2019 JAMB MATHEMATICS PAST QUESTIONS AND ANSWERS

To make q the subject, first multiply both sides by $b^2$ to eliminate the denominator. Then isolate terms with q and factor out q. Finally, divide both sides by p to get q alone.

Using tangent ratio in right-angled triangle: $\tan(78°) = \frac{\text{height}}{60}$. Therefore, height = $60 \tan(78°) = 282.4m$. Rounded to nearest whole number = 282m.

First calculate $2 \otimes 4 = (2 × 4) + 2 – 4 = 8 + 2 – 4 = 6$. Then calculate $3 \otimes 6 = (3 × 6) + 3 – 6 = 18 + 3 – 6 = 15$.

Total number of students = 100. For 8 years old: $\frac{13}{100} \times 360° = 46.8°$

Using the formula: $n(P \cup B) = n(P) + n(B) – n(P \cap B)$. Given total students = 50, so: $50 = 40 + 30 – n(P \cap B)$. Therefore $n(P \cap B) = 20$

Multiply numerator and denominator by $\sqrt{2} + \sqrt{3}$ to eliminate surd in denominator. This gives: $\frac{(\sqrt{2} + \sqrt{3})(\sqrt{2} + \sqrt{3})}{(\sqrt{2} – \sqrt{3})(\sqrt{2} + \sqrt{3})} = \frac{5 – 2\sqrt{6}}{1}$

Using the diagram and Pythagoras theorem in the right-angled triangle formed with the radius: If radius = 5cm and perpendicular distance = 4cm, then half chord length = $\sqrt{25-16} = 3$cm. Full chord = 4.2cm

Using the co-function relationship: $\sin \theta = \cos(90° – \theta)$. Therefore, $p = 90° – 58° = 32°$

First solve within parentheses: $\frac{1}{4} + \frac{3}{5} – \frac{1}{3} = \frac{31}{60}$. Then divide: $\frac{2}{3} \times \frac{5}{4} \times \frac{60}{31} = \frac{50}{31}$

Using polynomial long division, the remainder will be in the form ax + b. Substituting values shows remainder = 5x – 3

Total outcomes = 8. Favorable outcomes (HHH,HHT,HTH,THH) = 4. Probability = $\frac{4}{8} = \frac{1}{2}$

Using formula for repeated letters: $\frac{12!}{2!2!2!1!1!1!1!1!1!}$ = 2664910 ways

First find transpose, then add 2M. Result = $\begin{vmatrix} -6 & 0 & 14 \ 0 & -3 & 18 \ 13 & 18 & 9 \end{vmatrix}$

Using formula: $\frac{\sum fx}{\sum f}$ = $\frac{149}{46}$ = 3.26

Using formula: $\frac{\sum fx^2}{\sum f} – (\frac{\sum fx}{\sum f})^2$ = 4.85

The angle bisector of two intersecting lines contains all points equidistant from both lines

In cyclic quadrilateral, opposite angles are supplementary. Using given angles: $x + 141° = 180°$, therefore $x = 39°$

Solving: $4\sin^2 x = 3$, $\sin^2 x = \frac{3}{4}$, $\sin x = \frac{\sqrt{3}}{2}$, therefore $x = 60°$

$$\text{Area of rectangle ABCD} = \text{length} \times \text{breadth}$$ $$= 7 \times 4$$ $$= 28 \text{ cm}^2$$ $$\text{Area of triangle CDE} = \frac{1}{2} \times \text{base} \times \text{height}$$ $$= \frac{1}{2} \times 3 \times 4$$ $$= 6 \text{ cm}^2$$ $$\text{Area of the figure} = 28 \text{ cm}^2 + 6 \text{ cm}^2$$ $$= 34 \text{ cm}^2$$

Total marks = $\sum fx = (0×4 + 1×3 + 2×7 + 3×8 + 4×6 + 5×2) = 63$

Using point-slope form and gradient = $\frac{22-12}{5-3} = 5$, then y = 5x + 2

Using midpoint formula: $2 = \frac{m+n}{2}$ and $m = \frac{n-4}{2}$. Solving gives m = 2, n = 2

Using determinant formula: $2x(1) – (-4)(9) = 58$. Solving: $2x + 36 = 58$, $x = 14$

**Explanation**:

Given \( y = 6x^3 + 2x^{-2} – x^{-3} \), we differentiate each term with respect to \( x \):

\[
\frac{dy}{dx} = \frac{d}{dx}(6x^3) + \frac{d}{dx}(2x^{-2}) – \frac{d}{dx}(x^{-3}).
\]

1. For \( \frac{d}{dx}(6x^3) \):
\[
\frac{d}{dx}(6x^3) = 18x^2.
\]

2. For \( \frac{d}{dx}(2x^{-2}) \):
\[
\frac{d}{dx}(2x^{-2}) = -4x^{-3}.
\]

3. For \( \frac{d}{dx}(-x^{-3}) \):
\[
\frac{d}{dx}(-x^{-3}) = 3x^{-4}.
\]

Combine these results:

\[
\frac{dy}{dx} = 18x^2 – 4x^{-3} + 3x^{-4}.
\]

**Explanation**:

\[
\frac{d}{dx} \left[ \log(4x^3 – 2x) \right]
\tag{1}
\]

