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

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

Answer 1

Answer: (D)

We have $y'=\frac{1}{3\left(1-\left(f(x)^{2}\right)\right)}$ which is even
Hence $\int\limits_{-1}^{1} g'(x)=g(1)-g(-1)=2 g(1)$