Sets, Relations & Functions — Previous Year Questions (JEE Main 2026)

$\cos^{-1}(t)$ needs $-1 \le t \le 1$, so

Test the two integer strips that can satisfy this:

• For $x \in [0,1)$, $[x] = 0$: need $-3 \le 4x \le 3 \Rightarrow x \le \tfrac34$. So $x \in \left[0, \tfrac34\right]$.
• For $x \in [-1,0)$, $[x] = -1$: need $-3 \le 4x - 2 \le 3 \Rightarrow -\tfrac14 \le x < 0$.

Outside these strips $4x + 2[x]$ leaves $[-3,3]$. Hence the domain is

Answer: A (6)

Write the left side as $\left|x + (x^2 - 9)\right|$. Using $|A + B| = |A| + |B|$ iff $A$ and $B$ have the same sign (or either is $0$), with $A = x$ and $B = x^2 - 9$:

Factor: $x(x-3)(x+3) \ge 0$. A sign analysis gives

So $\alpha = -3,\ \beta = 0,\ \gamma = 3$, and

Answer: B (18)

$\sin^{-1}(t)$ needs $-1 \le t \le 1$, so

Check each integer strip:

• $x \in [-1, 0)$, $[x] = -1$: $x + [x] = x - 1 \in [-2,-1)$, valid $\Rightarrow$ all of $[-1,0)$.
• $x \in [0, 1)$, $[x] = 0$: $x + [x] = x \in [0,1)$, valid.
• $x \in [1, 2)$, $[x] = 1$: $x + [x] = x + 1 \in [2,3)$, valid $\Rightarrow$ all of $[1,2)$.
• $x \in [2, 3)$, $[x] = 2$: $x + [x] \ge 4 > 3$, invalid.
• $x < -1$: $x + [x] < -2$ soon drops below $-3$, invalid.

Combining the valid strips gives the domain $[-1, 2)$, so $\alpha = -1,\ \beta = 2$ and

Answer: B (5)

Common terms satisfy $5k + 1 = 7m + 9$. The smallest common term is $16$, and common terms recur with period $\operatorname{lcm}(5,7) = 35$:

The largest such term $\le \min(501, 499) = 499$ is $16 + 35(13) = 471$, so $t = 0,\dots,13$ (14 terms).

Divisible by 3: $16 + 35t \equiv 0 \pmod 3 \Rightarrow 1 + 2t \equiv 0 \Rightarrow t \equiv 1 \pmod 3$.

That is 5 elements.

Answer: B (5)

Two requirements:

1. The log argument must be defined and positive: $\left|\dfrac{2x-5}{x^2-4}\right| > 0$, so $x \ne 2, -2$ and $x \ne \tfrac52$ (numerator zero makes the value $0$).
2. The square root needs $\log_{0.6}(\cdot) \ge 0$. Since base $0.6 < 1$, this means $\left|\dfrac{2x-5}{x^2-4}\right| \le 1$, i.e. $|2x - 5| \le |x^2 - 4|$.

Square the last inequality:

Since $(x-1)^2 \ge 0$, this holds when $x^2 + 2x - 9 \ge 0$ (or $x = 1$). Roots of $x^2 + 2x - 9$ are $-1 \pm \sqrt{10}$, so

Now remove $x = \tfrac52$ (excluded above), which lies in the right branch and splits it:

Matching, $a = -1-\sqrt{10},\ b = 1,\ c = -1+\sqrt{10},\ d = \tfrac52,\ e = \tfrac52$:

Answer: 4

On $[1,\infty)$, $f$ is strictly increasing, so its inverse is $f^{-1}(x) = 1 + (x - 1)^{1/4}$.

Statement (I). For a strictly increasing $f$, $f(x) = f^{-1}(x)$ holds exactly at the fixed points $f(x) = x$:

Put $t = x - 1 \ge 0$: $t^4 = t \Rightarrow t(t^3 - 1) = 0 \Rightarrow t = 0,\,1$, giving $x = 1, 2$. So $S$ has exactly two elements — (I) is TRUE.

Statement (II). $f^{-1}(x+1) = 1 + (x+1-1)^{1/4} = 1 + x^{1/4}$. The equation is

At $x = 1$ the left side is $0$ and the right side is $1$ (left $<$ right); as $x$ grows, $(x-1)^4$ overtakes $x^{1/4}$, so by continuity a solution exists near $x \approx 2.05$. Thus $S$ is not empty — (II) is FALSE.

Only (I) is true.

Answer: A (only (I) is TRUE)

Put $x = 0$:

Check: $f(1) = 2 + 1 - 1 = 2$. ✓

Find $\alpha$ by expanding both sides with $f(t) = 2t^2 + t - 1$:

Comparing the $xy$ terms: $4xy = \alpha xy \Rightarrow \alpha = 4$.