Let \( u = 4x^3 – 2x \).

\[
\frac{d}{dx} \left[ \log(4x^3 – 2x) \right] = \frac{d}{du} (\log u) \cdot \frac{du}{dx}
\]

Now,

\[
\frac{d}{du} (\log u) = \frac{1}{u}
\]

and

\[
\frac{du}{dx} = 12x^2 – 2
\]

Thus,

\[
\frac{d}{dx} \left[ \log(4x^3 – 2x) \right] = \frac{12x^2 – 2}{u}
\]

Substitute \( u = 4x^3 – 2x \):

\[
\frac{d}{dx} \left[ \log(4x^3 – 2x) \right] = \frac{12x^2 – 2}{4x^3 – 2x}.
\]

Substitute x = -2: $3(-2)^3 + 4(-2)^2 + (-2) – 8 = -24 + 16 – 2 – 8 = -18$

Cross multiply and solve: $2(4x-6) \leq 3(3+2x)$. Simplify to get $x \leq \frac{21}{2}$

Solve both parts: -7 ≤ 9 – 8x gives x ≤ 2; 9 – 8x < 16 - x gives x > -1

Sum terms where n = 1,2,3,4: $2^1 + 2^3 + 2^5 + 2^7 = 2 + 8 + 32 + 128 = 170$

Using power rule: $[\frac{2x^3}{3} + \frac{x^2}{2}]_{-1}^2$ = $\frac{7}{2}$

Using $P = \frac{k}{\sqrt{q}}$, find k using given values, then solve for q when P = 4. q = 9

Cost = ₦125, Revenue from 170 mangoes = ₦204. Profit = ₦79. Percentage profit = 63.2%

Using SI formula: $I = \frac{PRT}{100}$. $4890 = \frac{8550 × 3 × x}{100}$. Solving gives x = 19%

Convert all to prime bases and simplify: $3^{-6} × 5 × 3^{\frac{10}{5}} = \frac{5}{3}$

Evaluate numerator and denominator separately, then divide: 431

Using product rule: $\frac{ds}{dt} = 4(t-2) + (4t+3)(1)$. At t = 5: 12 + 23 = 35

Sum of angles in quadrilateral = 360°. So 12x = 360°, x = 30°

Using formula: $\frac{\text{measured – actual}}{\text{actual}} × 100% = \frac{0.03}{0.18} × 100% = 16.7%$

Convert to standard form and evaluate: 3.2

Convert to simultaneous equations: x + y = 4 and 2(x-y) = -5. Solve to get x = $\frac{3}{4}$, y = $\frac{13}{4}$

Multiply numerator first, then divide by denominator: 0.098

Substitute y = 21 – 3x into xy = 30: 3x² – 21x + 30 = 0. Solve to get x = 5 or 2, y = 6 or 15

Using trigonometry in triangle: $\tan \theta = \frac{9}{20}$. θ = 156°

Solution:

We are given:

\[
P = \left( \frac{Q(R – T)}{15} \right)^{\frac{1}{3}}
\]

Step 1: Cube both sides to remove the cube root:

\[
P^3 = \frac{Q(R – T)}{15}
\]

Step 2: Multiply through by 15 to eliminate the denominator:

\[
15P^3 = Q(R – T)
\]

Step 3: Divide through by \( Q \) to isolate \( (R – T) \):

\[
R – T = \frac{15P^3}{Q}
\]

Step 4: Rearrange to solve for \( T \):

\[
T = R – \frac{15P^3}{Q}
\]

Final Answer:
\[
T = R – \frac{15P^3}{Q}
\]

This corresponds to option B.

< BAC = 13021302 (angle subtended at the centre)

< BAC = 65°

Also, x = 26° (theorem)

y = 65° – 26° = 39°

< AOC = 180° – (39° + 39°)

= 102°

For interior angle 140°, polygon has 9 sides. Sum = (n-2) × 180° = 1260°

Loss = ₦165,000. Percentage loss = $\frac{165000}{1250000} × 100% = 13.2%$

Solution:

We are given:

\[
V = \frac{\pi h}{3} \left( R^2 + Rr + r^2 \right).
\]

Step 1: Differentiate \( V \) with respect to \( R \):

\[
\frac{dV}{dR} = \frac{\pi h}{3} \cdot \frac{d}{dR} \left( R^2 + Rr + r^2 \right).
\]

Step 2: Differentiate each term inside the parentheses:

– The derivative of \( R^2 \) is \( 2R \),
– The derivative of \( Rr \) is \( r \) (since \( r \) is a constant with respect to \( R \)),
– The derivative of \( r^2 \) is \( 0 \) (since \( r^2 \) is independent of \( R \)).

So:

\[
\frac{d}{dR} \left( R^2 + Rr + r^2 \right) = 2R + r.
\]

Step 3: Multiply by the constant factor \(\frac{\pi h}{3}\):

\[
\frac{dV}{dR} = \frac{\pi h}{3} (2R + r).
\]

—
Final Answer:
\[
\frac{dV}{dR} = \frac{\pi h}{3} (2R + r)
\]

This corresponds to option A.

Convert decimals to fractions and divide: $\frac{12}{1000}$

Convert all to base 5 and equate powers: x = -2