Question

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 $g(0)=0$
$\int\limits_{-1}^1 g^{\prime}(x) d x=$

• A. $2 g(-1)$
• B. 0
• C. $-2 g(1)$
• D. $2 g(1)$

Answer 1

Answer: (D)

$\int\limits_{-1}^1 g^{\prime}(x) d x= {[g(x)]_{-1}^1=g(1)-g(-1)=2 g(1)} (\because g$ is odd)
Aliter : $\because (g(x))^3-3(g(x))=-x $
$ \because x \rightarrow-x $
$ (g(-x))^3-3(g(-x))=x $
$ (g(x))^3-3(g(x))=-(g(-x))^3+3 g(-x) $
$ (g(x))^3+(g(-x))^3-3(g(x)+g(-x))=0 $
So $ g(x)+g(-x)=0 $
$ g(x)=-g(-x) $
$ \because g(x) $ is odd so $g^{\prime}(x) $ is even
$ \therefore I=\int\limits_{-1}^1 g^{\prime}(x) d x=2 \int\limits_0^1 g^{\prime}(x) d x $
$=2(g(1)-g(0))=2 g(1)$