Further Mathematics Questions and Answers
Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.
Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.
If f(x)=mx2−6x−3 and f′(1)=12, find the value of the constant m.
9
3
-3
-4
Correct answer is A
f(x)=mx2−6x−3
f′(x)=dydx=2mx−6
f′(1)=2m−6=12
2m=18⟹m=9
4
3
2
0
Correct answer is A
\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3}
\frac{x^{2} - 2x - 3}{x - 3} = \frac{x^{2} - 3x + x - 3}{x - 3}
\frac{(x - 3)(x + 1)}{x - 3} = x + 1
\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3} \equiv \lim \limits_{x \to 3} (x + 1) (L'Hopital rule)
\lim \limits_{x \to 3} (x + 1) = 3 + 1 = 4
Calculate in surd form, the value of \tan 15°.
2 + \sqrt{3}
1 + \sqrt{3}
\sqrt{3} - 1
2 - \sqrt{3}
Correct answer is D
\tan 15 = \tan (60 - 45)
\tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}
\tan (60 - 45) = \frac{\tan 60 - \tan 45}{1 + \tan 60 \tan 45}
= \frac{\sqrt{3} - 1}{1 + (\sqrt{3} \times 1)}
= \frac{\sqrt{3} - 1}{1 + \sqrt{3}}
Rationalizing by multiplying denominator and numerator by 1 - \sqrt{3},
\tan 15 = 2 - \sqrt{3}