Question 6 - A-Level Maths

Edexcel C3 — Question 6 11 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Product & Quotient Rules
Type	Find equation of tangent
Difficulty	Standard +0.3 This is a straightforward multi-part question requiring standard techniques: proving a derivative formula using quotient rule (routine), finding a tangent at a given point using product rule (standard procedure), and solving f'(x)=0 numerically. All steps are textbook exercises with no novel insight required, making it slightly easier than average.
Spec	1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07m Tangents and normals: gradient and equations 1.07n Stationary points: find maxima, minima using derivatives 1.07q Product and quotient rules: differentiation 1.09d Newton-Raphson method

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

Show mark scheme Show mark scheme source

(a) $\frac{d}{dx}(\sec x) = \frac{d}{dx}[(\cos x)^{-1}]$

$= -(\cos x)^{-2} \times (-\sin x)$

Answer	Marks
$= \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \times \frac{\sin x}{\cos x}$	M1 A1 M1
$= \sec x \tan x$	A1
(b) $\frac{dy}{dx} = 2e^{2x} \times \sec x + e^{2x} \times \sec x \tan x = e^{2x}\sec x(2+\tan x)$	M1 A1
$x = 0, y = 1$, grad $= 2$	M1 A1
$\therefore y = 2x+1$	A1

(c) SP: $e^{2x}\sec x(2+\tan x) = 0$

$\tan x = -2$

Answer	Marks	Guidance
$x = -1.11$ (2dp)	M1 M1 A1	(11)

**(a)** $\frac{d}{dx}(\sec x) = \frac{d}{dx}[(\cos x)^{-1}]$

$= -(\cos x)^{-2} \times (-\sin x)$

$= \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \times \frac{\sin x}{\cos x}$ | M1 A1 M1

$= \sec x \tan x$ | A1

**(b)** $\frac{dy}{dx} = 2e^{2x} \times \sec x + e^{2x} \times \sec x \tan x = e^{2x}\sec x(2+\tan x)$ | M1 A1

$x = 0, y = 1$, grad $= 2$ | M1 A1

$\therefore y = 2x+1$ | A1

**(c)** SP: $e^{2x}\sec x(2+\tan x) = 0$

$\tan x = -2$

$x = -1.11$ (2dp) | M1 M1 A1 | **(11)**

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

\hfill \mbox{\textit{Edexcel C3  Q6 [11]}}

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Q1 6 Q2 7 Q3 7 Q4 9 Q5 10 Q6 11 Q7 12 Q8 13

Edexcel C3 — Question 6 11 marks

(a) \(\frac{d}{dx}(\sec x) = \frac{d}{dx}[(\cos x)^{-1}]\)

\(= -(\cos x)^{-2} \times (-\sin x)\)

(c) SP: \(e^{2x}\sec x(2+\tan x) = 0\)

\(\tan x = -2\)

This paper (8 questions)

Answer	Marks
\(= \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \times \frac{\sin x}{\cos x}\)	M1 A1 M1
\(= \sec x \tan x\)	A1
(b) \(\frac{dy}{dx} = 2e^{2x} \times \sec x + e^{2x} \times \sec x \tan x = e^{2x}\sec x(2+\tan x)\)	M1 A1
\(x = 0, y = 1\), grad \(= 2\)	M1 A1
\(\therefore y = 2x+1\)	A1