JEE Main 2026 (April 4th, Shift 1) Practice Questions & Answers

The domain of $\cos^{-1}(y)$cos−1(y) is $-1 \le y \le 1$−1≤y≤1.
So, $-1 \le \frac{4x + 2[x]}{3} \le 1 \implies -3 \le 4x + 2[x] \le 3$−1≤34x+2[x]​≤1⟹−3≤4x+2[x]≤3.
Let $x = I + f$x=I+f, where $I = [x]$I=[x] and $f = \{x\}$f={x} is the fractional part, so $0 \le f < 1$0≤f<1.
Then, $4x + 2[x] = 4(I + f) + 2I = 6I + 4f$4x+2[x]=4(I+f)+2I=6I+4f.
Therefore, $-3 \le 6I + 4f \le 3$−3≤6I+4f≤3.
For $I = 0$I=0: $-3 \le 4f \le 3 \implies 0 \le f \le \frac{3}{4}$−3≤4f≤3⟹0≤f≤43​. So $x \in \left[0, \frac{3}{4}\right]$x∈[0,43​].
For $I = -1$I=−1: $-3 \le -6 + 4f \le 3 \implies 3 \le 4f \le 9 \implies \frac{3}{4} \le f < 1$−3≤−6+4f≤3⟹3≤4f≤9⟹43​≤f<1. So $x \in \left[-\frac{1}{4}, 0\right)$x∈[−41​,0).
For any other integer $I$I, no such $f \in [0, 1)$f∈[0,1) satisfies the inequality.
Taking the union gives the domain $x \in \left[-\frac{1}{4}, \frac{3}{4}\right]$x∈[−41​,43​].
Thus, $\alpha = -\frac{1}{4}$α=−41​ and $\beta = \frac{3}{4}$β=43​.
$12(\alpha + \beta) = 12 \left( -\frac{1}{4} + \frac{3}{4} \right) = 12 \left( \frac{1}{2} \right) = 6$12(α+β)=12(−41​+43​)=12(21​)=6.

Observe that $(x) + (x^2 - 9) = x^2 + x - 9$(x)+(x2−9)=x2+x−9.
Let $A = x$A=x and $B = x^2 - 9$B=x2−9. The given equation is $|A + B| = |A| + |B|$∣A+B∣=∣A∣+∣B∣.
This property of absolute values holds true if and only if $A \cdot B \ge 0$A⋅B≥0.
Substituting the expressions back, we get: $x(x^2 - 9) \ge 0$x(x2−9)≥0.
$\implies x(x - 3)(x + 3) \ge 0$⟹x(x−3)(x+3)≥0.
The critical points are $x = -3, 0, 3$x=−3,0,3. Evaluating the sign of the product in the intervals formed by these points:
For $x \in [3, \infty)$x∈[3,∞), the product is $\ge 0$≥0.
For $x \in [-3, 0]$x∈[−3,0], the product is $\ge 0$≥0.
The solution set is $[-3, 0] \cup [3, \infty)$[−3,0]∪[3,∞).
Comparing this with $[\alpha, \beta] \cup [\gamma, \infty)$[α,β]∪[γ,∞), we get $\alpha = -3$α=−3, $\beta = 0$β=0, and $\gamma = 3$γ=3.
Therefore, $(\alpha^2 + \beta^2 + \gamma^2) = (-3)^2 + 0^2 + 3^2 = 9 + 0 + 9 = 18$(α2+β2+γ2)=(−3)2+02+32=9+0+9=18.

From $|z + 2| = |z - 2|$∣z+2∣=∣z−2∣, the locus of $z$z is the perpendicular bisector of the line segment joining $(-2, 0)$(−2,0) and $(2, 0)$(2,0), which is the imaginary axis. Let $z = iy$z=iy, where $y \in \mathbb{R}$y∈R.
Given $\arg \left( \frac{iy + 3}{iy - i} \right) = \frac{\pi}{4}$arg(iy−iiy+3​)=4π​.
Simplify the complex fraction: $\frac{3 + iy}{i(y - 1)} = \frac{(3 + iy)(-i)}{y - 1} = \frac{y - 3i}{y - 1} = \frac{y}{y - 1} + i \left( \frac{-3}{y - 1} \right)$i(y−1)3+iy​=y−1(3+iy)(−i)​=y−1y−3i​=y−1y​+i(y−1−3​).
Since the argument is $\frac{\pi}{4}$4π​, the real and imaginary parts must be equal and positive:
$\frac{y}{y - 1} = \frac{-3}{y - 1} > 0 \implies y = -3$y−1y​=y−1−3​>0⟹y=−3.
For $y = -3$y=−3, the real part is $\frac{-3}{-4} = \frac{3}{4} > 0$−4−3​=43​>0, satisfying the requirement.
Thus, $z = -3i$z=−3i.
Then, $|z|^2 = 0^2 + (-3)^2 = 9$∣z∣2=02+(−3)2=9.

The total number of functions from a set of 4 elements to a set of 3 elements is $3^4 = 81$34=81.
The number of onto (surjective) functions is given by the principle of inclusion-exclusion:
$\sum_{k=0}^{n} (-1)^k \binom{n}{k} (n-k)^m = \binom{3}{0} 3^4 - \binom{3}{1} 2^4 + \binom{3}{2} 1^4 - \binom{3}{3} 0^4$∑k=0n​(−1)k(kn​)(n−k)m=(03​)34−(13​)24+(23​)14−(33​)04
$= 1(81) - 3(16) + 3(1) - 0 = 81 - 48 + 3 = 36$=1(81)−3(16)+3(1)−0=81−48+3=36.
The number of functions that are NOT onto is equal to the total number of functions minus the number of onto functions:
$81 - 36 = 45$81−36=45.

