Mathematics Questions and Answers

y = x3 + x2 - x + 1

\(\frac{dy}{dx}\) = \(\frac{d(x^3)}{dx}\) + \(\frac{d(x^2)}{dx}\) - \(\frac{d(x)}{dx}\) + \(\frac{d(1)}{dx}\)

\(\frac{dy}{dx}\) = 3x2 + 2x - 1 = 0

\(\frac{dy}{dx}\) = 3x2 + 2x - 1

At the maximum point \(\frac{dy}{dx}\) = 0

3x2 + 2x - 1 = 0

(3x2 + 3x) - (x - 1) = 0

3x(x + 1) -1(x + 1) = 0

(3x - 1)(x + 1) = 0

therefore x = \(\frac{1}{3}\) or -1

For the maximum point

\(\frac{d^2y}{dx^2}\) < 0

\(\frac{d^2y}{dx^2}\) 6x + 2

when x = \(\frac{1}{3}\)

\(\frac{dx^2}{dx^2}\) = 6(\(\frac{1}{3}\)) + 2

= 2 + 2 = 4

\(\frac{d^2y}{dx^2}\) > o which is the minimum point

when x = -1

\(\frac{d^2y}{dx^2}\) = 6(-1) + 2

= -6 + 2 = -4

-4 < 0

therefore, \(\frac{d^2y}{dx^2}\) < 0

the minimum point is 1/3

Mode = L₁ + (\(\frac{D_1}{D_1 + D_2}\))C

D₁ = frequency of modal class - frequency of the class before it

D₁ = 5 - 2 = 3

D₂ = frequency of modal class - frequency of the class that offers it

D₂ = 5 - 3 = 2

L₁ = lower class boundary of the modal class

L₁ = 5 - 5

C is the class width = 8 - 5.5 = 3

Mode = L₁ + (\(\frac{D_1}{D_1 + D_2}\))C

= 5.5 + \(\frac{3}{2 + 3}\)C

= 5.5 + \(\frac{3}{5}\) x 3

= 5.5 + \(\frac{9}{5}\)

= 5.5 + 1.8

= 7.3 \(\approx\) = 7

2x + 1 \(\frac{d(3x + 1)}{\mathrm d x}\) + (3x + 1) \(\frac{d(2x + 1)}{\mathrm d x}\)

2x + 1 (3) + (3x + 1) (2)

6x + 3 + 6x + 2 = 12x + 5

tan\(\theta\) = \(\frac{100}{100}\) = 1

\(\theta\) = tan^-1(1) = 45^o

The bearing of x from z is N45^oE or 135^o

tan\(\theta\) = \(\frac{3}{4}\)

from Pythagoras tippet, the hypotenus is T

i.e. 3, 4, 5.

then sin \(\theta\) = \(\frac{3}{5}\) and cos\(\theta\) = \(\frac{4}{5}\)

cos\(\theta\) - sin\(\theta\)

\(\frac{4}{5}\) - \(\frac{3}{5}\) = \(\frac{1}{5}\)