Let `f,g:[-1,2]vecR` be continuous functions which are twice differentiable on the interval `(-1,2)dot` Let the values of `fa n dg` at the points `-1,0a n d2` be as given in the following table: , `x=-1` , `x=0` , `x=2` `f(x)` , 3, 6, 0 `g(x)` , 0, 1, `-1` In each of the intervals `(-1,0)a n d(0,2)` the function `(f-3g)' '` never vanishes. Then the correct statement(s) is (are) `f^(prime)(x)-3g^(prime)(x)=0` has exactly three solutions in `(-1,0)uu(0,2)dot` `f^(prime)(x)-3g^(prime)(x)=0` has exactly one solutions in `(-1,0)dot` `f^(prime)(x)-3g^(prime)(x)=0` has exactly one solutions in `(-1,2)dot` `f^(prime)(x)-3g^(prime)(x)=0` has exactly two solutions in `(-1,0)` and exactly two solutions in `(0,2)dot`

Options

A. `f'(x)-3g,(x)=0` has exctly three solution in `(-1, 0) uu (0,2)`
B. `f(x)-3g'(x)`=0 has exactly one solution in (-1,0)
C. f'(x)-3g'(x)=0 has exactly one solution in (0,2)
D. f'(x)-3g'(x)=0 has exactly one solution in (-1,0) and exactly one solution in (0,2)

Answer

D

Explanation

Let `F(x)=f(x)-3g(x) "for all " x in [-1,2]` We have,
`F(-1)=f(-1)-3g(-1)=3-3xx0=3`
`F(0)=f(0)-3g(0)=6-3xx1=3`
`F(2)=f(2)-3g(2)=0-3xx(-1)=3`
By Rolle's theorem, F'(x) vanishes at least once in each of the interva [-1,0] and [0,2]
Now, `F,(x)=f(x)=-3 g(x)`
`rArr F'(x)=f''(x)-3g''(x)=(f''-3g'')(x)=(f-3g)''(x)`
It is given that `(f-3g)''(x)i.e.,F'' (x)` never vanishes. Therefore, `F''(x)gt0 or, F'' (x) lt0 "for all" x in (-1,0) uu(0,2)`
`rArr F'(x)` is either increasing or decreasing in (-1,0( and (0,2) .
`rArr F'(x)` 0 has exactly on solution in (-1,0) and one solution in (0,2)
`rArr f'(x)-3g(x)=0` has exctly one solution in (-1,0)n adn exactly one solution in (0,2)