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Given that \( 6x + 3x^{\frac{5}{2}} \) can be written in the form \( 6x^p + 3x^q, \) (a) write down the value of \( p \) and the value of \( q \) - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 1

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Given-that-\(-6x-+-3x^{\frac{5}{2}}-\)-can-be-written-in-the-form-\(-6x^p-+-3x^q,-\)--(a)-write-down-the-value-of-\(-p-\)-and-the-value-of-\(-q-\)-Edexcel-A-Level Maths Pure-Question 8-2011-Paper 1.png

Given that \( 6x + 3x^{\frac{5}{2}} \) can be written in the form \( 6x^p + 3x^q, \) (a) write down the value of \( p \) and the value of \( q \). Given that \( \f... show full transcript

Worked Solution & Example Answer:Given that \( 6x + 3x^{\frac{5}{2}} \) can be written in the form \( 6x^p + 3x^q, \) (a) write down the value of \( p \) and the value of \( q \) - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 1

Step 1

(a) write down the value of p and the value of q.

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Answer

To express ( 6x + 3x^{\frac{5}{2}} ) in the form ( 6x^p + 3x^q ):

  • The term ( 6x ) corresponds to ( 6x^p ), hence ( p = 1 ).
  • The term ( 3x^{\frac{5}{2}} ) corresponds to ( 3x^q ), hence ( q = \frac{5}{2} ).

Thus, the values are:

  • ( p = 1 )
  • ( q = \frac{5}{2} )

Step 2

(b) find y in terms of x, simplifying the coefficient of each term.

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Answer

Starting with the equation:

[ \frac{dy}{dx} = \frac{6x + 3x^{\frac{5}{2}}}{\sqrt{x}} ]

We simplify this:

[ \frac{dy}{dx} = \frac{6x}{\sqrt{x}} + \frac{3x^{\frac{5}{2}}}{\sqrt{x}} = 6x^{\frac{1}{2}} + 3x^{2} ]

Next, we integrate both terms:

[ y = \int (6x^{\frac{1}{2}} + 3x^{2}) ; dx = 6 \cdot \frac{2}{3} x^{\frac{3}{2}} + 3 \cdot \frac{1}{3} x^{3} + C ]

This leads to:

[ y = 4x^{\frac{3}{2}} + x^{3} + C ]

Substituting the known values ( x = 4 ) and ( y = 90 ):

[ 90 = 4 \cdot (4)^{\frac{3}{2}} + (4)^{3} + C ]

Calculating ( (4)^{\frac{3}{2}} = 8 ) and ( (4)^{3} = 64 ):

[ 90 = 4 \cdot 8 + 64 + C ]
[ 90 = 32 + 64 + C ] [ 90 = 96 + C ]
[ C = 90 - 96 = -6 ]

Finally, substituting ( C ) back into the equation gives:

[ y = 4x^{\frac{3}{2}} + x^{3} - 6 ]

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