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 uniqu

Question

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 y = f ( x ) . If x ∈ ( − 2 , 2 ) , the equation implicitly defines a unique real valued differentiable function y = g ( x ) , satisfying g ( 0 ) = 0 . Question: ∫ − 1 1 ​ g ′ ( x ) d x is equal to

TestHub · Accepted Answer

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 , ( g ( 1 ) ) 3 − 3 g ( 1 ) + 1 At x ( g ( − 1 ) ) 3 − 3 g ( − 1 ) − 1 ​ = 0 = − 1 , = 0 ​ On adding Eqs. (i) and (ii), we get ⇒ ⇒ ⇒ ​ g ( 1 ) + g ( − 1 ) = 0 g ( 1 ) ∴ ​ ( 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 ) I ​ = g ( 1 ) − g ( − 1 ) = g ( 1 ) − ( − g ( 1 )) = 2 g ( 1 ) ​

Mathematics - Definite Integration Question with Solution | TestHub

Options:

Doubts & Discussion