Mathematics Syllabus for JEE Main

Mathematics Syllabus (JEE Main)

Algebra Calculus Coordinate Geometry Sets, Relations & Functions Statistics & Probability Trigonometry Vector & 3D Geometry

Q. For which value of a is the function f(x) = { ax + 1, x < 0; 2, x = 0; 3x - 1, x > 0 } continuous at x = 0?

A. -1
B. 0
C. 1
D. 2

Solution

Setting ax + 1 = 2 at x = 0 gives a = 2.

Correct Answer: A — -1

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Q. For which value of a is the function f(x) = { ax + 1, x < 0; 2x + a, x >= 0 } continuous at x = 0?

A. -1
B. 0
C. 1
D. 2

Solution

Setting the two pieces equal at x = 0 gives 1 = a, hence a = 1.

Correct Answer: A — -1

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Q. For which value of a is the function f(x) = { ax + 2, x < 1; 3, x >= 1 } continuous at x = 1?

A. 1
B. 2
C. 3
D. 4

Solution

Setting ax + 2 = 3 at x = 1 gives a = 1.

Correct Answer: B — 2

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Q. For which value of a is the function f(x) = { x^2 + a, x < 1; 3, x >= 1 } continuous at x = 1?

A. 1
B. 2
C. 3
D. 0

Solution

To ensure continuity at x = 1, we set limit as x approaches 1 from left (1 + a) equal to f(1) = 3, thus a = 2.

Correct Answer: B — 2

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Q. For which value of a is the function f(x) = { x^2 - a, x < 0; 2x + 1, x >= 0 } continuous at x = 0?

A. -1
B. 0
C. 1
D. 2

Solution

Setting the two pieces equal at x = 0 gives -a = 1, so a = -1.

Correct Answer: A — -1

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Q. For which value of a is the function f(x) = { x^2 - a, x < 1; 3x - 2, x >= 1 } continuous at x = 1?

A. 0
B. 1
C. 2
D. 3

Solution

Setting the two pieces equal at x = 1 gives 1 - a = 1. Thus, a = 0.

Correct Answer: C — 2

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Q. For which value of a is the function f(x) = { x^2, x < 1; ax + 1, x >= 1 } continuous at x = 1?

A. 0
B. 1
C. 2
D. 3

Solution

Setting the two pieces equal at x = 1 gives 1 = a(1) + 1, leading to a = 0.

Correct Answer: C — 2

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Q. For which value of b is the function f(x) = { 2x + 1, x < 1; b, x = 1; x^2 + 1, x > 1 continuous at x = 1?

A. 1
B. 2
C. 3
D. 4

Solution

Setting the left limit (2(1) + 1 = 3) equal to the right limit (1^2 + 1 = 2), we find b = 3.

Correct Answer: B — 2

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Q. For which value of b is the function f(x) = { x^2 - 1, x < 1; b, x = 1; 3x - 2, x > 1 continuous at x = 1?

A. 0
B. 1
C. 2
D. 3

Solution

Setting limit as x approaches 1 gives b = 2 for continuity.

Correct Answer: C — 2

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Q. For which value of b is the function f(x) = { x^2 - 4, x < 2; bx + 2, x >= 2 } continuous at x = 2?

A. 2
B. 4
C. 6
D. 8

Solution

Setting the two pieces equal at x = 2 gives us 0 = 2b + 2. Solving for b gives b = -1.

Correct Answer: B — 4

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Q. For which value of b is the function f(x) = { x^3 - 3x + b, x < 1; 2x + 1, x >= 1 continuous at x = 1?

A. 0
B. 1
C. 2
D. 3

Solution

Setting 1 - 3 + b = 2 gives b = 4 for continuity.

Correct Answer: A — 0

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Q. For which value of c is the function f(x) = { 3x + c, x < 1; 2x^2, x >= 1 continuous at x = 1?

A. -1
B. 0
C. 1
D. 2

Solution

Setting 3(1) + c = 2(1)^2 gives c = -1 for continuity.

Correct Answer: B — 0

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Q. For which value of c is the function f(x) = { x^2 - 4, x < c; 3x - 5, x >= c } continuous at x = c?

A. 1
B. 2
C. 3
D. 4

Solution

Setting the two pieces equal at x = c: c^2 - 4 = 3c - 5. Solving gives c = 3.

Correct Answer: C — 3

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Q. For which value of c is the function f(x) = { x^2 - c, x < 1; 2x + 1, x >= 1 continuous at x = 1?

A. 0
B. 1
C. 2
D. 3

Solution

Setting x^2 - c = 2x + 1 at x = 1 gives c = 2.

Correct Answer: C — 2

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Q. For which value of c is the function f(x) = { x^2, x < c; 2x + 1, x >= c continuous at x = c?

A. 0
B. 1
C. 2
D. 3

Solution

Setting x^2 = 2x + 1 at x = c gives c = 2.

