**Exam Body:**

Differentiate \((2x+5)^2(x-4)\) with respect to x.

A) \(4(2x+5)(x-4)\)

B) \(4(2x+5)(4x-3)\)

C) \((2x+5)(2x-13)\)

D) \((2x+5)(6x-11)\)

Exam Body: JAMB

To differentiate \((2x+5)^2(x-4),\) you first need to know that it is a product function.

Using the product rule you have \(\frac{dy}{dx}=\frac{udv}{dx}+\frac{vdu}{dx}\)

Let \((2x+5)^2\) be \(u\) and \((x-4)\) be \(v.\)

To find \(du\) we use chain rule which is \(\frac{du}{dx}=\frac{du}{dw} \times \frac{dw}{dx}\)

We say let \((2x+5)\) be \(w\)

We then have a new function \(u=w^2\)

\(\frac{du}{dw}=2w\)

\(\frac{dw}{dx}=2\)

So \(\frac{du}{dx}=2 \times 2w\) which equals \(4w\) and \(w\) was \((2x+5)\)

\(\frac{du}{dx}=4(2x+5)\)

\(\frac{dv}{dx}=1\)

Substituting everything into the product rule we have:

\(\frac{dy}{dx}=(2x+5)^2(1) + 4(2x+5)(x-4)\)

\(\frac{dy}{dx}=(4x^2 + 20x + 25) + (8x^2 - 12x - 80)\)

\(\frac{dy}{dx}=12x^2 + 8x - 55\)

\(\frac{dy}{dx}=(2x+5)(6x-11)\)

Explanation provided by Nigerian Scholars

Find the area bounded by the curves \(y = 4 - x^2\) and \(y = 2x + 1\)

A) \(20\frac{1}{3}\) sq. units

B) \(20\frac{2}{3}\) sq. units

C) \(10\frac{2}{3}\) sq. units

D) \(10\frac{1}{3}\) sq. units

Exam Body: JAMB

\(y = 4 - x^2\) and \(y = 2x + 1\)

\(\Rightarrow 4 - x^2 = 2x + 1\)

\(\Rightarrow x^2 + 2x - 3 = 0\)

\((x+3)(x-1) = 0\)

Thus, \(x = -3\) or \(x = 1.\)

Integrating \(x^2 + 2x - 3\) from \((-3\) to \(1)\) with respect to \(x\) will give \(\frac{31}{3} = 10\frac{1}{3}\)

Explanation provided by Nigerian Scholars

Find the rate of change of the volume, \(V\) of a sphere with respect to its radius, \(r\) when \(r = 1.\)

A) \(12\pi\)

B) \(4\pi\)

C) \(24\pi\)

D) \(8\pi\)

Exam Body: JAMB

Volume of sphere, \(V = \frac{4}{3} \times \pi r^3\)

Rate of change of \(V = \frac{dv}{dr}\)

Thus if, \(V = \frac{4}{3} \times \pi r^3,\)

\(\Rightarrow \frac{dv}{dr} = 4\pi^2\)

At \(r = 1,\) Rate \(= 4 \times \pi \times 1 = 4\pi\)

Explanation provided by Nigerian Scholars

If \(y = x \; \mathrm{sin} \; x,\) find \(\frac{dy}{dx}\) when \(x = \frac{\pi}{2}.\)

A) \(\frac{- \pi}{2}\)

B) \(-1\)

C) \(1\)

D) \(\frac{\pi}{2}\)

Exam Body: JAMB

\(y = x \; \mathrm{sin} \; x\)

\(\frac{dy}{dx} = 1 \times \mathrm{sin} \; x + x \; \mathrm{cos} \; x\)

\(= \mathrm{sin} \; x + x \; \mathrm{cos} \; x\)

At \(x = \frac{\pi}{2},\) \(= \mathrm{sin} \; \frac{\pi}{2} + \frac{\pi}{2} \; \mathrm{cos} \frac{π}{2}\)

\(= 1 + \frac{\pi}{2} \times (0) = 1\)

Explanation provided by Nigerian Scholars

Find the dimensions of a rectangle of greatest area which has a fixed perimeter \(p.\)

A) square of sides \(p\)

B) square of sides \(2p\)

C) square of sides \(\frac{p}{2}\)

D) square of sides \(\frac{p}{4}\)

Exam Body: JAMB

Let the rectangle be a square of sides \(\frac{p}{4}.\)

So that perimeter of square \(= 4p\)

\(4 \times \frac{p}{4} = p.\)

Explanation provided by Nigerian Scholars

Given the scores: \(4, 7, 8, 11, 13, 8\) with corresponding frequencies: \(3, 5, 2, 7, 2, 1\) respectively. Find the square of the mode.

