Let f(x)=∫10t|t−x|dt,xϵR The function is discontinuous at
A
x=0
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B
x=1
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C
x=12
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D
None of the above points
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Solution
The correct option is D None of the above points ∫10t|t−x|dx x<0⇒f(x)∫10t(t−x)dx=(t33−t22x)10=13−x2 x>0⇒f(x)∫10t(x−t)dt=(t2x2−t23)10=x2−13 0≤x≤1=∫x0(x−t)dt+∫x0t(x−t)dt (t2x2−t33)x0+(t33−xt22)1x =(x22−x33)+(13−x2)−(x33−x32) f(x)=x33−x2+13 Minimum value is possible at x=1√2 f(6)=2083