MTH 202 Business Mathematics II Chapter-wise Past Questions — TU BBA-F Semester 2

Asked on 2024 Exam[1 marks]2024 paper

$Integrate: \int \frac{x+2}{x-5} dx$

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Step-by-Step Solution:

Evaluate the indefinite integral:

I = \int \frac{x+2}{x-5} \, dx

Step 1: Perform algebraic division / rearrangement

Express the numerator in terms of $(x - 5)$ :

\frac{x + 2}{x - 5} = \frac{(x - 5) + 7}{x - 5} = \frac{x - 5}{x - 5} + \frac{7}{x - 5} = 1 + \frac{7}{x - 5}

Step 2: Integrate term by term

I = \int \left( 1 + \frac{7}{x - 5} \right) dx

I = \int 1 \, dx + 7 \int \frac{1}{x - 5} \, dx

I = x + 7 \ln|x - 5| + C

Final Answer: $x + 7 \ln|x - 5| + C$ .

Asked on 2024 Exam[1 marks]2024 paper

Find the area of the region bounded by the curve $y = \frac{3}{2}x^2$ , x axis and the two ordinates x=0 and x=2.

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Step-by-Step Solution:

Given:

Curve: $y = \frac{3}{2}x^2$
Bounded by the $x$ -axis and ordinates $x = 0$ and $x = 2$ .

Step 1: Set up the definite integral

Since $y \ge 0$ on $[0, 2]$ , the area $A$ is:

A = \int_0^2 \frac{3}{2}x^2 \, dx

Step 2: Evaluate the integral

A = \frac{3}{2} \left[ \frac{x^3}{3} \right]_0^2 = \frac{1}{2} \left[ x^3 \right]_0^2

A = \frac{1}{2} (2^3 - 0^3) = \frac{1}{2}(8) = 4 \text{ sq. units}

Final Answer: The area of the region is $4$ square units.

Asked on 2024 Exam[1 marks]2024 paper

The marginal cost function of manufacturing xxx units of a product is given by MC = $3x^2 - 10x + 3$ . The total cost of producing one unit of the product is Rs. 7. Find average cost function.

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Step-by-Step Solution:

Given:

Marginal Cost function: $MC = 3x^2 - 10x + 3$
Total cost of producing $1$ unit: $TC(1) = \text{Rs } 7$

Step 1: Find the Total Cost function $TC(x)$ by integration

TC(x) = \int MC \, dx = \int (3x^2 - 10x + 3) \, dx

TC(x) = 3\left(\frac{x^3}{3}\right) - 10\left(\frac{x^2}{2}\right) + 3x + C

TC(x) = x^3 - 5x^2 + 3x + C

Step 2: Determine the integration constant ( $C$ ) using $TC(1) = 7$ $7 = 1^3 - 5(1^2) + 3(1) + C$ $

7 = 1 - 5 + 3 + C

7 = -1 + C \implies C = 7 + 1 = 8

Thus, the Total Cost function is:

TC(x) = x^3 - 5x^2 + 3x + 8

Step 3: Find the Average Cost function $AC(x)$ $AC(x) = \frac{TC(x)}{x} = \frac{x^3 - 5x^2 + 3x + 8}{x}$ $

AC(x) = x^2 - 5x + 3 + \frac{8}{x}

Final Answer: The Average Cost function is $AC(x) = x^2 - 5x + 3 + \frac{8}{x}$ .

Asked on 2024 Exam[3 marks]2024 paper

Integrate the following:

(a) $\int \frac{\log x + 5}{x} dx$

(b) $\int (2x + 1)\sqrt{5x + 2} dx$

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Step-by-Step Solution:

(a) Evaluate $\int \frac{\log x + 5}{x} \, dx$ :

Let $u = \log x + 5 = \ln x + 5$ . Differentiating with respect to $x$ :

du = \frac{1}{x} \, dx

Substitute into the integral:

I = \int u \, du = \frac{u^2}{2} + C

Substitute back $u = \log x + 5$ :

I = \frac{(\log x + 5)^2}{2} + C

(Alternatively: $\int \frac{\ln x}{x} dx + 5 \int \frac{1}{x} dx = \frac{(\ln x)^2}{2} + 5\ln x + C_1$ , which is algebraically equivalent up to a constant).

