Mathematics 2019 Jamb Examination

To find the second derivative, differentiate twice: 1. First derivative \( \displaystyle \frac{dy}{dx} = 24x^2 - 6x + 7 \). 2. Second derivative \( \displaystyle \frac{d^2y}{dx^2} = 48x - 6 \).

Using the quotient rule \( \displaystyle \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \) where \( \displaystyle u = 2x \) and \( \displaystyle v = \sin x \): \[ \frac{(\sin x)(2) - (2x)(\cos x)}{\sin^2 x} = \frac{2\sin x - 2x\cos x}{\sin^2 x} \] \[ = \frac{2\sin x}{\sin^2 x} - \frac{2x\cos x}{\sin^2 x} = 2\csc x - 2x\csc x \cot x \] Factoring out \( \displaystyle 2\csc x \): \( \displaystyle 2\csc x(1 - x\cot x) \).

Using the distance formula \( \displaystyle d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \): \[ d = \sqrt{(5 - 2)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} \] \[ d = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ units} \]

First, find the gradient of the given line: \( \displaystyle y = \frac{7}{4}x + \frac{3}{4} \), so \( \displaystyle m_1 = \frac{7}{4} \). The perpendicular gradient \( \displaystyle m_2 = -\frac{4}{7} \). Using \( \displaystyle y - y_1 = m(x - x_1) \): \[ y - 1 = -\frac{4}{7}(x + 3) \implies 7y - 7 = -4x - 12 \] \[ 7y + 4x + 5 = 0 \]

Mean \( \displaystyle \bar{x} = \frac{\sum fx}{\sum f} = 3 \). Sum of frequencies (\( \displaystyle \sum f \)): \( (2y-2) + (y-1) + (3y-4) + (3-y) + (6-2y) = 3y + 2 \). Sum of \( \displaystyle fx \): \( 1(2y-2) + 2(y-1) + 3(3y-4) + 4(3-y) + 5(6-2y) \) \( = 2y-2 + 2y-2 + 9y-12 + 12-4y + 30-10y = -y + 26 \). Set up equation: \( \displaystyle \frac{-y + 26}{3y + 2} = 3 \implies -y + 26 = 9y + 6 \) \( 20 = 10y \implies y = 2 \).

Number of students that scored above 40 = 55 + 45 + 30 + 15 + 5

= 150 students.

The formula for arc length is \(\displaystyle L = \frac{\theta}{360} \times 2\pi r \): \[ L = \frac{57}{360} \times 2 \times \frac{22}{7} \times 7 \] \[ L = \frac{57}{360} \times 44 \approx 6.966 \approx 6.97\text{ cm} \]

|AC| = |DF| = 13 cm

Using Pythagoras theorem,

|AC|\(^2\) = |AB|\(^2\) + |BC|\(^2\)

13\(^2\) = 12\(^2\) + |BC|\(^2\)

|BC|\(^2\) = 169 - 144 = 25

|BC| = \(\sqrt{25}\)

= 5 cm

Volume of triangular prism = \(\frac{1}{2} \times base \times length \times height\)

= \(\frac{1}{2} \times 5 \times 12 \times 18\)

= 540 cm\(^3\)

Simplify the expression using identities: - \( 1 + \cot^2 x = \csc^2 x \) - \( \csc^2 x - 1 = \cot^2 x \) So, the expression is \( \frac{\csc^2 x}{\cot^2 x} = \frac{1/\sin^2 x}{\cos^2 x/\sin^2 x} = \frac{1}{\cos^2 x} = \sec^2 x \). If \( \sin x = \frac{4}{5} \), using a 3-4-5 triangle, \( \cos x = \frac{3}{5} \). \[ \sec^2 x = \left(\frac{5}{3}\right)^2 = \frac{25}{9} \]

Substitute values: \( \displaystyle \sin 30 = \frac{1}{2}, \tan 60 = \sqrt{3}, \sin 60 = \frac{\sqrt{3}}{2} \): \[ \frac{2(1/2) + 5\sqrt{3}}{\sqrt{3}/2} = \frac{1 + 5\sqrt{3}}{\sqrt{3}/2} = \frac{2(1 + 5\sqrt{3})}{\sqrt{3}} \] \[ = \frac{2 + 10\sqrt{3}}{\sqrt{3}} = \frac{2}{\sqrt{3}} + 10 = \frac{2\sqrt{3}}{3} + 10 \]