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 $y=f\left(x\right)$. If $x\in (-2,2)$, the equation implicitly defines a unique real valued differentiable function $y=g\left(x\right)$ satisfying $g\left(0\right)=0$.

${\int }_{-l}^1{g}^{\prime }\left(x\right)dx=$

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

Answer 1

Answer: (D)

$2g\left(1\right)$

For $x$ lies in $(-2,2)$

$\left(g\right(x){)}^3-g(x)+x=0$

$\left(g\right(-x){)}^3-g(-x)-x=0$

Adding we get $\left[g\right(x)+g(-x\left)\right]\left[g\right(x{)}^2+g(-x{)}^2-g(x\left)g\right(-x)-3]=0$

Suppose ${g\left(x\right)}^2+{g(-x)}^2-g\left(x\right)g(-x)-3=0$

Put$x=0$. Then we get ${g\left(0\right)}^2+{g\left(0\right)}^2-g\left(0\right)g\left(0\right)-3$

$0-3=0$, Not possible.

Thus,$g\left(x\right)+g(-x)=0$

Hence ${\int }_{-1}^1{g}^{\prime }\left(x\right)dx=g\left(1\right)-g(-1)=2g\left(1\right)$