OCR Further Pure Core AS 2019 June — Question 6 5 marks

Exam BoardOCR
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2019
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear transformations
TypeArea scale factor from determinant
DifficultyStandard +0.8 This question requires finding the determinant of a matrix with algebraic entries, recognizing that area scales by |det(T)|, then minimizing a quadratic expression. It combines multiple concepts (determinants, area scale factors, optimization) and requires algebraic manipulation beyond routine application, making it moderately challenging for Further Maths AS level.
Spec4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation
6 A transformation T is represented by the matrix \(\mathbf { T }\) where \(\mathbf { T } = \left( \begin{array} { c c } x ^ { 2 } + 1 & - 4 \\ 3 - 2 x ^ { 2 } & x ^ { 2 } + 5 \end{array} \right)\). A quadrilateral \(Q\), whose area is 12 units, is transformed by T to \(Q ^ { \prime }\). Find the smallest possible value of the area of \(Q ^ { \prime }\).
Question 6:
AnswerMarks Guidance
AnswerMarks Guidance
\((\Delta =) \begin{vmatrix} x^2+1 & -4 \\ 3-2x^2 & x^2+5 \end{vmatrix}\)M1 1.1 — Expanding the determinant of \(\mathbf{T}\). Condone sign error and/or one "\(x\)" rather than "\(x^2\)"
\(= (x^2+1)(x^2+5) - (-4)(3-2x^2)\)
\(= x^4 - 2x^2 + 17\)A1 1.1
\(\Delta = (x^2-1)^2 + 16\)M1* 3.1a — Attempt to find minimum value of quadratic in \(x^2\). Or from \(\dfrac{d\Delta}{dx} = 4x^3 - 4x = 0\)
\((x^2 = 1) \Rightarrow \Delta_{\min} = 16\)A1 1.1
So the smallest possible area is \(192\) (units)A1ft dep(*) [5] 3.2a — Their \(\Delta_{\min} \times 12\) 
# Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\Delta =) \begin{vmatrix} x^2+1 & -4 \\ 3-2x^2 & x^2+5 \end{vmatrix}$ | M1 | 1.1 — Expanding the determinant of $\mathbf{T}$. Condone sign error and/or one "$x$" rather than "$x^2$" |
| $= (x^2+1)(x^2+5) - (-4)(3-2x^2)$ | | |
| $= x^4 - 2x^2 + 17$ | A1 | 1.1 |
| $\Delta = (x^2-1)^2 + 16$ | M1* | 3.1a — Attempt to find minimum value of quadratic in $x^2$. Or from $\dfrac{d\Delta}{dx} = 4x^3 - 4x = 0$ |
| $(x^2 = 1) \Rightarrow \Delta_{\min} = 16$ | A1 | 1.1 |
| So the smallest possible area is $192$ (units) | A1ft dep(*) [5] | 3.2a — Their $\Delta_{\min} \times 12$ |
6 A transformation T is represented by the matrix $\mathbf { T }$ where $\mathbf { T } = \left( \begin{array} { c c } x ^ { 2 } + 1 & - 4 \\ 3 - 2 x ^ { 2 } & x ^ { 2 } + 5 \end{array} \right)$. A quadrilateral $Q$, whose area is 12 units, is transformed by T to $Q ^ { \prime }$.

Find the smallest possible value of the area of $Q ^ { \prime }$.

\hfill \mbox{\textit{OCR Further Pure Core AS 2019 Q6 [5]}}
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