arthham

JEE Advanced 2020 · Integrals

Let bb be a nonzero real number. Suppose f:R→Rf : \mathbb{R} \to \mathbb{R} is a differentiable function such that f(0)=1f(0) = 1. If the derivative f′f' of ff satisfies the equation f′(x)=f(x)b2+x2f'(x) = \frac{f(x)}{b^{2} + x^{2}} for all x∈Rx \in \mathbb{R}, then which of the following statements is/are True?

Answers: A. If b>0b > 0, then ff is an increasing function · C. f(x)f(−x)=1f(x)f(-x) = 1 for all x∈Rx \in \mathbb{R}

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Written solution

f′(x)=f(x)b2+x2,f(0)=1,b≠0f'(x) = \dfrac{f(x)}{b^2+x^2}, \quad f(0)=1, \quad b\neq 0

To find: which of statements (1)–(4) are True\text{To find: which of statements (1)--(4) are True}

dff=dxb2+x2\dfrac{df}{f} = \dfrac{dx}{b^2+x^2}

ln⁡f=1btan⁡−1 ⁣(xb)+C\ln f = \dfrac{1}{b}\tan^{-1}\!\left(\dfrac{x}{b}\right) + C

x=0, f=1: 0=0+C  ⇒  C=0x=0,\ f=1:\ 0 = 0 + C \;\Rightarrow\; C=0

f(x)=e1btan⁡−1(x/b)f(x) = e^{\frac{1}{b}\tan^{-1}(x/b)}

f>0, b2+x2>0  ⇒  f′(x)=f(x)b2+x2>0f>0,\ b^2+x^2>0 \;\Rightarrow\; f'(x)=\dfrac{f(x)}{b^2+x^2} > 0

⇒(1) is True\Rightarrow (1) \text{ is True}

f′>0 even when b<0  ⇒  (2) is Falsef' > 0 \ \text{even when } b<0 \;\Rightarrow\; (2) \text{ is False}

g(x)=1btan⁡−1 ⁣(xb) is odd  ⇒  f(−x)=e−g(x)=1f(x)g(x)=\dfrac{1}{b}\tan^{-1}\!\left(\dfrac{x}{b}\right) \text{ is odd} \;\Rightarrow\; f(-x)=e^{-g(x)}=\dfrac{1}{f(x)}

⇒f(x)f(−x)=1  ⇒  (3) is True\Rightarrow f(x)f(-x)=1 \;\Rightarrow\; (3) \text{ is True}

f(x)−f(−x)=eg(x)−e−g(x)=2sinh⁡(g(x))≠0f(x)-f(-x) = e^{g(x)}-e^{-g(x)} = 2\sinh(g(x)) \neq 0

⇒(4) is False\Rightarrow (4) \text{ is False}

True statements: (1), (3)\boxed{\text{True statements: (1), (3)}}

Answer: (1), (3)\text{Answer: (1), (3)}

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