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JEE MainMathematicsIndefinite Integration

Let g: (1, ) R be a differentiable function satisfying ( 1 x^3 - x - g(x) x^2 ) d x = g(x) x + C where C is an arbitrary constant. If g( 2 ) = 0 , then the value of g(2) is

Options

  1. A- 1 2 3
  2. B( 2 -1) + 1 2 3
  3. C( 2 +1) - 1 2 3
  4. D1 2 ( 3 2 )

Correct answer

C. ( 2 +1) - 1 2 3

Step-by-step solution

Differentiating both sides of the given integral equation with respect to x , we get: 1 x^3 - x - g(x) x^2 = d d x ( g(x) x ) Applying the quotient rule on the right-hand side: 1 x^3 - x - g(x) x^2 = x g'(x) - g(x) x^2 = g'(x) x - g(x) x^2 Canceling - g(x) x^2 from both sides gives: 1 x^3 - x = g'(x) x g'(x) = x x^3 - x = 1 x^2 - 1 Integrating both sides with respect to x : g(x) = 1 2 | x-1 x+1 | + K Using the given condition g( 2 ) = 0 : 1 2 ( 2 -1 2 +1 ) + K = 0 Rationalizing the argument of the logarithm: 2 -1 2

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