Answer:
using [hex]b^2 – 4ac < 0[/hex]
a = 6, b = 2a and c = b
[hex](2a)^2 – 4(6)(b) < 0 [/hex]
[hex]4(a)^2 – 24b < 0 [/hex]
[hex]4(a)^2 < 24b [/hex]
:- [hex](a)a^2 < 6b [/hex] - ANS
Explanation:
According to the equation given, the correct question is ‘The curve [hex]y = 2(x)^3 + a(x)^2 + bx – 30[/hex] has no stationary points. Show that a2 < 6b’
First Step
we have to differentiate the equation to get the equation of the gradient of the curve, (from here we will get an equation to find the stationary values) thought the question stated that the curve has no stationary points, the curve has stationary points actually, but they are in complex(unreal) form, so the curve has no real stationary point.
[hex]y = 2(x)^3 + a(x)^2 + bx – 30[/hex]
[hex]dy/dx = 6(x)^2 + 2ax + b[/hex]
The general form of a quadratic equation is [hex]a(x)^2 + bx + c[/hex].
So from the equation a = 6, b= 2a and c = b.
Second Step
Since the question stated that it has no stationary points, the discriminant formula ‘ [hex](b)^2 – 4ac < 0[/hex]’ will be used, this formula is used to test if a quadratic equation has unreal or no roots(same as stationary point) .
When testing for the type of roots a quadratic equation has,
if
[hex](b)^2 – 4ac > 0[/hex] the roots will be uequal and real
[hex](b)^2 – 4ac < 0[/hex] the roots will be unequal and unreal(complex number)
[hex](b)^2 – 4ac = 0[/hex] the roots will be equal and real
using [hex](b)^2 – 4ac < 0[/hex]
a = 6, b = 2a and c = b
[hex](2a)^2 – 4(6)(b) < 0 [/hex]
[hex]4(a)^2 – 24b < 0 [/hex]
[hex]4(a)^2 < 24b [/hex]
:- [hex](a)a^2 < 6b [/hex] - ANSWER