Correct Answer: C — 2

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Q. For which value of k does the equation x^2 + kx + 16 = 0 have real and distinct roots?

A. -8
B. -4
C. 0
D. 4

Solution

The discriminant must be positive: k^2 - 4*1*16 > 0 => k^2 > 64 => k > 8 or k < -8. Thus, k = -4 is valid.

Correct Answer: B — -4

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Q. For which value of k does the equation x^2 + kx + 4 = 0 have one root equal to 2?

A. -4
B. -2
C. 0
D. 2

Solution

Substituting x = 2 into the equation gives 2^2 + 2k + 4 = 0, leading to k = -4.

Correct Answer: B — -2

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Q. For which value of k does the equation x^2 - 4x + k = 0 have roots that differ by 2?

A. 2
B. 3
C. 4
D. 5

Solution

Let the roots be r and r+2. Then, r + (r+2) = 4 and r(r+2) = k leads to k = 4.

Correct Answer: C — 4

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Q. For which value of k does the equation x^2 - kx + 9 = 0 have roots that are both positive?

A. k < 6
B. k > 6
C. k = 6
D. k = 0

Solution

For both roots to be positive, k must be greater than 6, as the sum of the roots must be positive.

Correct Answer: B — k > 6

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Q. For which value of k does the quadratic equation x^2 - kx + 4 = 0 have no real roots?

A. k < 4
B. k = 4
C. k > 4
D. k ≤ 4

Solution

The discriminant must be negative: k^2 - 16 < 0, hence k > 4.

Correct Answer: C — k > 4

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Q. For which value of k is the function f(x) = kx^2 + 2x + 1 differentiable at x = -1?

A. 0
B. 1
C. -1
D. 2

Solution

f'(x) = 2kx + 2; f'(-1) = -2k + 2 must exist for any k.

Correct Answer: A — 0

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Q. For which value of m is the function f(x) = { 3x + m, x < 1; 2, x = 1; mx + 1, x > 1 continuous at x = 1?

A. -1
B. 0
C. 1
D. 2

Solution

Setting 3 + m = 2 and 2 = m + 1 gives m = 1 for continuity.

Correct Answer: B — 0

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Q. For which value of p is the function f(x) = { x^2 + p, x < 0; 3x - 1, x >= 0 } continuous at x = 0?

A. -1
B. 0
C. 1
D. 2

Solution

Setting the two pieces equal at x = 0 gives us p = -1.

Correct Answer: B — 0

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Q. From a group of 10 people, how many ways can a committee of 4 be formed?

A. 210
B. 120
C. 150
D. 180

Solution

The number of ways to choose 4 people from 10 is given by 10C4 = 210.

Correct Answer: A — 210

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Q. From a point A, the angle of elevation of the top of a building is 45 degrees. If the height of the building is 20 meters, how far is point A from the base of the building?

A. 10 m
B. 20 m
C. 30 m
D. 40 m

Solution

Using tan(45°) = height/distance, we have 1 = 20/distance. Therefore, distance = 20 m.

Correct Answer: B — 20 m

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Q. From a point on the ground, the angle of elevation of the top of a hill is 30 degrees. If the height of the hill is 50 meters, how far is the point from the base of the hill?

A. 50 m
B. 75 m
C. 100 m
D. 125 m

Solution

Using tan(30°) = height/distance, we have 1/√3 = 50/distance. Therefore, distance = 50√3 ≈ 86.6 m.

Correct Answer: C — 100 m

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Q. From a point on the ground, the angle of elevation of the top of a hill is 30 degrees. If the distance from the point to the base of the hill is 100 meters, what is the height of the hill?

A. 50 m
B. 60 m
C. 70 m
D. 80 m

Solution

Using tan(30°) = height/100, we have 1/√3 = height/100. Therefore, height = 100/√3 ≈ 57.74 m.

Correct Answer: A — 50 m

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Q. From a point on the ground, the angle of elevation to the top of a 40 m high building is 45 degrees. How far is the point from the base of the building?

A. 40 m
B. 20 m
C. 30 m
D. 50 m

Solution

Using tan(45°) = height/distance, we have distance = height/tan(45°) = 40/1 = 40 m.

Correct Answer: A — 40 m

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Q. From a point on the ground, the angle of elevation to the top of a 75 m high building is 45 degrees. How far is the point from the building?

A. 75 m
B. 50 m
C. 100 m
D. 25 m

Solution

Using tan(45°) = height/distance, we have distance = height/tan(45°) = 75/1 = 75 m.

Correct Answer: A — 75 m

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Q. From a point on the ground, the angle of elevation to the top of a building is 45 degrees. If the building is 50 meters tall, how far is the point from the base of the building?

A. 50 m
B. 25 m
C. 100 m
D. 75 m

Solution

Distance = height / tan(angle) = 50 / 1 = 50 m.

Correct Answer: A — 50 m

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Mathematics Syllabus (JEE Main)

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