Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be two non-constant differentiable functions. If $f^{\prime}(x)=\left(e^{(f(x)-g(x))}\right) g^{\prime}(x)$ for all $x \in \mathbb{R}$, and $f(1)=g(2)=1$, then which of the following statement(s) is (are) TRUE?

(B) $f(2)>1-\log _{\mathrm{e}} 2$; (C) $g(1)>1-\log _{\mathrm{e}} 2$

(A) $f(2)<1-\log _{\mathrm{e}} 2$

(D) $g(1)<1-\log _{\mathrm{e}} 2$

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Differential Equations

Let

f : R \to R

and

g : R \to R

be two non-constant differentiable functions. If

f^{'} (x) = (e^{(f (x) - g (x))}) g^{'} (x)

for all

x \in R

, and

f (1) = g (2) = 1

, then which of the following statement(s) is (are) TRUE?

Answer: (B), (C)

Rewrite the given equation:

e^{- f (x)} f^{'} (x) = e^{- g (x)} g^{'} (x)

.

Integrating:

- e^{- f (x)} = - e^{- g (x)} + c

, so

e^{- g (x)} - e^{- f (x)} = c

.

Using

f (1) = g (2) = 1

e^{- g (1)} - e^{- 1} = e^{- 1} - e^{- f (2)}

.

Thus

e^{- g (1)} + e^{- f (2)} = \frac{2}{e}

.

Since both terms are positive:

e^{- g (1)} < \frac{2}{e}

implies

- g (1) < ln (2/ e) = ln 2 - 1

, so

g (1) > 1 - ln 2

. (C) is TRUE.

Similarly,

e^{- f (2)} < \frac{2}{e}

implies

f (2) > 1 - ln 2

. (B) is TRUE.

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · One or more correct options