Paragraph: Consider the functions defined implicitly by the equation y^3-3 y+x=0 on various intervals in the real line. If x ∈(-∞,-2) (2, ∞) , the equation implicitly defines a unique real valued differentiable function

Correct answer: D. 2 g(1) Let I=∫_ -1 ^1 g^ (x) d x=[g(x)]_ -1 ^1=g(1)-g(-1) Since, y^3-3 y+x=0 and y=g(x) (g(x))^3-3 g(x)+x=0 [by Eq. (i)] At x=1 , aligned (g(1))^3-3 g(1)+1 & =0 At & =-1, (g(-1))^3-3 g(-1)-1 & =0 aligned On adding Eqs. (i) and (ii), we get array rlrl & & (g(1))^3+(g(-1))^3-3(g(1)+g(-1))=0 & & [g(1)+g(-1)][(g(1))^2+(g(-1))^2-g(1) g(-1)-3]=0 & g(1)+g(-1)=0 & g(1) & =-g(-1) & & I & =g(1)-g(-1)=g(1)-(-g(1))=2 g(1) array

JEE Advanced 2008Paper 1

Exam: JEE Advanced
Year: 2008
Paper: Paper 1
Subject: Mathematics
Chapter: Definite Integration

JEE Advanced 2008 · Paper 1 · Mathematics · Definite Integration

Paragraph: Consider the functions defined implicitly by the...

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

Options

A $2 g(-1)$
B0
C $-2 g(1)$
D $2 g(1)$

Correct answer

D. $2 g(1)$

Step-by-step solution

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 } \quadx & =-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}

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