A group of 40 students appeared in an examination of 3 subjects - Mathematics, Physics & Chemistry. It was found that all students passed in at least one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, at most 11 students passed in both Mathematics and Physics, at most 15 students passed in both Physics and Chemistry, at most 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is

Numerical Answer Type:

Enter a numerical value Answer: 10 to 10 +4 marks

Solution & Explanation

### Related Formula

n(M cup P cup C) = n(M) + n(P) + n(C) - n(M cap P) - n(P cap C) - n(M cap C) + n(M cap P cap C)

### Core Logic

Set Operations diagram for Q21 - JEE Main 2024 Morning

Let

x

be the number of students who passed in all three subjects, so

n(M cap P cap C) = x

. We are given:

n(M) = 20, n(P) = 25, n(C) = 16

n(M cup P cup C) = 40

Constraints:

n(M cap P) le 11

n(P cap C) le 15

n(M cap C) le 15

If we let

n(M cap P) = 11

, then the region representing only

M cap P

11 - x

. For this to be non-negative,

x le 11

. ### Step 1: Testing x = 11

Assume

x = 11

. To maximize, let's fix

n(M cap P) = 11

. This leaves

0

students in

(M cap P) - (M cap P cap C)

. Let

n(M cap C) = 11 + z

and

n(P cap C) = 11 + y

. Constraints limit these subsets:

11 + z le 15 Rightarrow z le 4

11 + y le 15 Rightarrow y le 4

Filling the Venn diagram nodes: Only

M = 20 - 11 - z - 0 = 9 - z

Only

P = 25 - 11 - y - 0 = 14 - y

Only

C = 16 - 11 - y - z = 5 - y - z

Summing all disjoint areas: Total

= (9 - z) + 0 + (14 - y) + z + 11 + y + (5 - y - z) = 40

39 - y - z = 40 Rightarrow y + z = -1

Since subsets cannot be negative,

x=11

is impossible. ### Step 2: Testing x = 10

Assume

x = 10

. Let's construct a valid distribution with

x = 10

. We can choose intersection values:

n(M cap P) = 10

, so only

M cap P

0

n(M cap C) = 11

, so only

M cap C

1

n(P cap C) = 10

, so only

P cap C

0

. This gives: Only

M = 20 - 10 - 0 - 1 = 9

Only

P = 25 - 10 - 0 - 0 = 15

Only

C = 16 - 10 - 1 - 0 = 5

Summing up =

9 + 15 + 5 + 0 + 1 + 0 + 10 = 40

. All conditions are satisfied perfectly. Hence, the maximum number is 10. ### Pattern Recognition In multi-constraint Venn diagram maximization, test the theoretical upper bound sequentially downward until a non-negative configuration for all disjoint subsets is achieved. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sets

Reference Study Guides

Sets

More Sets Previous-Year Questions — Page 3

Q54 jee_main_2025_28_jan_morning Functional Equations and Series

f(x) = frac2^x2^x + sqrt2

x in mathbbR

, then

sum_k=1^81 fleft(frack82right)

is equal to: (1) 41 (2)

frac812

(3) 82 (4)

81sqrt2

A. 41
B. $frac812$ $\frac{81}{2}$
C. 82
D. $81sqrt2$ $81\sqrt{2}$

Solution

### Related Formula Symmetric identity wrapper for matching indices:

f(x) + f(1-x) = 1

### Core Logic Let's evaluate

f(x) + f(1-x)

f(x) + f(1-x) = frac2^x2^x + sqrt2 + frac2^1-x2^1-x + sqrt2

= frac2^x2^x + sqrt2 + frac22 + sqrt2cdot 2^x = frac2^x + sqrt22^x + sqrt2 = 1

### Step 1: Expanding the Series Pairing matching terms from opposite ends of the summation:

sum_k=1^81 fleft(frack82right) = left[fleft(frac182right) + fleft(frac8182 ight)right] + dots + fleft(frac4182 ight)

There are 40 complete pairs matching the

f(x) + f(1-x) = 1

identity, plus one lone center term

fleft(frac12right)

