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Largest Interval for a Solution of y′=1+y² | JEE

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

Question

The largest value of

c

such that there exists a differentiable function

h(x)

for

-c < x < c

that is a solution of

y' = 1 + y^2

with

h(0) = 0

is:

2\pi

\pi

\dfrac{\pi}{2}

correct

\dfrac{\pi}{4}

Solution

Separate and integrate:

\frac{dy}{1 + y^2} = dx \Rightarrow \tan^{-1}y = x + c_1.

Apply

h(0) = 0

\tan^{-1}0 = 0 + c_1 \Rightarrow c_1 = 0

, so

y = \tan x

. The function

h(x) = \tan x

is differentiable only where

\tan^{-1}y = x

stays in the range

\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)

, i.e.

-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}

. Hence the largest such

c

\dfrac{\pi}{2}

. Correct answer: (3)

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