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N(A) E Math

PSLE

PSLE Foundation Mathematics

<p>If $a{{x}^{2}}+bx+c=0$ , where where $a,b$ and $c$ are real numbers and $a\ne 0$, then the solutions are given by </p>
<p style="text-align: center">$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$</p>
<p>This formula for solving quadratic equations is usually used when the <strong>quadratic expression cannot be factorised easily. </strong></p>
<p style="margin-top: 1.5rem"><strong>Tip: </strong></p>
<p>Always express the quadratic equation in the general form $a{{x}^{2}}+bx+c=0$ so that the correct value for $a,b$ and $c$ can be substituted into the formula.</p>

If $a{{x}^{2}}+bx+c=0$ , where where $a,b$ and $c$ are real numbers and $a\ne 0$, then the solutions are given by 

$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

This formula for solving quadratic equations is usually used when the **quadratic expression cannot be factorised easily. **

Always express the quadratic equation in the general form $a{{x}^{2}}+bx+c=0$ so that the correct value for $a,b$ and $c$ can be substituted into the formula.

Solving Quadratic Equations Using the Quadratic Formula

Solving Quadratic Equations by using Formula

Quadratic Functions, Equations and Inequalities

<p>Find the derivative of each of the following w.r.t $x$:</p>
<ol class="abc-list"><li data-list-item-id="item-1784097596054-iuq9f0oyv" data-list-item-path="item-1784097596054-iuq9f0oyv"><div class="list-item-paragraph"><p>$y = \ln{\left( (5x + 3)^2(3x^2 - 1) \right)} $</p></div></li><li data-list-item-id="item-1784097597557-iairj8pgi" data-list-item-path="item-1784097597557-iairj8pgi"><div class="list-item-paragraph"><p>$y = \ln \left( \frac{7x^2 - 8}{2 - x} \right) $</p></div></li><li data-list-item-id="item-1784097598086-2r5ar0fqh" data-list-item-path="item-1784097598086-2r5ar0fqh"><div class="list-item-paragraph"><p>$ y = \ln { \left( (x + 3)(x - 2)^2(5 - 3x^2) \right) }$</p></div></li></ol>

Find the derivative of each of the following w.r.t $x$:

Derivatives of Natural Logarithmic Functions

Chain Rule

Techniques of Differentiation

Differentiation

<p>Find the range of values of $x$ that satisfies the quadratic inequality $ 2x^2 - 4x + 7 &lt; 5x + 3$</p>

Find the range of values of $x$ that satisfies the quadratic inequality $ 2x^2 - 4x + 7 < 5x + 3$

Quadratic Inequalities

Solving Quadratic Inequalities

<p>Without expanding completely, prove that&nbsp;$ \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots + (-1)^n\binom{n}{n} = 0$ for every positive integer $n$.</p>

Without expanding completely, prove that $ \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots + (-1)^n\binom{n}{n} = 0$ for every positive integer $n$.

Sums of Coefficients in Polynomial Expansions

Binomial Theorem

<p>Suppose $\mathrm P(x) = ax^4 + bx^3 + cx^2 + dx + e$ satisfies $\mathrm P(1) = 1, \mathrm P(2) = 2, \mathrm P(3) = 3, \mathrm P(4) = 4$, it is also given&nbsp;$\mathrm P(5) = -19$.<br>Find $\mathrm P(6)$.</p>

Suppose $\mathrm P(x) = ax^4 + bx^3 + cx^2 + dx + e$ satisfies $\mathrm P(1) = 1, \mathrm P(2) = 2, \mathrm P(3) = 3, \mathrm P(4) = 4$, it is also given $\mathrm P(5) = -19$.
Find $\mathrm P(6)$.

Forming Polynomials from Roots and Factors

Polynomials, Cubic Equations and Partial Fractions

<p>A cubic polynomial satisfies $\mathrm P(1) = \mathrm P(2) =\mathrm P(3) = 7$. Also $ \mathrm P(0) = 1$<br>Find $ \mathrm P(4)$</p>


A cubic polynomial satisfies $\mathrm P(1) = \mathrm P(2) =\mathrm P(3) = 7$. Also $ \mathrm P(0) = 1$
Find $ \mathrm P(4)$

<ol class="abc-list"><li data-list-item-id="item-1"><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}}{\mathrm{d}x}\left[(2x)(3 - 5x)^{\frac{6}{5}}\right] = (6 - 22x)(3 - 5x)^{\frac{1}{5}}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-2"><div class="list-item-paragraph">Hence, evaluate $\int x(3 - 5x)^{\frac{1}{5}} dx$.<span class="inline-marks">[3]</span></div></li></ol>

Integration as an Inverse Process of Differentiation

Product Rule

2024 GESS PRELIMS P2 Q11

Integration

<p>The equation of a curve is $y = \frac{x}{e^{2x}}$.</p>
<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">Find an expression for&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">Given that&nbsp;$x$ is changing at a constant rate of $0.18$ units per second, find the rate of change of $y$ when&nbsp;$x = 1$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-3" data-list-item-path="item-3"><div class="list-item-paragraph">Explain if the curve has more than one stationary point.<span class="inline-marks">[2]</span></div></li></ol>

The equation of a curve is $y = \frac{x}{e^{2x}}$.

