IG MATH: A&G END OF TERM TESTS

Year 10 Mathematics Solutions

SCIMAIQ Solutions

End of Term 1 Examination (March 2026) - Solutions

Mathematics - Year 10

Question 1

1(a) Write down the inequality represented in the diagram. [2]

The diagram shows an open circle at \(-3\) (meaning \(x > -3\)) and a shaded circle at \(2\) (meaning \(x \le 2\)).

Inequality: \(-3 < x \le 2\)

1(b) Write down the integer values of \(x\) that satisfy the inequality \(-4 < 2x \le 8\). [2]

Step 1: Divide all parts by 2:

\[ \frac{-4}{2} < \frac{2x}{2} \le \frac{8}{2} \] \[ -2 < x \le 4 \]

Step 2: Identify integers greater than \(-2\) and less than or equal to \(4\):

Integer values: \(-1, 0, 1, 2, 3, 4\)

Question 2

Given \(y = 3x^2 - 12x + 7\)

2(a) Find the value of \(\frac{dy}{dx}\) when \(x=3\). [3]

Step 1: Find the derivative:

\[ \frac{dy}{dx} = 6x - 12 \]

Step 2: Substitute \(x=3\):

\[ \left.\frac{dy}{dx}\right|_{x=3} = 6(3) - 12 = 18 - 12 = 6 \]

Value: \(6\)

2(b) Find the coordinates of the point on the graph where the gradient is \(1\). [2]

Step 1: Set derivative equal to 1:

\[ 6x - 12 = 1 \implies 6x = 13 \implies x = \frac{13}{6} \approx 2.17 \]

Step 2: Find \(y\):

\[ y = 3\left(\frac{13}{6}\right)^2 - 12\left(\frac{13}{6}\right) + 7 = -\frac{59}{12} \approx -4.92 \]

Coordinates: \((2.17, -4.92)\)

Question 3

Given \(y = 2x^p - qx^2\) and \(\frac{dy}{dx} = 14x^6 + 6x\)

Step 1: Differentiate:

\[ \frac{dy}{dx} = 2p x^{p-1} - 2qx \]

Step 2: Compare with given derivative:

Exponents: \(p - 1 = 6 \implies p = 7\)
Coefficients: \(2p = 14\) ✓
Second term: \(-2q = 6 \implies q = -3\)

\(p = 7\)
\(q = -3\)

Question 4

Curve: \(y = x^3 + 8x^2 + 5x\)

4(a) Work out the coordinates of the two turning points. [6]

Step 1: Find derivative and set to zero:

\[ \frac{dy}{dx} = 3x^2 + 16x + 5 = 0 \]

Step 2: Solve using quadratic formula:

\[ x = \frac{-16 \pm \sqrt{256 - 60}}{6} = \frac{-16 \pm 14}{6} \] \[ x = -\frac{1}{3} \quad \text{or} \quad x = -5 \]

Step 3: Find y-coordinates:

\(x = -5\): \(y = (-5)^3 + 8(25) + 5(-5) = -125 + 200 - 25 = 50\)
\(x = -\frac{1}{3}\): \(y = -\frac{1}{27} + \frac{8}{9} - \frac{5}{3} = -\frac{22}{27}\)

Turning points: \((-5, 50)\) and \(\left(-\frac{1}{3}, -\frac{22}{27}\right)\)

4(b) Determine whether each turning point is a maximum or minimum. [3]

Step 1: Find second derivative:

\[ \frac{d^2y}{dx^2} = 6x + 16 \]

Step 2: Test each x-value:

At \(x = -5\): \(6(-5) + 16 = -14 < 0\) → Maximum
At \(x = -\frac{1}{3}\): \(6(-\frac{1}{3}) + 16 = 14 > 0\) → Minimum

\((-5, 50)\) is a maximum \(\left(\frac{d^2y}{dx^2} < 0\right)\)
\(\left(-\frac{1}{3}, -\frac{22}{27}\right)\) is a minimum \(\left(\frac{d^2y}{dx^2} > 0\right)\)

Question 5

Given \(y = x^3 - 2x + 3\)

5(a) Complete the table. [2]

For \(x = -1.5\): \(y = (-1.5)^3 - 2(-1.5) + 3 = -3.375 + 3 + 3 = 2.625\)

For \(x = 2\): \(y = 8 - 4 + 3 = 7\)

\(x = -1.5\) → \(y = 2.63\) (to 2 decimal places)
\(x = 2\) → \(y = 7\)

5(b) Draw the graph of \(y = x^3 - 2x + 3\) for \(-2 \le x \le 2\). [4]

(Plot points and draw a smooth curve on the provided grid.)

5(c) By drawing a suitable straight line, solve \(x^3 - 2.5x + 1 = 0\). [4]

Step 1: Rearrange to match the curve \(y = x^3 - 2x + 3\):

\[ x^3 - 2.5x + 1 = 0 \] \[ x^3 - 2x = 2.5x - 1 \]

Step 2: Add 3 to both sides to match the curve:

\[ x^3 - 2x + 3 = 2.5x - 1 + 3 \] \[ x^3 - 2x + 3 = 2.5x + 2 \]

So the line is \(y = 2.5x + 2\)

Step 3: Draw the line \(y = 2.5x + 2\). Solutions are the x-coordinates where this line intersects the curve \(y = x^3 - 2x + 3\).

Draw the line \(y = 2.5x + 2\). The solutions are the x-coordinates of the intersection points.

