JEE Main2009MathematicsApplication of DerivativesActual
Given P(x)=x^4+a x^3+b x^2+c x+d such that x=0 is the only real root of P^ (x)=0 . If P(-1) < P(1) , then in the interval [-1,1]
Options
- AP(-1) is the minimum and P(1) is the maximum of P
- BP(-1) is not minimum but P(1) is the maximum of P
- CP(-1) is the minimum and P(1) is not the maximum of P
- Dneither P(-1) is the minimum nor P(1) is the maximum of P
Correct answer
B. P(-1) is not minimum but P(1) is the maximum of P
Step-by-step solution
aligned & P(x)=x^4+a x^3+b x^2+c x+d & P^ (x)=4 x^3+3 a x^2+2 b x+c & x=0 is a solution for P^ (x)=0, c=0 & P(x)=x^4+a x^3+b x^2+d aligned Also, we have P(-1) 0 P^ (x)=0 , only when x=0 and P(x) is differentiable in (-1,1) , we should have the maximum and minimum at the points x=-1,0 and 1 only Also, we have P(-1) 0 aligned Thus, we have a>0 and b>0 P ^ ( x )=4 x ^3+3 a x ^2+2 bx >0, x (0,1) Hence P(x) is increasing in [0,1] Max. of P(x)=P(1) Similarly, P(x) is decreasing in [-1,0] Therefore Min. P(x) does not occu