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Algebra, Equations, Functions and Sequences Short Notes for GATE CS

Algebra, Equations, Functions and Sequences short notes for GATE CS: 8 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice

algebra equations functions and sequences short notes

Exam Readiness Checklist

Exam Readiness Checklist When you see a number properties or proportionality question, run through this sequence: Arbitrary Primes? Check parity. Product is never prime. Sum is even unless one prime is 2. Test with 2 and 3. Proportionality? Write it as an equation with a constant k . The ratio of the two quantities is fixed. Sums and Differences? If you see \frac{x + y}{x - y} = k, use componendo and dividendo to isolate \frac{x}{y} . Denominator Check? Before dividing, verify the denominator is not zero using "distinct" or "non-zero" constraints. Final check: Proportionality (A \propto B) means A = k B, not A = B . Never assume the constant of proportionality is 1 unless explicitly stated.

Identity for All x: Coefficient Matching

Identity for All x: Coefficient Matching THE PATTERN "If [equation] holds for all real values of x ..." The Technique: Rearrange equation to one side: f (x) = 0 Express as linear combination of independent functions Set each coefficient to zero EXAMPLE P e^{x} = Q e^{- x} for all x Rearrange: P e^{x} - Q e^{- x} = 0 Since e^{x} and e^{- x} are linearly independent: P = 0 and Q = 0

Symmetric Expressions: x/y + y/x

Symmetric Expressions: x/y + y/x THE PATTERN "Find the value of \frac{x}{y} + \frac{y}{x} " The Technique Let r = \frac{x}{y}, then: \frac{x}{y} + \frac{y}{x} = r + \frac{1}{r} 1. AM = GM condition (from log equations) Given equation implies x = y ⟹ r = 1 ⟹ Answer: 2 2. Quadratic in r Given equation becomes r^{2} - k r + 1 = 0 ⟹ r + \frac{1}{r} = k (Vieta's)

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Question 1

Level 1: Warm-up

How many pairs of positive integers (x, y) satisfy the equation lo g_{10} (\frac{x + y}{2}) = \frac{1}{2} (lo g_{10} x + lo g_{10} y) and the condition x + y \leq 12 ?

Question 2

Level 1: Warm-up

Consider the following statements: Assertion (A): If \frac{x + y}{x - y} = \frac{3}{1} for x \neq = y, then \frac{x}{y} = 2 . Reason (R): The componendo and dividendo rule states that if \frac{a}{b} = \frac{c}{d}, then \frac{a + b}{a - b} = \frac{c + d}{c - d}, which yields \frac{( x + y ) + ( x - y )}{( x + y ) - ( x - y )} = \frac{3 + 1}{3 - 1} ⟹ \frac{x}{y} = 2 . Which of the following is correct?

Question 3

Level 1: Warm-up

If ln (\frac{x + y}{2}) = \frac{1}{2} (ln x + ln y) for positive real numbers x and y, what is the value of \frac{x}{y} + \frac{y}{x} ?

Question 4

Level 1: Warm-up

Consider the following: <b>Assertion (A):</b> If lo g (x^{2} + y^{2}) = lo g x + lo g y + lo g 4 for positive real numbers x and y, then \frac{x}{y} + \frac{y}{x} = 4 . <b>Reason (R):</b> lo g x + lo g y = lo g (x + y) . Which of the following is correct?

Question 5

Level 1: Warm-up

Match the logarithmic equations in List I with their simplified algebraic forms in List II. <b>List I</b> P. lo g (a^{2} + b^{2}) = lo g a + lo g b + lo g 8 Q. lo g (a^{2} + b^{2}) = lo g a + lo g b + 2 lo g 2 R. lo g (a^{3} + b^{3}) = lo g a + lo g b + 2 lo g 3 <b>List II</b> a^{2} + b^{2} = 8 ab a^{2} + b^{2} = 4 ab a^{3} + b^{3} = 9 ab <b>Options:</b>

Question 6

Level 1: Warm-up

Consider the following Assertion and Reason: Assertion (A): If f (x) = g (y) for all real values of x and y, then f (x) must be a constant function. Reason (R): Because x and y are independent variables, changing x would change f (x) unless f (x) is constant, which would violate the equality since g (y) would remain unchanged. Which of the following is correct?

