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GNDU Question Paper-2023
BA 1
st
Semester
QANTITATIVE TECHNIQUES
(Quantitative Techniques - I)
Time Allowed: Three Hours Maximum Marks: 100
Note: Attempt Five questions in all, selecting at least One question from each section. The
Fifth question may be attempted from any section. All questions carry equal marks.
SECTION-A
(1) Solve :
(ii) Solve the equation: (x+1) (x+2)(x+3)(x+4) = 120
(iii) Find equilibrium price and quantity, given that demand D =
and supply,
2. (i) (Find sum of 12 terms of an AP, where n
th
term is 5n + 2
(ii) Sum up the following series,
upto 12 terms.
(iii). If a,b,c,d are in G.P .., show that :
SECTION-B
3. (i)Explain the concepts of permutations and combinations.
(ii) Find the equation of straight line passing through the points (3,5) and (4, 7).
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(iii) From the following data find linear demand curve:
Price Per Liter
(Milk)
Demand in Liters
Rs.1
100
Rs.2
50
Rs.3
0
4. (i) Define Set. Explain various types of Sets.
(ii) Explain Union, Intersection, difference and symmetric difference of Sets with the
help of Venn diagrams.
SECTION-C
5. (i) Explain the concepts of constant and variable.
(ii) Define function, explain various types of function.
(iii) Evaluate Limit
6. (i) Show that f(x) =
is discoutinuous at x = 0
(ii) Find the derivative of c
X
by first principle method.
SECTION-D
7. (i) Find derivative of
w.r.t. x.
(ii) Find :
if y= log
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(iii) Differentiate(1+x)
x
w.r.t. x.
8. (1) Find elasticity of demand, given the demand function
when q=9 and q=0
(ii) Demand Curve is given as p = (50 - x)/5 find MR at output = x. Also find MR when x = 0
and x = 25
GNDU Answer Paper-2023
BA 1
st
Semester
QANTITATIVE TECHNIQUES
(Quantitative Techniques - I)
Time Allowed: Three Hours Maximum Marks: 100
Note: Attempt Five questions in all, selecting at least One question from each section. The
Fifth question may be attempted from any section. All questions carry equal marks.
SECTION-A
(1) Solve :
(ii) Solve the equation: (x+1) (x+2)(x+3)(x+4) = 120
(iii) Find equilibrium price and quantity, given that demand D =
and supply,
Ans: Mathematics Through a Story: A Journey Through Equations
Imagine a university student named Ravi, studying mathematics. One day, Ravi sat down
with his math notebook and three challenging problems that his professor had assigned. He
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knew he couldn’t just solve them blindly; he had to understand every step deeply. Let’s walk
along with Ravi and explore these problems.
(1) Solving the System of Equations:
Problem:
Understanding the Equation:
Ravi read the first equation:
He noticed something interesting – both fractions have the same denominator x.
So he combined the numerators:
Now he had:
To solve for x, he multiplied both sides by x:
He found the value of x!
Now he substituted x = 5/18 into the second equation:
Plug in the value of x:
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To divide by a fraction, flip it and multiply:
Now subtract:
Now solve for y:
Simplify the fraction:
Final Answer:
(ii) Solving the Quartic Equation:
Problem:
Understanding the Equation:
Ravi saw that the equation had four consecutive linear terms multiplied together. He
remembered this is a quartic equation (degree 4).
He thought of a trick: use substitution.
Let’s make it symmetrical. Let:
Then the equation becomes:
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He grouped the pairs:
Apply identity:
So:
Now multiply them:
Multiply:
Bring 120 to LHS:
Now let u=z
2
:
Solve using quadratic formula:
Calculate:
Discriminant = 6.25+477.75=484
Two cases:
Recall x=z−2.5x = z - 2.5x=z−2.5
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(iii) Finding Equilibrium Price and Quantity:
Problem:
Ravi knew that the equilibrium point in economics is where demand equals supply.
So set:
Let’s Solve:
Bring all terms to one side:
Combine like terms:
Multiply through by 2 to eliminate fractions:
Use Quadratic Formula:
So,
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Now, find equilibrium quantity:
Use demand equation:
Or use supply:
Perfect! Demand = Supply = 42 when price = 4.
Final Answer:
Equilibrium Price: p=4p = 4p=4
Equilibrium Quantity: q=42q = 42q=42
Conclusion: What Did We Learn?
