Mathematics 2016 Jamb Examination

To find the mean, sum all the terms and divide by the total number of terms (5): \[ \text{Sum} = (2-t) + (4+t) + (3-2t) + (2+t) + (t-1) \] Group the constants and the \( t \) terms: \[ \text{Sum} = (2 + 4 + 3 + 2 - 1) + (-t + t - 2t + t + t) = 10 + 0t = 10 \] \[ \text{Mean} = \dfrac{10}{5} = 2 \]

Perform the multiplication in the numerator and denominator first: \[ \text{Numerator} = 12.02 \times 20.06 = 241.1212 \] \[ \text{Denominator} = 26.04 \times 60.06 = 1563.9624 \] Now, divide the two results: \[ \dfrac{241.1212}{1563.9624} \approx 0.154173... \] Rounding to three significant figures gives \( 0.154 \).

Differentiate each term with respect to \( x \) using the rule \( \frac{d}{dx}(ax^n) = anx^{n-1} \): \[ \dfrac{dy}{dx} = (2 \cdot 3)x^{3-1} + (6 \cdot 2)x^{2-1} + (6 \cdot 1)x^{1-1} + 0 \] \[ \dfrac{dy}{dx} = 6x^2 + 12x + 6 \]

This is a Geometric Progression (G.P.) where the first term \( a = 3 \) and the common ratio \( r = \dfrac{9}{3} = 3 \). The formula for the \( n^{\text{th}} \) term is: \[ T_n = ar^{n-1} \] \[ T_n = 3 \times 3^{n-1} \] Simplified, this is \( 3^1 \times 3^{n-1} = 3^n \). Choice B is the expanded form of this expression.

First, convert the mixed fraction and solve the bracket: \[ 3\dfrac{1}{2} = \dfrac{7}{2} \] \[ \text{Bracket: } \dfrac{1}{7} \times \dfrac{7}{2} = \dfrac{1}{2} \] Now, handle the division followed by subtraction (BODMAS): \[ \dfrac{1}{2} \div \dfrac{3}{4} = \dfrac{1}{2} \times \dfrac{4}{3} = \dfrac{2}{3} \] \[ 1 - \dfrac{2}{3} = \dfrac{1}{3} \]

This is a direct proportion problem. Let \( x \) be the required petrol: \[ \dfrac{120 \text{ km}}{45 \text{ litres}} = \dfrac{600 \text{ km}}{x \text{ litres}} \] Cross-multiply to solve for \( x \): \[ 120x = 600 \times 45 \] \[ x = \dfrac{600 \times 45}{120} = 5 \times 45 = 225 \; \text{litres} \]

Differentiate the polynomial term by term using the power rule: \[ \dfrac{d}{dx}(3x^3) = 9x^2 \] \[ \dfrac{d}{dx}(2x^2) = 4x \] \[ \dfrac{d}{dx}(3x) = 3 \] Summing these gives \( 9x^2 + 4x + 3 \).

Integrate each term separately: \[ \int \sin x dx = -\cos x \] \[ \int -5x^2 dx = -\dfrac{5x^3}{3} \] Combining the results and adding the constant of integration \( k \): \[ -\cos x - \dfrac{5x^3}{3} + k \]

Substitute the given values into the equation: \[ 12 = \dfrac{p}{2} \left( \dfrac{27 - 24}{27} \right) \] \[ 12 = \dfrac{p}{2} \left( \dfrac{3}{27} \right) \] \[ 12 = \dfrac{p}{2} \left( \dfrac{1}{9} \right) = \dfrac{p}{18} \] Multiply both sides by 18: \[ p = 12 \times 18 = 216 \]

Multiply the first equation by 2 to align the coefficients of \( x \): \[ (1) \quad 6x - 10y = 18 \] \[ (2) \quad 6x - 4y = 12 \] Subtract equation (2) from equation (1): \[ -6y = 6 \Rightarrow y = -1 \] Substitute \( y = -1 \) into the first original equation: \[ 3x - 5(-1) = 9 \Rightarrow 3x + 5 = 9 \Rightarrow 3x = 4 \Rightarrow x = \dfrac{4}{3} \]