Question 7

Level 1: Warm-up

If f (x) and g (y) are functions such that f (x) = g (y) for all real values of x and y, and f (1) = 7, what is the value of g (10) - f (- 5) ?

Question 8

Level 1: Warm-up

Consider the functions f (t) = t^{2} and g (t) = 100 t for t > 0 . Which of the following statements is true?

Question 9

Level 1: Warm-up

Let r be a root of the equation x^{2} - 3 x - 4 = 0 . What is the maximum possible value of the expression r^{3} - 3 r^{2} + 2 r ?

Question 10

Level 1: Warm-up

If 2 e^{x} = 8 e^{- x}, which of the following statements about x is true?

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Algebra, Equations, Functions and Sequences Short Notes for GATE CS

Algebra, Equations, Functions and Sequences short notes for GATE CS: 8 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Exam Readiness Checklist

When you see a number properties or proportionality question, run through this sequence:

Arbitrary Primes? Check parity. Product is never prime. Sum is even unless one prime is 2. Test with 2 and 3.
Proportionality? Write it as an equation with a constant $k$ . The ratio of the two quantities is fixed.
Sums and Differences? If you see $\frac{x + y}{x - y} = k$ , use componendo and dividendo to isolate $\frac{x}{y}$ .
Denominator Check? Before dividing, verify the denominator is not zero using "distinct" or "non-zero" constraints.

Final check: Proportionality (

A \propto B

) means

A = k B

, not

A = B

. Never assume the constant of proportionality is 1 unless explicitly stated.

Identity for All x: Coefficient Matching

THE PATTERN

"If [equation] holds for all real values of $x$ ..."

The Technique:

Rearrange equation to one side: $f (x) = 0$
Express as linear combination of independent functions
Set each coefficient to zero

EXAMPLE

$P e^{x} = Q e^{- x} for all x$

Rearrange: $P e^{x} - Q e^{- x} = 0$

Since $e^{x}$ and $e^{- x}$ are linearly independent:

$P = 0 and Q = 0$

Symmetric Expressions: x/y + y/x

THE PATTERN

"Find the value of $\frac{x}{y} + \frac{y}{x}$ "

The Technique

Let $r = \frac{x}{y}$ , then:

$\frac{x}{y} + \frac{y}{x} = r + \frac{1}{r}$

1. AM = GM condition (from log equations)

Given equation implies $x = y ⟹ r = 1 ⟹$ Answer: $2$

2. Quadratic in r

Given equation becomes $r^{2} - k r + 1 = 0 ⟹ r + \frac{1}{r} = k$ (Vieta's)

Complete Formula Sheet

Exponential

Product:

a^{x + y}

Quotient:

a^{x - y}

Power:

a^{x y}

Zero:

1

Negative:

1/ a^{x}

Logarithmic

Product:

lo g x + lo g y

Quotient:

lo g x - lo g y

Power:

n lo g x

Base:

1

One:

0

Natural Log & Change of Base

Natural log:

ln x = lo g_{e} x

Change of base:

\frac{l n b}{l n a}

Inverse 1:

e^{l n x} = x

Inverse 2:

ln (e^{x}) = x

Critical Reminders

Log arguments must be strictly positive
$lo g (a + b) \neq = lo g a + lo g b$
When equation holds for all $x$ , match coefficients
For $\frac{x}{y} + \frac{y}{x}$ , check if $x = y$

Algebra, Equations, Functions and Sequences: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Quantitative Aptitude MCQ

How many pairs of positive integers $(x, y)$ satisfy the equation $lo g_{10} (\frac{x + y}{2}) = \frac{1}{2} (lo g_{10} x + lo g_{10} y)$ and the condition $x + y \leq 12$ ?

A.
4
B.
5
C.
6
D.
11

Correct Answer:

Step-by-Step Solution

Key idea: Simplify the right side to recognize the AM-GM equality condition.

Step 1: Simplify the right side using log rules.

$\frac{1}{2} (lo g_{10} x + lo g_{10} y) = \frac{1}{2} lo g_{10} (x y) = lo g_{10} ((x y)^{1/2}) = lo g_{10} (x y)$

Step 2: Equate the arguments since the logs are equal.

$\frac{x + y}{2} = x y$

Step 3: Recognize this as the Arithmetic Mean (AM) equals Geometric Mean (GM) condition.

For positive real numbers, AM = GM if and only if $x = y$ .

Step 4: Apply the condition $x + y \leq 12$ .

