
A.
(frac{1}{3}) 
B.
(frac{1}{3}) 
C.
1 
D.
1
Correct Answer: Option D
Explanation
y = x^{3} + x^{2} – x + 1
(frac{dy}{dx}) = (frac{d(x^3)}{dx}) + (frac{d(x^2)}{dx}) – (frac{d(x)}{dx}) + (frac{d(1)}{dx})
(frac{dy}{dx}) = 3x^{2} + 2x – 1 = 0
(frac{dy}{dx}) = 3x^{2} + 2x – 1
At the maximum point (frac{dy}{dx}) = 0
3x^{2} + 2x – 1 = 0
(3x^{2} + 3x) – (x – 1) = 0
3x(x + 1) 1(x + 1) = 0
(3x – 1)(x + 1) = 0
therefore x = (frac{1}{3}) or 1
For the maximum point
(frac{d^2y}{dx^2})
(frac{d^2y}{dx^2}) 6x + 2
when x = (frac{1}{3})
(frac{dx^2}{dx^2}) = 6((frac{1}{3})) + 2
= 2 + 2 = 4
(frac{d^2y}{dx^2}) > o which is the minimum point
when x = 1
(frac{d^2y}{dx^2}) = 6(1) + 2
= 6 + 2 = 4
4
therefore, (frac{d^2y}{dx^2})
the maximum point is 1