Challenging +1.2 This question requires finding intersection points, setting up two separate integrals for regions above and below the x-axis, and solving a cubic equation. While it involves multiple steps and careful handling of absolute values in area calculations, the techniques are standard for A-level integration problems. The algebraic manipulation is straightforward once the integral is set up correctly.
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The diagram shows part of the curve \(y = ( 5 - x ) ( x - 1 )\) and the line \(x = a\).
Given that the total area of the regions shaded in the diagram is 19 units \({ } ^ { 2 }\), determine the exact value of \(a\).
Solve their cubic leading to an exact value for \(a\); dependent on both previous M marks
\(a \neq \dfrac{9-\sqrt{21}}{2} \because a > 5\)
B1
BC – must give a reason for rejection of this value of \(a\)
\(a = \dfrac{9+\sqrt{21}}{2}\) only
A1
BC
# Question 7:
$\dfrac{32}{3}$ | **B1** | Seen or implied by later working
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# Question 7 (integration part):
$\int(-x^2 + 6x - 5)\,dx = -\dfrac{x^3}{3} + 3x^2 - 5x$ | **M1\*, A1** | M1 attempt integration on 3-term quadratic in $x$; A1 ignore lack of $+c$
$-\dfrac{a^3}{3} + 3a^2 - 5a - \left(-\dfrac{5^3}{3} + 75 - 25\right)$ | **Dep\*M1** | $\pm(F(a) - F(5))$
$\dfrac{32}{3} + \dfrac{a^3}{3} - 3a^2 + 5a + \dfrac{25}{3} = 19$ | **A1** | oe
$a^3 - 9a^2 + 15a = 0 \Rightarrow a^2 - 9a + 15 = 0 \because a \neq 0$ | **M1** | Solve their cubic leading to an exact value for $a$; dependent on both previous M marks
$a \neq \dfrac{9-\sqrt{21}}{2} \because a > 5$ | **B1** | BC – must give a reason for rejection of this value of $a$
$a = \dfrac{9+\sqrt{21}}{2}$ only | **A1** | BC
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\includegraphics[max width=\textwidth, alt={}, center]{d6430776-0b87-4e5e-8f78-c6228ee163d5-5_647_741_260_260}
The diagram shows part of the curve $y = ( 5 - x ) ( x - 1 )$ and the line $x = a$.\\
Given that the total area of the regions shaded in the diagram is 19 units ${ } ^ { 2 }$, determine the exact value of $a$.
\hfill \mbox{\textit{OCR PURE Q7 [8]}}