A) \(49\)

B) \(121\)

C) \(25\)

D) \(64\)

Exam Body: JAMB

Mode = score with highest frequency \(= 11.\)

Square of \(11 = 121\)

Explanation provided by Nigerian Scholars

Teams \(P\) and \(Q\) are involved in a game of football. What is the probability that the game ends in a draw?

A) \(\frac{2}{3}\)

B) \(\frac{1}{2}\)

C) \(\frac{1}{3}\)

D) \(\frac{1}{4}\)

Exam Body: JAMB

\(P\) (games end in draw)

\(\Rightarrow\) Team \(P\) wins and \(Q\) wins

\(P\) (\(P\) wins) \(= \frac{1}{2}\)

\(P\) (\(Q\) wins) \(= \frac{1}{2}\)

Therefore, \(P\) (games ends in draw) \(= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)

Explanation provided by Nigerian Scholars

Given the scores: \(4, 7, 8, 11, 13, 8\) with corresponding frequencies: \(3, 5, 2, 7, 2, 1\) respectively. The mean score is

A) \(7.0\)

B) \(8.7\)

C) \(9.5\)

D) \(11.0\)

Exam Body: JAMB

Mean \(= \frac{\sum fx}{\sum f}\)

Mean \(= \frac{(4 \times 3) + (7 \times 5) + ... + (8 \times 1)}{20}\)

Mean \(= \frac{174}{20} = 8.7\)

Explanation provided by Nigerian Scholars

If \(^6\mathrm{P}_r = 6,\) find the value of \(^6\mathrm{P}_{r + 1}\)

A) \(30\)

B) \(33\)

C) \(35\)

D) \(15\)

Exam Body: JAMB

\(^6\mathrm{P}_r = 6\) Thus \(r = 1\)

N.B: \(^6\mathrm{P}_1 = \frac{6!}{(6 - 1)!}\)

\(= \frac{6!}{5!}\) \(= \frac{6 \times 5!}{5!} = 6\)

\(^6\mathrm{P}_{r + 1} = ^6\mathrm{P}_2\) \(= \frac{6!}{(6 - 2)!} = \frac{6!}{4!}\)

\(= \frac{6 \times 5 \times 4!}{4!} = 30\)

Explanation provided by Nigerian Scholars

Given distribution of color beads: blue, black, yellow, white and brown with frequencies \(1, 2, 3, 4,\) and \(5\) respectively. Find the probability that a bead picked at random will be blue or white.

A) \(\frac{7}{15}\)

B) \(\frac{2}{5}\)

C) \(\frac{1}{3}\)

D) \(\frac{1}{15}\)

Exam Body: JAMB

Total number of beads \(= 15.\)

Number of white beads \(= 4. \Rightarrow \mathrm{P}\)(white) \(= \frac{4}{15}.\)

Number of blue beads \(= 1. \Rightarrow \mathrm{P}\)(blue) \(= \frac{1}{15}.\)

\(\mathrm{P}\)(white or blue) \(= \mathrm{P}\)(white) \(+ \mathrm{P}\)(blue) \(= \frac{5}{15} = \frac{1}{3}\)

Explanation provided by Nigerian Scholars

Find the variance of \(2, 6, 8, 6, 2,\) and \(6.\)

A) \(6\)

B) \(5\)

C) \(\sqrt{6}\)

D) \(\sqrt{5}\)

Exam Body: JAMB

The mean \(\bar{x} = \frac{\sum x}{n}\) \(= \frac{30}{6} = 5\)

Variance \(= \frac{\sum (x - \bar{x})^2}{n}\) \(= \frac{30}{6} = 5\)

Explanation provided by Nigerian Scholars

Find the number of ways of selecting \(8\) subjects from \(12\) subjects for an examination.

A) \(490\)

B) \(495\)

C) \(496\)

D) \(498\)

Exam Body: JAMB

Combination is the \((n)\) number of ways of selecting a number of \((r)\) subjects from \(n.\)

\(^n\mathrm{C}_r = \frac{n!}{r!(n-r)!}\) \(= \frac{12!}{8! \times 4!} = 495\)

Explanation provided by Nigerian Scholars

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