P(x) is a fourth degree polynomial such that (1) P(-x)=P(x) x R , (2) P(x) 0 x R , (3) P(0)=1 , (4) P(x) has exactly two local minima at x₁ and x₂ such that |x₁-x₂ |=2 . The line y=1 touches the curve at a certain point Q and the enclosed area between the line and the curve is 8 2 15 . Let g(x)=A x^2+B x+C(A 0) such that _ x 0 P(x)-g(x)-g(-x) x^2 is finite and is equal to the slope of the tangent of g(x) at x=-1 . Al

Step-by-step solution

P(x) is an even function. P ( x )= ax ^4+b x ^2+1 and P ^ ( x )=4 ax ^3+2 bx =2 x (2 ax ^2+ b ) It has two minima. Hence, a and b 0 . So, at x= - b 2 a , P ( x ) has minima. 2 -b 2 a =2 b=-2 a Maximum at (0,1) . Also, 8 2 15 =2 ₀^ -b a (1- (a x^4+ bx ^2+1 ) ) dx b =-1 a = 1 2 Now, _ x 0 P(x)-(g(x)+g(-x)) x^2 is finite. C= 1 2 , B=-1 Also, y=1 is tangent to A x^2-x+ 1 2 =f(x) Ax - x + 1 2 =1 has equal roots A=- 1 2