Let `f:R to R` be a continuous odd function, which vanishes exactly at one point and `f(1)=(1)/(2)` Suppose that `F(x)=int_(-1)^(x) f(t)dt` for all `x in [-1,2]` and `G(x)=int_(-1)^(x) t|f(f(t))|dt` for all `x in [-1,2]`. If `lim_(x to x) (F(x))/(G(x))=(1)/(14)`, then the value of `f((1)/(2))`, is

Options

A. 7
B. 8
C. 9
D. 6

Answer

A

Explanation

It is given that f(x) is continuous on R.Therefore,
`F(x)=underset(-1)overset(x)int f(t)dt` and `G(x)=underset(-1)overset(x)int t|f(f(t))|dt` are also continous such that F'(x)=f(x) and G'(x)=x|f(f(x))|.
Since f(x) is an odd function. Therefore,
`F(1)=underset(-1)overset(1)int f(t)dt=0` and `G(1)=underset(-1)overset(1)int t|f(f(t))|dt=0`
Now,
`underset(x to 1)lim (F(x))/(G(x))=underset(x to 1)lim((F(x)-F(1))/(x-1))/((G(x)-G(1))/(x-1))=underset(x to 1)lim (F'(x))/(G'(x))`
`=underset(x to 1)lim(f(x))/(x|f(f(x))|)=(f(1))/(|f(f(1))|)=(1//2)/(|f(1//2)|)=(1)/(2|f(1//2)|)`
`:. underset(x to 1)lim(F(x))/(G(X))=(1)/(14)`
`rArr (1)/(2|f(1//2)|)=(1)/(14)rArr |f(1//2)|=7 rArr f(1//2)=+-7`
Note that `f(1//2)ne -7` as f(x) vanishes exactly at one point.