Since $x = y$ , we have $2 x \leq 12 ⟹ x \leq 6$ .

Step 5: Count the positive integer pairs.

$x$ can be $1, 2, 3, 4, 5, 6$ . Since $y = x$ , the pairs are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$ .

There are 6 such pairs.

Answer: 6

Question 2 · Quantitative Aptitude MCQ

Consider the following statements:

Assertion (A): If $\frac{x + y}{x - y} = \frac{3}{1}$ for $x \neq = y$ , then $\frac{x}{y} = 2$ .

Reason (R): The componendo and dividendo rule states that if $\frac{a}{b} = \frac{c}{d}$ , then $\frac{a + b}{a - b} = \frac{c + d}{c - d}$ , which yields $\frac{( x + y ) + ( x - y )}{( x + y ) - ( x - y )} = \frac{3 + 1}{3 - 1} ⟹ \frac{x}{y} = 2$ .

Which of the following is correct?

A.
Both A and R are true and R is the correct explanation of A
B.
Both A and R are true but R is NOT the correct explanation of A
C.
A is true but R is false
D.
Both A and R are false

Correct Answer:

Step-by-Step Solution

Insight: The componendo and dividendo rule directly transforms the given ratio into the target ratio.

Exam route: Apply the rule to $\frac{x + y}{x - y} = \frac{3}{1}$ to get $\frac{2 x}{2 y} = \frac{4}{2} ⟹ \frac{x}{y} = 2$ . Both A and R are true, and R explains A.

Learning route:

Step 1: Evaluate Assertion (A). We are given $\frac{x + y}{x - y} = 3$ . Cross-multiplying gives $x + y = 3 x - 3 y ⟹ 4 y = 2 x ⟹ \frac{x}{y} = 2$ . So, A is true.

Step 2: Evaluate Reason (R). The componendo and dividendo rule states that if $\frac{a}{b} = \frac{c}{d}$ , then $\frac{a + b}{a - b} = \frac{c + d}{c - d}$ .

Step 3: Apply the rule to the given equation: $\frac{( x + y ) + ( x - y )}{( x + y ) - ( x - y )} = \frac{3 + 1}{3 - 1}$ .

Step 4: Simplify: $\frac{2 x}{2 y} = \frac{4}{2} ⟹ \frac{x}{y} = 2$ . This matches A, and R is the direct method used to derive it. So, R is true and correctly explains A.

Trap warning: Do not assume the rule is invalid or miscalculate the sum/difference in the numerator and denominator.

Verification: Let $x = 2, y = 1$ . Then $\frac{2 + 1}{2 - 1} = \frac{3}{1}$ . And $\frac{x}{y} = \frac{2}{1} = 2$ . The rule holds perfectly.

Question 3 · Quantitative Aptitude MCQ

If $ln (\frac{x + y}{2}) = \frac{1}{2} (ln x + ln y)$ for positive real numbers $x$ and $y$ , what is the value of $\frac{x}{y} + \frac{y}{x}$ ?

A.
1
B.
4
C.
It cannot be determined.
D.
2

Correct Answer:

Step-by-Step Solution

Key idea: The right side is the log of the geometric mean, the left side is the log of the arithmetic mean.

Step 1: Simplify the right side: $\frac{1}{2} (ln x + ln y) = \frac{1}{2} ln (x y) = ln (x y)$ .

Step 2: Equate arguments: $\frac{x + y}{2} = x y$ .

Step 3: Recognize this as Arithmetic Mean = Geometric Mean. This equality holds only when $x = y$ .

Step 4: Substitute $x = y$ into the target expression: $\frac{x}{x} + \frac{x}{x} = 1 + 1 = 2$ .

Answer: D

Question 4 · Quantitative Aptitude MCQ

Consider the following:

Assertion (A): If $lo g (x^{2} + y^{2}) = lo g x + lo g y + lo g 4$ for positive real numbers $x$ and $y$ , then $\frac{x}{y} + \frac{y}{x} = 4$ .

Reason (R): $lo g x + lo g y = lo g (x + y)$ .

Which of the following is correct?

A.
Both A and R are true, and R is the correct explanation of A
B.
Both A and R are true, but R is NOT the correct explanation of A
C.
A is true, but R is false
D.
A is false, but R is true

Correct Answer:

Step-by-Step Solution

Key idea: Evaluate the assertion and reason separately.

