Substitution and Evaluation of Expressions
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Substitution and Evaluation of Expressions — Free MYP4 Mathematics (Standard) Practice Questions

1QuestionMultiple Substitutions in Complex ExpressionsConcept Practice
4 marks~6 minCriterion A
The graph shows a parabola modelled by y=ax2+bx+cy = ax^2 + bx + c, opening upwards, with vertex at (0,−1)(0, -1). The parabola passes through (−2,3)(-2, 3), (0,−1)(0, -1), and (2,3)(2, 3).
a
State the value of cc by identifying the yy-intercept from the graph. [1]
b
Substitute the points (−2,3)(-2, 3) and (2,3)(2, 3) into y=ax2+bx+cy = ax^2 + bx + c to construct two equations in aa and bb. [1]
c
Deduce the values of aa and bb by solving the system of equations from part (b). [1]
d
The parabola models the cross-sectional profile of a satellite dish. The dish is functional only if its profile is symmetric about the central axis. Justify whether this dish meets the symmetry requirement, using the values of aa, bb, and cc. [1]

Solutions

2QuestionMultiple Substitutions in Complex ExpressionsAssessment Practice
12 marks~18 minCriterion D
In electrical circuit analysis, the total resistance RTR_T (in ohms) of a parallel circuit with three resistors is given by:

1RT=1R1+1R2+1R3\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

A circuit is designed with R1=20 ΩR_1 = 20\ \Omega, R2=30 ΩR_2 = 30\ \Omega, R3=60 ΩR_3 = 60\ \Omega.
a
Calculate RTR_T for these values. [2]
b
Each resistor has a tolerance of ±5%\pm 5\%. Deduce the maximum and minimum possible values of RTR_T, showing clearly which resistor values you use in each case. [4]
c
A safety specification requires RTR_T to remain within ±5%\pm 5\% of its nominal value under all operating conditions. Temperature fluctuations cause each resistor's resistance to change independently of its stated tolerance. Advise the engineer whether the ±5%\pm 5\% tolerance alone is sufficient to guarantee the specification is met, and state what additional information is required to complete the analysis. [6]
Question diagram

Solutions

3QuestionEvaluating with Negative Numbers and FractionsAssessment Practice
6 marks~9 minCriterion C
A structural engineer models the flexibility of a beam joint using the expression (a+1a)−1\left(a + \dfrac{1}{a}\right)^{-1}, where aa is a dimensionless stiffness ratio. A negative value of aa indicates the joint resists in the opposite direction.
a
Evaluate (a+1a)−1\left(a + \dfrac{1}{a}\right)^{-1} and (a+1a)−2\left(a + \dfrac{1}{a}\right)^{-2} for each value of aa below. Show all working.

a=−2,a=−12,a=13,a=−3a = -2, \quad a = -\tfrac{1}{2}, \quad a = \tfrac{1}{3}, \quad a = -3 [2]
b
Show that (a+1a)−1=aa2+1\left(a + \dfrac{1}{a}\right)^{-1} = \dfrac{a}{a^2+1}, and hence justify why (a+1a)−1\left(a + \dfrac{1}{a}\right)^{-1} always has the same sign as aa, while (a+1a)−2\left(a + \dfrac{1}{a}\right)^{-2} is always positive. [2]
c
The engineer states: "For any negative stiffness ratio, the flexibility measure (a+1a)−1\left(a + \dfrac{1}{a}\right)^{-1} is always negative and its magnitude is always less than 12\tfrac{1}{2}." Evaluate (a+1a)−1\left(a + \dfrac{1}{a}\right)^{-1} for a=−23a = -\dfrac{2}{3} and a=−4a = -4, then critique the engineer's claim, using the expression aa2+1\dfrac{a}{a^2+1} to support your reasoning. [2]

Solutions

4QuestionEvaluating with Negative Numbers and FractionsAssessment Practice
6 marks~9 minCriterion B
A signal-processing engineer models the attenuation (reduction in strength) of a radio signal at stage nn of a transmission network using the expression (n+1n)−1\left(n + \dfrac{1}{n}\right)^{-1}.

The first five values are:

(1+11)−1=12,(2+12)−1=25,(3+13)−1=310,(4+14)−1=417,(5+15)−1=526\left(1 + \frac{1}{1}\right)^{-1} = \frac{1}{2}, \quad \left(2 + \frac{1}{2}\right)^{-1} = \frac{2}{5}, \quad \left(3 + \frac{1}{3}\right)^{-1} = \frac{3}{10}, \quad \left(4 + \frac{1}{4}\right)^{-1} = \frac{4}{17}, \quad \left(5 + \frac{1}{5}\right)^{-1} = \frac{5}{26}
a
Deduce a general formula for the attenuation at stage nn, expressing your answer in the form nf(n)\dfrac{n}{f(n)}. [2]
b
Calculate the attenuation at stage 1010 using your formula. [2]
c
The engineer claims that the attenuation will never fall below 0.090.09 for any stage in this network (n≤12n \leq 12). Justify whether this claim is correct. [2]

Solutions

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