1
Question
Let f(x)=⎧⎪
⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪
⎪⎩acot−1(b+x4),−23<x<02,x=0ln(1−cx)x,0<x<23.
If the function f is differentiable at x=0, then the value of b2−2a+c6 is
If the function f is differentiable at x=0, then the value of b2−2a+c6 is
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Solution
As f is differentiable at x=0, so f(x) is continuous also at x=0.
∴f(0+)=limx→0ln(1−cx)x (00) form
limx→0−c1−cx=−c
⇒−c=2
⇒c=−2 …(1)
Now, f′(0+)=limh→0+ln(1+2h)h−2h
=limh→0+ln(1+2h)−2hh2
=limh→0+(2h−(2h)22+⋯)−2hh2=f′(0+)=−2
Also, f′(0−)=limh→0+acot−1(b−h4)−2−h (00) form
=limh→0+−a1+(b−h4)2×(−14)−1=−4ab2+16
As f′(0−)=f′(0+)
∴−4ab2+16=−2
⇒2a=b2+16∴b2−2a=−16 …(2)
Hence, using equation (1) and equation (2),
b2−2a+c6=−16+64=48
∴f(0+)=limx→0ln(1−cx)x (00) form
limx→0−c1−cx=−c
⇒−c=2
⇒c=−2 …(1)
Now, f′(0+)=limh→0+ln(1+2h)h−2h
=limh→0+ln(1+2h)−2hh2
=limh→0+(2h−(2h)22+⋯)−2hh2=f′(0+)=−2
Also, f′(0−)=limh→0+acot−1(b−h4)−2−h (00) form
=limh→0+−a1+(b−h4)2×(−14)−1=−4ab2+16
As f′(0−)=f′(0+)
∴−4ab2+16=−2
⇒2a=b2+16∴b2−2a=−16 …(2)
Hence, using equation (1) and equation (2),
b2−2a+c6=−16+64=48
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