COMEDK2024MathematicsApplication of DerivativesActual
If f(x)= x+b x^2+a x, x 0 has extreme values (or turning points) at x=-1 and x=2 then the values of a and b are
Options
- Aa= 1 4 b=- 1 2
- Ba= 1 2 b=- 1 4
- Ca= 1 2 b= 1 4
- Da=- 1 2 b=- 1 4
Correct answer
B. a= 1 2 b=- 1 4
Step-by-step solution
Given f(x) = x + bx^2 + ax . The derivative of the function is f'(x) = 1 x + 2bx + a . Since the function has extreme values at x = -1 and x = 2 , f'(x) must be zero at these points. For x = -1 : f'(-1) = 1 -1 + 2b(-1) + a = 0 -1 - 2b + a = 0 a - 2b = 1 . For x = 2 : f'(2) = 1 2 + 2b(2) + a = 0 1 2 + 4b + a = 0 a + 4b = - 1 2 . Subtracting the first equation from the second: (a + 4b) - (a - 2b) = - 1 2 - 1 6b = - 3 2 b = - 3 12 = - 1 4 . Substituting b = - 1 4 into a - 2b = 1 : a - 2(- 1 4 ) = 1 a + 1 2 = 1 a = 1 2