Step 1: Check Reason (R)

The statement $lo g x + lo g y = lo g (x + y)$ is FALSE.

The correct identity is $lo g x + lo g y = lo g (x y)$ .

Step 2: Check Assertion (A)

Given: $lo g (x^{2} + y^{2}) = lo g x + lo g y + lo g 4$

Using the correct product rule on the right side:

$lo g x + lo g y + lo g 4 = lo g (x y) + lo g 4 = lo g (4 x y)$

Since logs are equal, arguments are equal:

$x^{2} + y^{2} = 4 x y$

Divide both sides by $x y$ :

$\frac{x ^{2}}{x y} + \frac{y ^{2}}{x y} = 4$

$\frac{x}{y} + \frac{y}{x} = 4$

So Assertion (A) is TRUE.

Step 3: Conclusion

A is true, R is false.

Answer: A is true, but R is false

Question 5 · Quantitative Aptitude MCQ

Match the logarithmic equations in List I with their simplified algebraic forms in List II.

List I

P. $lo g (a^{2} + b^{2}) = lo g a + lo g b + lo g 8$

Q. $lo g (a^{2} + b^{2}) = lo g a + lo g b + 2 lo g 2$

R. $lo g (a^{3} + b^{3}) = lo g a + lo g b + 2 lo g 3$

List II

$a^{2} + b^{2} = 8 ab$
$a^{2} + b^{2} = 4 ab$
$a^{3} + b^{3} = 9 ab$

Options:

A.
P-2, Q-1, R-3
B.
P-1, Q-2, R-3
C.
P-2, Q-3, R-1
D.
P-3, Q-1, R-2

Correct Answer:

Step-by-Step Solution

Key idea: Use log rules to combine the right side into a single log, then equate arguments.

Step 1: Simplify P.

$lo g a + lo g b + lo g 8 = lo g (ab) + lo g (8) = lo g (8 ab)$ .

Equating arguments: $a^{2} + b^{2} = 8 ab$ . (Matches 1)

Step 2: Simplify Q.

$lo g a + lo g b + 2 lo g 2 = lo g (ab) + lo g (2^{2}) = lo g (ab) + lo g (4) = lo g (4 ab)$ .

Equating arguments: $a^{2} + b^{2} = 4 ab$ . (Matches 2)

Step 3: Simplify R.

$lo g a + lo g b + 2 lo g 3 = lo g (ab) + lo g (3^{2}) = lo g (ab) + lo g (9) = lo g (9 ab)$ .

Equating arguments: $a^{3} + b^{3} = 9 ab$ . (Matches 3)

Step 4: Match the pairs.

P-1, Q-2, R-3.

Answer: P-1, Q-2, R-3

Question 6 · Quantitative Aptitude MCQ

Consider the following Assertion and Reason:

Assertion (A): If $f (x) = g (y)$ for all real values of $x$ and $y$ , then $f (x)$ must be a constant function.

Reason (R): Because $x$ and $y$ are independent variables, changing $x$ would change $f (x)$ unless $f (x)$ is constant, which would violate the equality since $g (y)$ would remain unchanged.

Which of the following is correct?

A.
Both A and R are true and R is the correct explanation of A.
B.
Both A and R are true but R is NOT the correct explanation of A.
C.
A is true but R is false.
D.
A is false but R is true.

Correct Answer:

Step-by-Step Solution

Key idea: If an equation holds for all independent values of two separate variables, both sides must be constant.

Step 1: Analyze Assertion (A). If $f (x) = g (y)$ for all $x, y$ , fix $y = y_{0}$ . Then $f (x) = g (y_{0})$ for all $x$ , meaning $f (x)$ is a constant. Thus, A is true.

Step 2: Analyze Reason (R). The reasoning correctly explains that since $x$ and $y$ vary independently, a non-constant $f (x)$ would change while $g (y)$ stays fixed, breaking the equality. Thus, R is true and correctly explains A.

Answer: A

Question 7 · Quantitative Aptitude MCQ

If $f (x)$ and $g (y)$ are functions such that $f (x) = g (y)$ for all real values of $x$ and $y$ , and $f (1) = 7$ , what is the value of $g (10) - f (- 5)$ ?

A.
15
B.
0
C.
-15
D.
12

Correct Answer:

Step-by-Step Solution

Key idea: If $f (x) = g (y)$ for all independent $x$ and $y$ , both functions must be the same constant.

