JAMB Mathematics Past Questions & Answers - Page 45

221.

Marks 1 2 3 4 5
Frequency 2y - 2 y - 1 3y - 4 3 - y 6 - 2y

The table above is the distribution of data with mean equals to 3. Find the value of y.

A.

5

B.

2

C.

3

D.

6

Correct answer is B

Marks (x) 1 2 3 4 5  
Frequency (f) 2y - 2 y - 1 3y - 4 3 - y 6 - 2y 3y + 2
fx 2y - 2 2y - 2 9y - 12 12 - 4y 30 - 10y 26 - y

Mean = \(\frac{\sum fx}{\sum f}\)

\(3 = \frac{26 - y}{3y + 2}\)

\(3(3y + 2) = 26 - y\)

\(9y + 6 = 26 - y\)

\(9y + y = 26 - 6\)

\(10y = 20 \implies y = 2\)

222.

Find the equation of a line perpendicular to the line 4y = 7x + 3 which passes through (-3, 1)

A.

7y + 4x + 5 = 0

B.

7y - 4x - 5 = 0

C.

3y - 5x + 2 = 0

D.

3y + 5x - 2 = 0

Correct answer is A

Equation: 4y = 7x + 3

\(\implies y = \frac{7}{4} x + \frac{3}{4}\)

Slope = coefficient of x = \(\frac{7}{4}\)

Slope of perpendicular line = \(\frac{-1}{\frac{7}{4}}\)

= \(\frac{-4}{7}\)

The perpendicular line passes (-3, 1)

\(\therefore\) Using the equation of line \(y = mx + b\)

m = slope and b = intercept.

\(y = \frac{-4}{7} x + b\)

To find the intercept, substitute y = 1 and x = -3 in the equation.

\(1 = \frac{-4}{7} (-3) + b\)

\(1 = \frac{12}{7} + b\)

\(b = \frac{-5}{7}\)

\(\therefore y = \frac{-4}{7} x - \frac{5}{7}\)

\(7y + 4x + 5 = 0\)

223.

Find the distance between the points C(2, 2) and D(5, 6).

A.

13 units

B.

7 units

C.

12 units

D.

5 units

Correct answer is D

= \(\sqrt{(5 - 2)^2 + (6 - 2)^2}\)

= \(\sqrt{3^2 + 4^2}\)

= \(\sqrt{9 + 16}\)

= \(\sqrt{25}\)

= 5 units

224.

Differentiate \(\frac{2x}{\sin x}\) with respect to x.

A.

\(2 \cot x \sec x (1 + \tan x)\)

B.

\(2 \csc x - x \cot x\)

C.

\(2x \csc x + \tan x\)

D.

\(2\csc x(1 - x\cot x)\)

Correct answer is D

No explanation has been provided for this answer.

225.

If y = 8x\(^3\) - 3x\(^2\) + 7x - 1, find \(\frac{\mathrm d^2 y}{\mathrm d x^2}\).

A.

48x - 6

B.

11x\(^2\) + 6x - 7

C.

32x + 7

D.

24x\(^2\) - 6x + 7

Correct answer is A

y = 8x\(^3\) - 3x\(^2\) + 7x - 1

\(\frac{\mathrm d^2 y}{\mathrm d x^2}  = \frac{\mathrm d}{\mathrm d x} (\frac{\mathrm d y}{\mathrm d x})\)

= \(\frac{\mathrm d}{\mathrm d x} (24x^2 - 6x + 7)\)

\(\frac{\mathrm d^2 y}{\mathrm d x^2} = 48x - 6\)