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A curve has equation y = \frac{2x + 3}{4x^2 + 7} 9 (a) (i) Find \frac{dy}{dx} 9 (a) (ii) Hence show that y is increasing when 4x^2 + 12x - 7 < 0 - AQA - A-Level Maths Mechanics - Question 9 - 2017 - Paper 1
Question 9
A curve has equation y = \frac{2x + 3}{4x^2 + 7} 9 (a) (i) Find \frac{dy}{dx} 9 (a) (ii) Hence show that y is increasing when 4x^2 + 12x - 7 < 0
Worked Solution & Example Answer:A curve has equation y = \frac{2x + 3}{4x^2 + 7} 9 (a) (i) Find \frac{dy}{dx} 9 (a) (ii) Hence show that y is increasing when 4x^2 + 12x - 7 < 0 - AQA - A-Level Maths Mechanics - Question 9 - 2017 - Paper 1
Step 1
Find \frac{dy}{dx}
Answer
To find the derivative of the function ( y = \frac{2x + 3}{4x^2 + 7} ), we will use the quotient rule. The quotient rule states that if ( y = \frac{u}{v} ), then ( \frac{dy}{dx} = \frac{u'v - uv'}{v^2} ).
Here, ( u = 2x + 3 ) and ( v = 4x^2 + 7 ).
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Calculate ( u' ) and ( v' ):
- ( u' = 2 )
- ( v' = 8x )
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Apply the quotient rule:
[ \frac{dy}{dx} = \frac{(2)(4x^2 + 7) - (2x + 3)(8x)}{(4x^2 + 7)^2} ] -
Expand and simplify the numerator:
[ \frac{dy}{dx} = \frac{8x^2 + 14 - 16x^2 - 24x}{(4x^2 + 7)^2} = \frac{-8x^2 - 24x + 14}{(4x^2 + 7)^2} ]
Step 2
Hence show that y is increasing when 4x^2 + 12x - 7 < 0
Answer
To show that ( y ) is increasing when ( \frac{dy}{dx} > 0 ), we need to evaluate when ( -8x^2 - 24x + 14 > 0 ).
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Rearranging yields:
[ -8x^2 - 24x + 14 > 0 ]\
[ 8x^2 + 24x - 14 < 0 ] -
Divide through by 2:
[ 4x^2 + 12x - 7 < 0 ] -
Now solve the quadratic inequality. First, find the roots using the quadratic formula:
[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]
with ( a = 4, b = 12, c = -7 ). [ x = \frac{-12 \pm \sqrt{12^2 - 4(4)(-7)}}{2(4)} = \frac{-12 \pm \sqrt{144 + 112}}{8} = \frac{-12 \pm \sqrt{256}}{8} = \frac{-12 \pm 16}{8} ] -
Thus, the roots are:
[ x = \frac{4}{8} = \frac{1}{2} \quad \text{and} \quad x = \frac{-28}{8} = -3.5 ] -
To determine the intervals:
Testing numbers in the intervals:- For x < -3.5, the expression is positive.
- For -3.5 < x < 0.5, the expression is negative.
- For x > 0.5, the expression is positive again.
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Therefore, ( 4x^2 + 12x - 7 < 0 ) in the interval ( (-3.5, 0.5) ).
Thus, ( y ) is increasing when ( 4x^2 + 12x - 7 < 0 ).