1.

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

Answer»

1
0
`(1)/(2)`
`-(1)/(2)`

Solution :`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'stheorem 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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