. ### Step 2: Computing Final Valuation

textSum = 40 + fleft(frac12right) = 40 + fracsqrt2sqrt2 + sqrt2 = 40 + frac12 = frac812

### Pattern Recognition When encountering fractional summation bounds, always check the sum of components

x + (1-x)

to find linear reduction templates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series Class 12 Maths: Relations and Functions

Q55 jee_main_2025_28_jan_morning Functional Relations and Properties

Let

f: mathbbR to mathbbR

be a function defined by

f(x) = (2 + 3a)x^2 + left( fraca + 2a - 1 right)x + b, a neq 1

. If

f(x + y) = f(x) + f(y) + 1 - frac27xy,

then the value of

28sum_i = 1^5|f(i)|

is: (1) 715 (2) 735 (3) 545 (4) 675

A. 715
B. 735
C. 545
D. 675

Solution

### Related Formula Given functional property equation:

f(x + y) = f(x) + f(y) + 1 - frac27xy

### Core Logic Substitute

x = y = 0

into the property equation:

f(0) = 2f(0) + 1 implies f(0) = -1

. Since

f(0) = b

, we instantly find

b = -1

. ### Step 1: Extracting Parameter Values Substitute

y = -x

into the property equation:

f(0) = f(x) + f(-x) + 1 + frac27x^2

-1 = 2(3a + 2)x^2 + 2b + 1 + frac27x^2

Matching coefficients for

x^2

gives:

6a + 4 + frac27 = 0 implies a = -frac57

Therefore, the absolute functional identity is:

f(x) = -frac17x^2 - frac34x - 1

### Step 2: Computing the Target Series Rewriting using common denominators:

|f(x)| = frac128|4x^2 + 21x + 28|

Evaluating for

i=1

5

28 sum_i = 1^5 |f(i)| = 675

### Pattern Recognition Substituting standard points like

0

and

-x

decouples symmetric multi-variable systems with maximum efficiency. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Relations and Functions

Q57 jee_main_2025_28_jan_morning Equivalence Relations

The relation

R = \(x, y) : x, y in mathbbZ text and x + y text is even\

is:

A. reflexive and transitive but not symmetric
B. reflexive and symmetric but not transitive
C. an equivalence relation
D. symmetric and transitive but not reflexive

Solution

### Related Formula An equivalence relation must be simultaneously reflexive, symmetric, and transitive. ### Core Logic Let's check each property sequentially: 1. **Reflexive:** For any

x in mathbbZ

x + x = 2x

, which is always even. Thus,

(x, x) in R

. 2. **Symmetric:** If

x + y

is even, then

y + x

must also be even due to commutative addition. Thus, if

(x, y) in R implies (y, x) in R

. 3. **Transitive:** If

x + y

is even and

y + z

is even, then adding them gives

(x + y) + (y + z) = x + 2y + z = texteven implies x + z = texteven - 2y = texteven

. Thus,

(x, z) in R

. ### Step 1: Final Property Summary Since all three criteria are satisfies simultaneously,

R

is an equivalence relation. ### Pattern Recognition Parity relation properties (even/odd checking sums) over integer sets universally form clean modular equivalence systems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Relations and Functions

Q55 jee_main_2025_03_april_morning Types of Relations

Let

A = \-3, -2, -1, 0, 1, 2, 3\

[cite: 555]. Let

R

be a relation on

A

defined by

xRy

if and only if

0 le x^2 + 2y le 4

[cite: 555]. Let

l

be the number of elements in

R

and

m

be the minimum number of elements required to be added in

R

to make it a reflexive relation[cite: 556, 557]. Then

l + m

is equal to[cite: 558]:

A. 19
B. 20
C. 17
D. 18

Solution

### Related Formula 1. Elements in a relation satisfy the exact range constraint. 2. Reflexive criteria: For every

x in A

(x, x) in R

. ### Core Logic Rewrite the inequality to isolate variables systematically [cite: 1235]:

