Question

Q1. Differentiate y = 12x⁵ + 3x⁴ + 7x³ + x² 

Q2. Find the distance between the points (2, 3) and (0, 6)

Q3. The compound interest and simple interest on a certain sum for 2 years is N1230 and N1200 respectively. The rate of interest is same for both compound interest and simple interest and it is compounded annually. What is the principal?

Answer 1

Solution. Q1 Using the rules of differentiation and the table of derivatives, we get

$$y'=12\times5\times x^4+3\times4\times x^3+7\times3\times x^2+2\times x$$

$$y'=60x^4+12x^3+21x^2+2x$$

Q2 We use the formula for the distance between two points with coordinates (x₁,y₁), (x₂,y₂)

$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$

Therefore, the distance between the points (2, 3) and (0, 6) is equal to

$$d=\sqrt{(0-2)^2+(6-3)^2}=\sqrt{4+9}=\sqrt{13}$$

Q3 Using Simple Interest Equation (Principal + Interest)

$$A=P(1+it)$$

where A is the total accrued amount (principal + interest), P is the principal amount, i is the rate of interest per year in decimal, t is time period involved in years.

The compound interest formula is as follows:

$$T=P(1+i)^t$$

where T is total accrued, including interest; P is the principal amount; i is the rate of interest per year in decimal, t is time period involved in years.

According to the condition of the problem t= 2years, T=1230, A=1200, and the rate of interest is the same for both compound interest and simple interest and it is compounded annually.

As result get

$$P(1+2i+i^2)=1230$$

$$P(1+2i)=1200$$

Subtracting the second from the first equation, we obtain

$$Pi^2=30.$$

Hence

$$P=\frac{30}{i^2}$$

Substituting the resulting expression into the second equation, we get

$$\frac{30}{i^2}(1+2i)=1200.$$

Therefore

$$1+2i=40i^2$$

$$40i^2-2i-1=0$$

$$D=(-2)^2-4\times40\times(-1)=4+160=164$$

$$i_1=\frac{2-\sqrt{164}}{80}<0$$

$$i_2=\frac{2+\sqrt{164}}{80}$$

As result the principal is equl to

$$P=\frac{30}{i^2}=\frac{30}{(\frac{2+\sqrt{164}}{80})^2}\approx 876$$

Answer. Q1

$$y'=60x^4+12x^3+21x^2+2x$$

Q2

$$d=\sqrt{13}$$

Q3

$$P= 876$$