Now

Answer: B (140)

The two conditions “$x \mid z$” and “$y \le w$” are independent, so

Divisibility pairs in $A = \{2,3,4,5,6\}$ (count $z$ that each $x$ divides):

• $x=2$: $z \in \{2,4,6\}$ → 3
• $x=3$: $z \in \{3,6\}$ → 2
• $x=4$: $z \in \{4\}$ → 1
• $x=5$: $z \in \{5\}$ → 1
• $x=6$: $z \in \{6\}$ → 1

Total $= 3+2+1+1+1 = 8$.

Ordered pairs with $y \le w$: out of $5 \times 5 = 25$, the number with $y \le w$ is $\dfrac{25 + 5}{2} = 15$.

Answer: 120

The denominator $3x^2 + x + 3$ has discriminant $1 - 36 < 0$, so it is always positive and $f$ is defined on all of $\mathbb{R}$.

Onto? Set $y = f(x)$ and clear denominators:

For a real $x$ (with $3y - 2 \ne 0$) the discriminant must be $\ge 0$:

The range is the bounded interval $\left[\tfrac17, \tfrac75\right] \ne \mathbb{R}$, so $f$ is not onto.

One-one? For a generic value $y$ in the range the quadratic in $x$ has two distinct roots, so different $x$ map to the same $y$ (e.g. $f(0) = \tfrac23 = f\!\left(\tfrac{?}{}\right)$ has a second preimage). Hence $f$ is not one-one.

$f$ is neither one-one nor onto.

Answer: D (neither one-one nor onto)

$\log_e(x+y) \le 2 \iff x + y \le e^2 \approx 7.389$. With $x, y \in \mathbb{N}$ (so $\ge 1$), this means

Counting by $s = x+y$ from $2$ to $7$: sizes $1,2,3,4,5,6$, so $|R| = 21$.

For transitivity we need: whenever $(x,y), (y,z) \in R$, also $(x,z) \in R$. The minimum superset that is transitive is the transitive closure of $R$.

$R$ contains $(1,1),\dots$ and links every $a \le 6$ to every $b \le 6$ eventually (e.g. $(1,6)$ and $(6,1)$ chain to force $(a,b)$ for all $a,b \in \{1,\dots,6\}$). Computing the transitive closure gives the full block $\{1,\dots,6\}\times\{1,\dots,6\}$ (size $36$) — every pair $(a,b)$ with $a,b \le 6$ becomes forced, while pairs involving a coordinate $\ge 7$ do not arise.

Answer: 15

Each element of $A \times B$ is a pair $(a, b)$; there are $9$ such pairs. We count quadruples where $(a_1 + b_2) \mid (a_2 + b_1)$.

Let $s = a_1 + b_2$ (from the first pair’s $a$ and the second pair’s $b$) and $t = a_2 + b_1$. Both $s$ and $t$ range over sums $a + b$ with $a \in \{1,4,7\}$, $b \in \{2,3,8\}$, i.e. values in $\{3,4,9,6,7,12,9,10,15\}$.

Enumerating all $9 \times 9 = 81$ ordered pairs $((a_1,b_1),(a_2,b_2))$ and checking $(a_1+b_2)\mid(a_2+b_1)$ gives exactly $18$ satisfying quadruples.

Answer: 18

On $S = \{-2,-1,0,1,2\}$, count pairs with $1 + ab > 0$, i.e. $ab > -1$.

• If either $a = 0$ or $b = 0$: $ab = 0 > -1$ → all such pairs qualify. That is $2\cdot 5 - 1 = 9$ pairs (rows/columns through $0$).
• Among nonzero $a, b \in \{-2,-1,1,2\}$: need $ab > -1$, i.e. $ab \ge 1$ (both same sign) — products of opposite signs give $ab \le -1$, excluded. Same-sign pairs: $\{-2,-1\}^2$ gives $4$, $\{1,2\}^2$ gives $4$ → $8$ pairs.

Total $= 9 + 8 = 17$. So Statement I is TRUE.

Equivalence? $R$ is reflexive ($1 + a^2 > 0$) and symmetric ($ab = ba$), but not transitive: $(-2, 0) \in R$ and $(0, 1) \in R$, yet $1 + (-2)(1) = -1 \not> 0$ so $(-2, 1) \notin R$. Hence $R$ is not an equivalence relation — Statement II is FALSE.

Only I is true.

Answer: A (Only I is true)

Factor the arguments:

So

Let $t = \log_{x+1}(2x+3)$, so $\log_{2x+3}(x+1) = \dfrac1t$. The equation becomes

• $t = 1$: $2x + 3 = x + 1 \Rightarrow x = -2$. But then base $x + 1 = -1 < 0$ — rejected.
• $t = 2$: $2x + 3 = (x+1)^2 = x^2 + 2x + 1 \Rightarrow x^2 = 2 \Rightarrow x = \pm\sqrt2$.
  • $x = \sqrt2$: bases $x+1 \approx 2.41 > 0,\ne 1$ and $2x+3 \approx 5.83$ — valid.
  • $x = -\sqrt2 \approx -1.41$: base $x + 1 < 0$ — rejected.

Only real solution: $x = \sqrt2$.

Answer: 2