Question

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 $g(0)=0$
$\underset{-1}{\overset{1}{\int }}{g}^{\prime }(x)dx=$

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

Answer 1

Answer: (D)

$\underset{-1}{\overset{1}{\int }}{g}^{\prime }(x)dx=[g(x){]}_{-1}^1=g(1)-g(-1)=2g(1)(\because g$ is odd)
Aliter : $\because (g(x){)}^3-3(g(x))=-x$
$\because x\to -x$
$(g(-x){)}^3-3(g(-x))=x$
$(g(x){)}^3-3(g(x))=-(g(-x){)}^3+3g(-x)$
$(g(x){)}^3+(g(-x){)}^3-3(g(x)+g(-x))=0$
So $g(x)+g(-x)=0$
$g(x)=-g(-x)$
$\because g(x)$ is odd so $g^{\prime }(x)$ is even
$\therefore I=\underset{-1}{\overset{1}{\int }}{g}^{\prime }(x)dx=2\underset{0}{\overset{1}{\int }}{g}^{\prime }(x)dx$
$=2(g(1)-g(0))=2g(1)$