Let `f: R->R` and `g: R->R` be two non-constant differentiable functions. If `f^(prime)(x)=(e^((f(x)-g(x))))g^(prime)(x)` for all `x in R` , and `f(1)=g(2)=1` , then which of the following statement(s) is (are) TRUE? `f(2)<1-(log)_e2` (b) `f(2)>1-(log)_e2` (c) `g(1)>1-(log)_e2` (d) `g(1)<1-(log)_e2`

Options

A. `f(2) lt 1 - log_(e)2`
B. `f(2) gt 1 - log_(e) 2`
C. `g(1) gt 1 - log_(e) 2`
D. `g(1) lt 1 - log_(e) 2`

Answer

B, C

Explanation

We have, `f'(x)=e^((f(x)-g(x)))g'(x)AA x in R`
`rArr" "f'(x)=(e^(f(x)))/(e^g(x))g'(x)`
`rArr" "(f'(x))/(e^(f(x)))=(g'(x))/(e^(g(x)))`
`rArr" "e^(-f(x))f'(x)=e^(-g(x))g'(x)`
On integrating both side, we get
`e^(-f(x))=e^(-g(x))+C`
At x 1
`{:(e^(-f(1))=e^(-g(1))+C,,),(e^(-1)=e^(-g(1))+C," "[therefore f(1)=1]," "...(i)):}`
At x = 2
`{:(e^(-f(2))=e^(-g(2))+C,,),(rArr" "e^(-f(2))=e^(-1)+C," "[therefore g(2)=1]," "...(ii)):}`
From Eqs. (i) and (ii)
`{:(e^(-f(2))=2e^(-1)-e^(-g(1)),),(e^(-f(2)) gt 2e^(-1)," "...(iii)):}`
We know that, `e^(-x)` is decreasing
`therefore" "-f(2) lt log_(e)2-1`
`" "f(2) gt 1-log_(e)2`
`rArr" "e^(-g(1))+e^(-f(2))=2e^(-1)" "["from Eq. (iii)"]`
`rArr" "e^(-g(1)) lt 2e^(-1)`
`" "-g(1) lt log_(e)2-1`
`rArr" "g(1) gt 1-log_(e)2`