Let f : R rightarrow R and g : R rightarrow 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 ?

Mathematics · Methods of Differentiation

JEE Advanced 2018 — Paper 1 — Question 31

Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ and $\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$ be two non-constant differentiable functions. If

$f^{\prime}(x)=\left(e^{(f(x)-g(x))}\right) g^{\prime}(x)$ for all $x \in R$ ,

and $f(1)=g(2)=1$ , then which of the following statement(s) is (are) TRUE ?

Option A:
$\mathrm{f}(2)<1-\log _{\mathrm{e}} 2$
Option B:
$\mathrm{f}(2)>1-\log _{\mathrm{e}} 2$
Correct
Option C:
$\mathrm{g}(1)>1-\log _{\mathrm{e}} 2$
Correct
Option D:
$\mathrm{g}(1)<1-\log _{\mathrm{e}} 2$

Answer: B, C

Step-by-step solution

\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})} \cdot \mathrm{g}^{\prime}(\mathrm{x})\\& \text { given } \mathrm{f}(1)=\mathrm{g}(2)=1 \\& \int-f^{\prime}(x) \cdot e^{-f(x)} d x=\int-e^{-g(x)} \cdot g^{\prime}(x) d x \\& \int \mathrm{~d}\left(\mathrm{e}^{-\mathrm{f}(\mathrm{x})}\right)=\int \mathrm{d}\left(\mathrm{e}^{-\mathrm{g}(\mathrm{x})}\right)+\mathrm{C} \\& \mathrm{e}^{-\mathrm{f}(\mathrm{x})}=\mathrm{e}^{-\mathrm{g}(\mathrm{x})}+\mathrm{C} \\& \text { put } \mathrm{x}=1 \Rightarrow \mathrm{C}=\frac{1}{\mathrm{e}}-\frac{1}{\mathrm{e}^{\mathrm{g}(1)}} \\& \mathrm{x}=2 \Rightarrow \mathrm{e}^{\mathrm{f}(2)}=\frac{\mathrm{e}^{1+\mathrm{g}(1)}}{2 \mathrm{e}^{\mathrm{g}(1)}-\mathrm{e}} \\ \end{aligned}

$\Rightarrow \mathrm{~g}(1)>1-\log _{\mathrm{e}} 2$

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

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 1
Subject: Mathematics
Chapter: Methods of Differentiation
Topic: Methods of Differentiation