BITSAT2015MathematicsApplication of DerivativesActual
Let f(x) be a polynomial of degree three satisfying f (0)=-1 and f (1)=0 . Also , 0 is a stationary point of f(x) If f(x) does not have an extremum at x=0 , then the value of f(x) x³-1 d x is
Options
- Ax² 2 +C
- Bx+C
- Cx³ 6 +C
- DNone of these
Correct answer
B. x+C
Step-by-step solution
Let f(x)=a x³+b x²+c x+d Put x=0 and x=1 Then, we get f(0)=-1 and f(1)=0 d =-1 and a + b + c + d =0 a + b + c =1 It is given that x=0 is a stationary point of f ( x ) , but it is not a point of extremum. Therefore, f(0)=0=f^ (0) and f^ (0)=0 Now, f ( x )= ax ³+ bx ²+ cx + d f(x)=3 a x²+2 b x+c array l f^ (x)=6 a x+2 b and f^ (x)=6 a f^ =0, f^ (0)=0 and f^ (0)=0 0 array c=0, b=0 and a 0 From Eqs. (i) and (ii), we get a =1, ~b = c =0 and d =-1 Put these values in f ( x ) we get f(x)=x³-1 Hence, f ( x ) x ³-1 dx = x ³