In this beautiful journey with Ravi, we explored:
How to solve systems of equations by simplifying and substituting variables step by step.
Solving high-degree equations using substitution and symmetry tricks.
Finding equilibrium in economics using algebraic methods and understanding how supply
and demand interact.
Each of these problems taught us an important lesson:
Equations are not scary if you break them into small parts.
Patterns and substitutions can turn complex problems into simple ones.
Math is a language – it expresses real-world ideas like price and quantity through numbers.
2. (i) (Find sum of 12 n
th
term is terms of an AP, where 5n + 2
(ii) Sum up the following series,
upto 12 terms.
(iii). If a,b,c,d are in G.P .., show that :
Ans: Understanding Arithmetic Progression, Geometric Progression, and Series
Summations
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Mathematics often introduces us to patterns. Among the most important and commonly
found patterns are Arithmetic Progressions (A.P.) and Geometric Progressions (G.P.). These
are essential concepts, not only for solving numerical problems but also for understanding
various real-life patterns — such as savings over time, depreciation of car value, or bacterial
growth.
Let us now take up each part of the question, solve it, and explore the concepts behind it
deeply and meaningfully.
Part (i): Find the sum of 12th nth term of an A.P. where nth term is 5n + 2
Step 1: Understand the nth term
We are given the nth term of an A.P. as:
Tₙ = 5n + 2
This means:
When n = 1: T₁ = 5(1) + 2 = 7
When n = 2: T₂ = 5(2) + 2 = 12
When n = 3: T₃ = 5(3) + 2 = 17
and so on…
So, the A.P. becomes:
7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62
We need to find the sum of the first 12 terms of this A.P.
Step 2: Use the formula of sum of A.P.
For an A.P., the sum of first n terms (Sₙ) is given by the formula:
Sₙ = (n / 2) × [2a + (n - 1)d]
Where:
n is the number of terms,
a is the first term,
d is the common difference.
From the series:
a = 7
d = 12 - 7 = 5
n = 12
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So,
S₁₂ = (12 / 2) × [2×7 + (12 - 1)×5]
S₁₂ = 6 × [14 + 55]
S₁₂ = 6 × 69 = 414
Final Answer:
The sum of first 12 terms of the A.P. where nth term is 5n + 2 is 414.
Part (ii): Sum up the series 1 - 2/3 + 4/9 - 8/27 + … up to 12 terms
This looks different from an A.P. Here we notice alternating signs and the numerators and
denominators follow patterns too.
Step 1: Observe the pattern
Let us write the terms:
T₁ = 1
T₂ = -2/3
T₃ = 4/9
T₄ = -8/27
T₅ = 16/81
T₆ = -32/243
...
Numerator: 1, 2, 4, 8, 16, 32 → this is a G.P. with first term = 1, common ratio = 2
Denominator: 1, 3, 9, 27, 81, 243 → this is a G.P. with first term = 1, common ratio = 3
So each term is:
Tₙ = (2ⁿ⁻¹)/(3ⁿ⁻¹) × (-1)ⁿ⁺¹ = (-1)ⁿ⁺¹ × (2/3)ⁿ⁻¹
Let’s write this series compactly:
S = ∑[n=1 to 12] (-1)ⁿ⁺¹ × (2/3)ⁿ⁻¹
So this is a geometric series with:
First term (a) = 1
Common ratio (r) = -2/3
Step 2: Use sum of G.P. formula
For a geometric progression:
Sₙ = a × [(1 - rⁿ) / (1 - r)], if |r| < 1
Here:
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a = 1
r = -2/3
n = 12
Let’s calculate:
S₁₂ = 1 × [(1 - (-2/3)¹²) / (1 - (-2/3))]
= [(1 - (4096 / 531441)) / (1 + 2/3)]
= [(1 - 4096/531441) / (5/3)]
Now, simplifying numerator:
Let’s write 1 as 531441/531441:
= [(531441 - 4096)/531441] ÷ (5/3)
= (527345 / 531441) × (3/5)
Now multiply:
= (527345 × 3) / (531441 × 5)
= 1582035 / 2657205
Now simplify the fraction:
Divide numerator and denominator by 15:
= 105469 / 177147
Final Answer:
The sum of the series 1 - 2/3 + 4/9 - 8/27 + … up to 12 terms is approximately:
≈ 0.5955 (approximately, after decimal evaluation).
Part (iii): If a, b, c, d are in G.P., show that:
1/(a² + b²), 1/(b² + c²), 1/(c² + d²) are in G.P.
This is a theoretical question and requires algebraic proof.
Step 1: Use the property of G.P.
Let a, b, c, d be in G.P.
So,
b = ar
c = ar²
d = ar³
(for some first term a and common ratio r)
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Let us now calculate the three terms:
T₁ = 1 / (a² + b²) = 1 / (a² + a²r²) = 1 / (a²(1 + r²))
T₂ = 1 / (b² + c²) = 1 / (a²r² + a²r⁴) = 1 / (a²r²(1 + r²))
T₃ = 1 / (c² + d²) = 1 / (a²r⁴ + a²r⁶) = 1 / (a²r⁴(1 + r²))
Step 2: Rewrite all three terms
Let us define K = 1 / (a²(1 + r²))
Then:
T₁ = K
T₂ = K / r²
T₃ = K / r⁴
Now we can write:
T₁ = K = K × r⁰
T₂ = K × r⁻²
T₃ = K × r⁻⁴
Clearly, these three terms form a G.P. because:
T₂² = T₁ × T₃
Let’s verify:
(K / r²)² = K × (K / r⁴)
(K² / r⁴) = (K² / r⁴)
Hence, proven.
Final Answer:
If a, b, c, d are in G.P., then
1 / (a² + b²), 1 / (b² + c²), 1 / (c² + d²)
are also in G.P.
Conclusion
In this complete solution, we have understood and applied the fundamental concepts of:
Arithmetic Progression (A.P.)
Formula for nth term: Tₙ = a + (n - 1)d
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Formula for sum of first n terms: Sₙ = (n/2)[2a + (n - 1)d]
Geometric Progression (G.P.)
nth term: Tₙ = arⁿ⁻¹
Sum of n terms: Sₙ = a(1 - rⁿ) / (1 - r), for |r| < 1
Alternating Series
Recognized patterns using powers of 2 and 3 and handled signs with (-1)ⁿ
Proof-based Algebra
Used substitution and algebraic identities to prove a statement
SECTION-B
3. (i)Explain the concepts of permutations and combinations.
(ii) Find the equation of straight line passing through the points (3,5) and (4, 7).
(iii) From the following data find linear demand curve:
Price Per Liter
(Milk)
Demand in Liters
Rs.1
100
Rs.2
50
Rs.3
0
Ans: 3(i) Concept of Permutations and Combinations
Let’s start with a simple story to understand permutations and combinations.
Understanding Through an Example
Imagine you are organizing a small contest at your college. You have 3 medals — gold, silver,
and bronze — and you have 5 participants. You need to decide who gets which medal.
Now comes the question: How many different ways can you distribute these 3 medals
among 5 participants?
This is where permutations come in.
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What is Permutation?
Permutation is a way of arranging or ordering things. When the order matters, we use
permutations.
Formula:
n = total number of objects
r = number of objects selected
! = factorial (e.g., 4!=4×3×2×1=24)
In Our Medal Example:
We have 5 people, and we want to give medals to 3 of them in order (Gold, Silver, Bronze):
So, there are 60 ways to distribute medals when the order matters.
What is Combination?
Combination is a way of selecting items where the order doesn’t matter.
Formula:
Used when order of selection is not important
Example:
Suppose you want to form a committee of 3 students out of 5, and you don’t care who is
first, second, or third, just the group matters.
So:
Thus, there are 10 ways to choose 3 members from 5 when the order is not important.
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Real-Life Application of Permutations & Combinations
Scenario
Permutation or Combination
Why?
Arranging books on a shelf
Permutation
Order matters
Choosing players for a team
Combination
Order doesn’t matter
Assigning ranks in a race
Permutation
1st, 2nd, 3rd matter
Selecting questions from a paper
Combination
No order required
Key Differences:
Permutations
Combinations
Order matters
Order doesn’t matter
More outcomes
Fewer outcomes
Used in arrangements
Used in selections
3(ii) Find the Equation of a Straight Line Passing Through the Points (3,5) and (4,7)
Let us now move to coordinate geometry, where we need to find the equation of a straight
line passing through two points.
Step 1: Understand What We’re Given
We are given two points:
A=(3,5)A = (3, 5)A=(3,5)
B=(4,7)B = (4, 7)B=(4,7)
We need to find the equation of the straight line that goes through both.
Step 2: Formula of a Line — Point-Slope Form
To find the equation, we first need the slope (m) of the line:
Slope Formula:
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Point-Slope Form of Line:
Using point (3, 5) and slope m=2m = 2m=2:
Now simplify:
Final Answer:
The equation of the straight line passing through the points (3,5) and (4,7) is:
3(iii) Find the Linear Demand Curve from the Given Data
Now let’s move into a real-world economic application: Demand Curve.
Given Data:
Price (Rs/Liter)
Demand (Liters)
Rs. 1
100
Rs. 2
50
Rs. 3
0
You are asked to find the linear demand curve, which means you must create an equation of
the form:
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Where:
Q is quantity demanded
P is price
a,b are constants
Step 1: Understand the Nature of the Data
From the table:
As Price increases, Demand decreases → This is a negative relationship, which is a typical
behavior in economics.
The demand falls at a constant rate: this implies a linear demand curve.
Step 2: Take Two Points From the Table
Let’s use two points to find the equation:
Point 1: (P1=1,Q1=100)(P_1 = 1, Q_1 = 100)(P1=1,Q1=100)
Point 2: (P2=2,Q2=50)(P_2 = 2, Q_2 = 50)(P2=2,Q2=50)
We are going to use the same two-point form as in coordinate geometry.
Find the Slope:
So, the slope a=−50a = -50a=−50
Step 3: Use Point-Slope Form Again
Let’s use:
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Final Answer:
The linear demand curve is:
This means:
At Rs. 1, Q=−50(1)+150=100Q = -50(1) + 150 = 100Q=−50(1)+150=100
At Rs. 2, Q=−50(2)+150=50Q = -50(2) + 150 = 50Q=−50(2)+150=50
At Rs. 3, Q=−50(3)+150=0Q = -50(3) + 150 = 0Q=−50(3)+150=0
So the equation perfectly fits the data.
Summary of All Three Parts
Part
Topic
Final Result
(i)
Permutations and Combinations
Explained with examples and formulas
(ii)
Line through (3,5) and (4,7)
y=2x−1y = 2x - 1y=2x−1
(iii)
Linear Demand Curve
Q=−50P+150Q = -50P + 150Q=−50P+150
Real-World Applications
Permutations & Combinations:
Used in probability, arrangements, lotteries, genetics, password generation, etc.
Straight Line Equation:
Used in geometry, physics (motion graphs), economics (cost/revenue functions), and data
analysis.
Linear Demand Curve:
Used in economics, marketing, price forecasting, and business planning to understand
consumer behavior.
Conclusion
In this lesson, we learned about three core mathematical and economic ideas:
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Permutations and combinations help us calculate the number of ways we can arrange or
select items — important for logic, counting, and probability.
The equation of a straight line lets us understand how two variables are related in a linear
way — useful in all sciences.
A linear demand curve represents the relation between price and quantity demanded —
crucial in economics and business.
Each of these topics is not only theoretical but also has practical uses in daily life, jobs, and
research. You’ll come across them in statistics, economics, computer science, and more.
4. (i) Define Set. Explain various types of Sets.
(ii) Explain Union, Intersection, difference and symmetric difference of Sets with the
help of Venn diagrams.
Ans: (i) Define Set. Explain various types of Sets
What is a Set?
Imagine you have a basket full of fruits. If someone asks you, “What fruits are in the
basket?” you might say, “Apple, Banana, Orange.” Now, you have just created a collection of
fruits.
In mathematics, such a collection is called a set.
Definition:
A set is a well-defined collection of distinct objects, considered as an object in its own right.
These objects are called the elements or members of the set.
The word well-defined means that it must be clear whether an object belongs to the set or
not.
How to represent a Set?
A set is usually represented by capital letters like A, B, C, and the elements are enclosed in
curly brackets { }.
Example:
Set of vowels in English = {a, e, i, o, u}
Set of natural numbers less than 5 = {1, 2, 3, 4}
If an element belongs to a set, we write:
a A (means a is an element of set A)
If it does not belong, we write:
b A (means b is not an element of set A)
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Types of Sets
Let’s now discuss the various types of sets in simple terms.
1. Empty Set (Null Set)
A set which has no elements is called an empty set.
Symbol: or {}
Example:
Set of natural numbers less than 1 =
Set of all square numbers between 2 and 3 =
2. Finite Set
If a set contains a countable number of elements, it is called a finite set.
Example:
A = {2, 4, 6, 8}
B = {January, February, March}
3. Infinite Set
If the number of elements in a set is uncountable or unlimited, it is called an infinite set.
Example:
Set of all natural numbers = {1, 2, 3, 4, …}
Set of points on a line = infinite
4. Equal Sets
Two sets are said to be equal if they contain exactly the same elements, regardless of order.
Example:
A = {1, 2, 3}
B = {3, 2, 1}
Here, A = B
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5. Equivalent Sets
Two sets are equivalent if they have the same number of elements, but not necessarily the
same elements.
Example:
A = {1, 2, 3}
B = {a, b, c}
A and B are equivalent but not equal.
6. Subset
A set A is a subset of set B if every element of A is also an element of B.
Symbol: A B
Example:
A = {2, 4}
B = {1, 2, 3, 4, 5}
A B
If A is a subset of B but not equal to B, it is called a proper subset.
7. Universal Set
A set which contains all the elements under consideration is called a universal set.
Symbol: Usually denoted by U
Example:
If U = {1, 2, 3, 4, 5, 6}, and A = {2, 4}, then A U
8. Power Set
The power set of a set is the set of all possible subsets, including the empty set and the set
itself.
Example:
A = {1, 2}
Power Set of A = {, {1}, {2}, {1, 2}}
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9. Singleton Set
A set with only one element is called a singleton set.
Example:
A = {5}
10. Disjoint Sets
Two sets are disjoint if they have no element in common.
Example:
A = {1, 2}, B = {3, 4}, then A ∩ B =
(ii) Explain Union, Intersection, Difference, and Symmetric Difference of Sets with Venn
Diagrams
Let’s explore the basic operations on sets with clear explanations and Venn diagrams to
visualize them.
1. Union of Sets
The union of two sets A and B is the set containing all elements that are in A or in B or in
both.
Symbol: A B
Example:
A = {1, 2, 3}, B = {3, 4, 5}
A B = {1, 2, 3, 4, 5}
Venn Diagram:
Imagine two overlapping circles. The union is the entire area covered by both circles.
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2. Intersection of Sets
The intersection of two sets A and B is the set containing only the elements that are
common to both A and B.
Symbol: A ∩ B
Example:
A = {1, 2, 3}, B = {2, 3, 4}
A ∩ B = {2, 3}
Venn Diagram:
The overlapping part between the two circles represents the intersection.
3. Difference of Sets
The difference of two sets A and B (A − B) is the set of elements that are in A but not in B.
Symbol: A − B
Example:
A = {1, 2, 3, 4}, B = {3, 4, 5}
A − B = {1, 2}
Venn Diagram:
Only the part of circle A that does not overlap with B.
Similarly, B − A would mean elements in B but not in A.
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4. Symmetric Difference of Sets
The symmetric difference of two sets A and B is the set of elements that are in either of the
sets but not in both.
Symbol: A B or A Δ B
Example:
A = {1, 2, 3}, B = {3, 4, 5}
A Δ B = {1, 2, 4, 5}
Venn Diagram:
The symmetric difference is represented by the two non-overlapping parts of the circles.
less
Why Study Sets?
Sets are the foundation of modern mathematics. Almost every mathematical structure is
built upon the concept of sets:
In Algebra, we talk about sets of numbers.
In Geometry, we deal with sets of points, lines, and shapes.
In Computer Science, we use sets for data structures, logic, and databases.
In Probability, we define events as sets of outcomes.
Understanding sets helps students develop logical thinking, understand classification, and
solve problems with clarity.
Conclusion
To summarize:
A Set is a collection of well-defined, distinct objects.
Sets are used to group, compare, and operate on elements.
There are many types of sets like empty set, finite set, infinite set, subset, power set, etc.
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We can perform operations on sets like union, intersection, difference, and symmetric
difference.
Venn diagrams help in visualizing these operations easily.
Mastering sets is essential because it acts like a gateway to higher concepts in mathematics
and computer science. If you understand sets clearly, you can explore other subjects like
relations, functions, logic, and programming more confidently.
SECTION-C
5. (i) Explain the concepts of constant and variable.
(ii) Define function, explain various types of function.
(iii) Evaluate Limit
Ans: (i) Explain the concepts of constant and variable
Let us imagine you are in a classroom. The teacher asks, “Who among you is always sitting
at the same place, no matter which class it is?” A student named Ravi raises his hand and
says, “Ma’am, I always sit at the first bench, first seat.” The teacher smiles and says, “You
are a constant.”
Now she turns to the rest of the students who change their seats every day. “And the rest of
you are variables because you keep changing your places.”
That’s the simplest way to understand constants and variables in mathematics.
What is a Constant?
In mathematics, a constant is a fixed value. It does not change regardless of the conditions
or operations performed.
Examples of constants:
Numbers like 2, -7, 3.14 (π), and 0 are constants.
In an equation like y = 5x + 3, the number 3 is a constant. It doesn’t change no matter what
value x takes.
Real-Life Examples:
The number of hours in a day (24).
The value of π (approximately 3.1416).
The boiling point of water at sea level (100°C).
What is a Variable?
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A variable is a symbol or letter that represents a value that can change. Variables are used
to generalize mathematical expressions or equations.
Examples of variables:
In the equation y = 5x + 3, x and y are variables.
If x = 1, y = 8; if x = 2, y = 13. The values are changing.
Real-Life Examples:
Your age (it changes every year).
The temperature of a room (it changes during the day).
The number of people visiting a website (it changes every hour).
Summary:
Term
Meaning
Changes?
Constant
Fixed value
No
Variable
Value that can change
Yes
(ii) Define Function and Explain Various Types of Functions
What is a Function?
Let’s go back to the classroom.
Imagine you are giving your notebook to a machine. The machine takes your notebook and
gives back a certain number depending on the number of pages.
If every student gives one notebook and gets one specific number back, then this is a
function.
Mathematical Definition:
A function is a relation between a set of inputs and a set of possible outputs where each
input is related to exactly one output.
It is usually written as:
f(x) = some expression
Where:
f is the name of the function.
x is the input variable.
f(x) is the output, also known as the value of the function at x.
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Example:
Let’s say:
f(x) = x²
If x = 2, then f(2) = 4
If x = 3, then f(3) = 9
So, this function takes a number and squares it.
Types of Functions
Let’s understand the most common types of functions in a friendly and simple way:
1. Constant Function
A constant function always gives the same output, no matter the input.
Example:
f(x) = 5
If x = 1, f(x) = 5
If x = 100, f(x) = 5
Think of it like a vending machine that always gives you a chocolate bar, no matter which
button you press!
2. Identity Function
It gives back the input as the output.
Example:
f(x) = x
If x = 2, f(x) = 2
If x = -5, f(x) = -5
It's like looking in a mirror—you see exactly what you are.
3. Linear Function
It’s a straight line when graphed.
Example:
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f(x) = 2x + 3
If x = 1, f(x) = 5
If x = 0, f(x) = 3
The equation of a straight line is a linear function.
4. Quadratic Function
It includes x² and makes a U-shaped graph (called a parabola).
Example:
f(x) = x² + 2x + 1
Used in physics and engineering (e.g., projectile motion).
5. Cubic Function
It includes x³ and makes an S-shaped curve.
Example:
f(x) = x³
The graph increases faster and has more twists.
6. Polynomial Function
A function made up of many powers of x.
Example:
f(x) = 3x⁴ + 2x² + 1
The general form is:
f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀
7. Rational Function
A function that has a polynomial in both the numerator and the denominator.
Example:
f(x) = (x² + 1)/(x + 1)
Defined only when the denominator is not zero.
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8. Exponential Function
The variable is in the exponent.
Example:
f(x) = 2^x
Used in compound interest, population growth, etc.
9. Logarithmic Function
The inverse of an exponential function.
Example:
f(x) = log(x)
10. Piecewise Function
Different expressions for different intervals of x.
Example:
Summary of Function Types:
Function Type
Example
Output Style
Constant
f(x) = 5
Same for all x
Identity
f(x) = x
Output = input
Linear
f(x) = 2x + 3
Straight line
Quadratic
f(x) = x²
U-shaped curve
Cubic
f(x) = x³
S-shaped curve
Polynomial
f(x) = x⁴ + 2x²
Mix of power terms
Rational
f(x) = (x²+1)/(x+1)
Division of polynomials
Exponential
f(x) = 2^x
Fast growth
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Function Type
Example
Output Style
Logarithmic
f(x) = log(x)
Slow growth
Piecewise
See above
Different rules for x intervals
(iii) Evaluate:
Lim x → 1 of (x³ - 1) / (x² - 1)
Let’s break this into a small problem-solving story.
We are given:
lim (x → 1) (x³ - 1)/(x² - 1)
First, try direct substitution:
(1³ - 1)/(1² - 1) = (1 - 1)/(1 - 1) = 0/0
Oops! This is an indeterminate form. We need to simplify.
Step 1: Factor numerator and denominator
We know:
x³ - 1 = (x - 1)(x² + x + 1)
x² - 1 = (x - 1)(x + 1)
So the expression becomes:
[(x - 1)(x² + x + 1)] / [(x - 1)(x + 1)]
Now cancel the common term (x - 1):
(x² + x + 1) / (x + 1)
Step 2: Now substitute x = 1
(1² + 1 + 1)/(1 + 1) = (1 + 1 + 1)/2 = 3/2
Final Answer:
lim (x → 1) (x³ - 1)/(x² - 1) = 3/2
Conclusion
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We have now understood:
Constants and Variables:
Constants are fixed values (like 3, 5).
Variables are changing values (like x, y).
Functions:
A function relates each input to one and only one output.
Types include: Constant, Linear, Quadratic, Rational, Exponential, and more.
Limit Evaluation:
The original form was 0/0 (indeterminate), but we simplified and got the answer 3/2.
6. (i) Show that f(x) =
is discoutinuous at x = 0
(ii) Find the derivative of c
X
by first principle method.
Ans: Understanding Continuity
Before we begin solving the problem, let's understand what continuity means in simple
terms. Suppose you are drawing the graph of a function without lifting your pen — that’s
continuity. Mathematically, a function f(x)f(x)f(x) is continuous at a point x=ax = ax=a if:
This means:
The function must be defined at x=ax = ax=a.
The limit of the function as x→ax \to ax→a must exist.
The limit must equal the value of the function at that point.
If any of these three conditions fail, the function is discontinuous at that point.
Our Function:
Given:
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We are told nothing about the function at x=0, so f(0)is not defined. Already, this violates
the first condition of continuity. But let's go deeper.
Check Left and Right Limits:
To check continuity at x=0, let us find the left-hand limit (LHL) and the right-hand limit (RHL)
as x→0
Case 1: As x→0+x \to 0^+x→0+ (approaching 0 from the right)
We know:
So:
Thus,
Case 2: As x→0 (approaching 0 from the left)
We know:
So:
Thus,
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Conclusion for Part (i):
We have:
Hence,
(ii) Derivative of f(x) = c
x
by First Principle
Understanding Derivative from First Principle
The derivative of a function f(x) at a point x using the first principle is given by:
This definition captures the instantaneous rate of change or the slope of the tangent to the
curve at a point.
Let’s Begin the Derivation
Let f(x)=c
x
. We want to find f′(x) using the first principle.
Step 1: Write the formula
Step 2: Use exponent laws
So,
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Important Observation:
Final Answer:
The derivative of c
x
by the first principle is:
Special Case: When c=e
Wrap-Up: Summary of Both Parts
Part (i): Discontinuity at x = 0
The left-hand limit and right-hand limit at x=0x = 0x=0 are not equal
Hence, the function is discontinuous at x=0x = 0x=0
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Part (ii): Derivative of c
x
Using the first principle:
Why This is Important for University Students
Understanding these topics deeply is crucial for success in higher mathematics:
Discontinuity tells us where functions break or fail to behave nicely — critical in physics,
engineering, and computer simulations.
First-principle differentiation is the foundation of calculus, which models change in
economics, biology, astronomy, and more.
Rather than memorizing formulas blindly, understanding the logic behind continuity and
differentiation will help you apply them better in real-world problems.
SECTION-D
7. (i) Find derivative of
w.r.t. x.
(ii) Find :
if y= log
(iii) Differentiate(1+x)
x
w.r.t. x.
Ans: Chapter: Understanding Differentiation Through Examples
Differentiation is one of the core concepts in calculus and is widely used in various branches
of science and engineering. It helps us understand how a function changes with respect to
its input. In simpler words, differentiation gives us the rate of change or slope of the curve
at any given point.
Let’s now understand the process of differentiation using the three problems given:
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Problem 7(i):
Find the derivative of
with respect to x
Step-by-Step Explanation:
This is a quotient of two functions:
Quotient Rule:
If
then the derivative is given by:
Apply the Quotient Rule:
Let:
Now use the formula:
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Let’s simplify each part carefully.
Numerator:
First Term:
Second Term:
So the full numerator becomes:
Final Answer for (i):
Problem 7(ii):
Find:
Step-by-Step Explanation:
This function involves a logarithmic function and a square root inside the log. We'll apply the
chain rule here.
Chain Rule:
If a function is in the form:
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Then,
Apply to our problem:
Let
Then,
Let’s differentiate the inner part:
So,
Therefore,
Final Answer for (ii):
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Problem 7(iii):
Differentiate (1+x)x with respect to x
Step-by-Step Explanation:
This is a unique type of function where both the base and the exponent contain variables.
Whenever we have a function like:
it is best to take logarithm on both sides and then use implicit differentiation.
Step 1: Take log on both sides
Let:
Step 2: Differentiate both sides
Differentiate implicitly:
Left side:
Right side:
Use product rule:
Why?
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Now equate both sides:
Multiply both sides by y:
But remember, y=(1+x)
x
Final Answer for (iii):
Summary of Key Concepts:
Concept
Description
Derivative
Measures the rate of change or slope of a function
Quotient Rule
Used when dividing two functions
Chain Rule
Used when a function is inside another function
Logarithmic Differentiation
Useful when both base and exponent are variables
Product Rule
Used when differentiating the product of two functions
Real-World Applications of Derivatives:
Physics: Finding velocity and acceleration
Economics: Analyzing cost and revenue functions
Biology: Growth rates of populations
Engineering: Studying stress-strain relationships
Medicine: Rate of spread of disease (epidemiology)
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Conclusion:
Differentiation is not just about solving mathematical expressions; it is a powerful tool that
explains how things change in the real world. Each of the problems we solved above shows
a different aspect of differentiation:
Quotient rule helps break down complex fractions
Chain rule shows how nested functions are treated
Logarithmic differentiation unlocks the secrets of variable powers
By understanding the rules and applying them step by step, any complex-looking problem
becomes manageable. Keep practicing these concepts with different types of functions to
master them!
8. (1) Find elasticity of demand, given the demand function
when q=9 and q=0
(ii) Demand Curve is given as p = (50 - x)/5 find MR at output = x. Also find MR when x = 0
and x = 25
Ans: Understanding Elasticity of Demand and Marginal Revenue
In economics, demand and revenue are two important concepts that help businesses make
decisions. Let's walk through them one by one and solve the two given problems in detail.
Part (1): Elasticity of Demand
What is Elasticity of Demand?
Imagine you're running a shop that sells cold drinks. If you lower the price of your drinks,
how much will your customers buy more? The answer to this question is called price
elasticity of demand. It tells us how sensitive the quantity demanded is to a change in price.
The formula for elasticity of demand is:
Given Demand Function:
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We need to find elasticity of demand (Ed) at:
q=9q = 9q=9
q=0q = 0q=0
Step 1: Find dp/dq
Step 2: Plug into the Elasticity Formula
When q = 9
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When q = 0
Conclusion for Part (1):
At q=9q = 9q=9, the elasticity is -0.75, meaning the demand is inelastic (price doesn’t affect
quantity much).
At q=0q = 0q=0, the elasticity is 0, meaning the demand is perfectly inelastic (no change in
quantity even if price changes).
Part (2): Marginal Revenue (MR)
What is Marginal Revenue?
Marginal Revenue is the extra revenue earned by selling one more unit of a product.
Where:
TR = Total Revenue = Price × Quantity
x = quantity/output
Given Demand Function:
We need to find:
The general Marginal Revenue (MR) formula
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MR when x=0x = 0x=0
MR when x=25x = 25x=25
Step 1: Find Total Revenue (TR)
Since:
So,
Step 2: Differentiate TR to get MR
Differentiate:
So the general formula for MR is:
Now calculate MR at given points
When x = 0:
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When x = 25:
Conclusion for Part (2):
The general MR function is 50−2x5\frac{50 - 2x}{5}550−2x
When x=0x = 0x=0, MR = 10
When x=25x = 25x=25, MR = 0
This means:
Initially, producing more units gives a high extra revenue.
But at some point (when x = 25), MR becomes zero, which means any further production
beyond this will not increase revenue—it might even reduce it.
Why This Matters
Let’s link this back to real business decisions:
Understanding elasticity helps in setting the right price. If the demand is inelastic, the seller
can raise prices without losing many customers.
Understanding marginal revenue helps a company to decide how much to produce. When
MR becomes zero or negative, producing more becomes unprofitable.
“This paper has been carefully prepared for educational purposes. If you notice any mistakes or
have suggestions, feel free to share your feedback.”