-2y le x^2 le 4-2y

[cite: 1235] Test every valid value of

y in A

to discover valid integer values for

x

[cite: 1237, 1238, 1241, 1259, 1260, 1261]: -

y = -3 implies 6 le x^2 le 10 implies x in \-3, 3\

y = -2 implies 4 le x^2 le 8 implies x in \-2, 2\

y = -1 implies 2 le x^2 le 6 implies x in \-2, 2\

y = 0 implies 0 le x^2 le 4 implies x in \-2, -1, 0, 1, 2\

y = 1 implies -2 le x^2 le 2 implies x in \-1, 0, 1\

y = 2 implies -4 le x^2 le 0 implies x in \0\

y = 3 implies -6 le x^2 le -2 implies textNo real x text exists

### Step 1: Listing set elements and counting Compile all distinct matching coordinate pairs

(x,y)

into set

R

[cite: 1264]:

R = \(-3,-3), (-3,3), (-2,-2), (-2,2), (-1,-2), (-1,2), (0,-2), (0,-1), (0,0), (0,1), (0,2), (1,-1), (1,0), (1,1), (2,0)\

[cite: 1264] Counting elements gives [cite: 1265]:

l = 15

[cite: 1265] To make the relation reflexive, the pairs

(-3,-3), (-2,-2), (-1,-1), (0,0), (1,1), (2,2), (3,3)

must all belong to

R

. Checking missing elements [cite: 1267]:

\(-1,-1), (2,2), (3,3)\ implies m = 3

[cite: 1267] Sum of variables [cite: 1268]:

l + m = 15 + 3 = 18

[cite: 1268] ### Pattern Recognition Isolating terms explicitly via a variable-by-variable bounded testing grid avoids missing distinct coordinate boundary values. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions

Q58 jee_main_2025_03_april_morning Domain of Functions

If the domain of the function

f(x) = log_e left(frac2x - 35 + 4xright) + sin^-1 left(frac4 + 3x2 - x ight)

[alpha, beta)

[cite: 598], then

alpha^2 + 4beta

is equal to[cite: 599]:

A. 5
B. 4
C. 3
D. 7

Solution

### Related Formula 1. For

log(g(x))

, we require

g(x) > 0

. 2. For

sin^-1(h(x))

, we require

-1 le h(x) le 1

. ### Core Logic Evaluate constraints independently [cite: 1307, 1309]: **Constraint 1 (Logarithmic Argument):** [cite: 1307]

frac2x-34x+5 > 0 implies x in left(-infty, -frac54right) cup left(frac32, inftyright)

[cite: 1309] **Constraint 2 (Arcsine Argument):** [cite: 1307]

-1 le frac3x+42-x le 1

[cite: 1309] ### Step 1: Solving the Arcsine inequalities Split inequality into separate conditional frames [cite: 1311]: Left frame:

frac3x+42-x + 1 ge 0 implies frac2x+62-x ge 0 implies fracx+3x-2 le 0 implies x in [-3, 2)

Right frame:

frac3x+42-x - 1 le 0 implies frac4x+22-x le 0 implies frac2x+1x-2 ge 0 implies x in left(-infty, -frac12right] cup (2, infty)

Intersecting both sets gives [cite: 1311]:

x in left[-3, -frac12right]

[cite: 1311] ### Step 2: Final Intersection and Value Solving Intersect Log constraint with Arcsine constraint solution range [cite: 1311]:

x in left[-3, -frac12right] cap left[left(-infty, -frac54right) cup left(frac32, inftyright)right] = left[-3, -frac54right)

[cite: 1311] Thus, identify parameters [cite: 1312]:

alpha = -3, quad beta = -frac54

[cite: 1312] Compute the requested expression value [cite: 1312]:

alpha^2 + 4beta = (-3)^2 + 4left(-frac54right) = 9 - 5 = 4

[cite: 1312] ### Pattern Recognition When dealing with fractional variables inside boundaries, flipping inequalities according to denominator signs prevents fatal zone misinterpretations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions

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