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AP EAMCET202017 Sep 2020Morning ShiftMathematicsIndefinite IntegrationActual

If (f(x) ) is a polynomial of the second degree in (x ) such that (f(0)=f(1)=3, f(2)=-3 ). Then, ( f(x) x^3-1 d x= )

Options

  1. A( ( x^2+x+1 |x-1| )+ 1 3 ⁻¹ ( 2 x+1 3 )+c )
  2. B( ( x^2+x+1 |x-1| )- 2 3 ⁻¹ ( 2 x+1 3 )+c )
  3. C( ( x^2+x+1 |x-1| )- 1 3 ⁻¹ ( 2 x+1 3 )+c )
  4. D( ( x^2+x+1 |x-1| )+ 2 3 ⁻¹ ( 2 x+1 3 )+c )

Correct answer

D. ( ( x^2+x+1 |x-1| )+ 2 3 ⁻¹ ( 2 x+1 3 )+c )

Step-by-step solution

(f(0)=f(1)=3 f(2)=-3 ) Let (f(x)=a x^2+b x+c ) ( aligned & c=-3, a+b+c=-3 (a+b=0) & and 4 a+2 b+c=-1 4 a+2 b=2 aligned ) and (4 a+2 b+c=-1 4 a+2 b=2 ) So, ( aligned a & =1, b=-1 f(x) x^3-1 d x & = x^2-x-3 x^3-1 d x x^2-x-3 x^3-1 & = A (x-1) + ( B x+C x^2+x+1 ) & = x^2(A+B)+x(A-B+C)+(A-C) (x^3-1 ) aligned ) On comparing (A+B=1, A-B+C=-1, A-C=-3 ) ( aligned & A=-1, B=2, C=2 & ( x^2-x-3 x^3-1 ) d x= -1 (x-1) d x+ (2 x+2) (x^2+x+1 ) d x & =- |x-1|+ (2 x+1) d x x^2+x+1 + 1 d x x^2+x+1 d x & =- x-1)+ (x^2+x+1 )+ d x-2)^2

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