Q. Find the equation of the line that is perpendicular to y = 5x + 2 and passes through (2, 3).
A.
y = -1/5x + 4
B.
y = 5x - 7
C.
y = -5x + 13
D.
y = 1/5x + 2
Show solution
Solution
The slope of the perpendicular line is -1/5. Using point-slope form: y - 3 = -1/5(x - 2) gives y = -1/5x + 13/5.
Correct Answer:
C
— y = -5x + 13
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Q. Find the equation of the line that is perpendicular to y = 5x + 2 and passes through the origin.
A.
y = -1/5x
B.
y = 5x
C.
y = -5x
D.
y = 1/5x
Show solution
Solution
The slope of the given line is 5. The slope of the perpendicular line is -1/5. Using y = mx + c, we get y = -1/5x.
Correct Answer:
C
— y = -5x
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Q. Find the equation of the line that passes through the origin and has a slope of -2.
A.
y = -2x
B.
y = 2x
C.
y = -x
D.
y = x
Show solution
Solution
Using the slope-intercept form: y = mx + b, where b = 0, we have y = -2x.
Correct Answer:
A
— y = -2x
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Q. Find the equation of the line that passes through the point (1, 2) and has a slope of 3.
A.
y = 3x + 1
B.
y = 3x - 1
C.
y = 3x + 2
D.
y = 3x - 2
Show solution
Solution
Using point-slope form: y - 2 = 3(x - 1) => y = 3x - 1.
Correct Answer:
C
— y = 3x + 2
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Q. Find the equation of the line that passes through the point (2, 3) and has a slope of -1.
A.
y = -x + 5
B.
y = -x + 3
C.
y = x + 1
D.
y = -x + 1
Show solution
Solution
Using point-slope form: y - 3 = -1(x - 2) => y = -x + 5.
Correct Answer:
A
— y = -x + 5
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Q. Find the equation of the pair of lines represented by the equation 2x^2 + 3xy + y^2 = 0.
A.
y = -2x, y = -x/3
B.
y = -3x/2, y = -x/2
C.
y = -x/3, y = -3x
D.
y = -x/2, y = -2x
Show solution
Solution
Using the quadratic formula for the slopes gives m1 = -2 and m2 = -1/3.
Correct Answer:
A
— y = -2x, y = -x/3
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Q. Find the equation of the pair of lines represented by the equation x^2 - 4y^2 = 0.
A.
x = 2y, x = -2y
B.
x = 4y, x = -4y
C.
x = 0, y = 0
D.
x = y, x = -y
Show solution
Solution
Factoring the equation gives (x - 2y)(x + 2y) = 0, which represents the lines x = 2y and x = -2y.
Correct Answer:
A
— x = 2y, x = -2y
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Q. Find the equation of the parabola that opens downwards with vertex at (0, 0) and passes through the point (2, -4).
A.
y = -x^2
B.
y = -2x^2
C.
y = -1/2x^2
D.
y = -4x^2
Show solution
Solution
Using the vertex form and substituting the point (2, -4), we find that the equation is y = -2x^2.
Correct Answer:
B
— y = -2x^2
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Q. Find the equation of the parabola with focus at (0, -3) and directrix y = 3.
A.
x^2 = -12y
B.
x^2 = 12y
C.
y^2 = -12x
D.
y^2 = 12x
Show solution
Solution
The distance from the focus to the directrix is 6, so p = -3. The equation is x^2 = 4py, which gives x^2 = -12y.
Correct Answer:
A
— x^2 = -12y
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Q. Find the equation of the parabola with focus at (0, 2) and directrix y = -2.
A.
x^2 = 8y
B.
y^2 = 8x
C.
y^2 = -8x
D.
x^2 = -8y
Show solution
Solution
The vertex is at (0, 0) and p = 2. The equation is y^2 = 4px, which gives y^2 = 8x.
Correct Answer:
A
— x^2 = 8y
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Q. Find the equation of the parabola with vertex at (2, 3) and focus at (2, 5).
A.
y = (1/4)(x - 2)^2 + 3
B.
y = (1/4)(x - 2)^2 - 3
C.
y = (1/4)(x + 2)^2 + 3
D.
y = (1/4)(x + 2)^2 - 3
Show solution
Solution
The vertex form of a parabola is given by (x - h)^2 = 4p(y - k). Here, h = 2, k = 3, and p = 1 (distance from vertex to focus). Thus, the equation is (x - 2)^2 = 4(1)(y - 3) or y = (1/4)(x - 2)^2 + 3.
Correct Answer:
A
— y = (1/4)(x - 2)^2 + 3
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Q. Find the equation of the tangent line to the curve y = x^2 + 2x at the point where x = 1.
