OCR C4 — Question 6 11 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeFind equation of tangent
DifficultyStandard +0.3 This is a slightly above-average C4 question. Part (i) is a standard proof from first principles using the quotient rule. Part (ii) requires applying the product rule to find dy/dx, evaluating at x=0 (straightforward), then writing the tangent equation. Part (iii) involves solving dy/dx=0 numerically. All techniques are routine for C4, though the combination of exponential and trig functions adds mild complexity.
Spec1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation

6. (i) Use the derivative of \(\cos x\) to prove that $$\frac { \mathrm { d } } { \mathrm {~d} x } ( \sec x ) = \sec x \tan x$$ The curve \(C\) has the equation \(y = \mathrm { e } ^ { 2 x } \sec x , - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }\).
(ii) Find an equation for the tangent to \(C\) at the point where it crosses the \(y\)-axis.
(iii) Find, to 2 decimal places, the \(x\)-coordinate of the stationary point of \(C\).

Question 6:
Part (i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{d}{dx}(\sec x) = \frac{d}{dx}[(\cos x)^{-1}]\)
\(= -(\cos x)^{-2} \times (-\sin x)\)M1 A1
\(= \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \times \frac{\sin x}{\cos x}\)M1
\(= \sec x \tan x\)A1
Part (ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{dy}{dx} = 2e^{2x} \times \sec x + e^{2x} \times \sec x \tan x = e^{2x}\sec x(2 + \tan x)\)M1 A1
\(x = 0,\ y = 1,\ \text{grad} = 2\)M1
\(\therefore\ y = 2x + 1\)A1
Part (iii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
SP: \(e^{2x}\sec x(2 + \tan x) = 0\)M1
\(\tan x = -2\)M1
\(x = -1.11\) (2dp)A1 (11)
# Question 6:

## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{dx}(\sec x) = \frac{d}{dx}[(\cos x)^{-1}]$ | | |
| $= -(\cos x)^{-2} \times (-\sin x)$ | M1 A1 | |
| $= \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \times \frac{\sin x}{\cos x}$ | M1 | |
| $= \sec x \tan x$ | A1 | |

## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 2e^{2x} \times \sec x + e^{2x} \times \sec x \tan x = e^{2x}\sec x(2 + \tan x)$ | M1 A1 | |
| $x = 0,\ y = 1,\ \text{grad} = 2$ | M1 | |
| $\therefore\ y = 2x + 1$ | A1 | |

## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| SP: $e^{2x}\sec x(2 + \tan x) = 0$ | M1 | |
| $\tan x = -2$ | M1 | |
| $x = -1.11$ (2dp) | A1 | **(11)** |

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6. (i) Use the derivative of $\cos x$ to prove that

$$\frac { \mathrm { d } } { \mathrm {~d} x } ( \sec x ) = \sec x \tan x$$

The curve $C$ has the equation $y = \mathrm { e } ^ { 2 x } \sec x , - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }$.\\
(ii) Find an equation for the tangent to $C$ at the point where it crosses the $y$-axis.\\
(iii) Find, to 2 decimal places, the $x$-coordinate of the stationary point of $C$.\\

\hfill \mbox{\textit{OCR C4  Q6 [11]}}