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9. (a) Show that f''(x) = \frac{(3 - x^3)^2}{x^2}, \ x \neq 0 where A and B are constants to be found - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 1
Question 11
9. (a) Show that f''(x) = \frac{(3 - x^3)^2}{x^2}, \ x \neq 0 where A and B are constants to be found. (b) Find f''(x). (c) Given that the point (-3, 10) li... show full transcript
Worked Solution & Example Answer:9. (a) Show that f''(x) = \frac{(3 - x^3)^2}{x^2}, \ x \neq 0 where A and B are constants to be found - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 1
Step 1
Show that f''(x) = 9x^2 + A + Bx^2.
Answer
To show that
can be expressed as
we start by expanding the numerator:
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Expand ((3 - x^3)^2: ) (= 9 - 6x^3 + x^6).
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Substitute back into the equation: (f''(x) = \frac{9 - 6x^3 + x^6}{x^2}).
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Now, divide each term by (x^2:\n ) (= \frac{9}{x^2} - 6x + x^4).
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Reorganizing gives: (= 9x^2 + A + Bx^2), where A = 0 and B = 0.
Step 2
Step 3
Given that the point (-3, 10) lies on the curve with equation y = f(x), find f(x).
Answer
Using the point (-3, 10) to find c:
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Substitute (x = -3) into (f(x) = -9x^3 - 6x^2 + \frac{x^3}{3} + c): (-10 = -9(-3)^3 - 6(-3)^2 + \frac{(-3)^3}{3} + c).
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Calculate: (-10 = -9(-27) - 6(9) + (-9) + c) (-10 = 243 - 54 - 9 + c) (-10 = 180 + c) (c = -10 - 180 = -190).
Hence, (f(x) = -9x^3 - 6x^2 + \frac{x^3}{3} - 190) after substituting for c.