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junior college 1 | H2 Maths
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I don’t understand how to do this question, the solution my school gave says discriminant is more than 0 because there is 2 stationary points. How is that even possible? I don’t understand :( can someone help step by step please thank you
Date Posted: 3 years ago
Views: 489
You can follow Yong Kc's working for the differentiation.
Explanation :
At any stationary point, dy/dx = 0
So if the curve C has stationary points,
αx² + 4αx + 1 = 0
(see his working)
Then look at the discriminant.
Recall that for real and distinct roots/solutions, discriminant > 0 (b² - 4ac > 0)
If C has 2 stationary points, there must be two distinct values of x, one for each point.
Which means :
(4α)² - 4(α)(1) > 0
16α² - 4α > 0
Divide by 4,
4α² - α > 0
Factorise,
α(4α - 1) > 0
This is an upward sloping curve (U-shaped) which cuts the horizontal axis at α = 0 and α = ¼
So for α(4α - 1) > 0,
α < 0 or 4a > 1
α < 0 or α > ¼
k = ¼
Explanation :
At any stationary point, dy/dx = 0
So if the curve C has stationary points,
αx² + 4αx + 1 = 0
(see his working)
Then look at the discriminant.
Recall that for real and distinct roots/solutions, discriminant > 0 (b² - 4ac > 0)
If C has 2 stationary points, there must be two distinct values of x, one for each point.
Which means :
(4α)² - 4(α)(1) > 0
16α² - 4α > 0
Divide by 4,
4α² - α > 0
Factorise,
α(4α - 1) > 0
This is an upward sloping curve (U-shaped) which cuts the horizontal axis at α = 0 and α = ¼
So for α(4α - 1) > 0,
α < 0 or 4a > 1
α < 0 or α > ¼
k = ¼
See 1 Answer
It should be > instead of ≥ since 2 stationary points implies they are distinct,
i.e real and distinct roots/solutions of x rather than real and repeated root.
i.e real and distinct roots/solutions of x rather than real and repeated root.
You are right. Overlooked that. Thanks.
No prob
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