In [0, 1] Lagrange's mean value theorem is NOT applicable to [IIT Screening 2003]
A
f(x)⎧⎪⎨⎪⎩12−x,x<12(12−x)2,x≥12
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B
f(x){sinxx,x≠01,x=0
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C
f(x) = x |x|
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D
f(x) = |x|
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Solution
The correct option is Af(x)⎧⎪⎨⎪⎩12−x,x<12(12−x)2,x≥12 f(x)⎧⎪⎨⎪⎩12−x,x<12(12−x)2,x≥12
On Verification you'll find that the function is not differentiable at x = 1/2.
RHD = 0 whereas LHD = -1 at x = 1/2