(b) Evaluate $\int (2x + 1)\sqrt{5x + 2} \, dx$ :

Let $u = 5x + 2 \implies 5x = u - 2 \implies x = \frac{u - 2}{5}$ . Differentiating gives:

dx = \frac{1}{5} \, du

Express the term $(2x + 1)$ in terms of $u$ :

2x + 1 = 2\left(\frac{u - 2}{5}\right) + 1 = \frac{2u - 4 + 5}{5} = \frac{2u + 1}{5}

Substitute into the integral:

I = \int \left( \frac{2u + 1}{5} \right) \sqrt{u} \cdot \left( \frac{1}{5} \, du \right) = \frac{1}{25} \int (2u + 1) u^{1/2} \, du

I = \frac{1}{25} \int \left( 2u^{3/2} + u^{1/2} \right) du

Integrate term by term:

I = \frac{1}{25} \left[ 2 \cdot \frac{u^{5/2}}{5/2} + \frac{u^{3/2}}{3/2} \right] + C

I = \frac{1}{25} \left[ \frac{4}{5} u^{5/2} + \frac{2}{3} u^{3/2} \right] + C

Factor out common terms

\frac{2}{15} u^{3/2}

:

I = \frac{1}{25} \cdot \frac{2}{15} u^{3/2} (6u + 5) + C = \frac{2}{375} u^{3/2} (6u + 5) + C

Substitute back $u = 5x + 2$ :

6u + 5 = 6(5x + 2) + 5 = 30x + 12 + 5 = 30x + 17

I = \frac{2}{375} (30x + 17)(5x + 2)^{3/2} + C

Final Answer:

(a) $\frac{(\log x + 5)^2}{2} + C$
(b) $\frac{2}{375}(30x + 17)(5x + 2)^{3/2} + C$

Asked on 2024 Exam[3 marks]2024 paper

The demand function for a product is $P_d = 8 - 2x$ and the supply function is $P_s = 2x$ under pure competition. Find the consumer’s surplus and producer’s surplus.

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Step-by-Step Solution:

Given market functions under pure competition:

Demand function: $P_d = 8 - 2x$
Supply function: $P_s = 2x$ where $x$ denotes quantity and $P$ denotes price.

Step 1: Find Market Equilibrium Quantity ( $x^$ ) and Price ( $P^$ )

At equilibrium, $P_d = P_s$ :

8 - 2x = 2x

8 = 4x \implies x^* = 2 \text{ units}

Substitute $x^* = 2$ into either equation:

P^* = 2(2) = 4

Step 2: Compute Consumer’s Surplus ( $CS$ )

Consumer’s surplus is the integral of $(P_d - P^*)$ from $0$ to $x^*$ :

CS = \int_0^{x^*} (P_d - P^*) \, dx = \int_0^2 (8 - 2x - 4) \, dx = \int_0^2 (4 - 2x) \, dx

CS = \left[ 4x - x^2 \right]_0^2 = 4(2) - (2)^2 = 8 - 4 = 4

(Geometrically: Triangle area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times (8 - 4) = 4$ ).

Step 3: Compute Producer’s Surplus ( $PS$ )

Producer’s surplus is the integral of $(P^* - P_s)$ from $0$ to $x^*$ :

PS = \int_0^{x^*} (P^* - P_s) \, dx = \int_0^2 (4 - 2x) \, dx

PS = \left[ 4x - x^2 \right]_0^2 = 4(2) - (2)^2 = 8 - 4 = 4

(Geometrically: Triangle area $= \frac{1}{2} \times 2 \times (4 - 0) = 4$ ).

Step 4: Total Surplus

TS = CS + PS = 4 + 4 = 8

Final Answer:

Equilibrium: $x^* = 2 \text{ units}$ , $P^* = 4$
Consumer’s Surplus: $4$
Producer’s Surplus: $4$

Asked on 2024 Exam[3 marks]2024 paper

Solve the differential equations: $\frac{dy}{dx} + \frac{1}{x} \cdot y = x^3$ .

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Step-by-Step Solution:

Given first-order linear differential equation:

\frac{dy}{dx} + \frac{1}{x} \cdot y = x^3

Step 1: Identify standard form and integrating factor

Standard linear form: $\frac{dy}{dx} + P(x)y = Q(x)$ Here:

P(x) = \frac{1}{x}, \quad Q(x) = x^3

The integrating factor $I(x)$ is:

I(x) = e^{\int P(x) \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x

Step 2: Multiply the differential equation by $I(x)$ $x \left( \frac{dy}{dx} + \frac{1}{x} y \right) = x(x^3)$ $

\frac{d}{dx}[y \cdot x] = x^4

Step 3: Integrate both sides with respect to $x$ $\int \frac{d}{dx}[y \cdot x] \, dx = \int x^4 \, dx$ $

y \cdot x = \frac{x^5}{5} + C

Step 4: Solve explicitly for $y$

Divide both sides by $x$ :

y(x) = \frac{x^4}{5} + \frac{C}{x}

Final Answer: $y(x) = \frac{x^4}{5} + \frac{C}{x}$ (where $C$ is an arbitrary constant of integration).

