Standard +0.3 This is a straightforward application of the quotient rule to find dy/dx, then solving dy/dx = 0. The algebra is clean since e^x(sin x + cos x) = 0 gives tan x = -1 immediately. Slightly above average difficulty due to requiring quotient rule with exponential/trig functions, but the solution path is direct and the equation solving is routine.
3 The curve \(y = \frac { \mathrm { e } ^ { x } } { \cos x }\), for \(- \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi\), has one stationary point. Find the \(x\)-coordinate of this point.
Obtain correctly the derivative in any form, e.g. \(\frac{e^x \cos x + e^x \sin x}{\cos^2 x}\)
A1
Equate derivative to zero and reach \(\tan x = k\)
M1*
Solve for \(x\)
M1(dep*)
Obtain \(x = -\frac{1}{4}\pi\) (or \(-0.785\)) only (accept \(x \in [-0.79, -0.78]\) but not in degrees)
A1
[5]
[The last three marks are independent. Fallacious log work forfeits the M1*. For the M1(dep*) the solution can lie outside the given range and be in degrees, but the mark is not available if \(k = 0\). The final A1 is only given for an entirely correct answer to the whole question.]
Use correct quotient or product rule | M1 |
Obtain correctly the derivative in any form, e.g. $\frac{e^x \cos x + e^x \sin x}{\cos^2 x}$ | A1 |
Equate derivative to zero and reach $\tan x = k$ | M1* |
Solve for $x$ | M1(dep*) |
Obtain $x = -\frac{1}{4}\pi$ (or $-0.785$) only (accept $x \in [-0.79, -0.78]$ but not in degrees) | A1 | [5] |
[The last three marks are independent. Fallacious log work forfeits the M1*. For the M1(dep*) the solution can lie outside the given range and be in degrees, but the mark is not available if $k = 0$. The final A1 is only given for an entirely correct answer to the whole question.] |