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.
12.2
12.27
12.9
13.4
Correct answer is B
\(\begin{array}{c|c} \text{Class intervals} & F & \text{Class boundary}\\ \hline 5 - 7 & 17 & 4.5 - 9.5\\ 10 - 14 & 32 & 9.5 - 14.5\\ 15 - 19 & 25 & 14.5 - 19.5\\ 20 - 24 & 24 & 19.5 - 24.5 \end{array}\)
mode = 9.5 + \(\frac{D_1}{D_2 + D_1}\) x C
= 9.5 + \(\frac{5(32 - 17)}{2(32) - 17 - 25}\)
= 9.5 + \(\frac{75}{27}\)
= 12.27
\(\approx\) 12.3
19
18
13
12
Correct answer is A
mean(x) = \(\frac{\sum x}{N}\)
= \(\frac{48}{8}\)
= 5.875
re-arranging the numbers;
2, 3, 5, 6, 2, 7, 8, 9
median = \(\frac{6 + 7}{2}\)
= \(\frac{1}{2}\)
= 6.5
m + 2n = 5.875 + (6.5)2
= 13 + 5.875
= 18.875
= \(\approx\) = 19
find the equation of the curve which passes through by 6x - 5
6x2 - 5x + 5
6x2 + 5x + 5
3x2 - 5x - 5
3x2 - 5x + 3
Correct answer is D
m = \(\frac{dy}{dv}\) = 6x - 5
∫dy = ∫(6x - 5)dx
y = 3x2 - 5x + C
when x = 2, y = 5
∴ 5 = 3(2)2 - 5(2) +C
C = 3
∴ y = 3x2 - 5x + 3
Evaluate ∫\(^{\pi}_{2}\)(sec2 x - tan2x)dx
\(\frac{\pi}{2}\)
\(\pi\) - 2
\(\frac{\pi}{3}\)
\(\pi\) + 2
Correct answer is B
∫\(^{\pi}_{2}\)(sec2 x - tan2x)dx
∫\(^{\pi}_{2}\) dx = [X]\(^{\pi}_{2}\)
= \(\pi\) - 2 + c
when c is an arbitrary constant of integration
Differentiate \(\frac{x}{cosx}\) with respect to x
1 + x sec x tan x
1 + sec2 x
cos x + x tan x
x sec x tan x + secx
Correct answer is D
let y = \(\frac{x}{cosx}\) = x sec x
y = u(x) v (x0
\(\frac{dy}{dx}\) = U\(\frac{dy}{dx}\) + V\(\frac{du}{dx}\)
dy x [secx tanx] + secx
x = x secx tanx + secx