Mathematics 2005 Jamb Examination

Write the quadratic in standard form. Since \(f(x)= -x^2 + x + 2\) is a downward-opening parabola (coefficient of \(x^2\) is negative), it has a maximum at its vertex.

Vertex abscissa: \[x_v = -\dfrac{b}{2a} \]\[= -\dfrac{1}{2(-1)} = \dfrac{1}{2}.\]

Evaluate \(f\) at \(x_v\):

\[f\left(\dfrac12\right) = 2 + \dfrac12 - \left(\dfrac12\right)^2 \]\[= 2 + 0.5 - 0.25 = 2.25 = \dfrac{9}{4}.\]

Therefore the maximum value is \(\dfrac{9}{4}\).

Find intersection points: solve \(x^2 - x + 3 = 3\) :

\[x^2 - x = 0 \Rightarrow x(x-1)=0 \Rightarrow x=0,\;1.\]

Area between curves (upper minus lower) on \([0,1]\):

\[A = \int_0^1 \bigl(3 - (x^2 - x + 3)\bigr)\,dx = \int_0^1 (-x^2 + x)\,dx.\]

Compute integral:

\[\int_0^1 (-x^2 + x)\,dx = \left[-\tfrac{x^3}{3} + \tfrac{x^2}{2}\right]_0^1 = -\tfrac{1}{3} + \tfrac{1}{2} = \tfrac{1}{6}.\]

Therefore the area is \(\tfrac{1}{6}\) square units, which matches option D.

We evaluate the integral:

\[\int \sin 2x\,dx = -\tfrac{1}{2}\cos 2x + C.\]

Now compute between \(\pi\) and \(\tfrac{\pi}{2}\):

\[\Bigl(-\tfrac{1}{2}\cos 2x\Bigr)\Big|_{\pi}^{\tfrac{\pi}{2}} = -\tfrac{1}{2}\Bigl(\cos(\pi) - \cos(2\pi)\Bigr).\]

\[= -\tfrac{1}{2}((-1) - 1) = -\tfrac{1}{2}(-2) = 1.\]

Therefore the correct answer is \(1\).

From the histogram: count all bars strictly to the right of the N700 mark.
Workers who earn above #700 = 4 + 6 + 2 = 12

Angle of Excellent
= 360 - (120+80+90)
= 360 - 290
= 70^o
If 360^o represents 36 students
1^o will represent ³⁶/₃₆₀
50^o will represent ³⁶/₃₆₀ * ⁷⁰/₁
= 7

Total students = \(1+4+5+6+x+2 = 18 + x\). Sum of scores = \(1\cdot1 + 2\cdot4 + 3\cdot5 + 4\cdot6 + 5\cdot x + 6\cdot2\).

Compute the numeric part: \(1+8+15+24+12 = 60\). So total sum = \(60 + 5x\).

 Average = \(\dfrac{\text{sum}}{\text{number}} = \dfrac{60+5x}{18+x} = 3.5 = \dfrac{7}{2}.\)

Solve: \[2(60+5x)=7(18+x) \Rightarrow 120+10x\]

\[= 126 +7x \Rightarrow 3x = 6 \Rightarrow x = 2.\]

Answer: \(x=2\).

1.35, 1.25, 1.35, 1.40, 1.35, 1.50, 1.35, 1.50, and 1.20
Modal height = 1.35
Range (r) = 1.50 - 1.20
= 0.30
∴m + 2r = 1.35 + 2(0.30)
= 1.35 + 0.60
= 1.95

\[mean(x) = \dfrac{20t}{5} = 4t\]

\[standard\hspace{1mm}deviation=\sqrt{\dfrac{10t^2}{5}}\]

\[\sqrt{2}=\sqrt{\dfrac{10t^2}{5}}\\\sqrt{2}\]

\[=\sqrt{2t^2}\\2 = 2t^2\\t^2=1\\t=1\]

Step 1 — model the problem. We have 6 objects where 2 are identical and the other 4 are distinct. The number of distinct permutations is:

\[\dfrac{6!}{2!} = \dfrac{720}{2} = 360.\]

The division by \(2!\) accounts for swapping the two identical chalks which does not create a new arrangement.

Answer: 360 (option D).

For circular arrangements of \(n\) distinct people, number of distinct seatings (up to rotation) is \((n-1)!\). Here \(n=7\), so:

\[(7-1)! = 6! = 720.\]

We regard two seatings that can be rotated to match as the same arrangement (standard circular-counting convention).

Answer: 720 (option B).