Step 1: Since $f (x) = g (y)$ for all $x, y$ , $f (x)$ cannot depend on $x$ . Thus, $f (x) = c$ for all $x$ .

Step 2: Given $f (1) = 7$ , the constant $c = 7$ . So $f (x) = 7$ and $g (y) = 7$ for all inputs.

Step 3: Calculate $g (10) - f (- 5) = 7 - 7 = 0$ .

Answer: B

Question 8 · Quantitative Aptitude MCQ

Consider the functions $f (t) = t^{2}$ and $g (t) = 100 t$ for $t > 0$ . Which of the following statements is true?

A.
$f (t) > g (t)$ for all $t > 0$
B.
$f (t) < g (t)$ for all $t > 0$
C.
$f (t) > g (t)$ for all $t > 100$
D.
$f (t) < g (t)$ for all $t > 100$

Correct Answer:

Step-by-Step Solution

Key idea: Quadratic growth eventually dominates linear growth, but you must find the exact crossover point.

Step 1: Set $f (t) = g (t)$ to find the crossover: $t^{2} = 100 t ⟹ t (t - 100) = 0$ .

Step 2: The positive crossover is at $t = 100$ .

Step 3: For $t > 100$ , $t^{2} > 100 t$ , so $f (t) > g (t)$ .

Answer: C

Question 9 · Quantitative Aptitude MCQ

Let $r$ be a root of the equation $x^{2} - 3 x - 4 = 0$ . What is the maximum possible value of the expression $r^{3} - 3 r^{2} + 2 r$ ?

A.
24
B.
-6
C.
4
D.
10

Correct Answer:

Step-by-Step Solution

Key idea: Since $r$ is a root of $x^{2} - 3 x - 4 = 0$ , we have $r^{2} - 3 r - 4 = 0$ , which means $r^{2} - 3 r = 4$ . We can factor the target expression to use this substitution.

Step 1: Factor the target expression: $r^{3} - 3 r^{2} + 2 r = r (r^{2} - 3 r) + 2 r$ .

Step 2: Substitute $r^{2} - 3 r = 4$ : $r (4) + 2 r = 6 r$ .

Step 3: Find the roots of $x^{2} - 3 x - 4 = 0$ . Factoring gives $(x - 4) (x + 1) = 0$ , so the roots are $r = 4$ and $r = - 1$ .

Step 4: Evaluate $6 r$ for both roots to find the maximum:

For $r = 4$ : $6 (4) = 24$ .
For $r = - 1$ : $6 (- 1) = - 6$ .

The maximum possible value is 24.

Answer: A

Question 10 · Quantitative Aptitude MCQ

If $2 e^{x} = 8 e^{- x}$ , which of the following statements about $x$ is true?

A.
$x > 1$
B.
$x = 0.5$
C.
$0 < x < 1$
D.
$x < 0$

Correct Answer:

Step-by-Step Solution

Key idea: Isolate the exponential term by multiplying both sides by $e^{x}$ .

Step 1: Multiply both sides by $e^{x}$ :

$2 e^{x} \cdot e^{x} = 8 e^{- x} \cdot e^{x}$

$2 e^{2 x} = 8$

Step 2: Isolate $e^{2 x}$ :

$e^{2 x} = 4$

Step 3: Take natural log of both sides:

$2 x = ln 4$

$x = \frac{l n 4}{2} = \frac{2 l n 2}{2} = ln 2$

Step 4: Estimate $ln 2 \approx 0.693$ , so $0 < x < 1$ .

Answer: $0 < x < 1$

algebra equations functions and sequences short notes

Exam Readiness Checklist

Exam Readiness Checklist

Identity for All x: Coefficient Matching

Identity for All x: Coefficient Matching

Symmetric Expressions: x/y + y/x

Symmetric Expressions: x/y + y/x

Try a question

Login to view the complete short notes

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Algebra, Equations, Functions and Sequences Short Notes for GATE CS

Exam Readiness Checklist

Exam Readiness Checklist

Identity for All x: Coefficient Matching

Identity for All x: Coefficient Matching

Symmetric Expressions: x/y + y/x

Symmetric Expressions: x/y + y/x

Complete Formula Sheet

Complete Formula Sheet

Algebra, Equations, Functions and Sequences: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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