JEE Advance - Mathematics (2011 - Paper 2 Offline - No. 1)
The number of distinct real roots of $${x^4} - 4{x^3} + 12{x^2} + x - 1 = 0$$
답변
2
설명
Let $$f(x) = {x^4} - 4{x^3} + 12{x^2} + x - 1 = 0$$
$$f'(x) = 4{x^3} - 12{x^2} + 24x + 1 = 4({x^3} - 3{x^2} + 6x) + 1$$
$$f''(x) = 12{x^2} - 24x + 24 = 12({x^2} - 2x + 2)$$
f''(x) has 0 real roots.
f(x) has maximum two distinct real roots as f(0) = $$-$$1.