IIT-JEE 2008 · Paper 1 · Mathematics Q20

MathematicsSingle correct+4 / -1Question 20 of 69

Passage

Paragraph for Question Nos. 18 to 20

Consider the functions defined implicitly by the equation $y^3-3y+x=0$ 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 $g(0)=0$ .

$\displaystyle\int_{-1}^1g'(x)\,dx=$

2g(-1)

-2g(1)

2g(1)

Answer and worked solution

Official answer: D

g is odd. If $(x,y)$ satisfies $y^3-3y+x=0$ , so does $(-x,-y)$ ; together with uniqueness and $g(0)=0$ , $g(-x)=-g(x)$ .

Integral.
$\int_{-1}^1g'(x)\,dx=g(1)-g(-1)=2g(1).$

(D)

Topics:Properties of definite integrals Types of functions

IIT-JEE 2008 · Paper 1, Mathematics question 20