11 The curve \(y = 4 x ^ { 2 } + \frac { a } { x } + 5\) has a stationary point. Find the value of the positive constant \(a\) given that the \(y\)-coordinate of the stationary point is 32 .
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Question 11:
Answer Marks
Guidance
Answer Marks
Guidance
\(y = 4x^2 + ax^{-1} + 5\) B1 \(ax^{-1}\) seen or implied
\(\frac{dy}{dx} = 8x - ax^{-2}\) M1 Attempt to differentiate – at least one non-zero term correct
A1 Fully correct
At stationary point, \(8x - ax^{-2} = 0\) M1 Sets their derivative to 0
\(a = 8x^3\) oe A1 Obtains expression for \(a\) in terms of \(x\), or \(x\) in terms of \(a\) www
When \(a = 8x^3\), \(y = 32\): \(32 = 4x^2 + 8x^2 + 5\) M1 Substitutes their expression and 32 into equation of the curve to form single variable equation
\(x = \frac{3}{2}\) oe A1 Obtains correct value for \(x\). Allow \(x = \sqrt{\frac{27}{12}}\). Ignore \(-\frac{3}{2}\) given as well.
\(a = 27\) A1 [8]Obtains correct value for \(a\). Ignore \(-27\) given as well.
OR (alternative method):
Answer Marks
Guidance
Answer Marks
Guidance
\(y = 4x^2 + ax^{-1} + 5\) B1 \(ax^{-1}\) seen or implied
\(\frac{dy}{dx} = 8x - ax^{-2}\) M1 Attempt to differentiate – at least one non-zero term correct
A1 Fully correct
\(32 = 4x^2 + ax^{-1} + 5\) M1 Substitutes 32 into equation of the curve to find expression for \(a\)
\(a = 27x - 4x^3\) A1 Obtains expression for \(a\) in terms of \(x\) www
At stationary point, \(8x - ax^{-2} = 0\): \(8x - (27x - 4x^3)x^{-2} = 0\) M1 Sets derivative to zero and forms single variable equation
\(x = \frac{3}{2}\) oe A1 Obtains correct value for \(x\). Allow \(x = \sqrt{\frac{27}{12}}\). Ignore \(-\frac{3}{2}\) given as well.
\(a = 27\) A1 Obtains correct value for \(a\). Ignore \(-27\) given as well.
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## Question 11:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 4x^2 + ax^{-1} + 5$ | **B1** | $ax^{-1}$ seen or implied |
| $\frac{dy}{dx} = 8x - ax^{-2}$ | **M1** | Attempt to differentiate – at least one non-zero term correct |
| | **A1** | Fully correct |
| At stationary point, $8x - ax^{-2} = 0$ | **M1** | Sets their derivative to 0 |
| $a = 8x^3$ oe | **A1** | Obtains expression for $a$ in terms of $x$, or $x$ in terms of $a$ **www** | $x = \frac{\sqrt[3]{a}}{2}$ oe, $a = 18x$ oe also fine |
| When $a = 8x^3$, $y = 32$: $32 = 4x^2 + 8x^2 + 5$ | **M1** | Substitutes their expression and 32 into equation of the curve to form single variable equation | |
| $x = \frac{3}{2}$ oe | **A1** | Obtains correct value for $x$. Allow $x = \sqrt{\frac{27}{12}}$. Ignore $-\frac{3}{2}$ given as well. | Or expression for $a$ e.g. $a^{\frac{2}{3}} = 9$ |
| $a = 27$ | **A1** [8] | Obtains correct value for $a$. Ignore $-27$ given as well. | |
**OR (alternative method):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 4x^2 + ax^{-1} + 5$ | **B1** | $ax^{-1}$ seen or implied |
| $\frac{dy}{dx} = 8x - ax^{-2}$ | **M1** | Attempt to differentiate – at least one non-zero term correct |
| | **A1** | Fully correct |
| $32 = 4x^2 + ax^{-1} + 5$ | **M1** | Substitutes 32 into equation of the curve to find expression for $a$ |
| $a = 27x - 4x^3$ | **A1** | Obtains expression for $a$ in terms of $x$ **www** |
| At stationary point, $8x - ax^{-2} = 0$: $8x - (27x - 4x^3)x^{-2} = 0$ | **M1** | Sets derivative to zero **and** forms single variable equation |
| $x = \frac{3}{2}$ oe | **A1** | Obtains correct value for $x$. Allow $x = \sqrt{\frac{27}{12}}$. Ignore $-\frac{3}{2}$ given as well. |
| $a = 27$ | **A1** | Obtains correct value for $a$. Ignore $-27$ given as well. |
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11 The curve $y = 4 x ^ { 2 } + \frac { a } { x } + 5$ has a stationary point. Find the value of the positive constant $a$ given that the $y$-coordinate of the stationary point is 32 .
\hfill \mbox{\textit{OCR C1 2016 Q11 [8]}}