ExamBro
ExamBro
JEE Advanced · Mathematics · 27. Definite Integration

Paragraph:
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 \(g(0)=0\).Question:
\(\int_{-1}^1 g^{\prime}(x) d x\) is equal to

  1. A
    \(2 g(-1)\)
  2. B
    0
  3. C
    \(-2 g(1)\)
  4. D
    \(2 g(1)\)
Verified Solution

Answer & Solution

Correct Answer

(D)
\(2 g(1)\)

Step-by-step Solution

Detailed explanation

Let \(I=\int_{-1}^1 g^{\prime}(x) d x=[g(x)]_{-1}^1=g(1)-g(-1)\)
Since, \(\quad y^3-3 y+x=0\) and \(\quad y=g(x)\)
\[
\therefore(g(x))^3-3 g(x)+x=0
\]
[by Eq. (i)]
At \(\quad x=1\),
\[
\begin{aligned}
(g(1))^3-3 g(1)+1 & =0 \\
\text { At } \quad x & =-1, \\
(g(-1))^3-3 g(-1)-1 & =0
\end{aligned}
\]
On adding Eqs. (i) and (ii), we get
\[
\begin{array}{rlrl}
& & (g(1))^3+(g(-1))^3-3(g(1)+g(-1))=0 \\
\Rightarrow & & {[g(1)+g(-1)]\left[(g(1))^2+(g(-1))^2-g(1) g(-1)-3\right]=0} \\
\Rightarrow & g(1)+g(-1)=0 \\
\Rightarrow & g(1) & =-g(-1) \\
& \therefore & I & =g(1)-g(-1)=g(1)-(-g(1))=2 g(1)
\end{array}
\]
Same subject
Explore more questions on app
From JEE Advanced
Explore more questions on app