Connected Rates of Change

Derivatives of Exponential Functions

Stationary Points and Their Nature

2024 GESS PRELIMS P2 Q8

<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">Prove the identity $\ \operatorname{cosec} 4x + \cot 4x = \cot 2x$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">Hence, find, for&nbsp;$-180^\circ \le \theta \le 180^\circ$, the values of $\theta$ for which $\ \operatorname{cosec} 2\theta + \cot 2\theta = -\sqrt{3}$.<span class="inline-marks">[4]</span></div></li></ol>

Trigonometric Equations

Compound and Double-Angle Identities

Proving and Simplifying Trigonometric Identities

2024 GESS PRELIMS P2 Q7

Trigonometry

<ol class="abc-list"><li data-list-item-id="item-1"><div class="list-item-paragraph">Solve the equation $15(16^x) - 4^x = 6$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-2"><div class="list-item-paragraph">Solve the equation $\lg(2y) + 2\ln(y) = 2$.<span class="inline-marks">[4]</span></div></li></ol>

Solving Exponential Equations Using Substitution

Solving Logarithm Equations

2024 GESS PRELIMS P2 Q5

Exponential and Logarithmic Functions

<ol class="abc-list"><li data-list-item-id="item-1"><div class="list-item-paragraph">Given that the expression $2x^3 + ax^2 + bx + c$ leaves the same remainder when divided by $(x - 5)$ and when divided by $(2x + 1)$, show that&nbsp;$9a + 2b = -91$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-2"><div class="list-item-paragraph">Given that the remainder is $-6$ when the expression $2x^3 + ax^2 + bx + c$ is divided by $(x - 1)$, and $(x + 1)$ is a factor, calculate the value of $c$.<span class="inline-marks">[4]</span></div></li></ol>

Remainder and Factor Theorems

2024 GESS PRELIMS P2 Q1

<p>The diagram shows a prism in which its cross-section is a regular hexagon of side $ \left(5 + \sqrt{3} \right)$ cm.</p>
<div class="graph-block" data-graph-config="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" data-width="300" data-height="200" style="width:300px;height:200px;max-width:100%;display:block;margin:0 auto;"></div>
<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">Find an expression for the cross-sectional area of the prism in the form $q\sqrt{3} + p$ cm$^2$, where&nbsp;$p$ and $q$ are integers.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">Given that the volume of the prism is $3 \left(32\sqrt{3} + 138 \right)$ cm$^3$, find the height of the prism in the form $ \left(a\sqrt{3} + b \right)$ cm, where&nbsp;$a$ and $b$ are integers.<span class="inline-marks">[3]</span></div></li></ol>

The diagram shows a prism in which its cross-section is a regular hexagon of side $ \left(5 + \sqrt{3} \right)$ cm.

Applications of Surds

2024 GESS PRELIMS P1 Q11

Surds

<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">Find the term independent of $x$ in the expansion of $\left(2x^2 + \frac{3}{x}\right)^9$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><ol class="i-list"><li data-list-item-id="subitem-1781511615480-i4ty918y5" data-list-item-path="item-2/subitem-1781511615480-i4ty918y5"><div class="list-item-paragraph">Find the first three terms of the expansion of $(2 - 3x)^5$ in ascending powers of $x$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1781511646569-ijhb25enu" data-list-item-path="item-2/subitem-1781511646569-ijhb25enu"><div class="list-item-paragraph">Hence, find the value of $n$ such that the coefficient of $x$ in the expansion of $ (2 - 3x)^5(1 - 3x)^n$ is $-720$<span class="inline-marks">[3]</span></div></li></ol></li></ol>

Binomial Expansion for Positive Integer Powers

Coefficients in Products of Binomial Expansions

General Terms and Coefficients in Binomial Expansions

Finding Parameters in Binomial Expansions

2024 GESS PRELIMS P1 Q5

<p>Express $\frac{x + 7}{(x^2 - 9)(x - 3)}$ as the sum of $3$ partial fractions.<span class="inline-marks">[6]</span></p>

Express $\frac{x + 7}{(x^2 - 9)(x - 3)}$ as the sum of $3$ partial fractions.

Partial Fractions

2024 GESS PRELIMS P1 Q3

<p>The equation of a curve is $y = -2x^2 + 3x + 5$.</p>
<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">Write $y = -2x^2 + 3x + 5$ in the form $y = a(x - h)^2 + k$, where&nbsp;$a$, $h$ and $k$ are constants.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">Using your results in part <strong>(a)</strong>, explain clearly why the maximum value of $y$ is $k$ when&nbsp;$x = h$.<span class="inline-marks">[2]</span></div></li></ol>

The equation of a curve is $y = -2x^2 + 3x + 5$.

