How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
2 cosx - x sinx
sinx + x cosx
sinx - x cosx
x sinx - 2 cosx
Correct answer is A
y=xsinx
dydx=xcosx+sinx
d2ydx2=x(−sinx)+cosx+cosx
= 2cosx−xsinx
7
2
3
4
Correct answer is B
\lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4}
\frac{(x - 2)(x^{2} + 3x - 2)}{x^{2} - 4} = \frac{(x - 2)(x^{2} + 3x - 2)}{(x - 2)(x + 2)}
= \frac{(x^{2} + 3x - 2)}{x + 2}
\therefore \lim \limits_{x \to 2} \frac{(x - 2)(x^2 + 3x - 2)}{x^2 - 4} = \lim \limits_{x \to 2} \frac{x^{2} + 3x - 2}{x + 2}
= \frac{2^{2} + 3(2) - 2}{2 + 2}
= \frac{4 + 6 - 2}{4} = 2
Given that \theta is an acute angle and sin \theta = \frac{m}{n}, find cos \theta
\frac{\sqrt{n^2 - m^2}}{m}
\frac{\sqrt{(n + m)(n - m)}}{n}
\frac{m}{\sqrt{n^2 - m^2}}
\sqrt{\frac{n}{n^2 - m^2}}
Correct answer is B
sin \theta = \frac{m}{n}
Opp = m; Hyp = n
Adj = \sqrt{n^{2} - m^{2}}
\cos \theta = \frac{\sqrt{n^{2} - m^{2}}}{n}
= \frac{\sqrt{(n + m)(n - m)}}{n}
Find then equation line through (5, 7) parallel to the line 7x + 5y = 12
5x + 7y = 120
7x + 5y = 70
x + y = 7
15x + 17y = 90
Correct answer is B
Equation (5, 7) parallel to the line 7x + 5y = 12
5Y = -7x + 12
y = \frac{-7x}{5} + \frac{12}{5}
Gradient = \frac{-7}{5}
∴ Required equation = \frac{y - 7}{x - 5} = \frac{-7}{5} i.e. 5y - 35 = -7x + 35
5y + 7x = 70