A.
y = 3x - 2
B.
y = 2x + 1
C.
y = 2x + 2
D.
y = x + 3
Show solution
Solution
f'(x) = 2x + 2. At x = 1, f'(1) = 4. The point is (1, 3). The tangent line is y - 3 = 4(x - 1) => y = 4x - 1.
Correct Answer:
A
— y = 3x - 2
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Q. Find the family of curves represented by the equation y = mx + c, where m and c are constants.
A.
Straight lines with varying slopes and intercepts
B.
Parabolas with varying vertices
C.
Circles with varying radii
D.
Ellipses with varying axes
Show solution
Solution
The equation y = mx + c represents straight lines where m is the slope and c is the y-intercept.
Correct Answer:
A
— Straight lines with varying slopes and intercepts
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Q. Find the focus of the parabola defined by the equation x^2 = 12y.
A.
(0, 3)
B.
(0, -3)
C.
(3, 0)
D.
(-3, 0)
Show solution
Solution
The equation x^2 = 12y can be rewritten as (y - 0) = (1/3)(x - 0)^2, indicating the focus is at (0, 3).
Correct Answer:
A
— (0, 3)
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Q. Find the focus of the parabola given by the equation y^2 = 12x.
A.
(3, 0)
B.
(0, 3)
C.
(0, 6)
D.
(6, 0)
Show solution
Solution
The standard form of a parabola is y^2 = 4px. Here, 4p = 12, so p = 3. The focus is at (p, 0) = (3, 0).
Correct Answer:
C
— (0, 6)
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Q. Find the general solution of the differential equation dy/dx = 2y.
A.
y = Ce^(2x)
B.
y = 2Ce^x
C.
y = Ce^(x/2)
D.
y = 2x + C
Show solution
Solution
This is a separable equation. Integrating gives ln|y| = 2x + C, hence y = Ce^(2x).
Correct Answer:
A
— y = Ce^(2x)
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Q. Find the general solution of the differential equation dy/dx = y.
A.
y = Ce^x
B.
y = Ce^(-x)
C.
y = Cx
D.
y = C/x
Show solution
Solution
This is a separable equation. Integrating gives ln|y| = x + C, hence y = Ce^x.
Correct Answer:
A
— y = Ce^x
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Q. Find the general solution of the differential equation y'' - 5y' + 6y = 0.
A.
y = C1 e^(2x) + C2 e^(3x)
B.
y = C1 e^(3x) + C2 e^(2x)
C.
y = C1 e^(x) + C2 e^(2x)
D.
y = C1 e^(4x) + C2 e^(5x)
Show solution
Solution
The characteristic equation is r^2 - 5r + 6 = 0, giving roots 2 and 3. Thus, y = C1 e^(2x) + C2 e^(3x).
Correct Answer:
B
— y = C1 e^(3x) + C2 e^(2x)
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Q. Find the general solution of the equation cos(2x) = 0.
A.
x = (2n+1)π/4
B.
x = nπ/2
C.
x = (2n+1)π/2
D.
x = nπ
Show solution
Solution
The general solution is x = (2n+1)π/4, where n is any integer.
Correct Answer:
A
— x = (2n+1)π/4
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Q. Find the general solution of the equation sin(x) + sin(2x) = 0.
A.
x = nπ
B.
x = nπ/2
C.
x = (2n+1)π/4
D.
x = nπ/3
Show solution
Solution
Factoring gives sin(x)(1 + 2cos(x)) = 0, leading to x = nπ or cos(x) = -1/2.
Correct Answer:
A
— x = nπ
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Q. Find the general solution of the equation sin(x) + √3 cos(x) = 0.
A.
x = (2n+1)π/3
B.
x = (2n+1)π/6
C.
x = nπ
D.
x = (2n+1)π/4
Show solution
Solution
The general solution is x = (2n+1)π/3, where n is an integer.
Correct Answer:
A
— x = (2n+1)π/3
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Q. Find the general solution of the equation sin(x) + √3cos(x) = 0.
A.
x = (2n+1)π/3
B.
x = nπ
C.
x = (2n+1)π/4
D.
x = nπ + π/6
Show solution
Solution
The general solution is x = (2n+1)π/3, where n is an integer.
Correct Answer:
A
— x = (2n+1)π/3
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Q. Find the general solution of the equation sin(x) = -1/2.
A.
x = 7π/6 + 2nπ
B.
x = 11π/6 + 2nπ
C.
x = 7π/6, 11π/6
D.
Both 1 and 2
Show solution
Solution
The general solutions are x = 7π/6 + 2nπ and x = 11π/6 + 2nπ.