Completing the Square and Vertex Form

Maximum and Minimum Values of General Quadratic Functions

2024 GESS PRELIMS P1 Q1

<p>The diagram above shows two rods $AB$ and $BC$ of length $4$ m and $9$ m respectively and $\angle ABC = 90^{\circ} $. Rollers are fixed at points $A$ and $B$ such that&nbsp;$A$ is able to move along the wall and $B$ is able to move along the celling. The horizontal distance of $C$ from the vertical wall is $d$ m.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1781508142282-maxm0gnyuj8.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODE1MDgxNDIyODItbWF4bTBnbnl1ajgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUxOTQ4MiwiZXhwIjoxNzkzNTE5NzgyfQ.mMs8a0SbujLrQ-EjBIE85-RwP3HU1Q8_e1lPHwCz85w" style="max-width: min(320px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">Show clearly that&nbsp;$d = 4\sin \theta + 9\cos \theta$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">Express $d$ in the form $R\cos(\theta - \alpha)$ where&nbsp;$R &gt; 0$ and $0^{\circ} &lt; \alpha &lt; 90^{\circ}$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-3" data-list-item-path="item-3"><div class="list-item-paragraph">Find the value of $\theta$ for which $d = 6\text{ m}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-4" data-list-item-path="item-4"><div class="list-item-paragraph">Find the maximum value of $d$ and the corresponding value of $\theta$.<span class="inline-marks">[2]</span></div></li></ol>

The diagram above shows two rods $AB$ and $BC$ of length $4$ m and $9$ m respectively and $\angle ABC = 90^{\circ} $. Rollers are fixed at points $A$ and $B$ such that $A$ is able to move along the wall and $B$ is able to move along the celling. The horizontal distance of $C$ from the vertical wall is $d$ m.

Trigonometric Modelling

R-Formula and Trigonometric Extrema

Applications of Trigonometric Ratios

2024 FMSS PRELIMS P2 Q9

<ol class="abc-list"><li data-list-item-id="item-1"><div class="list-item-paragraph">Write down the first three terms in the expansion, in descending powers of $x$, of $ \left(x^2 + \frac{m}{x} \right)^9$, where&nbsp;$m$ is an integer.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-2"><div class="list-item-paragraph">Given that the coefficient of $x^3$ in the expansion of $ \left(x^2 + \frac{m}{x} \right)^9$ is $-126$,</div><ol class="i-list"><li data-list-item-id="subitem-1781507291750-bcufrqavo"><div class="list-item-paragraph">show that&nbsp;$m = -1$,<span class="inline-marks">[3]</span></div></li><li data-list-item-id="subitem-1781507292113-kvd75ax8q"><div class="list-item-paragraph">hence, find the term independent of $x$ in the expression of $ \left(2 - \frac{1}{x^3} \right) \left(x^2 + \frac{m}{x} \right)^9$.<span class="inline-marks">[4]</span></div></li></ol></li></ol>

2024 FMSS PRELIMS P2 Q7

<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">$A$ and $B$ are acute angles such that&nbsp;$\sin{(A - B)} = \frac{3}{8}$ and $\sin{A}\cos{B} = \frac{5}{8} $.<br>Without using a calculator, find the value of $\cos A \sin B$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">Express $2\sin 2\theta(\sec \theta - \tan \theta)$ as a quadratic expression in $\sin \theta$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-3" data-list-item-path="item-3"><div class="list-item-paragraph">Use your answer to part <strong>(b)</strong> to find, for&nbsp;$0 \le \theta \le 2\pi$, the exact solutions of the equation $2\sin 2\theta(\sec \theta - \tan \theta) + 3 = 0$.<span class="inline-marks">[3]</span></div></li></ol>

Exact Trigonometric Values and the Unit Circle

2024 FMSS PRELIMS P2 Q5

<p>Given that&nbsp;$\mathrm{f}(x) = x^2 - ax + 3$, where&nbsp;$a$ is a constant,</p>
<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">find the range of values of $a$ for which $f(x) &gt; x - 1$, for all real values of $x$,<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">find the value(s) of $a$ for which the line $y = a + 4$ is a tangent to the curve $y = f(x)$.<span class="inline-marks">[3]</span></div></li></ol>

Given that $\mathrm{f}(x) = x^2 - ax + 3$, where $a$ is a constant,

Intersection of a Line and a Curve

Nature of Roots of Quadratic Equations

Quadratic Expressions Positive or Negative for All Real Values

2024 FMSS PRELIMS P2 Q4

<p>Given that&nbsp;$y = \frac{3\ln 2x}{x^2}$ for&nbsp;$x &gt; 0$.</p>
<ol class="abc-list"><li data-list-item-id="item-1781506047585-uk90uikz0"><div class="list-item-paragraph"> Show that&nbsp;$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{x^3}(1 - 2\ln 2x)$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1781506048125-wavvo5cvu"><div class="list-item-paragraph">Hence, find $\int \frac{\ln 2x}{x^3} \mathrm{d}x$.<span class="inline-marks">[3]</span></div></li></ol>

Given that $y = \frac{3\ln 2x}{x^2}$ for $x > 0$.

A Math (IGCSE) Question Bank