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Differential Equation from a Logarithmic Series | JEE

JEE Maths question with a full step-by-step solution.

Question

A function

y = f(x)

satisfies

x = \dfrac{dy}{dx} - \dfrac{1}{2}\left(\dfrac{dy}{dx}\right)^2 + \dfrac{1}{3}\left(\dfrac{dy}{dx}\right)^3 - \cdots\infty

, with

y(0) = 1

and

x \le \ln 2

. Then

f'\left(\ln\dfrac{1}{2}\right) + f\left(\ln\dfrac{1}{2}\right)

equals:

-\ln 2

\ln 2

correct

1 + \ln 2

1 - \ln 2

Solution

Step 1: The right side is the logarithm series

\ln(1 + u) = u - \dfrac{u^2}{2} + \dfrac{u^3}{3} - \cdots

with

u = \dfrac{dy}{dx}

, so

x = \ln\left(1 + \frac{dy}{dx}\right) \Rightarrow 1 + \frac{dy}{dx} = e^x \Rightarrow \frac{dy}{dx} = e^x - 1.

Step 2: Integrate and apply

y(0) = 1

y = e^x - x + c, \qquad 1 = 1 - 0 + c \Rightarrow c = 0 \Rightarrow f(x) = e^x - x, \quad f'(x) = e^x - 1.

Step 3: At

x = \ln\dfrac{1}{2}

e^x = \dfrac{1}{2}

and

x = -\ln 2

f\left(\ln\tfrac{1}{2}\right) = \frac{1}{2} + \ln 2, \qquad f'\left(\ln\tfrac{1}{2}\right) = \frac{1}{2} - 1 = -\frac{1}{2}.

\therefore f'\left(\ln\tfrac{1}{2}\right) + f\left(\ln\tfrac{1}{2}\right) = -\frac{1}{2} + \frac{1}{2} + \ln 2 = \ln 2.

Correct answer: (2)

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