2010 JAMB MATHEMATICS PAST QUESTIONS AND ANSWERS

To solve this, first calculate the binomial coefficient $\binom{81}{16}$, then raise it to -1/4 power, and multiply by $2^{-1}$. The binomial coefficient simplifies, and when raised to -1/4 and multiplied by 1/2, gives 1/6.

To rationalize, multiply both numerator and denominator by $\sqrt{5}+\sqrt{3}$. After simplification and combining like terms, we get $(3\sqrt{15}-11)/2$.

Multiply 0.21 × 0.34 = 0.0714. In standard form, this is $7.14 \times 10^{-2}$.

Using set theory, if x is the number reading both: 40 + 30 – x = 50 (total readers minus overlap). Solving: 70 – x = 50, therefore x = 20 people read both papers.

Factor numerator: $x(x^2+3x-10)$ = $x(x+5)(x-2)$. Factor denominator: $2(x^2-4)$ = $2(x+2)(x-2)$. After cancellation, get $\frac{x(x+5)}{2(x+2)}$.

Using direct variation: $y = k\sqrt{x}$. When x = 16, 3 = k\sqrt{16}$, so k = 3/4. When x = 64, y = (3/4)\sqrt{64} = 6.

First calculate 23 using the given operation: 23 = 2 + $3^2$ = 11. Then 11*5 = 11 + $5^2$ = 36.

Expand: 18p + 18q = 18q + pq. Cancel 18q from both sides: 18p = pq. Therefore: p(q-18) = 0. Since p≠0, q must equal 18.

For a 2×2 determinant, $\begin{vmatrix}x & 2 \ 3 & 7\end{vmatrix}$ = 7x – 6 = 15. Solving: 7x = 21, x = 3.

In a circle, angles in the same segment are equal. Additionally, angles in a cyclic quadrilateral sum to 360°. Using these properties, x = 65°.

In a cyclic quadrilateral, opposite angles are supplementary (sum to 180°). Using this property and the given angles, x = 26°.

Area of a triangle = $\frac{1}{2} × base × height$. Using the given sides: Area = $\frac{1}{2} × 4 × 5 = 10$ square cm.

Area of sector = $\frac{\theta}{360°} × \pi r^2$ = $\frac{50}{360} × \pi × 36$ = $\frac{100\pi}{7}$ $\text{cm}^2$

The locus is the perpendicular bisector of line PQ. Using the midpoint formula and perpendicular slope, we get y = -x + 6.

Using the area formula $\frac{1}{2} × base × height$ and the given 30° angle, solve: $12\sqrt{3} = \frac{1}{2} × q × q\sin(30°)$. Therefore q = 8 cm.

Using chain rule: $\frac{dy}{dx} = 3(2x+1)^2 × \frac{d}{dx}(2x+1) = 6(2x+1)^2$

Integrate term by term: ∫sin x dx = -cos x, ∫2 dx = 2x. Therefore, ∫(sin x + 2)dx = -cos x + 2x + C

Count students scoring below 5: 3 + 1 + 5 = 9 students failed.

Sum all frequencies: 2 + 2 + 8 + 4 + 4 = 20 students.

Calculate $\frac{\sum fx}{\sum f}$ = $\frac{(1×2)+(2×2)+(3×8)+(4×4)+(5×4)}{20}$ = 3.2

Raise both sides to power 72 (LCM of 8 and 9), simplify, and solve for r to get r = 3.

Convert to improper fractions and multiply: $(\frac{3}{4} × \frac{4}{9} \div \frac{19}{2}) \div \frac{24}{19}$ = $\frac{1}{4}$

Percentage error = $\frac{\text{measured – actual}}{\text{actual}} × 100%$ = $\frac{1.26-1.25}{1.25} × 100%$ = 0.8%

Using Simple Interest formula: I = PRT/100, 24 = 400 × R × 3/100. Solve for R to get 2%.

Convert to common denominator and simplify ratios to get 12 : 15 : 10.

Find common difference d using $a_7 – a_3 = 4d$, then use $a_{10} = a_3 + 7d$ to get -44.

Find derivative $\frac{dy}{dx} = -2 + 2x$, set equal to 0 and solve to get x = 1.

Parallel lines have same slope (-4). Use point-slope form: y – 3 = -4(x – 2) to get y – 4x – 11 = 0.

Multiply both sides by 5, subtract 5, subtract MX, divide by M to get $Q = \frac{5P-MX-5}{M}$

Divide $27x^3 – 8y^3$ by $9x^2 + 6xy + 4y^2$ to get 3x – 2y.

