Let f: R → R and g : R → R be two non-constant differentiable functions. If f'(x) = (e^(f(x)-g(x)))

Q: Wait, why do we need to separate the variables? Can't we just integrate both sides directly?

That is a great question! If you try to integrate directly, you will find that $f(x)$ and $g(x)$ are trapped inside the exponential function, making it impossible to integrate with respect to $x$ without knowing the relationship between $f$ and $g$. By separating the variables, we isolate the functions so that each side depends only on one variable, allowing us to use the chain rule in reverse to integrate.

Q: How do we know that $e^{-f(2)} + e^{-g(1)} = \frac{2}{e}$ implies $f(2) > 1 - \log_e 2$?

Remember that for any real number $y$, $e^y$ is always strictly positive. Since $e^{-g(1)} > 0$, if we remove it from the equation $e^{-f(2)} + e^{-g(1)} = \frac{2}{e}$, the remaining term $e^{-f(2)}$ must be strictly less than $\frac{2}{e}$. Taking the natural logarithm on both sides gives $-f(2) 1 - \log_e 2$.

Q: Is it possible that $C$ could be zero?

The problem states that $f$ and $g$ are non-constant differentiable functions. If $C$ were zero, we would have $e^{-f(x)} = e^{-g(x)}$, which implies $f(x) = g(x)$. While this is a possible solution, the problem asks for general properties that must be true for the given conditions. Our derivation holds regardless of the specific value of $C$ as long as it is consistent with the boundary conditions.

Question

Let $f: R o R$ and $g : R o R$ 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 ? Options: (A) $f(2) < 1- \log_e 2$ (B) $f(2) > 1- \log_e 2$ (C) $g(1) > 1- \log_e 2$ (D) $g(1) < 1- \log_e 2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing before a complex differential equation, a puzzle that seems to weave two different worlds—the world of $f(x)$ and the world of $g(x)$—into a single, tangled expression: $$f'(x) = e^{f(x)-g(x)} g'(x)$$ It looks intimidating, but in the realm of JEE Advanced, intimidation is often just a mask for elegance. Let us peel back that mask together. ### The Art of Separation Our first instinct, and our most powerful tool, is the art of separation. We see that exponential term $e^{f(x)-g(x)}$. By the laws of exponents, we know this is simply: $$\frac{e^{f(x)}}{e^{g(x)}}$$ Now, look at the equation again: $f'(x) = \frac{e^{f(x)}}{e^{g(x)}} g'(x)$. If we multiply both sides by $e^{-f(x)}$ and rearrange, we get a beautiful, clean separation: $$e^{-f(x)} f'(x) = e^{-g(x)} g'(x)$$ We have effectively divorced the two functions, allowing them to exist in their own independent integrals. ### The Integration Dance Now, we prepare for the integration. We are looking at: $$\int e^{-f(x)} f'(x) dx = \int e^{-g(x)} g'(x) dx$$ This is where the chain rule becomes our best friend. Recall that the derivative of $e^{-h(x)}$ is $-e^{-h(x)} h'(x)$. Our integrals are almost perfect matches for this! By integrating both sides, we get: $$-e^{-f(x)} = -e^{-g(x)} - C$$ Multiplying by $-1$ gives us the elegant relation: $$e^{-f(x)} = e^{-g(x)} + C$$ This constant $C$ is the bridge between our two functions, the hidden link that connects the behavior of $f$ at $x=1$ to the behavior of $g$ at $x=2$. ### The Constant of Mystery We are given $f(1) = 1$ and $g(2) = 1$. Let us use these keys to unlock the value of $C$. Substituting $x=1$ into our equation, we find: $$e^{-f(1)} = e^{-g(1)} + C \Rightarrow e^{-1} = e^{-g(1)} + C$$ Thus, $C = e^{-1} - e^{-g(1)}$. Now, let us look at $x=2$. Substituting this into our main equation, we get: $$e^{-f(2)} = e^{-g(2)} + C$$ Since $g(2) = 1$, this becomes $e^{-f(2)} = e^{-1} + C$. Substituting our expression for $C$, we arrive at: $$e^{-f(2)} = e^{-1} + (e^{-1} - e^{-g(1)}) \Rightarrow e^{-f(2)} + e^{-g(1)} = \frac{2}{e}$$ ### The Final Inequality We have arrived at the heart of the problem. We know that $e^{-f(2)} + e^{-g(1)} = \frac{2}{e}$. Because the exponential function $e^y$ is always strictly positive for any real $y$, we know that $e^{-g(1)} > 0$. This implies that $e^{-f(2)} < \frac{2}{e}$. Taking the natural logarithm of both sides, we get: $$-f(2) < \ln\left(\frac{2}{e}\right) \Rightarrow -f(2) < \ln 2 - 1$$ Multiplying by $-1$ flips the inequality, giving us **$f(2) > 1 - \ln 2$**. By symmetry, the same logic applies to $g(1)$, leading us to **$g(1) > 1 - \ln 2$**. We have navigated the complexity, solved the differential equation, and bounded the functions with precision. This is the power of mathematical reasoning—turning a terrifying equation into a clear, logical truth.

Animated Solution for Mathematics - 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 ?

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Visualized Solution

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

$Separating Variables$

$Integration Setup$

$Performing Integration$

$Substitute x = 1$

$Isolating C$

$Substitute x = 2$

$Eliminating C$

$Symmetric Relation$

$Bounding f (2)$

$Final Inequality for f (2)$

$Bounding g (1)$

$Conclusion$

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Integration Dance

The Constant of Mystery

The Final Inequality

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)=ef(x)−g(x)g′(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Separating Variables

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integration Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Performing Integration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitute x=1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitute x=2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Eliminating C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Symmetric Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bounding f(2)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Inequality for f(2)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bounding g(1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion