These essential MCQ questions are selected from the most important topics in JEE Main (Joint Entrance Examination) 2026 for admissions to IITs, NITs, IIITs and other engineering colleges. Sections include Algebra, Calculus, Coordinate Geometry, Trigonometry, Vectors and 3D Geometry as per the JEE Main 2026 syllabus. For truely unlimited daily MCQ practice, visit Vooo AI Education.
📐 JEE Main Mathematics
1The number of real roots of x² + 5|x| + 6 = 0 is:
Answer: A — 0
For x ≥ 0: x² + 5x + 6 = 0 → (x+2)(x+3) = 0 → x = -2 or -3 (both negative, rejected). For x < 0: x² - 5x + 6 = 0 → (x-2)(x-3) = 0 → x = 2 or 3 (both positive, rejected since x < 0). Therefore no real roots exist. The equation has 0 real solutions.
For x ≥ 0: x² + 5x + 6 = 0 → (x+2)(x+3) = 0 → x = -2 or -3 (both negative, rejected). For x < 0: x² - 5x + 6 = 0 → (x-2)(x-3) = 0 → x = 2 or 3 (both positive, rejected since x < 0). Therefore no real roots exist. The equation has 0 real solutions.
2If f(x) = x³ - 3x, then f'(x) at x = 2 is:
Answer: B — 9
f(x) = x³ - 3x. Using power rule: f'(x) = 3x² - 3. At x = 2: f'(2) = 3(4) - 3 = 12 - 3 = 9. The derivative represents the slope of the tangent to the curve at x = 2. Power rule: d/dx(xⁿ) = nxⁿ⁻¹.
f(x) = x³ - 3x. Using power rule: f'(x) = 3x² - 3. At x = 2: f'(2) = 3(4) - 3 = 12 - 3 = 9. The derivative represents the slope of the tangent to the curve at x = 2. Power rule: d/dx(xⁿ) = nxⁿ⁻¹.
3The value of sin 30° × cos 60° + cos 30° × sin 60° equals:
Answer: A — sin 90°
Using the sine addition formula: sin(A+B) = sinA cosB + cosA sinB. sin 30° cos 60° + cos 30° sin 60° = sin(30° + 60°) = sin 90° = 1. Standard values: sin 30° = 1/2, cos 60° = 1/2, cos 30° = √3/2, sin 60° = √3/2. So (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1.
Using the sine addition formula: sin(A+B) = sinA cosB + cosA sinB. sin 30° cos 60° + cos 30° sin 60° = sin(30° + 60°) = sin 90° = 1. Standard values: sin 30° = 1/2, cos 60° = 1/2, cos 30° = √3/2, sin 60° = √3/2. So (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1.
4The equation of a circle with centre (2, -3) and radius 5 is:
Answer: A — (x-2)² + (y+3)² = 25
Standard equation of circle: (x-h)² + (y-k)² = r², where (h,k) is centre and r is radius. Centre (2,-3) and radius 5: (x-2)² + (y-(-3))² = 5² → (x-2)² + (y+3)² = 25. Always verify: the sign inside brackets is opposite to the centre coordinates.
Standard equation of circle: (x-h)² + (y-k)² = r², where (h,k) is centre and r is radius. Centre (2,-3) and radius 5: (x-2)² + (y-(-3))² = 5² → (x-2)² + (y+3)² = 25. Always verify: the sign inside brackets is opposite to the centre coordinates.
5∫x²dx equals:
Answer: B — x³/3 + C
∫xⁿ dx = xⁿ⁺¹/(n+1) + C (for n ≠ -1). So ∫x² dx = x³/3 + C, where C is the constant of integration. Verification: d/dx(x³/3 + C) = 3x²/3 = x². Integration is the reverse of differentiation.
∫xⁿ dx = xⁿ⁺¹/(n+1) + C (for n ≠ -1). So ∫x² dx = x³/3 + C, where C is the constant of integration. Verification: d/dx(x³/3 + C) = 3x²/3 = x². Integration is the reverse of differentiation.
6If A and B are two events, P(A) = 0.4, P(B) = 0.5, P(A∩B) = 0.2, then P(A∪B) =
Answer: A — 0.7
Using the addition rule: P(A∪B) = P(A) + P(B) - P(A∩B) = 0.4 + 0.5 - 0.2 = 0.7. This formula avoids double-counting the intersection. If A and B were mutually exclusive, P(A∩B) = 0 and P(A∪B) = P(A) + P(B) = 0.9.
Using the addition rule: P(A∪B) = P(A) + P(B) - P(A∩B) = 0.4 + 0.5 - 0.2 = 0.7. This formula avoids double-counting the intersection. If A and B were mutually exclusive, P(A∩B) = 0 and P(A∪B) = P(A) + P(B) = 0.9.
7The slope of the line 3x - 4y + 8 = 0 is:
Answer: A — 3/4
Rewriting in slope-intercept form y = mx + c: -4y = -3x - 8 → y = (3/4)x + 2. The slope m = 3/4. General method: for line ax + by + c = 0, slope = -a/b = -3/(-4) = 3/4. The y-intercept is 2.
Rewriting in slope-intercept form y = mx + c: -4y = -3x - 8 → y = (3/4)x + 2. The slope m = 3/4. General method: for line ax + by + c = 0, slope = -a/b = -3/(-4) = 3/4. The y-intercept is 2.
8The sum of first n terms of an AP is Sn = n/2[2a + (n-1)d]. For a=2, d=3, n=5:
Answer: B — 35
Sn = n/2[2a + (n-1)d] = 5/2[2(2) + (5-1)(3)] = 5/2[4 + 12] = 5/2 × 16 = 40. Wait — let me recalculate: 5/2 × 16 = 40. Actually S₅ = 2+5+8+11+14 = 40. Correcting: the answer is 40. AP: a=2, a+d=5, a+2d=8, a+3d=11, a+4d=14. Sum = 40.
Sn = n/2[2a + (n-1)d] = 5/2[2(2) + (5-1)(3)] = 5/2[4 + 12] = 5/2 × 16 = 40. Wait — let me recalculate: 5/2 × 16 = 40. Actually S₅ = 2+5+8+11+14 = 40. Correcting: the answer is 40. AP: a=2, a+d=5, a+2d=8, a+3d=11, a+4d=14. Sum = 40.
9The vectors a⃗ = 2î + 3ĵ and b⃗ = î - ĵ. Their dot product a⃗·b⃗ is:
Answer: A — -1
Dot product: a⃗·b⃗ = (2)(1) + (3)(-1) = 2 - 3 = -1. For vectors a⃗ = a₁î + a₂ĵ + a₃k̂ and b⃗ = b₁î + b₂ĵ + b₃k̂, the dot product = a₁b₁ + a₂b₂ + a₃b₃. If the dot product is 0, the vectors are perpendicular.
Dot product: a⃗·b⃗ = (2)(1) + (3)(-1) = 2 - 3 = -1. For vectors a⃗ = a₁î + a₂ĵ + a₃k̂ and b⃗ = b₁î + b₂ĵ + b₃k̂, the dot product = a₁b₁ + a₂b₂ + a₃b₃. If the dot product is 0, the vectors are perpendicular.
10lim(x→0) sin(x)/x equals:
Answer: C — 1
The standard limit lim(x→0) sin(x)/x = 1 is a fundamental result in calculus. It can be proved geometrically (squeeze theorem) or using L'Hôpital's rule (0/0 form): lim sin(x)/x = lim cos(x)/1 = cos(0) = 1. This limit is used extensively in derivatives of trigonometric functions.
The standard limit lim(x→0) sin(x)/x = 1 is a fundamental result in calculus. It can be proved geometrically (squeeze theorem) or using L'Hôpital's rule (0/0 form): lim sin(x)/x = lim cos(x)/1 = cos(0) = 1. This limit is used extensively in derivatives of trigonometric functions.
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