Let b be a nonzero real number. Suppose f: ℝ → ℝ is a differentiable function such that f(0) = 1 . If the derivative f ^ ' of f satisfies the equation f ^ ' (x) = f(x) b ^ 2 + x ^ 2 for all x ∈ ℝ, then which of the

Correct answer: A. If b > 0, then f is an increasing function f ^ ' (x) = f(x) b ^ 2 + x ^ 2 ⇒ f ^ ' (x) f(x) = 1 b ^ 2 + x ^ 2 ⇒ lnf(x) = (1)/(b) ^ - 1 (x)/(b) + C ∴ f(0) = 1 ∴ 0 = 0 + C ⇒ C = 0 ∴ lnf(x) = (1)/(b) ^ - 1 (x)/(b) ⇒ f(x) = e ^ (1)/(b) ^ - 1 (x)/(b) ∴ f ^ ' (x) = e ^ (1)/(b) ^ - 1 (x)/(b) . 1 b ^ 2 + x ^ 2 > 0 for x ∈ R ⇒ f(x) is increasing for x ∈ R . f(x) . f(- x) = e ^ (1)/(b) ^ - 1 (x)/(b) · e ^ - (1)/(b) ^ - 1 (x)/(b) = e ^ 0 = 1 .

JEE Advanced 2020Paper 2

Exam: JEE Advanced
Year: 2020
Paper: Paper 2
Subject: Mathematics
Chapter: Differential Equations

JEE Advanced 2020 · Paper 2 · Mathematics · Differential Equations

Let b be a nonzero real number. Suppose f: ℝ → ℝ is a...

Let

b

be a nonzero real number. Suppose

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

is a differentiable function such that

f\left(0\right) = 1

. If the derivative

{f}^{'}

f

satisfies the equation

{f}^{'}\left(x\right) = \dfrac{f\left(x\right)}{{b}^{2} + {x}^{2}}

for all

x \in \mathbb{R},

then which of the following statements is/are TRUE?

Options

AIf $b > 0,$ then $f$ is an increasing function
BIf $b < 0,$ then $f$ is a decreasing function
C $f\left(x\right)f\left(- x\right) = 1$ for all $x \in \mathbb{R}$
D $f\left(x\right) - f\left(- x\right) = 0$ for all $x \in \mathbb{R}$

Correct answer

A. If $b > 0,$ then $f$ is an increasing function

Step-by-step solution

${f}^{'}\left(x\right) = \dfrac{f\left(x\right)}{{b}^{2} + {x}^{2}} ⇒ \dfrac{{f}^{'}\left(x\right)}{f\left(x\right)} = \dfrac{1}{{b}^{2} + {x}^{2}}$

$⇒ lnf\left(x\right) = \dfrac{1}{b}\tan^{- 1}\dfrac{x}{b} + C$

$∴ f\left(0\right) = 1 ∴ 0 = 0 + C ⇒ C = 0$

$∴ lnf\left(x\right) = \dfrac{1}{b}\tan^{- 1}\dfrac{x}{b}$

$⇒ f\left(x\right) = {e}^{\dfrac{1}{b}\tan^{- 1}\dfrac{x}{b}}$

$∴ {f}^{'}\left(x\right) = {e}^{\dfrac{1}{b}\tan^{- 1}\dfrac{x}{b}} . \dfrac{1}{{b}^{2} + {x}^{2}} > 0$ for $x \in R$

$⇒ f\left(x\right)$ is increasing for $x \in R$ .

$f\left(x\right) . f\left(- x\right) = {e}^{\dfrac{1}{b}\tan^{- 1}\dfrac{x}{b}} \cdot {e}^{- \dfrac{1}{b}\tan^{- 1}\dfrac{x}{b}} = {e}^{0} = 1$ .

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