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1 Unit 1: Pure Mathematics A

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Question 160

The curve C has equation y=f(x)y = \text{f}(x)y=f(x) where x>0x > 0x>0.

Given that

  • f′(x)=(x+2)2xx\text{f}'(x) = \dfrac{(x + 2)^2}{x\sqrt{x}}f′(x)=xx​(x+2)2​
  • the point P(9,30)P(9, 30)P(9,30) lies on C
a.

(i) Find the value of the gradient of the curve at PPP.

(ii) Hence find the equation of the tangent to CCC at PPP, giving your answer in the form ax+by+c=0ax + by + c = 0ax+by+c=0, where aaa, bbb and ccc are integers to be found.

[4]
b.

Find f(x)\text{f}(x)f(x), simplifying your answer.

[6]
Markscheme

1 Unit 1: Pure Mathematics A Questions

  1. A Level
  2. /Maths
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