Correct Answer:
D
— Both 1 and 2
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Q. Find the general solution of the equation sin(x) = sin(2x).
A.
x = nπ
B.
x = nπ/3
C.
x = nπ/2
D.
x = nπ/4
Show solution
Solution
Using the identity sin(a) = sin(b) gives x = nπ or x = (2n+1)π/3.
Correct Answer:
A
— x = nπ
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Q. Find the general solution of the equation sin(x) = sin(π/4).
A.
x = nπ + (-1)^n π/4
B.
x = nπ + π/4
C.
x = nπ + 3π/4
D.
x = nπ + π/2
Show solution
Solution
The general solution is x = nπ + (-1)^n π/4, where n is any integer.
Correct Answer:
A
— x = nπ + (-1)^n π/4
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Q. Find the general solution of the equation y' = 3y + 2.
A.
y = (C - 2/3)e^(3x)
B.
y = Ce^(3x) - 2/3
C.
y = 2/3 + Ce^(3x)
D.
y = 3x + C
Show solution
Solution
This is a first-order linear differential equation. The integrating factor is e^(-3x).
Correct Answer:
B
— y = Ce^(3x) - 2/3
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Q. Find the general solution of the equation y'' - 5y' + 6y = 0.
A.
y = C1 e^(2x) + C2 e^(3x)
B.
y = C1 e^(3x) + C2 e^(2x)
C.
y = C1 e^(x) + C2 e^(2x)
D.
y = C1 e^(4x) + C2 e^(5x)
Show solution
Solution
The characteristic equation is r^2 - 5r + 6 = 0, giving roots 2 and 3. Thus, y = C1 e^(2x) + C2 e^(3x).
Correct Answer:
B
— y = C1 e^(3x) + C2 e^(2x)
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Q. Find the integral of f(x) = 2x + 3.
A.
x^2 + 3x + C
B.
x^2 + 3x
C.
x^2 + 3
D.
2x^2 + 3x + C
Show solution
Solution
The integral ∫(2x + 3)dx = x^2 + 3x + C.
Correct Answer:
A
— x^2 + 3x + C
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Q. Find the integral of f(x) = 2x^3 - 4x + 1.
A.
(1/2)x^4 - 2x^2 + x + C
B.
(1/2)x^4 - 2x^2 + C
C.
(1/4)x^4 - 2x^2 + x + C
D.
(1/3)x^4 - 2x^2 + x + C
Show solution
Solution
The integral ∫(2x^3 - 4x + 1)dx = (1/2)x^4 - 2x^2 + x + C.
Correct Answer:
A
— (1/2)x^4 - 2x^2 + x + C
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Q. Find the integral ∫ (1/x) dx.
A.
ln
B.
x
C.
+ C
D.
x + C
.
1/x + C
.
e^x + C
Show solution
Solution
The integral of 1/x is ln|x| + C.
Correct Answer:
A
— ln
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Mathematics Syllabus (JEE Main) MCQ & Objective Questions
The Mathematics Syllabus for JEE Main is crucial for students aiming to excel in competitive exams. Understanding this syllabus not only helps in grasping key concepts but also enhances your ability to tackle objective questions effectively. Practicing MCQs and important questions from this syllabus is essential for solid exam preparation, ensuring you are well-equipped to score better in your exams.
What You Will Practise Here
Sets, Relations, and Functions
Complex Numbers and Quadratic Equations
Permutations and Combinations
Binomial Theorem
Sequences and Series
Limits and Derivatives
Statistics and Probability
Exam Relevance
The Mathematics Syllabus (JEE Main) is not only relevant for JEE but also appears in CBSE and State Board examinations. Students can expect a variety of question patterns, including direct MCQs, numerical problems, and conceptual questions. Mastery of this syllabus will prepare you for similar topics in NEET and other competitive exams, making it vital for your overall academic success.
Common Mistakes Students Make
Misinterpreting the questions, especially in word problems.
Overlooking the importance of units and dimensions in problems.
Confusing formulas related to sequences and series.
Neglecting to practice derivations, leading to errors in calculus.
Failing to apply the correct methods for solving probability questions.
FAQs
Question: What are the key topics in the Mathematics Syllabus for JEE Main? Answer: Key topics include Sets, Complex Numbers, Permutations, Binomial Theorem, and Calculus.
Question: How can I improve my performance in Mathematics MCQs? Answer: Regular practice of MCQs and understanding the underlying concepts are essential for improvement.
Now is the time to take charge of your exam preparation! Dive into solving practice MCQs and test your understanding of the Mathematics Syllabus (JEE Main). Your success is just a question away!
