Sets, Relations and Functions: JEE Main 2026 PYQs

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Key idea: Domain of composite functions involving logarithm and square root with rational expressions

Answer: Option (1)

Key idea: Trigonometric identities, Set theory, and Range analysis

1. Step 1: Simplify $\cos^8\theta - \sin^8\theta$ using factorization: $\cos^8\theta - \sin^8\theta = (\cos^4\theta - \sin^4\theta)(\cos^4\theta + \sin^4\theta) = \cos 2\theta \cdot \frac{1 + \cos^2 2\theta}{2}$
2. Step 2: Substitute into the given expression: $2(\cos^8\theta - \sin^8\theta) \sec 2\theta = 2 \cdot \cos 2\theta \cdot \frac{1 + \cos^2 2\theta}{2} \cdot \frac{1}{\cos 2\theta} = 1 + \cos^2 2\theta$
3. Step 3: Find the range: Since $0 \leq \cos^2 2\theta \leq 1$, we have $1 \leq a^2 \leq 2$
4. Step 4: Check integer solutions: Only $a^2 = 1$ (giving $a = \pm 1$) is a perfect square in this range
5. Step 5: Verify if $a^2 = 1$ is achievable: $a^2 = 1$ requires $\cos 2\theta = 0$, which gives $\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, ...$ where $\tan^2\theta = 1$
6. Step 6: Apply constraint: These values are excluded from $P$ since $\tan^2\theta \neq 1$ is required. Thus $S = \emptyset$ and $n(S) = 0$

Answer: 120

Key idea: Relations on Cartesian Products with Divisibility and Ordering Conditions

1. Understand that $R$ is a relation on $A \times A$, so elements are of the form $((x,y), (z,w))$ where $(x,y)R(z,w)$ means $x|z$ and $y \leq w$
2. Count valid $(x,z)$ pairs where $x|z$ and $x,z \in A$: For $x=2$: $z \in \{2,4,6\}$ (3 pairs); $x=3$: $z \in \{3,6\}$ (2 pairs); $x=4,5,6$: one pair each. Total: $3+2+1+1+1=8$ pairs
3. Count valid $(y,w)$ pairs where $y \leq w$ and $y,w \in A = \{2,3,4,5,6\}$: This gives $5+4+3+2+1=15$ pairs
4. Since the conditions on $(x,z)$ and $(y,w)$ are independent, multiply: Total elements $= 8 \times 15 = 120$

Answer: Option (1)

Key idea: Domain of inverse trigonometric functions with greatest integer function

1. For $\cos^{-1}\left(\frac{4x+2[x]}{3}\right)$ to be defined, we need $-1 \leq \frac{4x+2[x]}{3} \leq 1$, which gives $-3 \leq 4x+2[x] \leq 3$
2. Let $[x] = n$ (integer) and $x = n + \{x\}$ where $0 \leq \{x\} < 1$. Substituting: $-3 \leq 6n + 4\{x\} \leq 3$
3. For solutions to exist with $0 \leq \{x\} < 1$, we need $n \leq \frac{1}{2}$ and $n > -\frac{7}{6}$, so $n \in \{-1, 0\}$
4. Case 1: $n=0$ gives $0 \leq 4x \leq 3$ with $0 \leq x < 1$, yielding $x \in [0, \frac{3}{4}]$
5. Case 2: $n=-1$ gives $\frac{3}{4} \leq \{x\} < 1$ with $-1 \leq x < 0$, yielding $x \in [-\frac{1}{4}, 0)$
6. Domain is $[-\frac{1}{4}, \frac{3}{4}]$, so $\alpha = -\frac{1}{4}$, $\beta = \frac{3}{4}$
7. Therefore, $12(\alpha + \beta) = 12(\frac{1}{2}) = 6$

Answer: Option (2)

Key idea: Counting functions using inclusion-exclusion principle; Onto (surjective) functions

1. Calculate total functions from $\{1,2,3,4\}$ to $\{a,b,c\}$: Total = $3^4 = 81$
2. Use inclusion-exclusion to find non-onto functions. Functions missing at least one element in range.
3. Functions using only 2 elements: $\binom{3}{1} \cdot 2^4 = 3 \times 16 = 48$
4. Functions using only 1 element: $\binom{3}{2} \cdot 1^4 = 3 \times 1 = 3$
5. By inclusion-exclusion: Non-onto = $48 - 3 = 45$
6. Alternatively: Onto functions = $3^4 - 3 \cdot 2^4 + 3 \cdot 1^4 = 81 - 48 + 3 = 36$, so non-onto = $81 - 36 = 45$

Answer: Option (1)

Key idea: Inverse Functions and Fixed Points

1. Step 1: Find the inverse function. From $y = (x-1)^4 + 1$, we get $f^{-1}(x) = (x-1)^{1/4} + 1$
2. Step 2: For Statement (I), solve $(x-1)^4 = (x-1)^{1/4}$. Let $u = x-1$, then $u^4 = u^{1/4}$
3. Step 3: This gives $u^{1/4}(u^{15/4} - 1) = 0$, so $u = 0$ or $u = 1$, meaning $x = 1$ or $x = 2$
4. Step 4: Verify: $f(1) = f^{-1}(1) = 1$ and $f(2) = f^{-1}(2) = 2$. Statement (I) is TRUE
5. Step 5: For Statement (II), solve $(x-1)^4 = x^{1/4}$. Let $g(x) = (x-1)^4 - x^{1/4}$
6. Step 6: Check: $g(1) = -1 < 0$ and $g(16) = 50625 - 2 > 0$. By IVT, a solution exists
7. Step 7: Statement (II) is FALSE. Therefore, only (I) is TRUE

Answer: Option (2)

Key idea: Functional Equations and Quadratic Functions

1. Assume $f(x) = bx^2 + cx + d$ and use $f(0) = -1$ to get $d = -1$
2. Substitute into functional equation: $b(x+y)^2 + c(x+y) - 1 = bx^2 + cx - 1 + 2y^2 + y + axy$
3. Compare coefficients: $xy$ gives $2b = a$, $y^2$ gives $b = 2$, $y$ gives $c = 1$
4. Thus $a = 4$ and $f(x) = 2x^2 + x - 1$. Verify: $f(1) = 2(1) + 1 - 1 = 2$ ✓
5. Calculate: $f(1)=2, f(2)=9, f(3)=20, f(4)=35, f(5)=54$
6. Find $\sum_{n=1}^{5} (4 + f(n)) = 6 + 13 + 24 + 39 + 58 = 140$

Answer: Option (1)

Key idea: Counting non-negative integer solutions to linear Diophantine equations

1. Step 1: For the equation $a + b + 2c = 22$ with $a, b, c \geq 0$, find the range of $c$: Since $2c \leq 22$, we get $c \in \{0, 1, 2, ..., 11\}$
2. Step 2: For each fixed value of $c$, we need to count solutions to $a + b = 22 - 2c$ where $a, b \geq 0$
3. Step 3: The number of non-negative integer solutions to $a + b = k$ is $(k + 1)$
4. Step 4: For $c = 0, 1, 2, ..., 11$, the number of solutions are $23, 21, 19, ..., 3, 1$ respectively
5. Step 5: Total solutions: $n(A) = 23 + 21 + 19 + ... + 3 + 1 = \sum_{i=1}^{12}(2i-1) = 12^2 = 144$
6. Step 6: However, if we consider $c \in \{0, 1, ..., 10\}$ (11 values), we get $n(A) = 23 + 21 + ... + 3 = 11^2 = 121$

Answer: 72

Key idea: One-one functions with constraints, counting principle

1. From $f(2) + f(3) = 5$ and $f(3) \leq 4$, find possible pairs: $(f(2), f(3)) \in \{(4,1), (3,2), (2,3), (1,4)\}$
2. For each pair, determine valid values of $f(1)$ using $f(1) \geq 3$ and one-one property (all three values must be distinct)
3. In each case, $f(1)$ has exactly 3 valid choices from the remaining elements $\geq 3$
4. After fixing $f(1), f(2), f(3)$, the remaining three elements $\{4,5,6\}$ map to remaining three values in $3! = 6$ ways
5. Each of the 4 cases contributes $3 \times 6 = 18$ functions
6. Total number of functions = $4 \times 18 = 72$

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Key idea: Trigonometric equations using product-to-sum formulas

Answer: Option (2)

Key idea: Domain of inverse trigonometric functions with greatest integer function

1. For $\sin^{-1}\left(\frac{x + [x]}{3}\right)$ to be defined, we need $-1 \leq \frac{x + [x]}{3} \leq 1$, which gives $-3 \leq x + [x] \leq 3$
2. Write $x = [x] + \{x\}$ where $\{x\}$ is the fractional part with $0 \leq \{x\} < 1$. Then $x + [x] = 2[x] + \{x\}$
3. For $[x] = -1$: We get $-3 \leq -2 + \{x\} \leq 3$, giving $0 \leq \{x\} < 1$, so $x \in [-1, 0)$
4. For $[x] = 0$: We get $-3 \leq \{x\} \leq 3$, giving $0 \leq \{x\} < 1$, so $x \in [0, 1)$
5. For $[x] = 1$: We get $-3 \leq 2 + \{x\} \leq 3$, giving $0 \leq \{x\} < 1$, so $x \in [1, 2)$
6. For $[x] \geq 2$ or $[x] \leq -2$, the inequality fails. Domain is $[-1, 2)$, so $\alpha = -1, \beta = 2$
7. Therefore, $\alpha^2 + \beta^2 = (-1)^2 + 2^2 = 1 + 4 = 5$

Answer: Option (4)

Key idea: Testing injectivity and surjectivity of rational functions using derivatives and range analysis

1. Verify domain: Check $3x^2 + x + 3 = 0$. Discriminant $= 1 - 36 = -35 < 0$, so denominator is always positive. Function is well-defined on $\mathbb{R}$.
2. Test one-one: Find $f'(x) = \frac{11(x^2-1)}{(3x^2+x+3)^2}$ using quotient rule.
3. Analyze $f'(x)$: Since $f'(x) < 0$ for $x \in (-1,1)$ and $f'(x) > 0$ for $x \in (-\infty,-1) \cup (1,\infty)$, function is not monotonic. Hence NOT one-one.
4. Test onto: Find critical values $f(-1) = \frac{7}{5}$ (local min) and $f(1) = \frac{1}{7}$ (local max).
5. As $x \to \pm\infty$, $f(x) \to \frac{2}{3}$. Range is $[\frac{1}{7}, \frac{7}{5}] \neq \mathbb{R}$. Hence NOT onto.
6. Conclusion: $f$ is neither one-one nor onto.

