Solved Question Paper 2021 (2017 Syllabus) Mathematics for Economist

Answer:

A set is a collection of well-defined and distinct objects. The theory of set was developed by German mathematician Georg Cantor. The objects in a set are called the elements or the members of the set. The name of the set is written in upper case and the elements of the set are written in lower case. If x is an element of a set A, we say that x belongs to A or is a member of A, and is expressed symbolically as

$x\ \in \ A$

If y is not a member of A, then this is symbolically denoted as

$y\ \not\in \ A$

Let V be the set of all vowels. Then, V is written as

$V=\left \{ a,\ e,\ i,\ o,\ u \right \}$

Set Operations:

Let A and B be subsets of the universal set U.

$\\U=\left \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9 \right \}\\ A=\left \{1, 2, 3, 4, 5, 6 \right \}\\ B=\left \{3, 4, 5, 6, 7, 8 \right \}\\$

1. Union of Sets: The union of A and B, denoted by $A\cup B$ , is the set of all elements x in U such that is either in A or in B.

$A\cup B=\left \{ x \in U\ |\ x \in A\ or\ x \in B \right \}$

$\\A\cup B=\left \{1, 2, 3, 4, 5, 6, 7, 8 \right \}\\$

2. Intersection of Sets: The Intersection of A and B, denoted by $A\cap B$ , is the set of all elements x in U such that is in both A and B.

$A\cap B=\left \{ x \in U\ |\ x \in A\ and\ x \in B \right \}$

$\\A\cap B=\left \{3, 4, 5, 6 \right \}\\$

3. Difference of Sets: The Difference of B minus A, denoted by , is the set of all elements x in U such that is in B, but not in A.

$B-A=\left \{ x \in U\ |\ x \not\in A\ and\ x \in B \right \}$

$\\B-A=\left \{7,8 \right \}\\$

4. Complement of Sets: The complement of a set A, denoted as $A^{C}\ or\ A^{'}$ is the set of all elements in U such that is not in A.

$A^{C}\ or\ A^{'}=\left \{ x \in U\ |\ x \not\in A\ \right \}$

$A^{C}\ or\ A^{'}=\left \{0,7,8,9 \right \}$

$\\\mathbf{Solution:}\\ \\\mathrm{The\ De\ Morgan's\ law\ of\ set\ theory\ states\ that:}\\ \\\mathbf{For\ Union:}\ \left ( A\cup B \right )^{C}=A^{C}\cap B^{C}\\ \\\mathbf{For\ Intersection:}\ \left ( A\cap B \right )^{C}=A^{C}\cup B^{C}\\$

$\\Given: \\A=\left \{1, 2, 3, 4 \right \}\\ B=\left \{2,4,5,6 \right \}\\ C=\left \{0,3,4,7,8 \right \}\\ \therefore U=\left \{0, 1, 2, 3, 4, 5, 6, 7, 8 \right \}\\$

$\\\mathbf{For\ Union:}\ \left ( A\cup B \right )^{C}=A^{C}\cap B^{C}\\ \mathbf{LHS:}\ \left ( A\cup B \right )^{C}\\ A\cup B=\left \{1,2,3,4,5,6 \right \}\\ \left ( A\cup B \right )^{C}=\left \{0,7,8 \right \}\\ \mathbf{RHS:}\ A^{C}\cap B^{C}\\ A^{C}=\left \{0,5,6,7,8 \right \}\\ B^{C}=\left \{0,1,3,7,8 \right \}\\ A^{C}\cap B^{C}=\left \{0,7,8 \right \}\\ \because LHS=RHS\\ \therefore \left ( A\cup B \right )^{C}=A^{C}\cap B^{C}=\left \{0,7,8 \right \}$

$\\\mathbf{For\ Intersection:}\ \left ( A\cap B \right )^{C}=A^{C}\cup B^{C}\\ \mathbf{LHS:}\ \left ( A\cap B \right )^{C}\\ A\cap B=\left \{2,4 \right \}\\ \left ( A\cap B \right )^{C}=\left \{0,1,3,5,6,7,8 \right \}\\ \mathbf{RHS:}\ A^{C}\cup B^{C}\\ A^{C}=\left \{0,5,6,7,8 \right \}\\ B^{C}=\left \{0,1,3,7,8 \right \}\\ A^{C}\cup B^{C}=\left \{0,1,3,5,6,7,8 \right \}\\ \because LHS=RHS\\ \therefore \left ( A\cup B \right )^{C}=A^{C}\cap B^{C}=\left \{0,1,3,5,6,7,8 \right \}\\ Hence\ proved\\$

$\\\mathrm{Solution:}\\ \mathrm{Total\ no.\ of\ students\ =25}\\ \mathrm{Let}\\ E\rightarrow Students\ taking\ Economics\\ P \rightarrow Students\ taking\ Politics\\ By\ question,\\ A_{1}=12\\ A_{1}-A_{2}=8\\ \mathrm{To\ find:\ Number\ of\ students\ who\ have\ taken\ Economics\ and\ Politics,\ n(E\cap P)}\\ \- \ \ \- \- \ \- \ \- \ \- \ \ \ \ \ \- \ \mathrm{Number\ of\ students\ who\ have\ taken\ Politics\ but\ not\ Economics,\ n(P)-n(E\cap P)}\\$

$\\\mathrm{To\ find:\ Number\ of\ students\ who\ have\ taken\ Economics\ and\ Politics,\ n(E\cap P)}\\ \ \ \ \ \mathrm{Number\ of\ students\ who\ have\ taken\ Politics\ but\ not\ Economics,\ n(P)-n(E\cap P)}\\ A_{1}+A_{3}-A_{2}=25\ -----(1)\\ \ \ \mathrm{Also,}\ A_{1}-A_{2}=8\\ \ \ \mathrm{or,}\ 12-A_{2}=8\ \left ( \because A_{1}=12 \right )\\ \ \ \mathrm{or,}\ 12-8=A_{2}\\ \ \ \mathrm{or, }\ A_{2}=4\\ \ \ \mathrm{or, }\ A_{2}=n(E\cap P)=4\\ \mathrm{Hence,\ the\ number\ of\ students\ who\ have\ taken\ Economics\ and\ Politics,\ n(E\cap P)=4}\\ \mathrm{Next,\ the\ number\ of\ students\ who\ have\ taken\ Politics\ but\ not\ Economics}\\ \mathrm{From\ (i),}\\ A_{1}+A_{3}-A_{2}=25\\ 12+A_{3}-4=25\\ 8+A_{3}=25\\ A_{3}=25-8\\ A_{3}=17\\ \mathrm{Hence,\ the\ number\ of\ students\ who\ have\ taken\ Politics\ but\ not\ Economics}\ A_{3}=n(P)-n(E\cap P)=17-4=13\\$

Answer:

(i) Linear and quadratic functions:

Linear function: Linear functions are those whose graph is a straight line. A linear function has one independent variable and one dependent variable. A linear function has the following form:

y = f(x) = a+bx.

Quadratic function: A quadratic function is a polynomial function with one or more variables in which the highest-degree term is of the second degree. A quadratic function has the following form:

$\\y=f(x)=ax^{2}+bx+c$

(ii) Homogeneous and homothetic functions:

Homogeneous function: A function $f\left ( x_{1}, x_{2}, x_{3},..., x_{n} \right )$ is said to be homogenous of degree ‘k’, if multiplication of each of its independent variables by a constant $\lambda$ , will alter the value of the function by the proportion $\lambda^{k}$ .

$\\i.e.,\ f\left ( \lambda x_{1},\lambda x_{2},\lambda x_{3},...,\lambda x_{n} \right )=\lambda^{k}f\left ( x_{1}, x_{2}, x_{3},..., x_{n} \right )\\ \\or,\ f(\lambda x)=\lambda^{k}f(x)\\$

Note that ‘k’ may be negative

Homothetic Functions: A homothetic function is a monotonic transformation of a homogeneous function. A monotonic transformation is a non-negative transformation that preserves the ordinality of the function. In other words, given a set of numbers {1, 2, 3}, {3} should always be the biggest output followed by {2} then {1}. If this order changes, our transformation is non-monotonic.

For Example: Given a function, f(x) = 2x

A-monotonic transformation of this function would be:

$\\g\left ( f\left ( x \right ) \right )=2x+1\\ h\left ( f\left ( x \right ) \right )=2(2x)$

A non-monotonic transformation of this function would be:

$\\b\left ( f\left ( x \right ) \right )=-1(2x)\\$

This is because it would affect the order of our inputs.

Relationship between Homogeneity and homotheticity:

1.A homogeneous function is always a homothetic function.

2.A homothetic function may not be a homogeneous function.

(iii) Explicit and implicit functions:

An explicit function is one that is given in terms of the independent variable. It is a function that is represented in terms of an independent variable. For example, y = 3x+1 is explicit where y is a dependent variable and is dependent on the independent variable x.

Take the following function,

y =3x - 8

Here, y is the dependent variable and is given in terms of the independent variable x.

When in a function the dependent variable is not explicitly isolated on either side of the equation then the function becomes an implicit function. Implicit functions are usually given in terms of both dependent and independent variables.

Example:

$\\y + x^2 - 3x + 8 = 0\\ y-3x^2+2xy+5=0\\ x^3+2x^2y+3xy^2+y^3=0$

Sometimes, it is not convenient to express a function explicitly.

(iv) Domain and range of a function:

A function is defined as a relation with the added restriction that each value in the domain must have only one corresponding y-value in the range. A function is also called a mapping or transformation which implies the action of associating one thing with another. In the statement, y = f(x), the functional notation f may be interpreted to mean a rule by which the set x is mapped (transformed) into the set y. Thus we may write f: x -> y. The notation f: x -> y means that f is a function from x to y.

In the function,y = f(x), x is referred to as the argument of the function or the independent variable and y is called the value of the function or the dependent variable. The set of all permissible values that x can take in a given context is known as the domain of the function. The y value into which the x value is mapped is called the image of that x value. The set of all images is called the range of the function, which is the set of all values that the y variable can take.

In simple words, a function is nothing but a functional relationship between the independent variable x and the dependent variable y.

For example, let us take a function f(x) = 2x and $1\leq x\leq 4$

x	f(x)
1	2
2	4
3	6
4	8

$\\\mathbf{Solution:}\\ \\\mathrm{Let\ the\ two\ points\ be\ A(3,-2)\ and\ B=(-4,1)}\\ \\\mathrm{The\ equation\ of\ a\ straight\ line\ when\ two\ points\ are\ given\ can\ be\ written\ as:}\\ (y-y_{1})=\left ( \frac{y_{2}-y_{1}}{{x_{2}-x_{1}}} \right )\left ( x-x_{1} \right )\\ \\Here, \ \ x_{1}=3 \ \ \ \ \ \ y_{1}=-2\\ \- \- \- \- \- \- \ \ \ \ \ \ \ \ \ \ \ x_{2}=-4 \ \ \ y_{2}=1\\ (y-(-2))=\left ( \frac{1-(-2)}{{-4-3}} \right )\left ( x-3 \right )\\ (y+2)=\left ( \frac{1+2}{{-7}} \right )\left ( x-3 \right )\\ (y+2)=-\frac{3}{7} \left ( x-3 \right )\\ y+2=-\frac{3}{7}x+\frac{3}{7}.3\\ y+2=-\frac{3}{7}x+\frac{9}{7}\\ y=-\frac{3}{7}x+\frac{9}{7}-2\\ y=-\frac{3}{7}x+\frac{9-14}{7}\\ y=-\frac{3}{7}x-\frac{5}{7}\\ \therefore \mathrm{The\ required\ equation\ of\ a\ straight\ line\ is\ }\ y=-\frac{3}{7}x-\frac{5}{7}\ \ \mathrm{and\ the\ gradient\ of\ the\ line\ is\ -\frac{3}{7}.}\\$

A Matrix is defined as a rectangular array of numbers, parameters, or variables arranged in rows and columns. The members of the array referred to as the element of the matrix are usually enclosed in brackets [ ], in parentheses ( ) or with double vertical lines║ ║.

As a shorthand device, the array in matrix A can be simply written as:

$A=\left [ a_{ij} \right ]\ \begin{pmatrix} i=1,2,3,...,m\\ j=1,2,3,...,n\\ \end{pmatrix}$

For example, the matrix $\begin{bmatrix} 2 &3 &1 \\ 5 &7 &2 \end{bmatrix}$ is a 2x3 matrix that has 2 rows and 3 columns.

Some of the properties of matrices are discussed below:

Properties of matrix addition:

1. (A+B)+C = A+(B+C). This is the associative law for matrix addition.

2. A+B = B+A. This is the commutative law for matrix addition.

3. A + O = A = O + A. The zero matrix O, the same size as A, is the additive identity for matrices the same size as A.

4. A + (−A) = O = (−A) + A. The matrix −A is the unique additive inverse of A.

Properties of matrix multiplication:

1. The product (AB)C is defined precisely when the product A(BC) is defined, and when they are both defined (AB)C = A(BC). This is the associative law for matrix multiplication.

2. A(B + C) = AB + AC and (B + C)A = BA + CA when the products and sums are defined. These are the left and right distributivity laws, respectively, for matrix multiplication over matrix addition.

$\\\mathrm{Solution:}\\ \mathrm{Given:\ System\ of\ equations}\\ 2x_{1}+3x_{2}-x_{3}=15\\ 4x_{2}+2x_{3}=16\\ 3x_{1}+2x_{2}=18\\ or,\\ 2x_{1}+3x_{2}-x_{3}=15\\ 0.x_{1}+4x_{2}+2x_{3}=16\\ 3x_{1}+2x_{2}+0.x_{3}=18\\ \\\mathrm{To\ find:\ Solution\ to\ the\ given\ simultaneous\ equations\ by\ matrix\ inversion\ method}\\ \- \ \ \- \- \ \- \ \- \ \- \ \ \ \ \ \- \ \mathrm{i.e., value\ of}\ x_{1},\ x_{2}\ \mathrm{and}\ x_{3}\\$ $\\\mathrm{Rewriting\ the\ system\ of\ equations\ in\ matrix\ form:}\\ \\\begin{bmatrix} 2 & 3 & -1 \\ 0 & 4 & 2 \\ 3 & 2 & 0 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}=\begin{bmatrix} 15\\ 16\\ 18 \end{bmatrix}\\ \mathrm{Let}\\ A=\begin{bmatrix} 2 & 3 & -1 \\ 0 & 4 & 2 \\ 3 & 2 & 0 \end{bmatrix}\ \ X=\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}\ \ d=\begin{bmatrix} 15\\ 16\\ 18 \end{bmatrix}\\ \mathrm{We\ know\ that,}\\ X=A^{-1}.d\\ Also,\ A^{-1}=\frac{1}{\left | A \right |}.\ Adj A\\ {\left | A \right |}=2\begin{vmatrix} 4 &2\\ 2 &0 \end{vmatrix}-3\begin{vmatrix} 0 &2\\ 3 &0 \end{vmatrix}+(-1)\begin{vmatrix} 0 &4 \\ 3 &2 \end{vmatrix}\\ {\left | A \right |}=2(0-4)-3(0-6)-1(0-12)\\ {\left | A \right |}=2(-4)-3(-6)-1(-12)\\ {\left | A \right |}=-8+18+12\\ {\left | A \right |}=-8+30\\ {\left | A \right |}=22\\$

$\\C_{11}=\mathrm{Cofactor\ of\ a_{11}}=\begin{vmatrix} 4 &2 \\ 2 &0 \end{vmatrix}=0-4=-4\\ \\C_{12}=\mathrm{Cofactor\ of\ a_{12}}=-\begin{vmatrix} 0 &2 \\ 3 &0 \end{vmatrix}=-(0-6)=-(-6)=6\\ \\C_{13}=\mathrm{Cofactor\ of\ a_{13}}=\begin{vmatrix} 0 &4 \\ 3 &2 \end{vmatrix}=0-12=-12\\ \\C_{21}=\mathrm{Cofactor\ of\ a_{21}}=-\begin{vmatrix} 3 &-1 \\ 2 &0 \end{vmatrix}=-(0-(-2))=-(0+2)=-2\\ \\C_{22}=\mathrm{Cofactor\ of\ a_{22}}=\begin{vmatrix} 2 &-1\\ 3 &0 \end{vmatrix}=0-(-3)=0+3=3\\ \\C_{23}=\mathrm{Cofactor\ of\ a_{23}}=-\begin{vmatrix} 2 &3 \\ 3 &2 \end{vmatrix}=-(4-9)=-(-5)=5\\ \\C_{31}=\mathrm{Cofactor\ of\ a_{31}}=\begin{vmatrix} 3 &-1 \\ 4 &2 \end{vmatrix}=6-(-4)=6+4=10\\ \\C_{32}=\mathrm{Cofactor\ of\ a_{32}}=-\begin{vmatrix} 2 &-1 \\ 0 &2 \end{vmatrix}=-(4-0)=-4\\ \\C_{33}=\mathrm{Cofactor\ of\ a_{33}}=\begin{vmatrix} 2 &3 \\ 0 &4 \end{vmatrix}=8-0=8\\$

$\\\mathrm{Cofactor\ matrix,}C=\begin{bmatrix} -4 &\ 6 &-12 \\ -2 &\ 3 &5 \\ 10 &\ -4 &\ 8\ \end{bmatrix}\\ \\\mathrm{Adjoint\ matrix,}\ Adj. A=\begin{bmatrix} -4 &\ -2 &10 \\ 6 &\ 3 &-4 \\ -12 &\ 5 &\ 8\ \end{bmatrix}\\ \mathrm{Now,}\\ A^{-1}=\frac{1}{\left | A \right |}.\ Adj A\\ A^{-1}=\frac{1}{22}\begin{bmatrix} -4 &\ -2 &10 \\ 6 &\ 3 &-4 \\ -12 &\ 5 &\ 8\ \end{bmatrix}\\ \mathrm{Next,}\\ X=A^{-1}.d\\ X=\frac{1}{22}\begin{bmatrix} -4 &\ -2 &10 \\ 6 &\ 3 &-4 \\ -12 &\ 5 &\ 8\ \end{bmatrix}.\begin{bmatrix} 15\\ 16\\ 18 \end{bmatrix}\\$

$\\X=\frac{1}{22}\begin{bmatrix} -4*15+(-2)*16+10*18 \\ 6*15+3*16+(-4)*18 \\ -12*15+5*16+8*18\ \end{bmatrix}\\ X=\frac{1}{22}\begin{bmatrix} -60-32+180 \\ 90+48-72 \\ -180+80+144\ \end{bmatrix}\\ X=\frac{1}{22}\begin{bmatrix} -92+180 \\ 138-72 \\ -180+224\ \end{bmatrix}\\ X=\frac{1}{22}\begin{bmatrix} 88 \\ 66 \\ 44\ \end{bmatrix}\\ X=\begin{bmatrix} \frac{1}{22}\times 88 \\ \frac{1}{22}\times 66 \\ \frac{1}{22}\times 44\ \end{bmatrix}\\ X=\begin{bmatrix} 4 \\ 3 \\ 2 \ \end{bmatrix}\\ \\\mathrm{The\ Solution\ to\ the\ given\ simultaneous\ equations,}\\ \mathrm{i.e., value\ of}\ x_{1}=4,\ x_{2}=3\ \mathrm{and}\ x_{3}=2\\$

In some cases, the input-output model equation solution may give numbers expressed by negative numbers. However, in the economic sense, the output cannot be negative since a firm may produce some amount of a product or no product at all. It cannot produce a negative amount of a product, say, it cannot produce negative 2 units of bread.

If our solution gives negative output, it means that more than one unit of a product is used up in the production of every one unit of that product; it's definitely an unrealistic situation. Such a system is not viable.

The solution of the Leontief input-output model equation will yield a non-negative output if and only if it satisfies certain conditions. These conditions are known as the Hawkins-Simon condition. It ensures that Leontief input-output model equation does not give negative numbers as a solution.

For n-industries case:

Given the Leontief matrix (I - A):

$\begin{bmatrix} 1-a_{11} & -a_{12} &...... &-a_{1n} \\ -a_{21} & 1-a_{22} &...... &-a_{2n} \\ . &. &...... & .\\ . &. &...... & .\\ -a_{n1} & -a_{n2} &...... &1-a_{nn} \\ \end{bmatrix}$

The two conditions are:

1. The diagonal elements $(1-a_{11}),(1-a_{22}) ,.....,(1-a_{nn})\\$ should all be positive. In other words, elements $a_{11},a_{22} ,.....,a_{nn}\\$ should all be less than 1. Thus, the production of one unit of output of any sector should use not more than one unit of its own output, and

2. The determinant of the matrix must always be positive

These two conditions are known as the Hawkins-Simon condition.

Economic Interpretation of Hawkins-Simon conditions:

Let us consider a two-industry case. The Leontief Leontief matrix (I - A) can be written as:

$\begin{bmatrix} 1-a_{11} & -a_{12} \\ -a_{21} & 1-a_{22} \\ \end{bmatrix}\\$

1. The first condition requires that $1-a_{11}$ and $1-a_{22}\\$ must be positive or $a_{11}$ and $a_{22}$ must be less than 1. Economically, this implies that the amount of a commodity used in the production of a hundred rupee's worth of that commodity must be less than a hundred rupee.

In the case of the 1st commodity, the amount of the 1st commodity used in the production of a hundred rupee's worth of the 1st commodity must be less than a hundred rupee.

In the case of the 2nd commodity, the amount of the 2nd commodity used in the production of a hundred rupee's worth of the 2nd commodity must be less than a hundred rupee.

2. The second condition implies that the determinant must be positive, i.e., D>0 implies that $\begin{vmatrix} 1-a_{11} & -a_{12} \\ -a_{21} & 1-a_{22} \\ \end{vmatrix}>0$ .

$\\or,\ (1-a_{11})(1-a_{22})-a_{12}a_{21}>0\\ \\or,\ 1-a_{22}-a_{11}+a_{11}a_{22}-a_{12}a_{21}>0\\ \\\mathrm{Transfering\ all\ the\ terms\ involving\ a_{ij}\ to\ the\ right\ hand\ side,\ we\ have}\\ \\or,\ 1>a_{22}+a_{11}-a_{11}a_{22}+a_{12}a_{21}\\ \\or,\ a_{22}+a_{11}-a_{11}a_{22}+a_{12}a_{21}<1\\ \\or,\ a_{22}-a_{11}a_{22}+a_{11}+a_{12}a_{21}<1\\ \\or,\ (1-a_{11})a_{22}+a_{11}+a_{12}a_{21}<1\\ \\\mathrm{Since,\ (1-a_{11})a_{22}\ is\ positive.\ According\ to\ the\ first\ condition\ both\ a_{11}\ and\ a_{22}\ are\ less\ than\ one. }\\ \\a_{11}+a_{12}a_{21}<1\\$

Economically, the first term $a_{11}$ shows the direct use of the 1st commodity in the production of the 1st commodity itself. The second term shows indirect use. $a_{12}a_{21}\\$ shows that output of the first industry is used as input in the production of the second commodity which is in turn used as input in the production of the 1st commodity. Economically, the second condition implies that the direct and indirect requirement of any commodity to produce one unit of that commodity must also be less than one.

$\\\mathrm{Solution:}\\ \mathrm{Given:}\\ \\A=\begin{bmatrix} 2 &1 \\ 4 &2 \end{bmatrix}\ B=\begin{bmatrix} 3 &0 \\ 1 &5 \end{bmatrix}\ C=\begin{bmatrix} 1 &2 \\ 2 &3 \end{bmatrix}\\\ \\\mathrm{To\ prove:}\ \left ( ABC \right )^{T}=C^{T}B^{T}A^{T}\\ \mathrm{LHS:}\ \left ( ABC \right )^{T}\\ AB=\begin{bmatrix} 2 &1 \\ 4 &2 \end{bmatrix}.\begin{bmatrix} 3 &0 \\ 1 &5 \end{bmatrix}\\\ AB=\begin{bmatrix} 2*3+1*1 &\ 2*0+1*5 \\ 4*3+2*1 &\ 4*0+2*5 \\ \end{bmatrix}\\\\ AB=\begin{bmatrix} 6+1 &\ 0+5 \\ 12+2 &\ 0+10 \\ \end{bmatrix}\\\\ AB=\begin{bmatrix} 7 &\ 5 \\ 14 &\ 10 \\ \end{bmatrix}\\\\$

$\\ABC=\begin{bmatrix} 7 &\ 5 \\ 14 &\ 10 \\ \end{bmatrix}.\begin{bmatrix} 1 &2 \\ 2 &3 \end{bmatrix}\\ \\ABC=\begin{bmatrix} 7*1+5*2 &\ 7*2+5*3 \\ 14*1+10*2 &\ 14*2+10*3 \\ \end{bmatrix}\\ \\ABC=\begin{bmatrix} 7+10 &\ 14+15 \\ 14+20 &\ 28+30 \\ \end{bmatrix}\\ \\ABC=\begin{bmatrix} 17 &\ 29 \\ 34 &\ 58 \\ \end{bmatrix}\\ \\\therefore \left ( ABC \right )^{T}=\begin{bmatrix} 17 &\ 34 \\ 29 &\ 58 \\ \end{bmatrix}\\$

$\\\mathrm{RHS:}\ C^{T}B^{T}A^{T}\\ \\ C^{T}=\begin{bmatrix} 1 &\ 2 \\ 2 &\ 3 \end{bmatrix}\ \ \ B^{T}=\begin{bmatrix} 3 &\ 1 \\ 0 &\ 5 \end{bmatrix}\ \ \ A^{T}=\begin{bmatrix} 2 &\ 4 \\ 1 &\ 2 \end{bmatrix}\\ \\ C^{T} B^{T}=\begin{bmatrix} 1 &\ 2 \\ 2 &\ 3 \end{bmatrix}. \begin{bmatrix} 3 &\ 1 \\ 0 &\ 5 \end{bmatrix}\\ \\ C^{T} B^{T}=\begin{bmatrix} 1*3+2*0 &\ 1*1+2*5 \\ 2*3+3*0 &\ 2*1+3*5 \\ \end{bmatrix}\\ \\ C^{T} B^{T}=\begin{bmatrix} 3+0 &\ 1+10 \\ 6+0 &\ 2+15 \\ \end{bmatrix}\\ \\ C^{T} B^{T}=\begin{bmatrix} 3 &\ 11 \\ 6 &\ 17 \\ \end{bmatrix}\\$

$\\ C^{T} B^{T}A^{T}=\begin{bmatrix} 3 &\ 11 \\ 6 &\ 17 \\ \end{bmatrix}.\begin{bmatrix} 2 &\ 4 \\ 1 &\ 2 \end{bmatrix}\\ \\ C^{T} B^{T}A^{T}=\begin{bmatrix} 3*2+11*1 &\ 3*4+11*2\\ 6*2+17*1 &\ 6*4+17*2\\ \end{bmatrix}\\ \\ C^{T} B^{T}A^{T}=\begin{bmatrix} 6+11 &\ 12+22\\ 12+17 &\ 24+34\\ \end{bmatrix}\\ \\\therefore C^{T} B^{T}A^{T}=\begin{bmatrix} 17 &\ 34\\ 29 &\ 58\\ \end{bmatrix}\\ \\\because LHS=RHS\\ \\\therefore \left ( ABC \right )^{T}=C^{T}B^{T}A^{T}=\begin{bmatrix} 17 &\ 34\\ 29 &\ 58\\ \end{bmatrix}\\\\ \mathrm{Hence\ proved}\\$

$\\\mathrm{Solution:}\\ \mathrm{Given:}\\ \\A=\begin{bmatrix} 8 &4 \\ 3 &7 \end{bmatrix}\ B=\begin{bmatrix} 3 &-2 \\ 1 &5 \end{bmatrix}\\ \\\mathrm{To\ find:\ Matrix\ C\ such\ that\ 2A+4B-3C=0 }\\ \\2A=2.\begin{bmatrix} 8 &4 \\ 3 &7 \end{bmatrix}=\begin{bmatrix} 16 &8 \\ 6 &14 \end{bmatrix}\\ \\4B=4.\begin{bmatrix} 3 &-2 \\ 1 &5 \end{bmatrix}=\begin{bmatrix} 12 &-8 \\ 4 &20 \end{bmatrix}\\$

$\\\mathrm{Now,}\\ \\2A+4B-3C=0\\ \\\begin{bmatrix} 16 &8 \\ 6 &14 \end{bmatrix}+\begin{bmatrix} 12 &-8 \\ 4 &20 \end{bmatrix}-3C=0\\\ \\\begin{bmatrix} 16+12 &8+(-8) \\ 6+4 &14+20 \end{bmatrix}=3C\\ \\\mathrm{or,}\ 3C=\begin{bmatrix} 16+12 &8+(-8) \\ 6+4 &14+20 \end{bmatrix}\\ \\\mathrm{or,}\ 3C=\begin{bmatrix} 28 &8-8 \\ 10 &34 \end{bmatrix}\\ \\\mathrm{or,}\ 3C=\begin{bmatrix} 28 &0 \\ 10 &34 \end{bmatrix}\\ \\\mathrm{or,}\ C=\frac{1}{3}.\begin{bmatrix} 28 &0 \\ 10 &34 \end{bmatrix}\\$

$\\\ C=\begin{bmatrix} \frac{1}{3}*28 &\frac{1}{3}*0 \\ \frac{1}{3}*10 &\frac{1}{3}*34 \end{bmatrix}\\ \\\therefore C=\begin{bmatrix} \frac{28}{3} &0 \\ \frac{10}{3} &\frac{34}{3} \end{bmatrix}\\$

$\\\mathrm{Solution:}\\ \mathrm{Given:}\\ A=\begin{bmatrix} 2 &\ 3 &\ 8 \\ 4 &\ 5 &\ k \\ 2 &\ 2 &-2 \end{bmatrix}\\ \\\mathrm{To\ find:\ The\ value\ of\ \mathit{k},\ if\ A\ is\ a\ singular\ matrix}\\ \mathrm{We\ know\ that,\ matrix\ A\ is\ a\ singular\ matrix,\ if\ its\ determinant =0,\ i.e., \left | A \right |=0.}\\ \\\left | A \right |=2\begin{vmatrix} 5 &\ k\\ 2 & -2 \end{vmatrix}-3\begin{vmatrix} 4 &\ k\\ 2 & -2 \end{vmatrix}+8\begin{vmatrix} 4 &\ 5\\ 2 &\ 2 \end{vmatrix}=0\\ \\\left | A \right |=2(-10-2k)-3(-8-2k)+8(8-10)=0\\ \left | A \right |=-20-4k+24+6k+64-80=0\\ \left | A \right |=-20+24+64-80-4k+6k=0\\ \left | A \right |=-100+88-4k+6k=0\\ \left | A \right |=-100+88+2k=0\\ \left | A \right |=-12+2k=0\\ \mathrm{or,}\ -12+2k=0\\ \mathrm{or,}\ 2k=12\\ \mathrm{or,}\ 2k=12\\ \mathrm{or,}\ k=6\\ \therefore \mathrm{The\ value\ of\ \mathit{k}\ is\ 6.}\\$

$\\(i)\ \lim_{x\rightarrow 2}\ \frac{x^{2}-4}{x-2}\\ \mathrm{Solution:}\\ \\\lim_{x\rightarrow 2}\ \frac{x^{2}-4}{x-2}=\lim_{x\rightarrow 2}\ \frac{x^{2}-2^{2}}{x-2}\\ \\\lim_{x\rightarrow 2}\ \frac{x^{2}-4}{x-2}=\lim_{x\rightarrow 2}\ \frac{(x+2)(x-2)}{(x-2)}\left ( \because a^{2}-b^{2}=(a+b)(a-b) \right )\\ \\\lim_{x\rightarrow 2}\ \frac{x^{2}-4}{x-2}=\lim_{x\rightarrow 2}\ (x+2)\\ \\\lim_{x\rightarrow 2}\ \frac{x^{2}-4}{x-2}=2+2\\ \\\lim_{x\rightarrow 2}\ \frac{x^{2}-4}{x-2}=4\\$

$\\(ii)\ \lim_{x\rightarrow a}\ \frac{\sqrt{x}-\sqrt{a}}{x-a}\\ \mathrm{Solution:}\\ \\\lim_{x\rightarrow a}\ \frac{\sqrt{x}-\sqrt{a}}{x-a}=\lim_{x\rightarrow a}\ \frac{\sqrt{x}-\sqrt{a}}{x-a}.\frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}\\ \\\lim_{x\rightarrow a}\ \frac{\sqrt{x}-\sqrt{a}}{x-a}=\lim_{x\rightarrow a}\ \frac{\left ( \sqrt{x} \right )^2-\left (\sqrt{a} \right )^2}{\left ( x-a \right )\left (\sqrt{x}+\sqrt{a} \right )} \left ( \because a^{2}-b^{2}=(a+b)(a-b) \right )\\ \\\lim_{x\rightarrow a}\ \frac{\sqrt{x}-\sqrt{a}}{x-a}=\lim_{x\rightarrow a}\ \frac{\left ( x-a \right )}{\left ( x-a \right )\left (\sqrt{x}+\sqrt{a} \right )}\\ \\\lim_{x\rightarrow a}\ \frac{\sqrt{x}-\sqrt{a}}{x-a}=\lim_{x\rightarrow a}\ \frac{1}{\sqrt{x}+\sqrt{a}}\\ \\\lim_{x\rightarrow a}\ \frac{\sqrt{x}-\sqrt{a}}{x-a}=\frac{1}{\sqrt{a}+\sqrt{a}}\\ \\\lim_{x\rightarrow a}\ \frac{\sqrt{x}-\sqrt{a}}{x-a}=\frac{1}{2\sqrt{a}}\\$

$\\(iii)\ \lim_{x\rightarrow\ \infty }\ \frac{5x^{3}+2}{3x^{3}+x+1}\\ \mathrm{Solution:}\\ \\\lim_{x\rightarrow\ \infty }\ \frac{5x^{3}+2}{3x^{3}+x+1}=\lim_{x\rightarrow\ \infty }\ \frac{\frac{5x^{3}+2}{x^{3}}}{\frac{3x^{3}+x+1}{x^{3}}}\\\ \\\lim_{x\rightarrow\ \infty }\ \frac{5x^{3}+2}{3x^{3}+x+1}=\lim_{x\rightarrow\ \infty }\ \frac{\frac{5x^{3}}{x^{3}}+\frac{2}{x^{3}}}{\frac{3x^{3}}{x^{3}}+\frac{x}{x^{3}}+\frac{1}{x^{3}}}\\\ \\\lim_{x\rightarrow\ \infty }\ \frac{5x^{3}+2}{3x^{3}+x+1}=\lim_{x\rightarrow\ \infty }\ \frac{5+\frac{2}{x^{3}}}{3+\frac{1}{x^{2}}+\frac{1}{x^{3}}}\\\ \\\lim_{x\rightarrow\ \infty }\ \frac{5x^{3}+2}{3x^{3}+x+1}=\frac{5+\frac{2}{(\infty )^{3}}}{3+\frac{1}{(\infty )^{2}}+\frac{1}{(\infty )^{3}}}\\\ \\\lim_{x\rightarrow\ \infty }\ \frac{5x^{3}+2}{3x^{3}+x+1}=\frac{5+\frac{2}{\infty }}{3+\frac{1}{\infty }+\frac{1}{\infty }}\\\ \\\lim_{x\rightarrow\ \infty }\ \frac{5x^{3}+2}{3x^{3}+x+1}=\frac{5+0}{3+0+0}\\\ \\\lim_{x\rightarrow\ \infty }\ \frac{5x^{3}+2}{3x^{3}+x+1}=\frac{5}{3}\\\$

$\\(iv)\ \lim_{x\rightarrow\ 2}\ \frac{x^{2}-6x+8}{x^{2}-7x+10}\\ \mathrm{Solution:}\\ \lim_{x\rightarrow\ 2}\ \frac{x^{2}-6x+8}{x^{2}-7x+10}=\lim_{x\rightarrow\ 2}\ \frac{x^{2}-(4+2)x+8}{x^{2}-(5+2)x+10}\\ \lim_{x\rightarrow\ 2}\ \frac{x^{2}-6x+8}{x^{2}-7x+10}=\lim_{x\rightarrow\ 2}\ \frac{x^{2}-4x-2x+8}{x^{2}-5x-2x+10}\\ \lim_{x\rightarrow\ 2}\ \frac{x^{2}-6x+8}{x^{2}-7x+10}=\lim_{x\rightarrow\ 2}\ \frac{x(x-4)-2(x-4)}{x(x-5)-2(x-5)}\\ \lim_{x\rightarrow\ 2}\ \frac{x^{2}-6x+8}{x^{2}-7x+10}=\lim_{x\rightarrow\ 2}\ \frac{(x-2)(x-4)}{(x-2)(x-5)}\\ \lim_{x\rightarrow\ 2}\ \frac{x^{2}-6x+8}{x^{2}-7x+10}=\lim_{x\rightarrow\ 2}\ \frac{x-4}{x-5}\\ \lim_{x\rightarrow\ 2}\ \frac{x^{2}-6x+8}{x^{2}-7x+10}=\frac{2-4}{2-5}\\ \lim_{x\rightarrow\ 2}\ \frac{x^{2}-6x+8}{x^{2}-7x+10}=\frac{-2}{-3}\\ \lim_{x\rightarrow\ 2}\ \frac{x^{2}-6x+8}{x^{2}-7x+10}=\frac{2}{3}\\$

$\\(v)\&space;\lim_{x\rightarrow\&space;0}\&space;\frac{\sqrt{1+x}-\sqrt{1-x}}{x}\\&space;\mathrm{Solution:}\\&space; \lim_{x\rightarrow\&space;0}\&space;\frac{\sqrt{1+x}-\sqrt{1-x}}{x}=\lim_{x\rightarrow\&space;0}\&space;\frac{\sqrt{1+x}-\sqrt{1-x}}{x}\times&space;\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\\ \lim_{x\rightarrow\&space;0}\&space;\frac{\sqrt{1+x}-\sqrt{1-x}}{x}=\lim_{x\rightarrow\&space;0}\&space;\frac{\left ( \sqrt{1+x} \right )^2-\left ( \sqrt{1-x} \right )^2}{x\left ( \sqrt{1+x}+\sqrt{1-x} \right )}\\ \lim_{x\rightarrow\&space;0}\&space;\frac{\sqrt{1+x}-\sqrt{1-x}}{x}=\lim_{x\rightarrow\&space;0}\&space;\frac{\left ( 1+x \right )-\left (1-x \right )}{x\left ( \sqrt{1+x}+\sqrt{1-x} \right )}\\ \lim_{x\rightarrow\&space;0}\&space;\frac{\sqrt{1+x}-\sqrt{1-x}}{x}=\lim_{x\rightarrow\&space;0}\&space;\frac{1+x-1+x}{x\left ( \sqrt{1+x}+\sqrt{1-x} \right )}\\ \lim_{x\rightarrow\&space;0}\&space;\frac{\sqrt{1+x}-\sqrt{1-x}}{x}=\lim_{x\rightarrow\&space;0}\&space;\frac{2x}{x\left ( \sqrt{1+x}+\sqrt{1-x} \right )}\\ \lim_{x\rightarrow\&space;0}\&space;\frac{\sqrt{1+x}-\sqrt{1-x}}{x}=\lim_{x\rightarrow\&space;0}\&space;\frac{2}{\sqrt{1+x}+\sqrt{1-x}}\\$

$\\ \lim_{x\rightarrow\&space;0}\&space;\frac{\sqrt{1+x}-\sqrt{1-x}}{x}=\frac{2}{\sqrt{1+0}+\sqrt{1-0}}\\ \\ \lim_{x\rightarrow\&space;0}\&space;\frac{\sqrt{1+x}-\sqrt{1-x}}{x}=\frac{2}{\sqrt{1}+\sqrt{1}}\\ \\ \lim_{x\rightarrow\&space;0}\&space;\frac{\sqrt{1+x}-\sqrt{1-x}}{x}=\frac{2}{1+1}\\ \\ \lim_{x\rightarrow\&space;0}\&space;\frac{\sqrt{1+x}-\sqrt{1-x}}{x}=\frac{2}{2}\\ \\ \lim_{x\rightarrow\&space;0}\&space;\frac{\sqrt{1+x}-\sqrt{1-x}}{x}=1\\$

Answer:

Continuity of a function can be defined as the ability to draw a graph without having to lift our pencil. A more formal definition of continuity is that continuity, at a point a, is defined when the limit of the function from the left equals the limit from the right and this value is also equal to the value of the function. Using notation, for all points a where

$\lim_{x\rightarrow\ a^{+}} f(x)=\lim_{x\rightarrow\ a^{-}} f(x)=f(a)\\\\$

the function is said to be continuous.

Conditions for continuity of a function:

A function f is said to be continuous at a point ‘a’ if the following conditions are satisfied:

a) f(a) is defined

b) $\lim_{x\rightarrow\ a}f(x)\\$ exists

c) $\lim_{x\rightarrow\ a}f(x)=f(a)\\\\$

If one of the conditions is not satisfied, then the function f(x) is said to be discontinuous at x = a.

$\\\mathrm{Solution:}\\ \mathrm{Given:\ The\ function}\\ f(x)=3+2x\ \ \ for\ -3/2< x\leq 0\\ \-\-\ \ \ \ \ \ =3-2x\ \ \ for\ \ \ \ \ \ \ \ 0< x < 3/2\\ \-\-\ \ \ \ \ \ =-3+2x\ for\ \ \ \ \ \ \ x\geq 3/2\\ \\\mathrm{To\ find:\ Whether\ the\ function\ is\ continuous\ at\ x=0}\\ \mathrm{For\ the\ continuity\ of\ a\ function:}\\ a)\ f(a)\ \mathrm{ is\ defined.}\\ b) \lim_{x\rightarrow\ a}f(x)\mathrm{exists}\\ c) \lim_{x\rightarrow\ a}f(x)=f(a)\\\\$

$\\\mathbf{ To\ find\ \mathit{f(a)}:}\\ f(0)=3+2x=3+2(0)=3+0=3\\\\ \mathbf{ LHL:}\\ \lim_{x\rightarrow\ 0^{-}}f(x)=\lim_{x\rightarrow\ 0^{-}}(3+2x)\\ \lim_{x\rightarrow\ 0^{-}}f(x)=3+2(0)\\ \lim_{x\rightarrow\ 0^{-}}f(x)=3+0\\ \lim_{x\rightarrow\ 0^{-}}f(x)=3\\\\ \mathbf{ RHL:}\\ \lim_{x\rightarrow\ 0^{+}}f(x)=\lim_{x\rightarrow\ 0^{+}}(3-2x)\\ \lim_{x\rightarrow\ 0^{+}}f(x)=3-2(0)\\ \lim_{x\rightarrow\ 0^{+}}f(x)=3-0\\ \lim_{x\rightarrow\ 0^{+}}f(x)=3\\ \because f(0)=\lim_{x\rightarrow\ 0^{-}}f(x)=\lim_{x\rightarrow\ 0^{+}}f(x)=3\\ \mathrm{\therefore\ the\ function\ is\ continuous\ at\ x=0. }$

$\\(i)\ y=\left ( 2x-5 \right )\left ( x^{2}+x+1 \right )\\ \mathrm{Solution:}\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x} \left ( 2x-5 \right )\left ( x^{2}+x+1 \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}= \left ( 2x-5 \right )\frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{2}+x+1 \right )+\left ( x^{2}+x+1 \right )\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2x-5 \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}= \left ( 2x-5 \right )\left (\frac{\mathrm{d} }{\mathrm{d} x} x^{2}+\frac{\mathrm{d} }{\mathrm{d} x}x+\frac{\mathrm{d} }{\mathrm{d} x}1 \right )+\left ( x^{2}+x+1 \right )\left ( \frac{\mathrm{d} }{\mathrm{d} x}2x-\frac{\mathrm{d} }{\mathrm{d} x}5 \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}= \left ( 2x-5 \right )\left (2x^{2-1}+1+0 \right )+\left ( x^{2}+x+1 \right )\left (2 \frac{\mathrm{d} }{\mathrm{d} x}x-0 \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}= \left ( 2x-5 \right )\left (2x \right )+\left ( x^{2}+x+1 \right )\left (2.1 \right )\\$

$\\\frac{\mathrm{d} y}{\mathrm{d} x}= \left ( 2x-5 \right ).2x+\left ( x^{2}+x+1 \right ).2\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}= \left ( 4x^{2}-10x \right )+\left ( 2x^{2}+2x+2 \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=4x^{2}-10x+2x^{2}+2x+2\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=6x^{2}-8x+2\\$

$\\(ii)\ y=\left ( 2x^{2}+7 \right )^{10}\\ \mathrm{Solution:}\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x} \left ( 2x^{2}+7 \right )^{10}\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=10 \left ( 2x^{2}+7 \right )^{10-1}\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2x^{2}+7 \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=10 \left ( 2x^{2}+7 \right )^{9}\left (\frac{\mathrm{d} }{\mathrm{d} x} 2x^{2}+\frac{\mathrm{d} }{\mathrm{d} x}7 \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=10 \left ( 2x^{2}+7 \right )^{9}\left (2\frac{\mathrm{d} }{\mathrm{d} x} x^{2}+0 \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=10 \left ( 2x^{2}+7 \right )^{9}\left (2.2 x^{2-1} \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=10 \left ( 2x^{2}+7 \right )^{9}.4x\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=40x\left ( 2x^{2}+7 \right )^{9}\\$

$\\(iii)\ y=\sqrt{x^{2}+5x}\\ \mathrm{Solution:}\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x} \sqrt{x^{2}+5x}\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2\sqrt{x^{2}+5x}}\ \frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{2}+5x \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2\sqrt{x^{2}+5x}}\ \left ( \frac{\mathrm{d} }{\mathrm{d} x}x^{2}+\frac{\mathrm{d} }{\mathrm{d} x}5x \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2\sqrt{x^{2}+5x}}\ \left (2x^{2-1}+5\frac{\mathrm{d} }{\mathrm{d} x}x \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2\sqrt{x^{2}+5x}}\ \left (2x+5.1 \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2\sqrt{x^{2}+5x}}\ \left (2x+5 \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2x+5}{2\sqrt{x^{2}+5x}}\\$

$\\(iv)\ y=\frac{x^{2}+7}{x^{2}-7}\\ \mathrm{Solution:}\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{x^{2}+7}{x^{2}-7}\right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{\left (x^{2}-7 \right )\frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{2}+7 \right )-\left (x^{2}+7 \right )\frac{\mathrm{d} }{\mathrm{d} x}\left (x^{2}-7 \right )}{\left ( x^{2}-7 \right )^{2}}\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{\left (x^{2}-7 \right )\left (\frac{\mathrm{d} }{\mathrm{d} x} x^{2}+\frac{\mathrm{d} }{\mathrm{d} x}7 \right )-\left (x^{2}+7 \right )\left (\frac{\mathrm{d} }{\mathrm{d} x}x^{2}-\frac{\mathrm{d} }{\mathrm{d} x}7 \right )}{\left ( x^{2}-7 \right )^{2}}\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{\left (x^{2}-7 \right )\left (2x^{2-1}+0 \right )-\left (x^{2}+7 \right )\left (2x^{2-1}-0 \right )}{\left ( x^{2}-7 \right )^{2}}\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{\left (x^{2}-7 \right )\left (2x \right )-\left (x^{2}+7 \right )\left (2x \right )}{\left ( x^{2}-7 \right )^{2}}\\$

$\\\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{\left (x^{2}-7 \right ).2x-\left (x^{2}+7 \right ).2x}{\left ( x^{2}-7 \right )^{2}}\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{\left (2x^{3}-14x \right )-\left (2x^{3}+14x \right )}{\left ( x^{2}-7 \right )^{2}}\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{2x^{3}-14x-2x^{3}-14x}{\left ( x^{2}-7 \right )^{2}}\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{-28x}{\left ( x^{2}-7 \right )^{2}}\\$

$\\(v)\ x^{3}+3xy+5y-6=0\\ \mathrm{Solution:}\\ \mathrm{Differentiating\ both\ sides\ with\ respect\ to\ }\ x\\ \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{3}+3xy+5y-6 \right )=\frac{\mathrm{d} }{\mathrm{d} x}0\\ \frac{\mathrm{d} }{\mathrm{d} x}x^{3}+\frac{\mathrm{d} }{\mathrm{d} x}3xy+\frac{\mathrm{d} }{\mathrm{d} x}5y-\frac{\mathrm{d} }{\mathrm{d} x}6=0\\ 3x^{3-1}+3\frac{\mathrm{d} }{\mathrm{d} x}xy+5\ \frac{\mathrm{d} }{\mathrm{d} x}y-0=0\\ 3x^{2}+3\left (x \frac{\mathrm{d} }{\mathrm{d} x}y+y\frac{\mathrm{d} }{\mathrm{d} x}x \right )+5\ \frac{\mathrm{d} y}{\mathrm{d} x}=0\\ 3x^{2}+3\left (x\ \frac{\mathrm{d} y}{\mathrm{d} x}+y.1 \right )+5\ \frac{\mathrm{d} y}{\mathrm{d} x}=0\\ 3x^{2}+3\left (x\ \frac{\mathrm{d} y}{\mathrm{d} x}+y \right )+5\ \frac{\mathrm{d} y}{\mathrm{d} x}=0\\ 3x^{2}+3 x\ \frac{\mathrm{d} y}{\mathrm{d} x}+3y+5\ \frac{\mathrm{d} y}{\mathrm{d} x}=0\\$

$\\3 x\ \frac{\mathrm{d} y}{\mathrm{d} x}+5\ \frac{\mathrm{d} y}{\mathrm{d} x}=-3x^{2}-3y\\ \\\left (3 x+5 \right )\ \frac{\mathrm{d} y}{\mathrm{d} x}=-(3x^{2}+3y)\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{(3x^{2}+3y)}{\left (3 x+5 \right )}\\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{3x^{2}+3y}{3 x+5}\\$

$\\(vi)\ y=x^{x}\\ \\\mathrm{Solution:}\\ \mathrm{Taking\ log\ on\ both\ sides,\ we\ have}\\ \log\ y=log\ x^{x}\\ \log\ y=x\ log\ x\ \left ( \because log\ x^{m}=m\ log\ x \right )\\ \mathrm{Differentiating\ both\ sides\ with\ respect\ to\ }\ x\\ \frac{\mathrm{d} }{\mathrm{d} x}\ log\ y=\frac{\mathrm{d} }{\mathrm{d} x}(x\ log\ x)\\ \frac{1}{y}\ \frac{\mathrm{d} }{\mathrm{d} x}y=x\ \frac{\mathrm{d} }{\mathrm{d} x} log\ x+ log\ x\ \frac{\mathrm{d} }{\mathrm{d} x} x \\ \frac{1}{y}\ \frac{\mathrm{d} y}{\mathrm{d} x}=x\ .\ \frac{1}{x}+ log\ x\ .1 \\ \frac{1}{y}\ \frac{\mathrm{d} y}{\mathrm{d} x}=1+ log\ x\\ \frac{\mathrm{d} y}{\mathrm{d} x}=\left ( 1+ log\ x \right )\ .\ y\\ \frac{\mathrm{d} y}{\mathrm{d} x}=\left ( 1+ log\ x \right ) \ x^{x}\\$

$\\\mathrm{Solution:}\\ \mathrm{ Given:}\\ z=2x^{3}+5x^{2}y+xy^{2}+y^{2}\\ \mathrm{ To\ find:\ First\ and\ second-order\ partial\ derivatives\ and\ also\ to\ verify\ that}\\ \frac{\partial^2 z}{\partial x \partial y}=\frac{\partial^2 z}{\partial y \partial x}\\\\ \mathrm{First-order\ partial\ derivatives}\\ \frac{\partial z}{\partial x}=\frac{\partial }{\partial x}\left ( 2x^{3}+5x^{2}y+xy^{2}+y^{2} \right )\\ \frac{\partial z}{\partial x}= \frac{\partial }{\partial x}2x^{3}+\frac{\partial }{\partial x}5x^{2}y+\frac{\partial }{\partial x}xy^{2}+\frac{\partial }{\partial x}y^{2}\\ \frac{\partial z}{\partial x}=2 \frac{\partial }{\partial x}x^{3}+5y\frac{\partial }{\partial x}x^{2}+y^{2}\frac{\partial }{\partial x}x+0\\ \frac{\partial z}{\partial x}=2.3x^{3-1}+5y.2x^{2-1}+y^{2}.1+0\\ \frac{\partial z}{\partial x}=6x^{2}+10xy+y^{2}\\\\$

$\\\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}\left ( 2x^{3}+5x^{2}y+xy^{2}+y^{2} \right )\\ \frac{\partial z}{\partial y}= \frac{\partial }{\partial y}2x^{3}+\frac{\partial }{\partial y}5x^{2}y+\frac{\partial }{\partial y}xy^{2}+\frac{\partial }{\partial y}y^{2}\\ \frac{\partial z}{\partial y}= 0 +5x^{2}\frac{\partial }{\partial y}y+x\frac{\partial }{\partial y}y^{2}+2y^{2-1}\\ \frac{\partial z}{\partial y}= 5x^{2}.1+x.2y^{2-1}+2y\\ \frac{\partial z}{\partial y}= 5x^{2}+2xy+2y\\$

$\\\mathrm{Second-order\ partial\ derivatives}\\ \frac{\partial^2 z}{\partial^2 x}=\frac{\partial }{\partial x}\left (\frac{\partial z}{\partial x} \right )\\ \frac{\partial^2 z}{\partial^2 x}=\frac{\partial }{\partial x}\left (6x^{2}+10xy+y^{2} \right )\\ \frac{\partial^2 z}{\partial^2 x}=\frac{\partial }{\partial x}6x^{2}+\frac{\partial }{\partial x}10xy+\frac{\partial }{\partial x}y^{2}\\ \frac{\partial^2 z}{\partial^2 x}=6\frac{\partial }{\partial x}x^{2}+10y\frac{\partial }{\partial x}x+ 0\\ \frac{\partial^2 z}{\partial^2 x}=6.2x^{2-1}+10y.1\\ \frac{\partial^2 z}{\partial^2 x}=12x+10y$

$\\\frac{\partial^2 z}{\partial^2 y}=\frac{\partial }{\partial y}\left (\frac{\partial z}{\partial y} \right )\\ \frac{\partial^2 z}{\partial^2 y}=\frac{\partial }{\partial y}\left (5x^{2}+2xy+2y \right )\\ \frac{\partial^2 z}{\partial^2 y}=\frac{\partial }{\partial y}5x^{2}+\frac{\partial }{\partial y}2xy+\frac{\partial }{\partial y}2y\\ \frac{\partial^2 z}{\partial^2 y}=0+2x\frac{\partial }{\partial y}y+2\frac{\partial }{\partial y}y\\ \frac{\partial^2 z}{\partial^2 y}=2x.1+2.1\\ \frac{\partial^2 z}{\partial^2 y}=2x+2\\$

$\\\mathrm{Next,}\\ \\LHS:\\ \frac{\partial^2 z}{\partial x \partial y}=\frac{\partial }{\partial x}\left (\frac{\partial z}{\partial y} \right )\\ \frac{\partial^2 z}{\partial x \partial y}=\frac{\partial }{\partial x}\left (5x^{2}+2xy+2y \right )\\ \frac{\partial^2 z}{\partial x \partial y}=\frac{\partial }{\partial x}5x^{2}+\frac{\partial }{\partial x}2xy+\frac{\partial }{\partial x}2y\\ \frac{\partial^2 z}{\partial x \partial y}=5\frac{\partial }{\partial x}x^{2}+2y\frac{\partial }{\partial x}x+0\\ \frac{\partial^2 z}{\partial x \partial y}=5.2x^{2-1}+2y.1\\ \frac{\partial^2 z}{\partial x \partial y}=10x+2y\\$

$\\RHS: \\\frac{\partial^2 z}{\partial y \partial x}=\frac{\partial }{\partial y}\left (\frac{\partial z}{\partial x} \right )\\ \frac{\partial^2 z}{\partial y \partial x}=\frac{\partial }{\partial y}\left (6x^{2}+10xy+y^{2} \right )\\ \frac{\partial^2 z}{\partial y \partial x}=\frac{\partial }{\partial y}6x^{2}+\frac{\partial }{\partial y}10xy+\frac{\partial }{\partial y}y^{2}\\ \frac{\partial^2 z}{\partial y \partial x}=0+10x\frac{\partial }{\partial y}y+2y^{2-1}\\ \frac{\partial^2 z}{\partial y \partial x}=10x.1+2y\\ \frac{\partial^2 z}{\partial y \partial x}=10x+2y\\ \\\because LHS=RHS\\ \\\therefore \frac{\partial^2 z}{\partial^2 y}=\frac{\partial^2 z}{\partial y \partial x}=10x+2y\\ \\Hence\ verified\\$

$\\\mathrm{Solution:}\\ \mathrm{ Given:}\\ u=\left ( 3x^{2}+5y^{2} \right )^{5}\\ \mathrm{ To\ find:\ du}\\ \mathrm{The\ total\ differential\ formula\ for\ the\ given\ function\ is}\\ du=f_{x}.dx+f_{y}.dy\\ or\ du=\frac{\partial u}{\partial x}.dx+\frac{\partial u}{\partial y}.dy\\ \\\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}\left ( 3x^{2}+5y^{2} \right )^{5}\\ \\\frac{\partial u}{\partial x}=5\left ( 3x^{2}+5y^{2} \right )^{5-1}\frac{\partial }{\partial x}\left ( 3x^{2}+5y^{2} \right )\\ \frac{\partial u}{\partial x}=5\left ( 3x^{2}+5y^{2} \right )^{4}\left ( \frac{\partial }{\partial x}3x^{2}+\frac{\partial }{\partial x}5y^{2} \right )\\ \frac{\partial u}{\partial x}=5\left ( 3x^{2}+5y^{2} \right )^{4}\left (3 \frac{\partial }{\partial x}x^{2}+0 \right )\\ \frac{\partial u}{\partial x}=5\left ( 3x^{2}+5y^{2} \right )^{4}.3.2x^{2-1}\\ \frac{\partial u}{\partial x}=5\left ( 3x^{2}+5y^{2} \right )^{4}.6x\\ \frac{\partial u}{\partial x}=30x\left ( 3x^{2}+5y^{2} \right )^{4}\\$

$\\\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}\left ( 3x^{2}+5y^{2} \right )^{5}\\ \\\frac{\partial u}{\partial y}=5\left ( 3x^{2}+5y^{2} \right )^{5-1}\frac{\partial }{\partial y}\left ( 3x^{2}+5y^{2} \right )\\ \frac{\partial u}{\partial y}=5\left ( 3x^{2}+5y^{2} \right )^{4}\left ( \frac{\partial }{\partial y}3x^{2}+\frac{\partial }{\partial y}5y^{2} \right )\\ \frac{\partial u}{\partial y}=5\left ( 3x^{2}+5y^{2} \right )^{4}\left ( 0+5\frac{\partial }{\partial y}y^{2} \right )\\ \frac{\partial u}{\partial y}=5\left ( 3x^{2}+5y^{2} \right )^{4}. 5.2y^{2-1}\\ \frac{\partial u}{\partial y}=5\left ( 3x^{2}+5y^{2} \right )^{4}. 10y\\ \frac{\partial u}{\partial y}=50y\left ( 3x^{2}+5y^{2} \right )^{4}\\ \\\mathrm{Now,}\\ \ du=\frac{\partial u}{\partial x}.dx+\frac{\partial u}{\partial y}.dy\\ \ du=\left (30x\left ( 3x^{2}+5y^{2} \right )^{4} \right ).dx+\left (50y\left ( 3x^{2}+5y^{2} \right )^{4} \right ).dy\\$

In economics, we are always faced with the choice of maximising or minimising something. We may face with the choice of maximising a firm's profit, a consumer's utility (satisfaction), the rate of growth of a firm or of a country or with minimising the cost of production of a given output. In economics, problems of maximisation and minimisation are known as optimisation. However, the collective term for maximisation and minimisation in mathematics is extremum or extreme values.

Maxima (Maximum Values)

The function y = f(x) has a relative maximum value at x = a, if f(a) is greater than any value immediately preceding or following. We call it a "relative" maximum because other values of the function may in fact be greater. At a point of maximum, at A, the graph is concave downward.

At the point of maximum A, the f '(x) changes sign from + to −.

Slope of the line at point C: f '(x) > 0

Slope of the line at point D: f '(x) < 0

Minima (Minimum Values)

The function y = f(x) has a relative minimum value at x = b, if f(b) is less than any value immediately preceding or following. We call it a "relative" minimum because other values of the function may in fact be less. At a point of minimum, at B, the graph is concave upward.

At the point of minimum B, the f '(x) changes sign from - to +.

Slope of the line at point E: f '(x) < 0

Slope of the line at point F: f '(x) > 0

Conditions for the existence of maximum and minimum:

There are two conditions. They are:

1.Necessary Condition/First-Order Condition

2.Sufficient Condition/Second-Order Condition

Necessary Condition/First-Order Condition:

The first-order condition states that the first-order derivative should be equal to zero, i.e., f(x) = 0

Sufficient Condition/Second-Order Condition:

For the existence of Maximum:

For the existence of maximum, the second-order condition states that second-order derivative should be negative, i.e.,

$f^{''}(x)\equiv \frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}<0$

For the existence of Minimum:

For the existence of minimum, the second-order condition states that second-order derivative should be positive, i.e.,

$f^{''}(x)\equiv \frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}>0$

$\\\mathrm{Solution:}\\ \mathrm{ Given:}\\ y= \frac{1}{3}x^{3}-3x^{2}+8x+10\\ \mathrm{ To\ find:\ The\ maximum\ and\ minimum\ values\ of\ the\ function}\\ \mathbf{First-order\ derivative:}\\ \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{1}{3}x^{3}-3x^{2}+8x+10 \right )\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\frac{1}{3}x^{3}-\frac{\mathrm{d} }{\mathrm{d} x}3x^{2}+\frac{\mathrm{d} }{\mathrm{d} x}8x+\frac{\mathrm{d} }{\mathrm{d} x}10\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{3}\frac{\mathrm{d} }{\mathrm{d} x}x^{3}-3\frac{\mathrm{d} }{\mathrm{d} x}x^{2}+8\frac{\mathrm{d} }{\mathrm{d} x}x+0\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{3}.3x^{3-1}-3.2x^{2-1}+8.1\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=x^{2}-6x+8\\$

$\\\mathbf{First-order\ Condition:}\\ \\\frac{\mathrm{d} y}{\mathrm{d} x}=0\ \ \ \ \ i.e.,\ \ \ \ x^{2}-6x+8=0\\\\ x^{2}-6x+8=0\\ \mathrm{Applying\ the\ quadratic\ formula,\ we\ have}\\ x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ a^{2}+bx+c=0\\ \mathrm{Here,}\\ a=1\ \ \ b=-6\ \ \ c=8\\\\ x=\frac{-(-6)\pm \sqrt{(-6)^{2}-4(1)(8)}}{2(1)}\\ x=\frac{6\pm \sqrt{36-32}}{2}\\ x=\frac{6\pm \sqrt{4}}{2}\\ x=\frac{6\pm 2}{2}\\ x=\frac{6+ 2}{2},\ \frac{6- 2}{2}\\ x=\frac{8}{2},\ \frac{4}{2}\\ x=4,\ 2\\ \therefore \mathrm{Stationary\ points\ are\ at\ x=4\ and\ x=2.}$

$\\\mathbf{Second-order\ derivative:}\\ \\\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )\\ \\\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=\frac{\mathrm{d} }{\mathrm{d} x}\left (x^{2}-6x+8 \right )\\ \\\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=\frac{\mathrm{d} }{\mathrm{d} x}x^{2}-\frac{\mathrm{d} }{\mathrm{d} x}6x+\frac{\mathrm{d} }{\mathrm{d} x}8\\ \\\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=2x^{2-1}-6\frac{\mathrm{d} }{\mathrm{d} x}x+0\\ \\\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=2x-6.1\\ \\\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=2x-6\\$

$\\\mathbf{Second-order\ Condition:}\\ \\\mathrm{At\ x=4}\\ \\\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=2x-6=2(4)-6=8-6=2>0\\ \\\mathrm{At\ x=2}\\ \\\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=2x-6=2(2)-6=4-6=-2<0\\ \\\therefore \mathrm{The\ funcion\ is\ maximum\ at\ x=2\ and\ minimum\ at\ x=4.}\\$ $\\y_{max}=\frac{1}{3}x^{3}-3x^{2}+8x+10\\ y_{max}=\frac{1}{3}(2)^{3}-3(2)^{2}+8(2)+10\\ y_{max}=\frac{1}{3}.8-3.4+16+10\\ y_{max}=\frac{8}{3}-12+26\\ y_{max}=\frac{8}{3}+14\\ y_{max}=\frac{8+42}{3}\\ y_{max}=\frac{50}{3}\\ \\y_{min}=\frac{1}{3}x^{3}-3x^{2}+8x+10\\ y_{min}=\frac{1}{3}(4)^{3}-3(4)^{2}+8(4)+10\\ y_{min}=\frac{1}{3}.64-3.16+32+10\\ y_{min}=\frac{64}{3}-48+42\\ y_{min}=\frac{64}{3}-6\\ y_{min}=\frac{64-18}{3}\\ y_{min}=\frac{46}{3}\\ \\\therefore \mathrm{The\ maximum\ and\ minimum\ values\ of\ the\ function\ are\ y_{max}=\frac{50}{3}\ and\ y_{min}=\frac{46}{3}\ respectively.}$

$\\\mathrm{Solution:}\\ \mathrm{ Given:}\\ \mathrm{Demand\ function:}\ P=500-2q\\ \mathrm{Average\ Cost\ function:}\ AC=0.25q+4+\frac{400}{q}\\ \mathrm{ To\ find:}\\ \mathrm{(i)\ The\ level\ of\ output\ at\ which\ profit \ is\ maximised} \\ \mathrm{(ii)\ the\ price\ at\ which\ profit \ is\ maximised} \\ \mathrm{(iii)\ the\ maximum\ profit} \\ \mathrm{We \ Know \ that,}\\ \mathrm{Profit=Revenue-Cost}\\ \mathrm{or, \ \pi =R-C}\\ \mathrm{Also, \ Revenue=Price \times Quantity}\\ \mathrm{or, \ R=P\times q}\\ R=\left ( 500-2q \right )q\\ R=500q-2q^{2}\\ \mathrm{Next,\ Total-cost \ function=Average-cost \ function \times q}\\ \mathrm{or, \ C=AC \times q}\\ C=\left ( 0.25q+4+\frac{400}{q} \right )\times q\\ C=0.25q^{2}+4q+\frac{400}{q}.q \\ C=0.25q^{2}+4q+400 \\$

$\\\mathrm{Now, \ Profit=Revenue-Cost}\\ \mathrm{or, \ \pi =R-C}\\ \pi =\left ( 500q-2q^{2} \right )-\left ( 0.25q^{2}+4q+400 \right )\\ \pi =500q-2q^{2}-0.25q^{2}-4q-400\\ \pi =-2q^{2}-0.25q^{2}+500q-4q-400\\ \pi =-2.25q^{2}+496q-400\\ \mathrm{which \ is \ the \ profit \ function.}\\ \mathbf{First-order\ derivative:}\\ \frac{\mathrm{d} \pi}{\mathrm{d} q}=\frac{\mathrm{d} }{\mathrm{d} q}\left (-2.25q^{2}+496q-400 \right )\\ \frac{\mathrm{d} \pi}{\mathrm{d} q}=\frac{\mathrm{d} }{\mathrm{d} q}-2.25q^{2}+\frac{\mathrm{d} }{\mathrm{d} q} 496q-\frac{\mathrm{d} }{\mathrm{d} q}400\\ \frac{\mathrm{d} \pi}{\mathrm{d} q}=-2.25\frac{\mathrm{d} }{\mathrm{d} q}q^{2}+ 496\frac{\mathrm{d} }{\mathrm{d} q}q-0\\ \frac{\mathrm{d} \pi}{\mathrm{d} q}=-2.25(2)q^{2-1}+ 496(1)\\ \frac{\mathrm{d} \pi}{\mathrm{d} q}=-4.5q+ 496\\$

$\\\mathbf{First-order\ Condition:}\\ \frac{\mathrm{d} \pi}{\mathrm{d} q}=0\ \ \ \ \ i.e.,\ \ \ \ -4.5q+ 496=0\\ or, \ -4.5q+ 496=0\\ or, \ -4.5q=- 496\\ or, \ 4.5q=496\\ or, \ q=\frac{496}{4.5}=110.22\\ \therefore \mathrm{Stationary\ point\ is\ at\ q=\frac{496}{4.5}=110.22.}\\ \\\mathbf{Second-order\ derivative:}\\ \frac{\mathrm{d^{2}} \pi}{\mathrm{d} q^{2}}=\frac{\mathrm{d} }{\mathrm{d} q}\left ( \frac{\mathrm{d} \pi}{\mathrm{d} q} \right )\\ \\\frac{\mathrm{d^{2}} \pi}{\mathrm{d} q^{2}}=\frac{\mathrm{d} }{\mathrm{d} q}\left ( -4.5q+ 496 \right )\\ \\\frac{\mathrm{d^{2}} \pi}{\mathrm{d} q^{2}}=\frac{\mathrm{d} }{\mathrm{d} q} -4.5q+\frac{\mathrm{d} }{\mathrm{d} q} 496\\ \\\frac{\mathrm{d^{2}} \pi}{\mathrm{d} q^{2}}=-4.5\frac{\mathrm{d} }{\mathrm{d} q}q+0\\ \\\frac{\mathrm{d^{2}} \pi}{\mathrm{d} q^{2}}=-4.5(1)\\ \\\frac{\mathrm{d^{2}} \pi}{\mathrm{d} q^{2}}=-4.5<0\\ \\\therefore \mathrm{The\ \ profit \ funcion\ is\ maximum\ at\ the\ level\ of\ output\ q=\frac{496}{4.5}=110.22}\\$

$\\\mathrm{Next, \ the\ price\ at\ which\ profit \ is\ maximum,}\\\ P=500-2q \\ P=500-2\left (\frac{496}{4.5} \right ) \\ P=500-\frac{992}{4.5} \\ P=500-220.44\\ P=279.56\\ \\\mathrm{Also,}\\ \pi_{max}=-2.25q^{2}+496q-400\\ \pi_{max}=-2.25\left ( \frac{496}{4.5} \right )^{2}+496\left ( \frac{496}{4.5} \right )-400\\ \pi_{max}=-2.25\left ( \frac{246,016}{20.25} \right )+\frac{246,016}{4.5}-400\\ \pi_{max}=-\frac{553,536}{20.25}+\frac{246,016}{4.5}-400\\ \pi_{max}=-27,335.11+54,670.22-400\\ \pi_{max}=-27,735.11+54,670.22\\ \pi_{max}=26,935.11\\ \therefore \mathrm{The\ maximum\ profit\ of\ the\ firm\ is\ \pi_{max}=26,935.11}\\$

$\\\mathrm{Solution:}\\ \mathrm{ Given:}\\ \mathrm{Total-cost\ function:}\ C=0.005x^{3}-0.02x^{2}-30x+3000\\ \mathrm{ To\ find:\ AC\ at\ 10\ units\ of\ output \ and\ MC \ at \ 3\ units\ of\ output.}\\ \mathrm{ Average-cost \ function=\frac{Total-cost \ function}{x}}\\ or, \ AC=\frac{C}{x}\\ AC=\frac{0.005x^{3}-0.02x^{2}-30x+3000}{x}\\ AC=\frac{0.005x^{3}}{x}-\frac{0.02x^{2}}{x}-\frac{30x}{x}+\frac{3000}{x}\\ AC=0.005x^{2}-0.02x-30+\frac{3000}{x}\\ \mathrm{At \ 10 \ units \ of \ output,\ i.e.,\ at }\ x=10,\\ AC=0.005(10)^{2}-0.02(10)-30+\frac{3000}{10}\\ AC=0.005(100)-0.2-30+300\\ AC=0.5-0.2-30+300\\ AC=300.5-30.2\\ AC=270.3\\$

$\\\mathrm{ Marginal-cost \ function=\frac{\Delta Total-cost \ function}{\Delta Units \ of \ output \ produced}}\\ MC=\frac{\mathrm{d} C}{\mathrm{d} x}\\ MC=\frac{\mathrm{d} }{\mathrm{d} x}\left ( 0.005x^{3}-0.02x^{2}-30x+3000 \right )\\ MC=\frac{\mathrm{d} }{\mathrm{d} x}0.005x^{3}-\frac{\mathrm{d} }{\mathrm{d} x}0.02x^{2}-\frac{\mathrm{d} }{\mathrm{d} x}30x+\frac{\mathrm{d} }{\mathrm{d} x}3000\\ MC=0.005\frac{\mathrm{d} }{\mathrm{d} x}x^{3}-0.02\frac{\mathrm{d} }{\mathrm{d} x}x^{2}-30\frac{\mathrm{d} }{\mathrm{d} x}x+0\\ MC=0.005(3)x^{3-1}-0.02(2)x^{2-1}-30(1)\\ MC=0.015x^{2}-0.04x-30\\ \mathrm{At \ 3\ units \ of \ output,\ i.e.,\ at }\ x=3,\\ MC=0.015(3)^{2}-0.04(3)-30\\ MC=0.015(9)-0.12-30\\ MC=0.135-0.12-30\\ MC=0.135-30.12\\ MC=-29.985\\ \therefore \mathrm{ AC=270.3\ units\ at\ 10\ units\ of\ output \ and\ MC=-29.985\ units \ at \ 3\ units\ of\ output.}\\$

$\\\mathrm{Solution:}\\ \mathrm{ Given:\ Demand\ function,}\ q=\frac{20}{p+3}\\ \mathrm{ To\ find:\ The\ price\ elasticity\ of\ demand\ at\ price\ p=2\ and\ also\ interpret\ the\ result}\\ \mathrm{ We\ know\ that,}\\ \mathrm{ Price\ elasticity\ of\ demand,}\ \epsilon_{d}=\frac{Marginal\ function}{Average\ function}=\frac{\frac{dq}{dp}}{\frac{d}{p}}\\ \mathbf{Marginal\ function:}\\ \frac{dq}{dp}=\frac{d}{dp}\left (\frac{20}{p+3} \right )\\ \frac{dq}{dp}=\frac{\left ( p+3 \right )\frac{d}{dp}20-20\frac{d}{dp}\left ( p+3 \right )}{\left ( p+3 \right )^2}\\ \frac{dq}{dp}=\frac{\left ( p+3 \right ).0-20\left ( \frac{d}{dp}p+\frac{d}{dp}3 \right )}{\left ( p+3 \right )^2}\\ \frac{dq}{dp}=\frac{0-20\left ( 1+0 \right )}{\left ( p+3 \right )^2}\\ \frac{dq}{dp}=\frac{-20 (1)}{\left ( p+3 \right )^2}\\ \frac{dq}{dp}=\frac{-20}{\left ( p+3 \right )^2}\\$

$\\\mathbf{Average\ function:}\\ \frac{q}{p}=\frac{\frac{20}{p+3}}{p}\\\\ \frac{q}{p}=\frac{20}{p+3}\times \frac{1}{p}\\ \frac{q}{p}=\frac{20}{p^2+3p}\\ \mathrm{ Price\ elasticity\ of\ demand,}\ \epsilon_{d}=\frac{Marginal\ function}{Average\ function}=\frac{\frac{dq}{dp}}{\frac{d}{p}}\\ \mathrm{ Price\ elasticity\ of\ demand,}\ \epsilon_{d}=\frac{\frac{-20}{\left ( p+3 \right )^2}}{\frac{20}{p^2+3p}}\\\\ \mathrm{ Price\ elasticity\ of\ demand,}\ \epsilon_{d}=\frac{-20}{\left ( p+3 \right )^2}\times \frac{p^2+3p}{20}\\ \mathrm{ Price\ elasticity\ of\ demand,}\ \epsilon_{d}=\frac{-\left ( p^2+3p \right )}{\left ( p+3 \right )^2}\\ \mathrm{ Price\ elasticity\ of\ demand,}\ \epsilon_{d}=\frac{-p^2-3p}{\left ( p+3 \right )^2}\\$

$\\\mathrm{ At\ price\ p=2,}\\ \\\mathrm{ Price\ elasticity\ of\ demand,}\ \epsilon_{d}=\frac{-(2)^2-3(2)}{\left ( 2+3 \right )^2}\\ \\\mathrm{ Price\ elasticity\ of\ demand,}\ \epsilon_{d}=\frac{-4-6}{\left ( 5 \right )^2}\\ \\\mathrm{ Price\ elasticity\ of\ demand,}\ \epsilon_{d}=-\frac{10}{25}\\ \\\mathrm{ Price\ elasticity\ of\ demand,}\ \epsilon_{d}=-\frac{2}{5}\\ \\\mathrm{ Price\ elasticity\ of\ demand,}\ \epsilon_{d}=-0.4\\\\ \mathrm{ Ignoring\ the\ sign,\ we\ have}\\ \\\mathrm{ Price\ elasticity\ of\ demand,}\ \epsilon_{d}=0.4<1\\ \therefore \mathrm{The\ price\ elasticity\ of\ demand\ at\ price\ p=2\ is \ \epsilon_{d}=0.4\ and\ since \ \epsilon_{d}<1, it\ is\ inelastic.}\\$

Integration is the inverse process of differentiation. Instead of differentiating a function, we are given the derivative of a function and asked to find its primitive, i.e., the original function. Such a process is called integration or anti-differentiation. Suppose we differentiate the function $y=x^{2}$ . We obtain $\frac{\mathrm{d} y}{\mathrm{d} x}=2x$ . Integration reverses this process and we say that the integral of 2x is .

The situation is just a little more complicated because there are lots of functions we can differentiate to give 2x. Here are some of them: $x^2+4,\ x^2-15,\ x^2+0.5\$ . All these functions have the same derivative, 2x, because when we differentiate the constant term we obtain zero. Consequently, when we reverse the process, we have no idea what the original constant term might have been. So we include in our answer an unknown constant, c say, called the constant of integration. When we want to integrate a function we use a special notation: $\int f(x)\ dx$ . The symbol for integration, ʃ, is known as an integral sign.

To integrate 2x we write $\int 2x\ dx=x^2+c$ . Note that along with the integral sign, ʃ there is a term of the form dx, which must always be written, and which indicates the variable involved, in this case x. We say that 2x is being integrated with respect to x. The function, 2x being integrated is called the integrand. Technically, integrals of this sort are called indefinite integrals, to distinguish them from definite integrals. When finding an indefinite integral, the answer should always contain a constant of integration.

$\\(i)\ \int \left ( 4x^3+\frac{1}{\sqrt{x}}-3 \right )dx\\ Solution:\\ \int \left ( 4x^3+\frac{1}{\sqrt{x}}-3 \right )dx=\int 4x^3dx+\int \frac{1}{\sqrt{x}}dx-\int 3 dx\\ \int \left ( 4x^3+\frac{1}{\sqrt{x}}-3 \right )dx=4\int x^3dx+\int \frac{1}{x^{1/2}}dx-3\int dx\\ \int \left ( 4x^3+\frac{1}{\sqrt{x}}-3 \right )dx=4\frac{x^{3+1}}{3+1}+\int x^{-1/2}dx-3x\\ \int \left ( 4x^3+\frac{1}{\sqrt{x}}-3 \right )dx=4\frac{x^{4}}{4}+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}-3x+C\\ \int \left ( 4x^3+\frac{1}{\sqrt{x}}-3 \right )dx=x^{4}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}-3x+C\\ \int \left ( 4x^3+\frac{1}{\sqrt{x}}-3 \right )dx=x^{4}+2x^{\frac{1}{2}}-3x+C\\ \int \left ( 4x^3+\frac{1}{\sqrt{x}}-3 \right )dx=x^{4}+2\sqrt{x}-3x+C\\$

$\\(ii)\ \int 4 \left ( e^{2x}+x \right )\left ( e^{2x}+x^2 \right )^2dx\\ Solution:\\ \int \left ( 4e^{2x}+4x \right )\left ( e^{2x}+x^2 \right )^2dx\\ Let\ u=e^{2x}+x^2\\ \frac{\mathrm{d} u}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\left (e^{2x}+x^2 \right )\\ \frac{\mathrm{d} u}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}e^{2x}+\frac{\mathrm{d} }{\mathrm{d} x}x^2\\ \frac{\mathrm{d} u}{\mathrm{d} x}=2e^{2x}+2x\\ du=\left ( 2e^{2x}+2x \right )dx\\ 2du=2\left ( 2e^{2x}+2x \right )dx\\ 2du=\left ( 4e^{2x}+4x \right )dx\\ \int \left ( 4e^{2x}+4x \right )\left ( e^{2x}+x^2 \right )^2dx=\int 2u^2du\\ \int \left ( 4e^{2x}+4x \right )\left ( e^{2x}+x^2 \right )^2dx=2 \int u^2du\\ \int \left ( 4e^{2x}+4x \right )\left ( e^{2x}+x^2 \right )^2dx=2\frac{u^{2+1}}{2+1}+C\\ \int \left ( 4e^{2x}+4x \right )\left ( e^{2x}+x^2 \right )^2dx=2\frac{u^{3}}{3}+C\\ \int \left ( 4e^{2x}+4x \right )\left ( e^{2x}+x^2 \right )^2dx=\frac{2}{3} u^{3}+C\\ \int \left ( 4e^{2x}+4x \right )\left ( e^{2x}+x^2 \right )^2dx=\frac{2}{3} (e^{2x}+x^2)^{3}+C\\$

$\\(iii)\ \int \frac{8x}{\left ( 2x^2+1 \right )}dx\\ Solution:\\ Let \ u=2x^2+1 \\ \frac{\mathrm{d} u}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\left (2x^2+1 \right )\\ \frac{\mathrm{d} u}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}2x^2+\frac{\mathrm{d} }{\mathrm{d} x}1\\ \frac{\mathrm{d} u}{\mathrm{d} x}=2\frac{\mathrm{d} }{\mathrm{d} x}x^2+0\\ \frac{\mathrm{d} u}{\mathrm{d} x}=2(2)x\\ \frac{\mathrm{d} u}{\mathrm{d} x}=4x\\ du=4xdx\\ 2du=2.4xdx\\ 2du=8xdx\\\ \int \frac{8x}{\left ( 2x^2+1 \right )}dx=\int \frac{2}{u}du\\ \int \frac{8x}{\left ( 2x^2+1 \right )}dx=2 \int \frac{1}{u}du\\ \int \frac{8x}{\left ( 2x^2+1 \right )}dx=2\ log u\ +\ C\\ \int \frac{8x}{\left ( 2x^2+1 \right )}dx=2\ log (2x^2+1)\ +\ C\\$

$\\(iv)\ \int x^{2}e^xdx\\ Solution:\\ Integration\ by\ parts\ formula:\\ \int udv=uv-\int vdu\\ u=x^2 \ \ \ \ \ \ \ \ \ \ \ \ v=e^x\\ du=2xdx \ \ \ \ \ \ \ dv=e^xdx\\ \int udv=uv-\int vdu\\ \int x^2e^x dx=x^2e^x-\int e^x2x\ dx\\ \int x^2e^x dx=x^2e^x-2\int e^xx\ dx\\ u=x \ \ \ \ \ \ \ \ \ \ \ \ v=e^x\\ du=dx \ \ \ \ \ \ \ dv=e^xdx\\ \int udv=uv-\int vdu\\ \int xe^xdx=xe^x-\int e^xdx\\ \int xe^xdx=xe^x-\int e^xdx\\ \int xe^xdx=xe^x-e^x\\ \int x^2e^x dx=x^2e^x-2(xe^x-e^x)+C\\ \int x^2e^x dx=x^2e^x-2xe^x+2e^x+C\\ \int x^2e^x dx=e^x(x^2-2x+2)+C\\$

$\\(v)\ \int x.logx\ dx\\ Solution:\\ Integration\ by\ parts\ formula:\\ \int udv=uv-\int vdu\\ u=logx\ \ \ \ \ \ \ \ \ \ \ \ \ v=\frac{1}{2}x^2\\ du=\frac{1}{x}dx \ \ \ \ \ \ \ dv=x dx\\ \int udv=uv-\int vdu\\ \int logx\ x dx=logx\ \frac{1}{2}x^2-\int \frac{1}{2}x^2\frac{1}{x}dx\\ \int logx\ x dx=logx\ \frac{1}{2}x^2-\frac{1}{2}\int x^2\frac{1}{x}dx\\ \int logx\ x dx=logx\ \frac{1}{2}x^2-\frac{1}{2}\int x\ dx\\ \int logx\ x dx=\frac{1}{2}x^2logx\ -\frac{1}{2}\frac{1}{2}x^2+C\\ \int logx\ x dx=\frac{1}{2}x^2logx\ -\frac{1}{4}x^2+C\\ \int logx\ x dx=\frac{1}{2}x^2\left (logx\ -\frac{1}{2} \right )+C\\$

