Mathematics Syllabus for JEE Main

Mathematics Syllabus (JEE Main)

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

Q. Find the weighted mean of the numbers 10, 20, and 30 with weights 1, 2, and 3 respectively.

A. 20
B. 25
C. 30
D. 35

Solution

Weighted mean = (10*1 + 20*2 + 30*3) / (1 + 2 + 3) = (10 + 40 + 90) / 6 = 140 / 6 = 23.33.

Correct Answer: B — 25

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Q. Find the x-coordinate of the point where the function f(x) = 2x^3 - 9x^2 + 12x has a local maximum.

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

Solution

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. f''(1) < 0 indicates a local maximum at x = 1.

Correct Answer: B — 2

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Q. Find the x-coordinate of the point where the function f(x) = x^2 - 4x + 5 has a local minimum.

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

Solution

The vertex occurs at x = -b/(2a) = 4/2 = 2. This is where the local minimum occurs.

Correct Answer: B — 2

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Q. Find the x-coordinate of the point where the function f(x) = x^2 - 4x + 5 has a minimum.

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

Solution

The vertex occurs at x = -b/(2a) = 4/2 = 2, which is the x-coordinate of the minimum point.

Correct Answer: A — 2

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Q. Find the y-intercept of the line represented by the equation 5x - 2y = 10.

A. 5
B. 2
C. 0
D. 10

Solution

Set x = 0: -2y = 10 => y = -5. The y-intercept is (0, -5).

Correct Answer: B — 2

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Q. For f(x) = { x^2, x < 1; 2x - 1, x ≥ 1 }, is f differentiable at x = 1?

A. Yes
B. No
C. Only left
D. Only right

Solution

f'(1) from left = 2, from right = 2; hence f is differentiable at x = 1.

Correct Answer: B — No

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Q. For the data set 10, 20, 30, 40, 50, what is the mean deviation?

A. 10
B. 15
C. 20
D. 25

Solution

Mean = 30; Mean Deviation = (|10-30| + |20-30| + |30-30| + |40-30| + |50-30|) / 5 = 10.

Correct Answer: B — 15

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Q. For the data set {10, 12, 23, 23, 16, 23, 21}, what is the mode?

A. 10
B. 12
C. 23
D. 21

Solution

The mode is the number that appears most frequently, which is 23.

Correct Answer: C — 23

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Q. For the data set {12, 15, 20, 22, 25}, what is the mode?

A. 12
B. 15
C. 20
D. No mode

Solution

There is no mode as all values appear only once.

Correct Answer: D — No mode

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Q. For the data set {2, 4, 6, 8, 10}, what is the mean deviation?

A. 2
B. 1.6
C. 3
D. 2.5

Solution

Mean = 6; Mean deviation = (|2-6| + |4-6| + |6-6| + |8-6| + |10-6|)/5 = (4 + 2 + 0 + 2 + 4)/5 = 12/5 = 2.4.

Correct Answer: B — 1.6

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Q. For the data set {4, 8, 6, 5, 3}, what is the mean?

A. 4.5
B. 5.5
C. 6.0
D. 5.0

Solution

Mean = (4 + 8 + 6 + 5 + 3) / 5 = 26 / 5 = 5.0.

Correct Answer: D — 5.0

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Q. For the data set: 1, 2, 3, 4, 5, what is the interquartile range?

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

Solution

Q1 = 2, Q3 = 4; Interquartile Range = Q3 - Q1 = 4 - 2 = 2.

Correct Answer: B — 2

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Q. For the data set: 5, 7, 8, 9, 10, what is the mean absolute deviation?

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

Solution

Mean = 7.5; MAD = (|5-7.5| + |7-7.5| + |8-7.5| + |9-7.5| + |10-7.5|) / 5 = 1.

Correct Answer: B — 2

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Q. For the data set: 5, 7, 8, 9, 10, what is the standard deviation?

A. 1.5
B. 2
C. 2.5
D. 3

Solution

Mean = 7.5; Variance = [(5-7.5)^2 + (7-7.5)^2 + (8-7.5)^2 + (9-7.5)^2 + (10-7.5)^2] / 5 = 2; Standard Deviation = sqrt(2) = 1.41

Correct Answer: B — 2

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Q. For the ellipse defined by the equation 9x^2 + 16y^2 = 144, what are the lengths of the semi-major and semi-minor axes?

A. 3, 4
B. 4, 3
C. 6, 8
D. 8, 6

Solution

The semi-major axis is 4 and the semi-minor axis is 3 after rewriting the equation in standard form.

