Question

Let f:R→R and g:R→R be the two non constant differentiable functions. if f'(x) = $\sf e^{[f(x) - g(x)]}$g'(x) for all values of x belonging to real numbers and f(1) = g(2) = 1. Which of the following statements are true?
(a) f(2) < 1 - log(2)
(b) g(2) < 1 - log(2)
(c) g(2) > 1 - log(2)
(d) f(2) > 1 - log(2)

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Answer 1

$\rm \longrightarrow f'(x) = e^{ [f(x) - g(x)]}g'(x) \\ \\ \rm \longrightarrow f'(x) = e^{f(x)} \times e^{ - g(x)} \times g'(x)\\ \\ \rm \longrightarrow \dfrac{f'(x) }{e^{f(x)}} = e^{ - g(x)} \times g'(x)e^{f(x)}\\ \\ \rm \longrightarrow f'(x) \times e^{ - f(x)}= e^{ - g(x)} \times g'(x) \\ \\ \rm \longrightarrow f'(x) \: e^{ - f(x)} - e^{ - g(x)} \: g'(x) = 0$

Now we will integrate :-

$\rm \longrightarrow \displaystyle \int \rm f'(x) \: e^{ - f(x)} - e^{ - g(x)} \: g'(x) =\displaystyle \int \rm 0 \\ \\ \\ \rm \longrightarrow \displaystyle \int \rm f'(x) \: e^{ - f(x)} - \displaystyle \int \rm e^{ - g(x)} \: g'(x) = constant(c) \\ \\ \\ \rm let \: \: g(x) = t \: and \: f(x) = m\: \\ \rm \: g'(x)dx = dt \: and \: f'(x) = dm$

$\longrightarrow \displaystyle \int \rm {e}^{ - m} dm - \displaystyle \int \rm {e}^{ - t} dt = c \\ \\ \longrightarrow \displaystyle \rm - {e}^{ - m} + c_1 + \displaystyle \rm {e}^{ - t} + c_2 = c \\ \\ \rm \: c - c_1 - c_2 = c_t \: (c_t \: is \: constant)\\ \\ \longrightarrow \displaystyle \rm - {e}^{ - m} + \displaystyle \rm {e}^{ - t} =c_t$

$\longrightarrow \displaystyle \rm - {e}^{ - f(x)} + \displaystyle \rm {e}^{ - g(x)} =c_t \\ \\ \boxed{ \bf f(1) = g(2) = 1} \\ \\ \\ \longrightarrow \displaystyle \rm - {e}^{ - f(1)} + \displaystyle \rm {e}^{ - g(1)} =c_t\\ \\ \\ \longrightarrow \displaystyle \rm - {e}^{ - f(2)} + \displaystyle \rm {e}^{ - g(2)} =c_t\\ \\ \\ \longrightarrow \displaystyle \rm - {e}^{ - f(1)} + \displaystyle \rm {e}^{ - g(1)} =\rm - {e}^{ - f(2)} + \displaystyle \rm {e}^{ - g(2)} =c_t\\ \\ \\ \longrightarrow \displaystyle \rm - {e}^{ - f(1)} + \displaystyle \rm {e}^{ - g(1)} =\rm - {e}^{ - f(2)} + \displaystyle \rm {e}^{ - g(2)} \\ \\ \\ \longrightarrow \displaystyle \rm - {e}^{ - (1)} + \displaystyle \rm {e}^{ - g(1)} =\rm - {e}^{ - f(2)} + \displaystyle \rm {e}^{ - 1} \\ \\ \\ \longrightarrow \displaystyle \rm \displaystyle \rm {e}^{ - g(1)} + {e}^{ - f(2)} =\rm \displaystyle \rm 2{e}^{ - 1} \\ \\ \\ \longrightarrow \displaystyle \rm \displaystyle \rm {e}^{ - g(1)} + {e}^{ - f(2)} =\rm \displaystyle \rm \frac{2}{{e} }$

Now , we know that :-

If a+b = p and a , b and p are positive numbers then a< p and b <p

$\therefore \: \rm {e}^{ - f(2)} < \dfrac{2}{e} and \: {e}^{ - g(1)} < \dfrac{2}{e}$

$\therefore \rm {e}^{ - f(2)} < \dfrac{2}{e} \\\rightarrow \rm ln \: {e}^{ - f(2)} < ln \dfrac{2}{e} \\ \rightarrow\rm { - f(2)} \: ln \: {e} < ln{2}-ln \: e \\ \rightarrow\rm { - f(2)} < ln{2}-1\\ \boxed{\rightarrow\rm { f(2)} > 1 - ln2}$

$\rm \rightarrow {e}^{ - g(1)} < \dfrac{2}{e} \\\rightarrow \rm ln \: {e}^{ - g(1)} < ln \dfrac{2}{e} \\ \rightarrow\rm { - g(1)} \: ln \: {e} < ln{2}-ln \: e \\ \rightarrow\rm { - g(1)} < ln{2}-1\\ \boxed{\rightarrow\rm { g(1)} > 1 - ln2}$

Therefore , option (d) is correct

Answer 2

Answer:

f′(x)=e(f(x)−g(x))g′(x)∀xϵR

⇒e−f(x)⋅f′(x)−e−g(x)g′(x)=0

⇒∫(e−f(x)f′(x)−e−g(x)⋅g′(x))dx=C

⇒−e−f(x)+e−g(x)=C

⇒−e−f(1)+e−g(1)=−e−f(2)+e−g(2)

⇒−e1+e−g(1)=−e−f(2)+e1

⇒e−f(2)+e−g(1)=e2

∴ e−f(2)<e2 and e−g(1)<e2

⇒−f(2)<loge2−1 and −g(1)<loge2−1

⇒f(2)>1−loge2  and  g(1)>1−loge2.