JEE Main 2026 April 4 Shift 1 mathematics question paper is available here with answer key and solutions. NTA conducted the first shift of the day on April 4, 2026, from 9:00 AM to 12:00 PM.
- The JEE Main Mathematics Question Paper contains a total of 25 questions.
- Each correct answer gets you 4 marks while incorrect answers gets you a negative mark of 1.
Candidates can download the JEE Main 2026 April 4 Shift 1 mathematics question paper along with detailed solutions to analyze their performance and understand the exam pattern better.
JEE Main 2026 April 4 Shift 1 Mathematics Question Paper with Solution PDF
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Let \(A\) is a matrix of order 3 such that \(|A| = -4\), then the value of \(|adj(adj(2adj A)^{-1})|\) is
If \(A = \frac{\sin 3^\circ}{\cos 9^\circ} + \frac{\sin 9^\circ}{\cos 27^\circ} + \frac{\sin 27^\circ}{\cos 81^\circ}\) and \(B = \tan 81^\circ - \tan 3^\circ\), find \(\frac{B}{A}\)
If \(f(x) = \min \{2x^2 + 3, 6x\} + |x-1| \cos(x^2 - \frac{1}{4})\), then the number of points of non derivability of \(f(x)\) is/are
If \(\int_{-2}^2 ([\sin x] + |x \sin x|) dx = 2 \sin 2 - 4 \cos 2 - \beta\), then the value of \(|\beta|\) where \([\cdot]\) is GIF is
Consider the differential equation \(\frac{dy}{dx} = (x^2 + x + 1)(y^2 - y + 1)\). If \(y(0) = \frac{1}{2}\), then the value of \(2(y(1)) - 1\) is
The domain of \(f(x) = \cos^{-1} \left( \frac{4x + 2[x]}{3} \right)\) (where \([x]\) denotes greatest integer function) is
If \(Z\) be a complex number such that \(|Z + 2| = |Z - 2|\) and \(arg \left( \frac{Z - 3}{Z + 1} \right) = \frac{\pi}{4}\), then the value of \(|Z|^2\) is
If coefficient of \(x^3\) in \((1+x)^3 + (1+x)^4 + \dots + (1+x)^{99} + (1+kx)^{100}\) is \(\binom{100}{3} \left( \frac{101}{4} - 43n \right)\), then the value of \((k^3 + 43n)\) is
If \(S = \{ A = \begin{bmatrix} a & b
c & d \end{bmatrix}, A^2 - 4A + 3I = Null matrix, a, b, c, d \in \{0, 1, 2, 3, 4\} and Tr(A) = 4 \}\). Then the number of elements of set \(S\) is/are
Let \(y = \tan^{-1}\left( \frac{3\cos x - 4\sin x}{4\cos x + 3\sin x} \right) + \tan^{-1}\left( \frac{x}{1 + \sqrt{1 + x^2}} \right)\). Then the value of \(\frac{dy}{dx}\) at \(x = \frac{\sqrt{3}}{2}\) is :
Let mean and median of 9 observations 8, 13, a, 17, 21, 51, 103, b, 67 are 40 and 21 respectively where a \(>\) b. If mean deviation about median is 26 then 2a is :-
A circle \(x^2 + y^2 + x - 3y = 0\) passes through \(P(1, 2)\). If 2 chords (PS \& PR) drawn from P are bisected by \(y\)-axis, then mid point of RS is \((\alpha, \beta)\), find \(6(\alpha + \beta)\)
Let \(A = \{1, 2, 3, 4, 5\}\) and \(B = \{a, b, c\}\) then total number of functions from \(A\) to \(B\) which are not onto are
Find number of ways of arranging 4 Boys \& 3 Girls such that all girls are not together :
If Latus rectum of parabola \(y^2 = 4kx\) and ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) coincide then the value of \(e^2 + 2\sqrt{2}\) is, where \(e\) is eccentricity of ellipse
If \(L_1 : x - y = 0\), \(L_2 : y = -3x\), \(L_3\) is obtuse angle bisector of \(L_1\) \& \(L_2\) and \(L_4\) is \(x + 3 = 0\). \(A\) is point of intersection of \(L_4\) \& \(L_1\), \(B\) is point of intersection of \(L_4\) \& \(L_2\), \(C\) is point of intersection of \(L_4\) \& \(L_3\). Then the value of \(\frac{(BC)^2}{(AC)^2}\)
In an A.P. first term is \(\frac{10}{3}\) and first 30 terms are non-negative such that sum of first 30 terms = \((T_{30})^3\), then d is equal to (where \(T_n\) is \(n^{th}\) term of A.P., and d is the common difference of A.P.)