Question 6

Given: \(f(x) = 5^x\), \(g(x) = 3x - 2\), \(h(x) = x^2 + 1\)

6(a) Find \(f(5)\). [1]

\(f(5) = 5^5 = 3125\)

6(b) Find \(g(-2x)\). [1]

\(g(-2x) = 3(-2x) - 2 = -6x - 2\)

6(c) Find \(g^{-1}(x)\). [2]

Let \(y = 3x - 2\). Swap \(x\) and \(y\): \(x = 3y - 2\)

Solve for \(y\): \(3y = x + 2 \implies y = \frac{x+2}{3}\)

\(g^{-1}(x) = \frac{x+2}{3}\)

6(d) Find \(h(1)\). [1]

\(h(1) = 1^2 + 1 = 2\)

6(e) Simplify \(g(x) + h(x)\). [2]

\(g(x) + h(x) = (3x - 2) + (x^2 + 1) = x^2 + 3x - 1\)

6(f) Find \(h(2-x)\). [2]

\(h(2-x) = (2-x)^2 + 1 = (4 - 4x + x^2) + 1 = x^2 - 4x + 5\)

Question 7

By drawing a suitable tangent, estimate the gradient of the curve at point \(P\). [3]

Draw a tangent line at point \(P\). Choose two points on this tangent line and calculate:

Gradient \(m = \frac{y_2 - y_1}{x_2 - x_1}\)

Question 8

Curve: \(y = x^3 + 3x^2 - 13x\)

8(a) Find \(\frac{dy}{dx}\). [2]

\(\frac{dy}{dx} = 3x^2 + 6x - 13\)

8(b) Find the gradient at \(x = 3\). [2]

\[ \left.\frac{dy}{dx}\right|_{x=3} = 3(9) + 6(3) - 13 = 27 + 18 - 13 = 32 \]

Gradient: \(32\)

Question 9

9(a) Sketch the graph of \(y = \frac{1}{x}\). [2]

Reciprocal graph with asymptotes at \(x=0\) and \(y=0\). Two branches in quadrants I and III.

9(b) Sketch the graph of \(y = 2^{-x}\). [2]

Exponential decay graph. Passes through \((0, 1)\). As \(x \to \infty\), \(y \to 0\). Horizontal asymptote at \(y=0\).

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30‑minute test · SOLUTIONS

📘 SOLUTIONS · 30‑minute test

Inequalities · Derivatives · Gradients · Turning points

1 [4]

(a) Number line shows an open circle at 1 and a closed circle at 4. The interval is highlighted between them.
Inequality: 1 < x ≤ 4.

(b) −5 ≤ 3x + 1 < 10
Subtract 1 from all parts: −6 ≤ 3x < 9
Divide by 3: −2 ≤ x < 3
Integer values: −2, −1, 0, 1, 2.

1(a) 1 < x ≤ 4 1(b) −2, −1, 0, 1, 2

2 [5]

y = 2x³ − 5x² + 3x − 4

(a) dy/dx = 6x² − 10x + 3

(b) At x = 2: 6(4) − 10(2) + 3 = 24 − 20 + 3 = 7

(c) Set dy/dx = 1: 6x² − 10x + 3 = 1 → 6x² − 10x + 2 = 0
Divide 2: 3x² − 5x + 1 = 0
Using quadratic formula: x = [5 ± √(25 − 12)] / 6 = [5 ± √13] / 6
Numerical values: x ≈ (5+3.606)/6 = 1.434, x ≈ (5−3.606)/6 = 0.232
Find corresponding y:
For x = 1.434: y = 2(1.434)³ −5(1.434)² +3(1.434)−4 ≈ 2(2.95) −5(2.056) +4.302−4 = 5.90 −10.28 +0.302 ≈ −4.08
For x = 0.232: y = 2(0.0125) −5(0.0538) +0.696−4 ≈ 0.025 −0.269 +0.696−4 = −3.548
Coordinates (to 3 s.f.): (1.43, −4.08) and (0.232, −3.55)

2(a) 6x²−10x+3 2(b) 7 2(c) (1.43,−4.08) and (0.232,−3.55)

3 [3]

y = 4xᵖ − q x³
dy/dx = 4p·xᵖ⁻¹ − 3q x²

Given dy/dx = 20x⁴ + 6x²
Compare powers:
x⁴ term → p−1 = 4 ⇒ p = 5. Then 4p = 4·5 = 20, matches coefficient.
x² term → −3q = 6 ⇒ q = −2.

Thus p = 5, q = −2.

p = 5 , q = −2

4 [6]

y = x³ − 6x² + 5x

(a) dy/dx = 3x² − 12x + 5
Set dy/dx = 0: 3x² − 12x + 5 = 0
Quadratic formula: x = [12 ± √(144 − 60)] / 6 = [12 ± √84] / 6 = [12 ± 2√21] / 6 = 2 ± (√21)/3
√21 ≈ 4.583, so (√21)/3 ≈ 1.528
x₁ = 2 + 1.528 = 3.528, x₂ = 2 − 1.528 = 0.472

Find y‑coordinates:
y(3.528) = (3.528)³ −6(3.528)² +5(3.528)
≈ 43.91 −6(12.45) +17.64 = 43.91 −74.70 +17.64 = −13.15
y(0.472) = (0.472)³ −6(0.472)² +5(0.472)
≈ 0.105 −6(0.223) +2.36 = 0.105 −1.338 +2.36 = 1.127

Turning points: (3.53, −13.2) and (0.472, 1.13) (3 s.f.)

(b) Second derivative: d²y/dx² = 6x − 12
At x = 3.53: 6(3.53)−12 = 21.18−12 = 9.18 > 0 → minimum
At x = 0.472: 6(0.472)−12 = 2.83−12 = −9.17 < 0 → maximum

4(a) (3.53,−13.2) and (0.472,1.13) 4(b) (3.53,−13.2) min, (0.472,1.13) max