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Let f(x) be a function such that its der...

Let f(x) be a function such that its derovative f'(x) is continuous in [a, b] and differentiable in (a, b). Consider a function `phi(x)=f(b)-f(x)-(b-x)f'(x)-(b-x)^(2)`A. If Rolle's theorem is applicable to `phi(x)` on, [a,b], answer following questions.
If there exists some unmber c(a lt c lt b) such that `phi'(c)=0 and f(b)=f(a)+(b-a)f'(a)+lambda(b-a)^(2)f''(c)`, then `lambda` is

A

1

B

0

C

`(1)/(2)`

D

`-(1)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`phi(x)=f(b)-f(x)-(b-x)f'(x)-(b-x)^(2)A`
`phi(b)=0`
`phi(a)=f(b)-f(a)-(b-a)f'(a)-(b-a)6(2)A`
Since Rolle's theorem is applicable
`therefore" "A=(f(b)-f(a)-(b-a)f'(a))/((b-a)^(2))" (i)"`
`phi'(x)=-f'(x)+f'(x)-(b-x)f''(x)+2(b-x)A`
`therefore" There exists some number " cin (a,b)` such that
`0=phi'(c)=-(b-c)f''(c)+2(b-c)A`
`"i.e. A=(1)/(2)f''(c)" (ii)"`
From (i) and (ii), `f(b)-f(a)-(b-a)f'(a)=(1)/(2)(b-a)^(2)f''(c)`
`"i.e. "f(b)=f(a)+(b-a)f'(a)+(1)/(2)(b-a)^(2)f''(c)`
`therefore" "lambda=(1)/(2)`
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