Asked on 2023 Exam[2 marks]2023 paper

Integrate: $\int \left( x^2 + e^x + \frac{1}{x} + 1 \right) dx$

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Step-by-Step Solution:

Evaluate the indefinite integral:

I = \int \left( x^2 + e^x + \frac{1}{x} + 1 \right) dx

Step 1: Apply the sum rule of integration

I = \int x^2 dx + \int e^x dx + \int \frac{1}{x} dx + \int 1 dx

Step 2: Integrate each term individually

$\int x^2 dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3}$
$\int e^x dx = e^x$
$\int \frac{1}{x} dx = \ln|x|$
$\int 1 dx = x$

Step 3: Combine terms and add constant of integration ( $C$ )

I = \frac{x^3}{3} + e^x + \ln|x| + x + C

Final Answer: $\frac{x^3}{3} + e^x + \ln|x| + x + C$ .

Asked on 2023 Exam[2 marks]2023 paper

Find the area of curve bounded by x-axis and ordinates of $y = 7x^2$ from $x_1 = 0$ and $x_2 = 2.$

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Step-by-Step Solution:

Given:

Curve: $y = 7x^2$
Bounded by the $x$ -axis ( $y = 0$ ) and the vertical lines (ordinates) $x_1 = 0$ and $x_2 = 2$ .

Step 1: Set up the definite integral for area

Since $y = 7x^2 \ge 0$ on the interval $[0, 2]$ , the area $A$ is:

A = \int_0^2 y \, dx = \int_0^2 7x^2 dx

Step 2: Evaluate the definite integral

A = 7 \left[ \frac{x^3}{3} \right]_0^2 = 7 \left( \frac{2^3}{3} - \frac{0^3}{3} \right)

A = 7 \left( \frac{8}{3} \right) = \frac{56}{3} \approx 18.67 \text{ sq. units}

Final Answer: The area is $\frac{56}{3}$ square units (or approximately $18.67$ square units).

Asked on 2023 Exam[5 marks]2023 paper

Integrate the following:

(a) $\int \frac{\log x}{x} dx$

$(b) \int \frac{3x + 1}{\sqrt{5x + 2}} dx$

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Step-by-Step Solution:

(a) Evaluate $\int \frac{\log x}{x} dx$ :

Let $u = \log x = \ln x$ . Differentiating:

du = \frac{1}{x} dx

Substitute $u$ into the integral:

\int \frac{\log x}{x} dx = \int u \, du = \frac{u^2}{2} + C

Substitute back $u = \ln x$ :

\int \frac{\log x}{x} dx = \frac{(\ln x)^2}{2} + C

(b) Evaluate $\int \frac{3x + 1}{\sqrt{5x + 2}} dx$ :

Use algebraic substitution: Let $u = 5x + 2 \implies 5x = u - 2 \implies x = \frac{u - 2}{5}$ . Differentiating gives:

dx = \frac{1}{5} du

Express numerator in terms of $u$ :

3x + 1 = 3\left(\frac{u - 2}{5}\right) + 1 = \frac{3u - 6 + 5}{5} = \frac{3u - 1}{5}

Substitute into the integral:

I = \int \frac{\frac{3u - 1}{5}}{\sqrt{u}} \cdot \left(\frac{1}{5} du\right) = \frac{1}{25} \int \frac{3u - 1}{u^{1/2}} du

I = \frac{1}{25} \int \left( 3u^{1/2} - u^{-1/2} \right) du

Integrate term by term:

I = \frac{1}{25} \left[ 3 \cdot \frac{u^{3/2}}{3/2} - \frac{u^{1/2}}{1/2} \right] + C

I = \frac{1}{25} \left[ 2u^{3/2} - 2u^{1/2} \right] + C = \frac{2}{25} \sqrt{u}(u - 1) + C

Substitute back $u = 5x + 2$ :

I = \frac{2}{25} \sqrt{5x + 2} \left( (5x + 2) - 1 \right) + C

I = \frac{2}{25} (5x + 1) \sqrt{5x + 2} + C

Final Answer:

(a) $\frac{(\ln x)^2}{2} + C$
(b) $\frac{2}{25}(5x + 1)\sqrt{5x + 2} + C$

Asked on 2023 Exam[5 marks]2023 paper

The annual rate of repair of a machine is given by $\frac{dc}{dt} = 5t + 12.5$ , where “t” is the age of machine in years, $\frac{dc}{dt}$ is Rs per year. Find total cost after 4 years.

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Step-by-Step Solution:

Given:

Rate of repair cost: $\frac{dC}{dt} = 5t + 12.5$ (in Rs per year)
Age of machine: $t$ in years
We want to find the total repair cost incurred over the first 4 years ( $t = 0$ to $t = 4$ ).

