2026 WAEC Further mathematics like questions and Answers

QUESTION

$$ \text{10a. The gradient of a tangent to the curve } y = 4x^3 \text{ at points } P \text{ and } Q \text{ is } 108. \text{ Find the coordinates of } P \text{ and } Q. $$ $$ \text{b(i). Given } \hat{A}=45^\circ,\ \hat{B}=30^\circ, $$ $$ \sin(A+B)=\sin A \cos B+\cos A \sin B $$ $$ \text{and} $$ $$ \cos(A+B)=\cos A \cos B-\sin A \sin B $$ $$ \text{Show that} $$ $$ \sin 15^\circ=\frac{\sqrt{6}-\sqrt{2}}{4} $$ $$ \text{and} $$ $$ \cos 15^\circ=\frac{\sqrt{6}+\sqrt{2}}{4} $$ $$ \text{(ii). Hence, find } \tan 15^\circ $$

SOLUTION

$$ \textbf{10a. Given the curve } y = 4x^3 $$ $$ \text{The gradient of the tangent is obtained by differentiation.} $$ $$ \frac{dy}{dx} = \frac{d}{dx}(4x^3) $$ $$ \frac{dy}{dx} = 12x^2 $$ $$ \text{Since the gradient is } 108, $$ $$ 12x^2 = 108 $$ $$ x^2 = \frac{108}{12} $$ $$ x^2 = 9 $$ $$ x = \pm 3 $$ $$ \text{When } x = 3: $$ $$ y = 4(3)^3 $$ $$ y = 4(27) $$ $$ y = 108 $$ $$ P = (3,108) $$ $$ \text{When } x = -3: $$ $$ y = 4(-3)^3 $$ $$ y = 4(-27) $$ $$ y = -108 $$ $$ Q = (-3,-108) $$ $$ \boxed{P=(3,108)\quad \text{and} \quad Q=(-3,-108)} $$

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$$ \textbf{b(i). Given } A=45^\circ,\ B=30^\circ $$ $$ \sin(A-B)=\sin A \cos B-\cos A \sin B $$ $$ \text{Since } 15^\circ = 45^\circ - 30^\circ, $$ $$ \sin 15^\circ = \sin(45^\circ-30^\circ) $$ $$ = \sin45^\circ \cos30^\circ - \cos45^\circ \sin30^\circ $$ $$ = \left(\frac{\sqrt2}{2}\right)\left(\frac{\sqrt3}{2}\right) - \left(\frac{\sqrt2}{2}\right)\left(\frac12\right) $$ $$ = \frac{\sqrt6}{4} - \frac{\sqrt2}{4} $$ $$ = \frac{\sqrt6-\sqrt2}{4} $$ $$ \boxed{\sin15^\circ=\frac{\sqrt6-\sqrt2}{4}} $$

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$$ \text{Using } \cos(A-B)=\cos A \cos B+\sin A \sin B $$ $$ \cos15^\circ = \cos(45^\circ-30^\circ) $$ $$ = \cos45^\circ\cos30^\circ + \sin45^\circ\sin30^\circ $$ $$ = \left(\frac{\sqrt2}{2}\right)\left(\frac{\sqrt3}{2}\right) + \left(\frac{\sqrt2}{2}\right)\left(\frac12\right) $$ $$ = \frac{\sqrt6}{4} + \frac{\sqrt2}{4} $$ $$ = \frac{\sqrt6+\sqrt2}{4} $$ $$ \boxed{\cos15^\circ=\frac{\sqrt6+\sqrt2}{4}} $$

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$$ \textbf{b(ii). Hence, find } \tan15^\circ $$ $$ \tan15^\circ = \frac{\sin15^\circ}{\cos15^\circ} $$ $$ = \frac{\frac{\sqrt6-\sqrt2}{4}} {\frac{\sqrt6+\sqrt2}{4}} $$ $$ = \frac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2} $$ $$ \text{Rationalizing the denominator:} $$ $$ = \frac{(\sqrt6-\sqrt2)(\sqrt6-\sqrt2)} {(\sqrt6+\sqrt2)(\sqrt6-\sqrt2)} $$ $$ = \frac{6-2\sqrt{12}+2}{6-2} $$ $$ = \frac{8-4\sqrt3}{4} $$ $$ = 2-\sqrt3 $$ $$ \boxed{\tan15^\circ = 2-\sqrt3} $$