatest integer function}, t...
If −1≤x≤1, then 0≤xsinπx≤1/2
∴ f(x)=[xsinπx]=0, for −1≤x≤1
If 1<x<1+h, where h is a small positive real number, then
π<πx<π+πh
⇒−1<sinπx<0
⇒−1<xsinπx<0
∴ f(x)=[xsinπx]=−1 in the right neighbourhood of x=1.
Thus, f(x) is constant and equal to zero in [−1,1] and so f(x) is differentiable and hence continuous on (−1,1).
∴ f(x)=[xsinπx]=0, for −1≤x≤1
If 1<x<1+h, where h is a small positive real number, then
π<πx<π+πh
⇒−1<sinπx<0
⇒−1<xsinπx<0
∴ f(x)=[xsinπx]=−1 in the right neighbourhood of x=1.
Thus, f(x) is constant and equal to zero in [−1,1] and so f(x) is differentiable and hence continuous on (−1,1).
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