Mathematics Exam > Mathematics Questions > In [0,1] .Lagrange’s mean value theorem...
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In [0,1] .Lagrange’s mean value theorem is NOT .applicable to
- a)
- b)
- c)f(x) = X |X|
- d)f(x) = |X|
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
In [0,1] .Lagrange’s mean value theorem is NOT .applicable toa)b...
There is only one function in option(A), where critical point 
but in other part critical point 
. then we can say that function in option (B),(C) and (D) are continuous on [0,1] and differentiable in (0,1).
but in other part critical point . then we can say that function in option (B),(C) and (D) are continuous on [0,1] and differentiable in (0,1).now for
here
⇒ f is not differentiable at 
f is not differentiable at 
⇒ LMVT is not applicable to f(x) in [0,11.
Most Upvoted Answer
In [0,1] .Lagrange’s mean value theorem is NOT .applicable toa)b...
There is only one function in option(A), where critical point 
but in other part critical point 
. then we can say that function in option (B),(C) and (D) are continuous on [0,1] and differentiable in (0,1).
but in other part critical point . then we can say that function in option (B),(C) and (D) are continuous on [0,1] and differentiable in (0,1).now for
here
⇒ f is not differentiable at 
f is not differentiable at 
⇒ LMVT is not applicable to f(x) in [0,11.
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In [0,1] .Lagrange’s mean value theorem is NOT .applicable toa)b)c)f(x) = X |X|d)f(x) = |X|Correct answer is option 'A'. Can you explain this answer?
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