If f(x), g(x) be twice differentiable functions on [0,2] satisfying `f''(x) = g''(x)` ,`f'(1) = 2g'(1) = 4` and `f(2) = 3 g(2) = 9`, then `f(x)-g(x)` at x = 4 equals (A) 0 (B) 10 (C) 8 (D) 2

Options

A. `f(4)-g(4)=10`
B. `|f(x)-g(x)|lt 2 rArr -2 lt x lt 0`
C. `f(2) = g(2) rArr x=-1`
D. `f(x)-g(x) = 2x` has real root

Answer

A, B, C

Explanation

We have `f^('')(x)=g^('')(x)`. On integration, we get
`f^(')(x)=g^(')(x)+C`……….(1)
Putting x=1, we get
`f^(')(1)=g^(')(x)+C` or `4=2+C` or `C=2`
`therefore f^(')(x)=g^(')(x)+2`
Integrating w.r.t. x, we get `f(x) =g(x)+2x+c_(1)` ..........(2)
Putting, `x=2,` we get
`f(2)=g(2)+4+c_(1)` or `0=3+4+c_(1)` or `c_(1)=2`
`therefore f(x) =g^(')(x)+2x+2`
Putting `x=4` we get
`|f(x)-g(x)| lt 2` or `|2x+2| lt 2`
or `|x+1|lt 1` or `-2 lt x lt 0`
Also, `f(2)=g(2)` or `x=-1`
`f(x)-g(x)=2x` has no solution.