| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Inverse function graphs and properties |
| Difficulty | Standard +0.8 This Further Maths question requires knowledge of inverse trig derivatives (non-standard at A-level), verification by substitution, finding gradients using chain rule, and testing perpendicularity. While systematic once the derivatives are known, it combines multiple techniques and requires careful algebraic manipulation of the gradient product, placing it moderately above average difficulty. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07l Derivative of ln(x): and related functions |
| Answer | Marks | Guidance |
|---|---|---|
| For 2nd curve \(\tan^{-1}\left(\sqrt{2} \times \frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}\) | B1, B1 | Alt: M1 Set up quadratic in sin or cos and solve; A1 Both values correct |
| (ii) For 1st curve \(y = \cos^{-1}x\), \(\frac{dy}{dx} = \frac{-1}{\sqrt{1-x^2}}\) | B1 | soi |
| For 2nd curve \(y = \tan^{-1}x\), \(\frac{dy}{dx} = \frac{\sqrt{2}}{1+2x^2}\) | B1 | soi |
| For 1st curve, when \(x = \frac{1}{\sqrt{2}}\), \(\frac{dy}{dx} = \frac{-1}{\sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^2}} = \frac{-1}{\sqrt{\frac{1}{2}}} = -\sqrt{2}\) | M1 | Substituting value into their derivatives and using \(m_1 \times m_2 = (1)\) (i.e. evidence of finding the product of gradients) |
| For 2nd curve, when \(x = \frac{1}{\sqrt{2}}\), \(\frac{dy}{dx} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\) | A1 | Depends on exact correct numerical values being seen |
| Since \(m_1 \times m_2 = -1\) then Yes | [4] | Acceptable reason: One the negative reciprocal of the other. Condone: One the negative inverse of the other. |
**(i)** For 1st curve $\cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$
For 2nd curve $\tan^{-1}\left(\sqrt{2} \times \frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$ | B1, B1 | Alt: M1 Set up quadratic in sin or cos and solve; A1 Both values correct
**(ii)** For 1st curve $y = \cos^{-1}x$, $\frac{dy}{dx} = \frac{-1}{\sqrt{1-x^2}}$ | B1 | soi
For 2nd curve $y = \tan^{-1}x$, $\frac{dy}{dx} = \frac{\sqrt{2}}{1+2x^2}$ | B1 | soi
For 1st curve, when $x = \frac{1}{\sqrt{2}}$, $\frac{dy}{dx} = \frac{-1}{\sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^2}} = \frac{-1}{\sqrt{\frac{1}{2}}} = -\sqrt{2}$ | M1 | Substituting value into their derivatives and using $m_1 \times m_2 = (1)$ (i.e. evidence of finding the product of gradients)
For 2nd curve, when $x = \frac{1}{\sqrt{2}}$, $\frac{dy}{dx} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$ | A1 | Depends on exact correct numerical values being seen
Since $m_1 \times m_2 = -1$ then Yes | [4] | Acceptable reason: One the negative reciprocal of the other. Condone: One the negative inverse of the other.
---4 The curves $y = \cos ^ { - 1 } x$ and $y = \tan ^ { - 1 } ( \sqrt { 2 } x )$ intersect at a point $A$.\\
(i) Verify that the coordinates of $A$ are $\left( \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { 4 } \pi \right)$.\\
(ii) Determine whether the tangents to the curves at $A$ are perpendicular.
\hfill \mbox{\textit{OCR FP2 2014 Q4 [6]}}