Let f: ℝ → ℝandg: ℝ → ℝ be two non-constant differentiable functions. If f ^ ′ (x) = ( e ^ (f(x)) - g(x) ) g ^ ′ (x)forallx ∈ ℝ , and f(1) = g(2) = 1 then which of the following statement(s) is (are) TRUE?

Correct answer: B. f(2) > 1 - _ e 2 f ^ ′ (x) = e ^ (f(x) - g(x)) g ^ ′ (x) x ∈ ℝ ⇒ 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) ⇒ - (1)/(e) + e ^ g(1) = - e ^ - f(2) + (1)/(e) ⇒ e ^ - f(2) + e ^ - g(1) = (2)/(e) ∴ e ^ - f(2) ⇒ - f(2) ⇒ f(2) > 1 - 2andg(1) > 1 - 2

JEE Advanced 2018Paper 1

Exam: JEE Advanced
Year: 2018
Paper: Paper 1
Subject: Mathematics
Chapter: Definite Integration

JEE Advanced 2018 · Paper 1 · Mathematics · Definite Integration

Let f: ℝ → ℝandg: ℝ → ℝ be two non-constant differentiable...

Let

f: \mathbb{R} \to \mathbb{R}andg: \mathbb{R} \to \mathbb{R}

be two non-constant differentiable functions. If

{f}^{′}\left(x\right) = \left({e}^{\left(f\left(x\right)\right) - g\left(x\right)}\right){g}^{′}\left(x\right)forallx \in \mathbb{R}

, and

f\left(1\right) = g\left(2\right) = 1

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

Options

A $f\left(2\right) < 1 - \log_{e} 2$
B $f\left(2\right) > 1 - \log_{e} 2$
C $g\left(1\right) > 1 - \log_{e} 2$
D $g\left(1\right) < 1 - \log_{e} 2$

Correct answer

B. $f\left(2\right) > 1 - \log_{e} 2$

Step-by-step solution

{f}^{&prime;}\left(x\right) = {e}^{\left(f\left(x\right) - g\left(x\right)\right)}{g}^{&prime;}\left(x\right) \forall x \in \mathbb{R}

⇒ {e}^{- f\left(x\right)} . {f}^{&prime;}\left(x\right) - {e}^{- g\left(x\right)}{g}^{&prime;}\left(x\right) = 0

⇒ ∫ \left({e}^{- f\left(x\right)}{f}^{&prime;}\left(x\right) - {e}^{- g\left(x\right)} . {g}^{&prime;}

\left(x\right)\right)dx = C

<br >

⇒ - {e}^{- f\left(x\right) + {e}^{- g\left(x\right)}} = C

<br >

⇒ - {e}^{- f\left(1\right)} + {e}^{- g\left(1\right)} = - {e}^{- f\left(2\right)} + {e}^{- g\left(2\right)}

<br >

⇒ - \dfrac{1}{e} + {e}^{g\left(1\right)} = - {e}^{- f\left(2\right)} + \dfrac{1}{e}

<br >

⇒ {e}^{- f\left(2\right)} + {e}^{- g\left(1\right)} = \dfrac{2}{e}

<br >

∴ {e}^{- f\left(2\right)} < \dfrac{2}{e}and{e}^{- g\left(1\right)} < \dfrac{2}{e}

<br >

⇒ - f\left(2\right) < \ln 2 - 1and - g\left(1\right) < \ln 2 - 1

<br >

⇒ f\left(2\right) > 1 - \ln 2andg\left(1\right) > 1 - \ln 2

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