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3. (a) Show that \( \frac{dy}{dx} = \frac{A}{(x + 1)^{n}} \) where A and n are constants to be found - Edexcel - A-Level Maths Pure - Question 5 - 2019 - Paper 2
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3. (a) Show that \( \frac{dy}{dx} = \frac{A}{(x + 1)^{n}} \) where A and n are constants to be found. (b) Hence deduce the range of values for x for which \( \frac... show full transcript
Worked Solution & Example Answer:3. (a) Show that \( \frac{dy}{dx} = \frac{A}{(x + 1)^{n}} \) where A and n are constants to be found - Edexcel - A-Level Maths Pure - Question 5 - 2019 - Paper 2
Step 1
Show that \( \frac{dy}{dx} = \frac{A}{(x + 1)^{n}} \)
Answer
To find ( \frac{dy}{dx} ), we will use the quotient rule for differentiation. The given function is:
[ y = \frac{5x^{2} + 10x}{(x + 1)^{2}} ]
By applying the quotient rule, we have:
[ \frac{dy}{dx} = \frac{(x + 1)^{2}(10x + 10) - (5x^{2} + 10x) \cdot 2(x + 1)}{(x + 1)^{4}} ]
Now, simplifying the expression:
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First, calculate the derivative of the numerator:
- The product terms are: ( (x + 1)^{2}(10x + 10) ) and ( (5x^{2} + 10x) \times 2 ) contributing towards the numerator.
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Combine and simplify:
- Resulting in:
- ( 10(x + 1)^{2} - 2(5x^{2} + 10x) )
- Factor and combine like terms to achieve: [ A = \text{resulting constant from simplification,}\ n = \text{power of } (x + 1)\ ]
Identifying constants A and n from the final expression gives the required form.
Step 2
Hence deduce the range of values for x for which \( \frac{dy}{dx} < 0 \)
Answer
Given that:
[ \frac{dy}{dx} = \frac{A}{(x + 1)^{n}} ]
For ( \frac{dy}{dx} < 0 ), it implies that A must be negative because ( (x + 1)^{n} ) will always be positive for ( x > -1 ). Hence:
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The condition becomes:
- ( A < 0 )
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Since we know A is a constant derived from step (a), we deduce the range of x values by ensuring that:\
- ( (x + 1) > 0 ) which simplifies to ( x > -1 ).
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Since we're interested in where ( \frac{dy}{dx} < 0 ):
- The ranges for x fall in: ( x < -1 ) must be satisfied to ensure negativity under the established conditions.