Answer: 276

Key idea: Transitive Relations and Ordered Pairs

1. Step 1: From $\log_5(x + y) \leq 2$, we get $x + y \leq 25$. So $R = \{(x, y) \in \mathbb{N} \times \mathbb{N} : x + y \leq 25\}$.
2. Step 2: For transitivity, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c)$ must be in $R$. This fails when $a + b \leq 25$, $b + c \leq 25$, but $a + c > 25$.
3. Step 3: A pair $(a, c)$ with $a + c > 25$ needs to be added if there exists $b \in \mathbb{N}$ such that $a + b \leq 25$ and $b + c \leq 25$.
4. Step 4: This requires $b \leq \min(25-a, 25-c)$ and $b \geq 1$, which is possible when $\max(a, c) \leq 24$.
5. Step 5: Count pairs $(a, c)$ where $a, c \in \{1, 2, ..., 24\}$ and $a + c > 25$. For each $a \in \{2, ..., 24\}$, we have $c \in \{26-a, ..., 24\}$.
6. Step 6: For $a = 2$: 1 pair, $a = 3$: 2 pairs, ..., $a = 24$: 23 pairs. Total = $1 + 2 + ... + 23 = \frac{23 \times 24}{2} = 276$.

Answer: Option (1)

Key idea: Relations: Counting elements and checking equivalence relation properties (reflexive, symmetric, transitive)

1. The relation $R$ is defined by $(a,b) \in R$ iff $1 + ab > 0$, i.e., $ab > -1$
2. Check all 25 ordered pairs from $\{-2,-1,0,1,2\} \times \{-2,-1,0,1,2\}$ systematically
3. Count pairs satisfying $ab > -1$: For $a=0$, all 5 pairs work. For other values, check each product. Total = 17 elements. Statement I is **true**
4. Check reflexivity: $(a,a) \in R$ requires $a^2 > -1$, which holds for all $a$. ✓
5. Check symmetry: If $ab > -1$, then $ba > -1$ (commutative). ✓
6. Check transitivity: Counterexample: $(-2,0) \in R$ and $(0,1) \in R$, but $(-2,1) \notin R$ since $1+(-2) = -1 \not> 0$. ✗
7. $R$ is not an equivalence relation. Statement II is **false**. Answer: Only I is true

Answer: 2

Key idea: Logarithmic equations with variable bases and change of base formula

1. Factor the arguments: $2x^2+5x+3 = (2x+3)(x+1)$ and $x^2+2x+1 = (x+1)^2$
2. Apply logarithm properties to get: $\log_{(x+1)}(2x+3) + 1 = 4 - 2\log_{(2x+3)}(x+1)$
3. Let $t = \log_{(x+1)}(2x+3)$, then $\log_{(2x+3)}(x+1) = \frac{1}{t}$, giving $t = 3 - \frac{2}{t}$
4. Solve $t^2 - 3t + 2 = 0$ to get $t = 1$ or $t = 2$
5. For $t=1$: $x+1 = 2x+3 \Rightarrow x = -2$ (invalid: base $x+1 = -1 < 0$)
6. For $t=2$: $(x+1)^2 = 2x+3 \Rightarrow x^2 = 2 \Rightarrow x = \pm\sqrt{2}$. Only $x = \sqrt{2}$ is valid (check bases are positive and $\neq 1$)
7. Sum of squares: $(\sqrt{2})^2 = 2$

Answer: 395

Key idea: Functional equations with logarithms, polynomial functions, and summation formulas

1. Evaluate the geometric series: $2 + \frac{2}{3} + \frac{2}{9} + ... = 3$
2. Simplify the equation using $\log_2(3) = \frac{1}{\log_3(2)}$ to get $f(x) = 1 + \frac{f(x)}{f(1/x)}$
3. Rearrange to obtain $(f(x) - 1)(f(1/x) - 1) = 1$
4. For polynomial $f(x)$, this implies $f(x) = x^k + 1$ for some integer $k$
5. Use $f(6) = 37$ to find $6^k = 36$, so $k = 2$ and $f(x) = x^2 + 1$
6. Calculate $\sum_{n=1}^{10} f(n) = \sum_{n=1}^{10} (n^2 + 1) = \frac{10 \cdot 11 \cdot 21}{6} + 10 = 385 + 10 = 395$