If _ x 0 a+ b+x^2 - 3 x^2 = 1 24 3 , where a and b are real constants, then the value of a+b is :

Step-by-step solution

For the limit to exist and be finite, the numerator must approach 0 as x 0 . a+ b - 3 = 0 a+ b = 3 Now, rationalizing the numerator: _ x 0 a+ b+x^2 - 3 x^2( a+ b+x^2 + 3 ) Substitute a - 3 = - b : _ x 0 b+x^2 - b x^2( a+ b+x^2 + 3 ) Rationalizing the numerator again: _ x 0 b+x^2 - b x^2( a+ b+x^2 + 3 )( b+x^2 + b ) _ x 0 1 ( a+ b+x^2 + 3 )( b+x^2 + b ) Substituting x = 0 : 1 ( a+ b + 3 )( b + b ) = 1 ( 3 + 3 )(2 b ) = 1 4 3b Given that the limit is 1 24 3 : 1 4 3b = 1 24 3 4 b = 24 b = 6 b = 36 From a+ b = 3 , we g