JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Definite Integrals

Consider the functions defined implicitly by the equation

y^{3} - 3 y + x = 0

on various intervals in the real line.
If

x \in (- \infty, - 2) \cup (2, \infty)

, the equation implicitly defines a unique real valued differentiable function

y = f (x)

.
If

x \in (- 2, 2)

, the equation implicitly defines a unique real valued differentiable function

y = g (x)

satisfying

g (0) = 0

\int_{- 1}^{1} g^{'} (x) d x =

Answer: (D)

Solution

By the fundamental theorem of calculus:

\int_{- 1}^{1} g^{'} (x) d x = g (1) - g (- 1)

From the implicit equation

y^{3} - 3 y + x = 0

and the symmetry of the function:

If

g (x)

satisfies the equation, then for

y^{3} - 3 y + x = 0

, we can check if there's a relationship between

g (x)

and

g (- x)

.

From

y^{3} - 3 y + x = 0

, replacing

x

with

- x

and

y

with

- y

(- y)^{3} - 3 (- y) + (- x) = - y^{3} + 3 y - x = 0

y^{3} - 3 y + x = 0

This shows

g (- x) = - g (x)

, i.e.,

g

is an odd function.

Therefore:

g (- 1) = - g (1)

So:

\int_{- 1}^{1} g^{'} (x) d x = g (1) - g (- 1) = g (1) - (- g (1)) = 2 g (1)

Answer: (D)

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option