The characteristic equation for a $2 \times 2$2×2 matrix $A$A is $A^2 - \text{Tr}(A)A + |A|I = 0$A2−Tr(A)A+∣A∣I=0.
Given $A^2 - 4A + 3I = 0$A2−4A+3I=0 and the trace (sum of diagonal elements) $\text{Tr}(A) = a + d = 4$Tr(A)=a+d=4. By comparing, $|A| = ad - bc = 3$∣A∣=ad−bc=3.
If a matrix is not a scalar multiple of the identity, it must satisfy its characteristic polynomial exactly matching the given relation.
Possibilities for $(a, d)$(a,d) such that $a + d = 4$a+d=4 from $\{0, 1, 2, 3, 4\}${0,1,2,3,4} are $(0, 4), (4, 0), (1, 3), (3, 1), (2, 2)$(0,4),(4,0),(1,3),(3,1),(2,2).
Now evaluate $bc = ad - 3$bc=ad−3 for each case:

• $(a, d) = (0, 4) \implies bc = 0 - 3 = -3$(a,d)=(0,4)⟹bc=0−3=−3 (Not possible as $b, c \ge 0$b,c≥0)
• $(a, d) = (4, 0) \implies bc = -3$(a,d)=(4,0)⟹bc=−3 (Not possible)
• $(a, d) = (1, 3) \implies bc = 3 - 3 = 0$(a,d)=(1,3)⟹bc=3−3=0. Pairs $(b, c)$(b,c) from $\{0,1,2,3,4\}${0,1,2,3,4} such that $bc = 0$bc=0 are $(0,0),(0,1),(0,2),(0,3),(0,4),(1,0),(2,0),(3,0),(4,0)$(0,0),(0,1),(0,2),(0,3),(0,4),(1,0),(2,0),(3,0),(4,0) $\implies 9$⟹9 matrices.
• $(a, d) = (3, 1) \implies bc = 3 - 3 = 0$(a,d)=(3,1)⟹bc=3−3=0. Same $9$9 pairs $\implies 9$⟹9 matrices.
• $(a, d) = (2, 2) \implies bc = 4 - 3 = 1$(a,d)=(2,2)⟹bc=4−3=1. Pairs $(b, c)$(b,c) must be $(1, 1)$(1,1) $\implies 1$⟹1 matrix.
  Could $A$A be a scalar matrix? $A = xI$A=xI implies $x^2 - 4x + 3 = 0$x2−4x+3=0, giving $x=1$x=1 or $x=3$x=3. For $x=1$x=1, trace is $2$2 (discard). For $x=3$x=3, trace is $6$6 (discard).
  Total matrices $= 9 + 9 + 1 = 19$=9+9+1=19.

First find the determinant of $A$A:
$|A| = 1(0 - 3) - 1(-10 - 1) + 2(-6 - 0) = -3 + 11 - 12 = -4$∣A∣=1(0−3)−1(−10−1)+2(−6−0)=−3+11−12=−4.
Recall that $(\text{adj} A)^{-1} = \frac{A}{|A|}$(adjA)−1=∣A∣A​. So, let $M = 2(\text{adj} A)^{-1} = 2 \frac{A}{-4} = -\frac{1}{2}A$M=2(adjA)−1=2−4A​=−21​A.
We want $\text{adj}(\text{adj} M)$adj(adjM). For a $3 \times 3$3×3 matrix $M$M, $\text{adj}(\text{adj} M) = |M|^{3-2} M = |M|M$adj(adjM)=∣M∣3−2M=∣M∣M.
Find the determinant of $M$M:
$|M| = \left| -\frac{1}{2} A \right| = \left( -\frac{1}{2} \right)^3 |A| = -\frac{1}{8}(-4) = \frac{1}{2}$∣M∣=​−21​A​=(−21​)3∣A∣=−81​(−4)=21​.
So, $\text{adj}(\text{adj} M) = \frac{1}{2} M = \frac{1}{2} \left( -\frac{1}{2} A \right) = -\frac{1}{4} A$adj(adjM)=21​M=21​(−21​A)=−41​A.
The sum of all elements of $A$A is $1 + 1 + 2 - 2 + 0 + 1 + 1 + 3 + 5 = 12$1+1+2−2+0+1+1+3+5=12.
Therefore, the sum of all elements of $-\frac{1}{4} A$−41​A is $-\frac{1}{4} \times 12 = -3$−41​×12=−3.

Let the first term be $a = \frac{10}{3}$a=310​ and the last term be $l$l.
The sum of the A.P. of $30$30 terms is $S_{30} = \frac{30}{2}(a + l) = 15\left( \frac{10}{3} + l \right) = 50 + 15l$S30​=230​(a+l)=15(310​+l)=50+15l.
Given $S_{30} = l^3$S30​=l3, we have the equation $l^3 = 50 + 15l \implies l^3 - 15l - 50 = 0$l3=50+15l⟹l3−15l−50=0.
Testing integer values, we find that for $l = 5$l=5: $5^3 - 15(5) - 50 = 125 - 75 - 50 = 0$53−15(5)−50=125−75−50=0. So $l = 5$l=5 is a root.
(The quadratic factor $l^2 + 5l + 10$l2+5l+10 has no real roots, so $l = 5$l=5 is the only real root).
The $30\text{th}$30th term is $a + 29d = l$a+29d=l.
$\frac{10}{3} + 29d = 5 \implies 29d = 5 - \frac{10}{3} = \frac{5}{3} \implies d = \frac{5}{87}$310​+29d=5⟹29d=5−310​=35​⟹d=875​.