1.

Let f: Rto R be by f(x)=(x-1)(x-2)(x-5). Define F(x)=int_(0)^(x)f(t)dt,xgt0 Then which of the following options is/are correct ?

Answer»

`F(x) ne 0` for all `x in (0,5)`
F has a local maximum at `x=2`
F has two local MAXIMA and ONE local minimum is `(0,oo)`
F has a local minimum at `x=1`

Solution :Given , `f:R toR and f(x)=(x-1)(x-2)(x-5)`
Since , `F(x)-int_(0)^(x) f(t)dt,x gt 0`
So, `F'(x)=(x-2)(x-5)`
Accroding to wavy curve method
`(-""+""-""+)/(1"2"5)`
F'(x) chages , it's sign form negative to positive at `x =1` and 5, So, F(x) has minima at ` x=1` and 5 and as F'(x) changes, it's sign from positive to negative at` x=2` so F(x) has maxima at ` x=2`
` because F(2)=int_(0)^(2) f(t)dt= int_(0)^(2)(t^2-8t^2+17t-10)dt`
` =[(t^4)/(4)-8(t^3)/(3)+17(t^2)/(2)-10T]_(0)^(2)`
` =4-(64)/(3)+34-20=38-(124)/(3)=-(10)/(3)`
`because` At the point of maxima `x=2`, the functional value `F(2)=-(10)/(3)`, negative for the internal `x in (0,5)`, so `F(x)ne 0` for any value of `x in (0,5)`,
Hence , OPTIONS (a),(b) and (d) are correct.


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