Mathematics 2007 Jamb Examination

Use the product rule: if y = x\cos x then \(\dfrac{dy}{dx} = 1\cdot\cos x + x\cdot(-\sin x) = \cos x - x\sin x\).

Compute stationary points: \(f'(x)=6x^2-2x-4\).
Solve \(6x^2-2x-4=0\) divide by 2
We have \(3x^2-x-2=0\).
Roots: $$x=\dfrac{1\pm\sqrt{1+24}}{6}=\dfrac{1\pm5}{6}$$ Giving \(x=1\) and \(x=-\tfrac{2}{3}\).
Second derivative $$f''(x)=12x-2$$ At \(x=1\), \(f''(1)=10>0\) → local minimum.
At \(x=-\tfrac{2}{3}\), \(f''(-\tfrac{2}{3})=-10<0\) → local maximum.
So the function has its maximum at \(x=-\tfrac{2}{3}\).

$$ \int_{0}^{\dfrac{\pi}{2}} -\left(2\cos x\right)\, dx $$ Factor out the $-2$: $$ = -2 \int_{0}^{\dfrac{\pi}{2}} \cos x \, dx $$ Integrate $\cos x$: $$ \int \cos x \, dx = \sin x $$ Apply bounds: $$ -2 \big[\sin x \big]_{0}^{\pi/2} = -2 \left( \sin\left(\tfrac{\pi}{2}\right) - \sin(0) \right) $$ $$ = -2 (1 - 0) = -2 $$

Compute $(-\tfrac{3}{4})\oplus 6 $
$= (-\tfrac{3}{4})\cdot 6 + (-\tfrac{3}{4}) + 6 $
$= -\tfrac{18}{4} -\tfrac{3}{4} + 6 $
$= -\tfrac{21}{4} + \tfrac{24}{4} = \tfrac{3}{4}$.

From the graph, the x-intercepts are $x=-2, -1,$ and $1$.
The cubic with these roots is $(x+2)(x+1)(x-1)=x^3+2x^2-x-2$.
This matches option B exactly, with the correct y-intercept at $y=-2$.

Answer: B. $\dfrac{42w}{5d^2}$. Step 1 — square both sides to remove the square root: \[ d^2 = \dfrac{42w}{5l}.\] Step 2 — multiply both sides by $5l$ to isolate $l$ in the denominator: \[5l d^2 = 42w.\] Step 3 — solve for $l$ by dividing both sides by $5d^2$: \[ l = \dfrac{42w}{5d^2}.\] So $l=\dfrac{42w}{5d^2}$ (option B).

Answer: D. $x\ge 3$ or $x\le -4$. Step 1 — factor the quadratic: \[x^2+x-12=(x+4)(x-3).\] Step 2 — find the roots: $x=-4$ and $x=3$. The parabola $y=x^2+x-12$ opens upward (coefficient of $x^2$ is positive), so the quadratic is \(\ge 0\) outside the interval between the roots. Step 3 — therefore the solution set is \[x\le -4\quad\text{or}\quad x\ge 3.\] This matches option D.

Answer: B. $(t+3)(2t-5)$. Step 1 — multiply $2\times(-15)=-30$. We seek two integers whose product is $-30$ and whose sum is the middle coefficient $+1$. Step 2 — $6$ and $-5$ work because $6\times(-5)=-30$ and $6+(-5)=1$. Step 3 — split the middle term and factor by grouping: \[2t^2+6t-5t-15=2t(t+3)-5(t+3)=(t+3)(2t-5).\] So the factorization is $(t+3)(2t-5)$ (option B).

Answer: A. $x>12$. Step 1 — expand both sides: \[-3x+6 < -2x-6.\] Step 2 — collect like terms. Add $3x$ to both sides: \[6 < x - 6.\] Step 3 — add $6$ to both sides: \[12 < x\quad\text{or}\quad x>12.\] So the solution is $x>12$ (option A).

Write the proportionality as an equation: $W=kL^2$ for some constant $k$.
Use $W=6$ at $L=4$ to find $k$: \[6 = k(4^2)=16k\Rightarrow k=\dfrac{6}{16}=\dfrac{3}{8}.\] Compute W at $L=\sqrt{17}$: \[W = \dfrac{3}{8}\cdot(\sqrt{17})^2 = \dfrac{3}{8}\cdot17 = \dfrac{51}{8}.\]