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Let f:R rarr(0,oo) and g:R rarr R be twi...

Let `f:R rarr(0,oo) and g:R rarr` R be twice differntiable function such that f'' and g'' ar continous fucntion on R. Suppose `f(2)=g(2)=0,f''(2)ne0and g''(2)ne0.If lim_(xrarr2) (f(X)g(x))/(f'(x)g'(x))=1` then

A

`f` has a local minimum at ` x = 2 `

B

`f` has a local maximum at `x = 2`

C

`f'' (2) gt f(2)`

D

`f (x) - f '' (x) = 0`, for atleast one ` x in R`

Text Solution

Verified by Experts

The correct Answer is:
A, D
Here, ` underset ( x to 2) ( lim )= (f (x)* g(x)) /( f ' (x) * g' (x)) = 1 `
`rArr underset ( x to 2) ( lim) (f (x) g' (x) + f ' (x) g(x))/( f' ' (x) g ' (x) + f' (x) g'' (x)) = 1 `
`" "` [ using L' Hospital's rule]
`rArr " " (f (2) g' (2)+ f ' (2) g (2))/( f ' ' (2) g' (2) + f ' (2) g'' (2)) = 1 `
`rArr " " (f (2) g' (2))/( f'' (2) g ' (2)) = 1 " " [ because f ' (2) = g (2) = 0 ]`
`rArr f (x) - f '' (x)= 0`, for atleast one ` x in R`
`rArr ` Option (d) is correct.
Also, `f : R to (0, oo)`
`rArr " " f(2) gt 0 `
`therefore " " f'' (2) = f(2) gt 0" " ` [ from Eq. (i)]`
Since, ` f' (2) = 0 and f ''(2) gt 0`
`therefore f(x) ` attains local minimum at ` x =2`.
`rArr ` Option (a) is correct.
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