$\\(vi)\ \int 10^{-x}\ dx\\ Solution:\\ Let\ u=-x\\ \frac{\mathrm{d} u}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(-x)\\ \frac{\mathrm{d} u}{\mathrm{d} x}=-1\frac{\mathrm{d} }{\mathrm{d} x}x\\ \frac{\mathrm{d} u}{\mathrm{d} x}=-1.1\\ \frac{\mathrm{d} u}{\mathrm{d} x}=-1\\ du=-1\ dx\\ \frac{du}{-1}=\frac{-1}{-1} dx\\\\ -1du=dx\\ \int 10^{-x}\ dx=\int 10^u-1du\\ \int 10^{-x}\ dx=-1\int 10^udu\\ \int 10^{-x}\ dx=-1\frac{10^u}{log(10)}+C\\ \int 10^{-x}\ dx=-\frac{10^u}{log(10)}+C\\ \int 10^{-x}\ dx=-\frac{10^{-x}}{log(10)}+C\\$

$\\Solution:\\ \int_{2}^{4} 3x^2(x^2+1)dx\\ \int_{2}^{4} \left (3x^4+3x^2 \right ) dx\\ \mathbf{Indefinite\ integral:}\\ \int \left (3x^4+3x^2 \right ) dx=\int 3x^4 dx + \int 3x^2 dx\\ \int \left (3x^4+3x^2 \right ) dx=3\int x^4 dx + 3\int x^2 dx\\ \int \left (3x^4+3x^2 \right ) dx=3\frac{x^{4+1}}{4+1} + 3\frac{x^{2+1}}{2+1}+C\\ \int \left (3x^4+3x^2 \right ) dx=3\frac{1}{5}x^{5} + 3\frac{1}{3}x^{3}+C\\ \int \left (3x^4+3x^2 \right ) dx=\frac{3}{5}x^{5} +x^{3}+C\\$

$\mathbf{Definite\ integral:}\\ \int_{2}^{4} \left (3x^4+3x^2 \right )dx=\left [ \frac{3}{5}x^{5} +x^{3} \right ]_{2}^{4}\\ \int_{2}^{4} \left (3x^4+3x^2 \right )dx=\left [ \frac{3}{5}(4)^{5} +(4)^{3} \right ]-\left [\frac{3}{5}(2)^{5} +(2)^{3} \right ]\\ \int_{2}^{4} \left (3x^4+3x^2 \right )dx=\left [ \frac{3}{5}*1024 + 64 \right ]-\left [\frac{3}{5}*32 +8 \right ]\\ \int_{2}^{4} \left (3x^4+3x^2 \right )dx=\left [ \frac{3072}{5} + 64 \right ]-\left [\frac{96}{5}+8 \right ]\\ \int_{2}^{4} \left (3x^4+3x^2 \right )dx=\left [ \frac{3072+320}{5} \right ]-\left [\frac{96+40}{5}\right ]\\ \int_{2}^{4} \left (3x^4+3x^2 \right )dx=\left (\frac{3392}{5} \right )-\left (\frac{136}{5}\right )\\ \int_{2}^{4} \left (3x^4+3x^2 \right )dx=678.4-27.2\\ \int_{2}^{4} \left (3x^4+3x^2 \right )dx=651.2\\$

Answer:

The producers may be willing to sell their products for a lower price than the prevailing market price. If the market price is higher than where producers expect to price their items, then the difference is called the producers’ surplus. In the sketch shown below, the shaded region represents the producers’ surplus.

The difference between the amount received by the producers (i.e., ) and the amount they would have received at the price level determined by the supply curve (the area under the supply curve from zero to ) is called the producer surplus.

Producer surplus = area above the supply curve and below the line, .

The producers’ surplus is given by:

$P.S=Q_eP_e-\int_{0}^{Q_e}S(x)\ dx$

In this formula, is the supply function, represents the unit market price and represents the quantity supplied. The producer’s surplus is the area of the region bounded above by the line that represents the price and below by the supply curve.

$\\\mathrm{Solution:}\\ \mathrm{Given:\ Production\ function\ or,\ Supply\ function:}\ Q=\sqrt{-4+4p} \ \ \mathrm{and\ market\ price}\ p=10\\ \mathrm{To\ find:\ Producer's\ surplus\ at\ market\ price,}\ p=10\\ \mathrm{We\ know\ that,}\\ P.S=Q_eP_e-\int_{0}^{Q_e}S(Q)\ dQ\\ \mathrm{From\ the\ supply\ function:}\ Q=\sqrt{-4+4p} \\ Squaring\ both\ sides,\ we\ have\\ Q^2=\left ( \sqrt{-4+4p} \right )^2 \\ Q^2=-4+4p\\ Q^2+4=4p\\ 4p=Q^2+4\\ p=\frac{Q^2+4}{4}\\ p=\frac{Q^2}{4}+\frac{4}{4}\\ p=\frac{1}{4}Q^2+1\\$

$\\\mathrm{At\ market\ price,}\ p=10\\ p=\frac{1}{4}Q^2+1\\ 10=\frac{1}{4}Q^2+1\\ 10-1=\frac{1}{4}Q^2\\ 9=\frac{1}{4}Q^2\\ 4*9=Q^2\\ Q^2=4*9\\ Q^2=36\\ Q=\sqrt{36}\\ Q=6\\\\ \mathrm{Now,}\\ P.S=Q_eP_e-\int_{0}^{Q_e}S(Q)\ dQ\\ P.S=6*10-\int_{0}^{6} \left ( \frac{1}{4}Q^2+1 \right )\ dQ\\$

$\\Indefinite\ Integral:\\ \int \left ( \frac{1}{4}Q^2+1 \right )\ dQ=\int \frac{1}{4}Q^2\ dQ+\int 1 \ dQ\\ \int \left ( \frac{1}{4}Q^2+1 \right )\ dQ=\frac{1}{4}\int Q^2\ dQ+1\int dQ\\ \int \left ( \frac{1}{4}Q^2+1 \right )\ dQ=\frac{1}{4}.\frac{Q^{2+1}}{2+1}+1.Q+C\\ \int \left ( \frac{1}{4}Q^2+1 \right )\ dQ=\frac{1}{4}.\frac{1}{3}Q^{3}+Q+C\\ \int \left ( \frac{1}{4}Q^2+1 \right )\ dQ=\frac{1}{12}Q^{3}+Q+C\\ P.S=60-\int_{0}^{6} \left ( \frac{1}{4}Q^2+1 \right )\ dQ\\ P.S=60-\left [ \frac{1}{12}Q^{3}+Q \right ]_{0}^{6}\\ P.S=60-\left [ \left ( \frac{1}{12}(6)^{3}+6 \right )-\left ( \frac{1}{12}(0)^{3}+0 \right ) \right ]\\ P.S=60-\left [ \left ( \frac{1}{12}\times 216+6 \right )-0 \right ]\\ P.S=60-\left ( 18+6 \right )\\ P.S=60-24\\ P.S=36\\ \therefore \mathrm{the\ producer's\ surplus\ at\ market\ price,}\ p=10\ is\ 36\ units.\\$

$\\\mathrm{Solution:}\\ \mathrm{Given:\ Demand\ function\ and\ supply\ function:}\\ P_d=(6-q)^2=36-12q+q^2\ \left [ Using\ (a-b)^2=a^2-2ab+b^2\ \ \ (6-q)^2=6^2-2(6)q+q^2=36-12q+q^2 \right ]\\ P_s=14+q\\ \mathrm{To\ find:\ Consumer's\ surplus\ under\ perfect\ competition}\\ \mathrm{We\ know\ that,}\\ C.S=\int_{0}^{Q_e}D(q)\ dq-Q_eP_e\\ \mathrm{At\ market\ equilibrium,}\\ \mathrm{Market\ demand=Market\ supply}\\ P_d=P_s\\ 36-12q+q^2=14+q\\ 36-12q+q^2-14-q=0\\ q^2-12q-q+36-14=0\\ q^2-13q+22=0\\$

$\\\mathrm{Applying\ quadratic\ formula,}\\ q=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ a^{2}+bx+c=0\\ q^2-13q+22=0\\ \mathrm{Here,}\\ a=1\ \ \ b=-13\ \ \ c=22\\ q=\frac{-(-13)\pm \sqrt{(-13)^{2}-4(1)(22)}}{2(1)}\\ q=\frac{13 \pm \sqrt{169-88}}{2}\\ q=\frac{13 \pm \sqrt{81}}{2}\\ q=\frac{13\pm 9}{2}\\ q=\frac{13+9}{2},\ \frac{13-9}{2}\\ q=\frac{22}{2},\ \frac{4}{2}\\ q=11,\ 2\\$

$\\\mathrm{At}\ q=11 \ \mathrm{using\ the\ demand\ function,}\\ P_d=36-12q+q^2\ P_e=36-12(11)+(11)^2=36-132+121=157-132=25\\ \mathrm{At}\ q=2 \ \mathrm{using\ the\ demand\ function,}\\ P_d=36-12q+q^2\ P_e=36-12(2)+(2)^2=36-24+4=40-24=16\\ \mathrm{For\ Q_e=11, P_e=25}\\ C.S=\int_{0}^{Q_e}D(q)\ dq-Q_eP_e\\ C.S=\int_{0}^{11} (36-12q+q^2)\ dq-11*25\\ \\Indefinite\ Integration\\ \int \left ( 36-12q+q^2 \right ) dq=\int 36 dq-\int 12q dq+\int q^2 dq\\ \int \left ( 36-12q+q^2 \right ) dq=36 \int dq-12\int q dq+\frac{q^{2+1}}{2+1}\\ \int \left ( 36-12q+q^2 \right ) dq=36q-12\frac{q^{1+1}}{1+1}+\frac{1}{3}q^{3}\\ \int \left ( 36-12q+q^2 \right ) dq=36q-12.\frac{1}{2}q^{2}+\frac{1}{3}q^{3}\\ \int \left ( 36-12q+q^2 \right ) dq=36q-6 q^{2}+\frac{1}{3}q^{3}\\$

$\\C.S=\int_{0}^{11}\left ( 36-12q+q^2 \right )dq-275\\ C.S=\left [ 36q-6 q^{2}+\frac{1}{3}q^{3} \right ]_{0}^{11}-275\\ C.S=\left [ \left ( 36(11)-6 (11)^{2}+\frac{1}{3}(11)^{3} \right )-\left ( 36(0)-6 (0)^{2}+\frac{1}{3}(0)^{3} \right ) \right ]-275\\ C.S=\left [ \left ( 396-6 (121)+\frac{1}{3}(1331) \right )-\left (0 \right ) \right ]-275\\ C.S=\left ( 396-726+\frac{1331}{3} \right )-275\\ C.S=\left (\frac{1331}{3} + 396-726\right )-275\\ C.S=\left (\frac{1331}{3}-330 \right )-275\\ C.S=\left (\frac{1331-330*3}{3} \right )-275\\ C.S=\left (\frac{1331-990}{3} \right )-275\\ C.S=\frac{341}{3}-275\\ C.S=113.67-275\\ C.S=-161.33\\$

$\\\mathrm{For\ Q_e=2, P_e=16}\\ C.S=\int_{0}^{Q_e}D(q)\ dq-Q_eP_e\\ C.S=\int_{0}^{2} (36-12q+q^2)\ dq-2*16\\ \\Indefinite\ Integration\\ \int \left ( 36-12q+q^2 \right ) dq=36q-6 q^{2}+\frac{1}{3}q^{3}\\ \\C.S=\left [ 36q-6 q^{2}+\frac{1}{3}q^{3} \right ]_{0}^{2}-32\\ \\C.S=\left [\left ( 36(2)-6 (2)^{2}+\frac{1}{3}(2)^{3} \right )-\left ( 36(0)-6 (0)^{2}+\frac{1}{3}(0)^{3} \right ) \right ]-32\\ \\C.S=\left [\left ( 72-6(4)+\frac{1}{3}.8 \right )-0 \right ]-32\\ \\C.S=\left ( 72-24+\frac{8}{3} \right )-32\\ \\C.S=\left ( 48+\frac{8}{3} \right )-32\\ \\C.S=\frac{48*3+8}{3}-32\\ \\C.S=\frac{144+8}{3}-32\\ \\C.S=\frac{152}{3}-32\\ \\C.S=50.67-32\\ \\C.S=18.67\\ \because C.S=-161.33\ units\ \mathrm{ which\ implies\ a\ negative\ consumer's\ surplus.}\\ \therefore \mathrm{Consumer's\ surplus\ under\ perfect\ competition\ is} \ C.S=18.67\ units\\$

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