Correct Answer: A — 3, 4

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Q. For the function f(x) = 2x^3 - 9x^2 + 12x, find the inflection point.

A. (1, 1)
B. (2, 2)
C. (3, 3)
D. (4, 4)

Solution

f''(x) = 12x - 18. Setting f''(x) = 0 gives x = 1.5. The inflection point is (1.5, f(1.5)).

Correct Answer: B — (2, 2)

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Q. For the function f(x) = 2x^3 - 9x^2 + 12x, find the intervals where the function is increasing.

A. (-∞, 1)
B. (1, 3)
C. (3, ∞)
D. (0, 3)

Solution

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 3. Testing intervals shows f is increasing on (1, 3).

Correct Answer: B — (1, 3)

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Q. For the function f(x) = 2x^3 - 9x^2 + 12x, find the local maxima.

A. (1, 5)
B. (2, 0)
C. (3, 0)
D. (0, 0)

Solution

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. f(1) = 5 is a local maximum.

Correct Answer: A — (1, 5)

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Q. For the function f(x) = 3x^2 - 12x + 7, find the coordinates of the vertex.

A. (2, -5)
B. (2, -1)
C. (3, -2)
D. (1, 1)

Solution

The vertex is at x = -b/(2a) = 12/(2*3) = 2. f(2) = 3(2^2) - 12(2) + 7 = -1. So, the vertex is (2, -1).

Correct Answer: B — (2, -1)

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Q. For the function f(x) = 3x^3 - 12x^2 + 9, find the x-coordinates of the inflection points.

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

Solution

f''(x) = 18x - 24. Setting f''(x) = 0 gives x = 4/3. This is the inflection point.

Correct Answer: B — 2

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Q. For the function f(x) = 3x^3 - 12x^2 + 9x, the number of local maxima and minima is:

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

Solution

Finding f'(x) = 9x^2 - 24x + 9 and solving gives two critical points. The second derivative test confirms one maximum and one minimum.

Correct Answer: C — 2

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Q. For the function f(x) = e^x - x^2, the point of inflection occurs at:

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

Solution

To find the point of inflection, we compute f''(x) = e^x - 2. Setting f''(x) = 0 gives e^x = 2, leading to x = ln(2). The closest integer is x = 1.

Correct Answer: B — x = 1

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Q. For the function f(x) = ln(x), find the point where it is not differentiable.

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

Solution

f(x) = ln(x) is not defined for x ≤ 0, hence not differentiable at x = 0.

Correct Answer: A — x = 0

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Q. For the function f(x) = sin(x) + cos(x), find the x-coordinate of the maximum point in the interval [0, 2π].

A. π/4
B. 3π/4
C. 5π/4
D. 7π/4

Solution

f'(x) = cos(x) - sin(x). Setting f'(x) = 0 gives tan(x) = 1, so x = π/4 + nπ. In [0, 2π], the maximum occurs at x = 3π/4.

Correct Answer: B — 3π/4

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Q. For the function f(x) = x^2 + 2x + 1, what is f'(x)?

A. 2x + 1
B. 2x + 2
C. 2x
D. x + 1

Solution

f'(x) = 2x + 2.

Correct Answer: B — 2x + 2

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Q. For the function f(x) = x^2 + 2x + 3, find the point where it is not differentiable.

A. x = -1
B. x = 0
C. x = 1
D. It is differentiable everywhere

Solution

The function is a polynomial and is differentiable everywhere.

Correct Answer: D — It is differentiable everywhere

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Q. For the function f(x) = x^2 + kx + 1 to be differentiable at x = -1, what must k be?

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

Solution

Setting the derivative f'(-1) = 0 gives k = 1 for differentiability.

Correct Answer: C — 1

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Q. For the function f(x) = x^2 - 2x + 1, find the slope of the tangent line at x = 1.

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

Solution

f'(x) = 2x - 2. Thus, f'(1) = 2(1) - 2 = 0.

Correct Answer: A — 0

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Q. For the function f(x) = x^2 - 4x + 4, find the point where it is not differentiable.

A. x = 0
B. x = 2
C. x = 4
D. It is differentiable everywhere

Solution

As a polynomial, f(x) is differentiable everywhere, including at x = 2.

Correct Answer: D — It is differentiable everywhere

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Q. For the function f(x) = x^2 - 4x + 5, find the minimum value.

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

Solution

The vertex occurs at x = 2. f(2) = 2^2 - 4*2 + 5 = 1, which is the minimum value.

Correct Answer: B — 2

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Mathematics Syllabus (JEE Main)

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