If \(f(x)\) is a non-constant polynomial satisfying \(f(x) = f'(x)f''(x)\) and \(f(0) = 0\). Then the value of \(\int_0^2 f(x) dx + f'(2) + f''(2)\) is :
Let in a \(\triangle ABC\), given that \(A \equiv (1, 2)\), mid-point of \(AB\) is \((-5, -1)\) and centroid is \((3, 4)\) then circumcentre is \((\alpha, \beta)\), then the value of \(21(\alpha + \beta)\) is :
If a coin is tossed 8 times and probability of getting exactly 3 heads in first 6 tosses and exactly 2 tails in last 5 tosses is \(p\) then \(64p\) is
View Solution
Step 1: Understanding the Concept:
We define events on specific tosses. The tosses 4, 5, and 6 are common to both "first 6" and "last 5".
Step 2: Key Formula or Approach:
Split the 8 tosses into three groups: \(G_1 = \{1, 2, 3\}\), \(G_2 = \{4, 5, 6\}\), \(G_3 = \{7, 8\}\).
Total outcomes \(= 2^8 = 256\).
Step 3: Detailed Explanation:
1. Define Heads count in each group: \(h_1, h_2, h_3\).
Condition A (3 heads in tosses 1-6): \(h_1 + h_2 = 3\).
Condition B (2 tails in tosses 4-8 \(\implies\) 3 heads): \(h_2 + h_3 = 3\).
2. Analyze possible values for \(h_2\) (size of group 2 is 3):
- If \(h_2 = 1\): \(h_1 = 2\) (3 ways), \(h_3 = 2\) (1 way). Ways \(= \binom{3}{1} \cdot 3 \cdot 1 = 9\).
- If \(h_2 = 2\): \(h_1 = 1\) (3 ways), \(h_3 = 1\) (2 ways). Ways \(= \binom{3}{2} \cdot 3 \cdot 2 = 18\).
- If \(h_2 = 3\): \(h_1 = 0\) (1 way), \(h_3 = 0\) (1 way). Ways \(= \binom{3}{3} \cdot 1 \cdot 1 = 1\).
- \(h_2 = 0\) is impossible because then \(h_3 = 3\), but group 3 only has 2 tosses.
3. Total favorable outcomes:
\(9 + 18 + 1 = 28\).
4. Probability \(p\):
\(p = \frac{28}{256} = \frac{7}{64}\).
5. Find \(64p\):
\(64 \cdot \frac{7}{64} = 7\).
Step 4: Final Answer:
The value of \(64p\) is 7.
Quick Tip: When events overlap in probability, partition the trials into disjoint sets based on the overlap. This allows you to treat the outcomes in each part independently.
JEE Main 2026 Mathematics Exam Pattern
| Particulars | Details |
|---|---|
| Exam Mode | Online (Computer-Based Test) |
| Paper | B.E./B.Tech |
| Medium of Exam | 13 languages: English, Hindi, Gujarati, Bengali, Tamil, Telugu, Kannada, Marathi, Malayalam, Odia, Punjabi, Assamese, Urdu |
| Type of Questions | Multiple Choice Questions (MCQs) + Numerical Value Questions |
| Total Marks | 100 marks |
| Marking Scheme | +4 for correct answer & -1 for incorrect MCQ and Numerical Value-based Questions |
| Total Questions | 25 Questions |
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