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.
If y = 243(4x + 5)-2, find \(\frac{dy}{dx}\) when x = 1
\(\frac{-8}{3}\)
\(\frac{3}{8}\)
\(\frac{9}{8}\)
-\(\frac{8}{9}\)
Correct answer is A
y = 243(4x + 5)-2, find \(\frac{dy}{dx}\)
= -1944(4x + 5)-3
= 1944(9)-3
\(\frac{dy}{dx}\) when x = 1
= -\(\frac{1944}{9^3}\)
= -\(\frac{1944}{729}\)
= \(\frac{-8}{3}\)
150(1 + \(\sqrt{3}\))m
50( \(\sqrt{3}\) - \(\sqrt{3}\))m
150 \(\sqrt{3}\)m
\(\frac{50}{\sqrt{3}}\)m
Correct answer is B
\(\frac{150}{Z}\) = tan 60o,
Z = \(\frac{150}{tan 60^o}\)
= \(\frac{150}{3}\)
= 50\(\sqrt{3}\)cm
\(\frac{150}{X x Z}\) = tan45o = 1
X + Z = 150
X = 150 - Z
= 150 - 50\(\sqrt{3}\)
= 50( \(\sqrt{3}\) - \(\sqrt{3}\))m
\(\frac{\pi}{2}\), \(\frac{3\pi}{2}\)
\(\frac{\pi}{3}\), \(\frac{2\pi}{3}\)
0, \(\frac{\pi}{3}\)
0, \(\pi\)
Correct answer is D
cos x + sin x \(\frac{1}{cos x - sinx}\)
= (cosx + sinx)(cosx - sinx) = 1
= cos2x + sin2x = 1
cos2x - (1 - cos2x) = 1
= 2cos2x = 2
cos2x = 1
= cosx = \(\pm\)1 = x
= cos-1x (\(\pm\), 1)
= 0, \(\pi\) \(\frac{3}{2}\pi\), 2\(\pi\)
(possible solution)
(\(\frac{3}{2}\), \(\frac{3}{2}\))
(\(\frac{2}{3}\), \(\frac{3}{2}\))
(\(\frac{3}{8}\), \(\frac{3}{2}\))
(-\(\frac{3}{8}\), \(\frac{3}{2}\))
Correct answer is D
y = 4x + 3
when x = 0, y = 3 \(\to\) (0, 3)
when y = 0, x = -\(\frac{3}{4}\) \(\to\) (\(\frac{3}{4}\), 0)
mid-point \(\frac{0 + (-{\frac{3}{4}})}{2}\), \(\frac{3 + 0}{4}\)
-\(\frac{3}{8}\), \(\frac{3}{2}\)
If the distance between the points (x, 3) and (-x, 2) is 5. Find x
6.0
2.5
\(\sqrt{6}\)
\(\sqrt{3}\)
Correct answer is C
d2 = (y - y)2 + (x - x)2
5 = 4x2 + 1 = 25= 4x2 + 1
= 4x2 = 25 - 1= 24
x2 = \(\frac{24}{4}\)
x = \(\sqrt{6}\)