Easy -1.2 This is a straightforward linear equation problem disguised as an inequality question. Students need to find equations for both firms (Firm A: find gradient and y-intercept from two points; Firm B: already given) then solve when they're equal. It requires basic coordinate geometry and solving a simple linear equation, which are routine skills well below average A-level difficulty.
1 In this question you must show detailed reasoning.
Andrea is comparing the prices charged by two different taxi firms.
Firm A charges \(\pounds 20\) for a 5 mile journey and \(\pounds 30\) for a 10 mile journey, and there is a linear relationship between the price and the length of the journey.
Firm B charges a pick-up fee of \(\pounds 3\) and then \(\pounds 2.40\) for each mile travelled.
Find the length of journey for which both firms would charge the same amount.
\(2 + 5 + x + x + (x + 2) + (x + 5) < 44\) or equivalent
B1
Correct inequality
[1]
Must be < only
Question 2(a)(ii):
Answer
Marks
Guidance
\(x(x + 2) + 10 \geq 45\) or equivalent
B1
Correct inequality relating to area
[1]
Must be \(\geq\) only
Question 2(b):
Answer
Marks
Guidance
\(x < 7.5\)
B1FT
Obtain \(x < 7.5\) from linear inequality
FT their linear inequality in (a)
\(x^2 + 2x - 35 \geq 0\)
M1
Attempt to solve three term quadratic
Critical values are \(-7\) and \(5\)
BC: Both values of \(x\) needed
\(x \leq -7, x \geq 5\)
A1FT
Choose 'outside' region for inequality
but \(x\) is a length so \(x \geq 5\)
FT their quadratic inequality in (b), as long as one positive root and one negative root
\(\{x: 5 \leq x < 7.5\}\) or \([5, 7.5)\)
B1
Single correct interval – any correct notation
[4]
Condone \(5 \leq x < 7.5\)
B1M1A0B1 possible if no reason for rejecting \(-7\)
Question 3(a):
Answer
Marks
Guidance
DR
\(\cos\alpha = \pm\frac{1}{3}\sqrt{5}\)
M1
Attempt Pythagoras on correct right-angled triangle
or \(\frac{9}{4} + \cos^2\alpha = 1\) or \(\cos^2\alpha = \frac{5}{9}\)
\(\pm\frac{1}{3}\sqrt{5}\) or eg \(\pm\sqrt{\frac{5}{9}}\)
A1
Obtain \(\pm\frac{1}{3}\sqrt{5}\) or eg \(\pm\frac{\sqrt{5}}{3}\)
[2]
A0 for answer only as DR
Question 3(b):
Answer
Marks
Guidance
\(2\sec^2\beta - 2 - 7\sec\beta + 5 = 0\)
M1
Attempt to use \(\tan^2\beta = \sec^2\beta - 1\)
Allow equiv methods involving \(\cos\beta\)
\(2\sec^2\beta - 7\sec\beta + 3 = 0\)
A1
Obtain correct equation
\((\sec\beta - 3)(2\sec\beta - 1) = 0\)
M1
Attempt to solve quadratic
\(\sec\beta = 3, \sec\beta = 0.5\)
\(\sec\beta \geq 1\) for \(0° \leq \beta < 90°\)
A1
Obtain \(\sec\beta = 3\) only
hence \(\sec\beta = 3\)
\(\sec\beta = 0.5\) must be seen and discarded with a reason
[4]
Allow more general reason such as \(
\sec\beta
$A: C = 2m + 10$ | M1 | Either correct equation
$B: C = 2.4m + 3$ | A1 | Both equations correct and variables defined
$2.4m + 3 = 2m + 10$ | M1 | Attempt to solve simultaneously
$0.4m = 7$ | | Attempt to solve their two equations
$m = 17.5$ | A1 | Obtain 17.5 miles
Same cost for a journey of 17.5 miles | [4] | Units needed
## Question 2(a)(i):
$2 + 5 + x + x + (x + 2) + (x + 5) < 44$ or equivalent | B1 | Correct inequality
| [1] | Must be < only
## Question 2(a)(ii):
$x(x + 2) + 10 \geq 45$ or equivalent | B1 | Correct inequality relating to area
| [1] | Must be $\geq$ only
## Question 2(b):
$x < 7.5$ | B1FT | Obtain $x < 7.5$ from linear inequality
| | FT their linear inequality in (a)
$x^2 + 2x - 35 \geq 0$ | M1 | Attempt to solve three term quadratic
Critical values are $-7$ and $5$ | | BC: Both values of $x$ needed
$x \leq -7, x \geq 5$ | A1FT | Choose 'outside' region for inequality
but $x$ is a length so $x \geq 5$ | | FT their quadratic inequality in (b), as long as one positive root and one negative root
$\{x: 5 \leq x < 7.5\}$ or $[5, 7.5)$ | B1 | Single correct interval – any correct notation
| [4] | Condone $5 \leq x < 7.5$
| | B1M1A0B1 possible if no reason for rejecting $-7$
## Question 3(a):
| DR | |
$\cos\alpha = \pm\frac{1}{3}\sqrt{5}$ | M1 | Attempt Pythagoras on correct right-angled triangle
| | or $\frac{9}{4} + \cos^2\alpha = 1$ or $\cos^2\alpha = \frac{5}{9}$
$\pm\frac{1}{3}\sqrt{5}$ or eg $\pm\sqrt{\frac{5}{9}}$ | A1 | Obtain $\pm\frac{1}{3}\sqrt{5}$ or eg $\pm\frac{\sqrt{5}}{3}$
| [2] | A0 for answer only as DR
## Question 3(b):
$2\sec^2\beta - 2 - 7\sec\beta + 5 = 0$ | M1 | Attempt to use $\tan^2\beta = \sec^2\beta - 1$
| | Allow equiv methods involving $\cos\beta$
$2\sec^2\beta - 7\sec\beta + 3 = 0$ | A1 | Obtain correct equation
$(\sec\beta - 3)(2\sec\beta - 1) = 0$ | M1 | Attempt to solve quadratic
$\sec\beta = 3, \sec\beta = 0.5$ | |
$\sec\beta \geq 1$ for $0° \leq \beta < 90°$ | A1 | Obtain $\sec\beta = 3$ only
hence $\sec\beta = 3$ | | $\sec\beta = 0.5$ must be seen and discarded with a reason
| [4] | Allow more general reason such as $|\sec\beta| \geq 1$
1 In this question you must show detailed reasoning.
Andrea is comparing the prices charged by two different taxi firms.\\
Firm A charges $\pounds 20$ for a 5 mile journey and $\pounds 30$ for a 10 mile journey, and there is a linear relationship between the price and the length of the journey.\\
Firm B charges a pick-up fee of $\pounds 3$ and then $\pounds 2.40$ for each mile travelled.\\
Find the length of journey for which both firms would charge the same amount.
\hfill \mbox{\textit{OCR H240/01 2018 Q1 [4]}}