Step 1: Set up the Definite Integral

The total accumulated cost $C$ is the definite integral of the marginal rate of repair:

C = \int_0^4 \frac{dC}{dt} \, dt = \int_0^4 (5t + 12.5) \, dt

Step 2: Evaluate the Integral

C = \left[ \frac{5t^2}{2} + 12.5t \right]_0^4

C = \left( \frac{5(4^2)}{2} + 12.5(4) \right) - \left( \frac{5(0)}{2} + 12.5(0) \right)

C = \left( \frac{5(16)}{2} + 50 \right) - 0

C = 40 + 50 = 90

Final Answer: The total cost of repair after 4 years is Rs 90.

Chapter-wise questions

Unit 1: Integration and applications

Step-by-Step Solution:

Step 1: Perform algebraic division / rearrangement

Step 2: Integrate term by term

Step-by-Step Solution:

Given:

Step 1: Set up the definite integral

Step 2: Evaluate the integral

Step-by-Step Solution:

Given:

Step 1: Find the Total Cost function TC(x)TC(x)TC(x) by integration

Step 2: Determine the integration constant (CCC) using TC(1)=7TC(1) = 7TC(1)=77=13−5(12)+3(1)+C7 = 1^3 - 5(1^2) + 3(1) + C7=13−5(12)+3(1)+C$

Step 3: Find the Average Cost function AC(x)AC(x)AC(x)AC(x)=TC(x)x=x3−5x2+3x+8xAC(x) = \frac{TC(x)}{x} = \frac{x^3 - 5x^2 + 3x + 8}{x}AC(x)=xTC(x)​=xx3−5x2+3x+8​$

Step-by-Step Solution:

(a) Evaluate ∫log⁡x+5x dx\int \frac{\log x + 5}{x} \, dx∫xlogx+5​dx:

(b) Evaluate ∫(2x+1)5x+2 dx\int (2x + 1)\sqrt{5x + 2} \, dx∫(2x+1)5x+2​dx:

Step-by-Step Solution:

Step 1: Find Market Equilibrium Quantity (x∗x^*x∗) and Price (P∗P^*P∗)

Step 2: Compute Consumer’s Surplus (CSCSCS)

Step 3: Compute Producer’s Surplus (PSPSPS)

Step 4: Total Surplus

Step-by-Step Solution:

Step 1: Identify standard form and integrating factor

Step 2: Multiply the differential equation by I(x)I(x)I(x)x(dydx+1xy)=x(x3)x \left( \frac{dy}{dx} + \frac{1}{x} y \right) = x(x^3)x(dxdy​+x1​y)=x(x3)$

Step 3: Integrate both sides with respect to xxx∫ddx[y⋅x] dx=∫x4 dx\int \frac{d}{dx}[y \cdot x] \, dx = \int x^4 \, dx∫dxd​[y⋅x]dx=∫x4dx$

Step 4: Solve explicitly for yyy

Step-by-Step Solution:

Step 1: Apply the sum rule of integration

Step 2: Integrate each term individually

Step 3: Combine terms and add constant of integration (CCC)

Step-by-Step Solution:

Given:

Step 1: Set up the definite integral for area

Step 2: Evaluate the definite integral

Step-by-Step Solution:

(a) Evaluate ∫log⁡xxdx\int \frac{\log x}{x} dx∫xlogx​dx:

(b) Evaluate ∫3x+15x+2dx\int \frac{3x + 1}{\sqrt{5x + 2}} dx∫5x+2​3x+1​dx:

Step-by-Step Solution:

Given:

Step 1: Set up the Definite Integral

Step 2: Evaluate the Integral

Step 1: Find the Total Cost function $TC(x)$ by integration

Step 2: Determine the integration constant ( $C$ ) using $TC(1) = 7$ $7 = 1^3 - 5(1^2) + 3(1) + C$ $

Step 3: Find the Average Cost function $AC(x)$ $AC(x) = \frac{TC(x)}{x} = \frac{x^3 - 5x^2 + 3x + 8}{x}$ $

(a) Evaluate $\int \frac{\log x + 5}{x} \, dx$ :

(b) Evaluate $\int (2x + 1)\sqrt{5x + 2} \, dx$ :

Step 1: Find Market Equilibrium Quantity ( $x^$ ) and Price ( $P^$ )

Step 2: Compute Consumer’s Surplus ( $CS$ )

Step 3: Compute Producer’s Surplus ( $PS$ )

Step 2: Multiply the differential equation by $I(x)$ $x \left( \frac{dy}{dx} + \frac{1}{x} y \right) = x(x^3)$ $

Step 3: Integrate both sides with respect to $x$ $\int \frac{d}{dx}[y \cdot x] \, dx = \int x^4 \, dx$ $

Step 4: Solve explicitly for $y$

Step 3: Combine terms and add constant of integration ( $C$ )

(a) Evaluate $\int \frac{\log x}{x} dx$ :

(b) Evaluate $\int \frac{3x + 1}{\sqrt{5x + 2}} dx$ :