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f(x) is cubic polynomial with f(x)=18a n...

`f(x)` is cubic polynomial with `f(x)=18a n df(1)=-1` . Also `f(x)` has local maxima at `x=-1a n df^(prime)(x)` has local minima at `x=0` , then the distance between `(-1,2)a n d(af(a)),` where `x=a` is the point of local minima is `2sqrt(5)` `f(x)` is increasing for `x in [1,2sqrt(5])` `f(x)` has local minima at `x=1` the value of `f(0)=15`

A

the distance between `(-1, 2) and (a, f(a))`, where ` x = a` is the point of local minima, is `2sqrt 5 `

B

`f(x) ` is increasing for ` x in [ 1, 2 sqrt 5] `

C

` f(x)` has local minima at ` x = 1 `

D

the value of ` f(0) = 5 `

Text Solution

Verified by Experts

The correct Answer is:
B, C
Since, `f(x) ` has local maxima at ` x =-1 and f' (x) ` has local minima at ` x = 0`
`therefore " " f'' (x) = lamda x`
On integrating, we get
`" " f' (x) = lamda (x^(2))/( 2) + c " " [ because f' (1) = 0 ]`
`rArr " " ( lamda)/(2) + c =0 rArr lamda = - 2 c " " ` ... (i)
Again, integrating on both sides, we get
`" " f (x) = lamda (x ^(3))/( 6) + cx + d `
`rArr " " f(2) = lamda ((8)/(6)) + 2c + d = 18 " " `... (ii)
and `" " f(1) = ( lamda)/( 6) + c + d = - 1 " " `... (iii)
From Eqs. (i), (ii) and (iii),
`" " f(x) = (1)/(4) (19x ^(3) - 57x + 34)`
`therefore f ' (x) = (1)/(4) ( 57 x ^(2) - 57) = ( 57)/(4) (x - 1) ( x + 1)`
For maxima or minima, put ` f' (x) = 0 rArr x =1, - 1 `
Now,`" " f'' (x) = (1)/(4) (114 x )`
At `" " x = 1 , f'' (x) gt 0`, minima
At ` " " x = -1, f'' (x) lt 0 `, maxima
`therefore f(x)` is increasing for `[ 1, 2 sqrt 5]`
`therefore f(x)` has local maxima at ` x = -1 and f(x)` has local minima at ` x = 1`.
Also, `" " f(0) = 34/4 `
Hence (b) and (c) are the correct answers.
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