JAMB Mathematics Past Questions & Answers - Page 369

1,841.

If y = x cos x, find dy/dx

A.

sin x - x cos x

B.

sin x + x cos x

C.

cos x + x sin x

D.

cos x - x sin x

Correct answer is D

y = x cos x
dy/dx = 1. cos x + x (-sin x)
= cos x - x sin x

1,842.

Integrate \(\frac{x^2 -\sqrt{x}}{x}\) with respect to x

A.

\(\frac{x^2}{2}-2\sqrt{x}+K\)

B.

\(\frac{2(x^2 - x)}{3x}+K\)

C.

\(\frac{x^2}{2}-\sqrt{x}+K\)

D.

\(\frac{(x^2 - x)}{3x}+K\)

Correct answer is A

\(\int \frac{x^2 -\sqrt{x}}{x} = \int \frac{x^2}{x} - \frac{x^{\frac{1}{2}}}{x}\\
\int x - x^{\frac{-1}{2}}\\
=\left(\frac{1}{2}\right)x^2 - \frac{x^{\frac{1}{2}}}{\frac{1}{2}}+K\\
=\frac{x^2}{2}-2x^{\frac{1}{2}}+K\\
=\frac{x^2}{2}-2\sqrt{x}+K\)

1,843.

In how many ways can 6 subjects be selected from 10 subjects for an examination

A.

218

B.

216

C.

215

D.

210

Correct answer is D

\(^{10}C_6 = \frac{10!}{(10-6)!6!}=\frac{10!}{4!6!}\\
=\frac{(10\times 9\times 8\times 7 \times 6!)}{4\times 3\times 2\times 1\times 6!}\\
=210\)

1,844.

Marks 3 4 5 6 7 8
Frequency 5 y - 1 y 9 4 1

The table above gives the frequency distribution of marks obtained by a group of students in a test. If the total mark scored is 200, calculate the value of y

A.

15

B.

13

C.

11

D.

8

Correct answer is C

Total mark scored = 200
∴200 = 15 + 4y - 4 + 5y + 54 + 28 + 8
200 = 9y + 101
200 - 101 = 9y
99 = 9y
∴y = 11

1,845.

What is the mean deviation of 3, 5, 8, 11, 12 and 21?

A.

4.7

B.

60

C.

3.7

D.

10

Correct answer is A

No explanation has been provided for this answer.