Factor $x^2 – y^2 = (x+y)(x-y)$. Substitute x-y = 2 to get x+y = 4. Solve system to get (3,1).

Use x = k/y. Find k using given values, then solve for new x to get $1\frac{1}{4}$.

Solve inequality: $\frac{1}{2}x – \frac{1}{3}x > \frac{1}{2} – \frac{1}{4}$, simplify to get x < $\frac{3}{2}$.

Solve both parts: -6 ≤ 4 – 2x and 4 – 2x < 5 - x. Combine to get -1 ≤ x < 6.

To find the sum to infinity of the given series, consider the series:

\[

0.5 + 0.05 + 0.005 + 0.0005 + \dots

\]

 Step 1: Identify the first term and common ratio

The series is a **geometric series**, where:

– The first term \( a = 0.5 \)

– The common ratio \( r \) is:

\[

r = \frac{\text{second term}}{\text{first term}} = \frac{0.05}{0.5} = 0.1

\]

 Step 2: Formula for the sum to infinity of a geometric series

For a geometric series with \( |r| < 1 \), the sum to infinity is given by:

\[

S_{\infty} = \frac{a}{1 – r}

\]

where \( a \) is the first term and \( r \) is the common ratio.

 Step 3: Calculate the sum to infinity

Substitute the values of \( a \) and \( r \) into the formula:

\[

S_{\infty} = \frac{0.5}{1 – 0.1} = \frac{0.5}{0.9}

\]

Simplify the fraction:

\[

S_{\infty} = \frac{5}{9}

\]

 Step 4: Conclusion

Thus, the sum to infinity of the series is \( \frac{5}{9} \).

The correct answer is Option D.

Use the 3×3 determinant formula: $2(0×1 – 9×3) – 8(4×1 – 9×5) + 6(4×3 – 0×5)$ = 102

To find the inverse of matrix \( P \), we will use the formula for the inverse of a 2×2 matrix:

\[

P^{-1} = \frac{1}{\text{det}(P)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

\]

where the matrix \( P = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \).

 Step 1: Identify the elements of \( P \)

From the matrix \( P = \begin{pmatrix} 2 & 1 \\ -3 & 1 \end{pmatrix} \), we have:

– \( a = 2 \)

– \( b = 1 \)

– \( c = -3 \)

– \( d = 1 \)

 Step 2: Calculate the determinant of \( P \)

The determinant of \( P \) is given by:

\[

\text{det}(P) = ad – bc = 2(1) – 1(-3) = 2 + 3 = 5

\]

 Step 3: Apply the formula for the inverse

Substitute the values into the formula for \( P^{-1} \):

\[

P^{-1} = \frac{1}{5} \begin{pmatrix} 1 & -1 \\ 3 & 2 \end{pmatrix}

\]

Thus, the inverse of \( P \) is:

\[

P^{-1} = \begin{pmatrix} \frac{1}{5} & -\frac{1}{5} \\ \frac{3}{5} & \frac{2}{5} \end{pmatrix}

\]

 Step 4: Final Answer

So, the inverse of \( P \) is:

\[

\boxed{\begin{pmatrix} \frac{1}{5} & -\frac{1}{5} \\ \frac{3}{5} & \frac{2}{5} \end{pmatrix}}

\]

 Option List:

1. \( \begin{pmatrix} -\frac{1}{5} & -\frac{1}{5} \\ -\frac{3}{5} & -\frac{2}{5} \end{pmatrix} \)  
2. \( \begin{pmatrix} \frac{1}{5} & \frac{1}{5} \\ \frac{3}{5} & \frac{2}{5} \end{pmatrix} \)  
3. \( \begin{pmatrix} -\frac{1}{5} & -\frac{1}{5} \\ \frac{3}{5} & \frac{2}{5} \end{pmatrix} \)  
4. \( \begin{pmatrix} \frac{1}{5} & -\frac{1}{5} \\ \frac{3}{5} & \frac{2}{5} \end{pmatrix} \)  

Thus, the correct answer is D.

Sum of interior angles = 360°. Solve equation: (x + 15) + (2x – 45) + (x – 30) + (x + 10) = 360. Find x and substitute to find smallest angle.

Surface area = lateral area + area of one end = $2\pi r h + \pi r^2$ = $2\pi(7)(50) + \pi(7^2)$ = $700\pi + 49\pi$ = $749\pi$ $\text{m}^2$

Use distance formula: $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ = $\sqrt{(-\frac{1}{2}-\frac{1}{2})^2 + (-\frac{1}{2}-\frac{1}{2})^2}$ = $\sqrt{2}$