Q. Find the value of cos(tan^(-1)(1)).
A.
1/√2
B.
1/2
C.
√2/2
D.
√3/2
Show solution
Solution
cos(tan^(-1)(1)) = 1/√2
Correct Answer:
C
— √2/2
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Q. Find the value of cos(tan^(-1)(3)).
A.
3/√10
B.
1/√10
C.
√10/10
D.
1/3
Show solution
Solution
cos(tan^(-1)(3)) = 3/√10
Correct Answer:
A
— 3/√10
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Q. Find the value of cos(tan^(-1)(3/4)).
A.
4/5
B.
3/5
C.
5/4
D.
3/4
Show solution
Solution
Using the identity, cos(tan^(-1)(3/4)) = 4/5
Correct Answer:
A
— 4/5
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Q. Find the value of cos^(-1)(-1/2).
A.
2π/3
B.
π/3
C.
π/2
D.
π
Show solution
Solution
cos^(-1)(-1/2) = 2π/3, since cos(2π/3) = -1/2.
Correct Answer:
A
— 2π/3
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Q. Find the value of cos^(-1)(0).
Show solution
Solution
cos^(-1)(0) = π/2, since cos(π/2) = 0.
Correct Answer:
B
— π/2
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Q. Find the value of i^4.
Show solution
Solution
i^4 = (i^2)^2 = (-1)^2 = 1.
Correct Answer:
A
— 1
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Q. Find the value of k for which the equation x^2 + kx + 16 = 0 has no real roots.
A.
k < 8
B.
k > 8
C.
k = 8
D.
k < 0
Show solution
Solution
For no real roots, the discriminant must be less than 0: k^2 - 4*1*16 < 0, which gives k < 8.
Correct Answer:
A
— k < 8
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Q. Find the value of k for which the equation x^2 + kx + 9 = 0 has roots that are both negative.
Show solution
Solution
For both roots to be negative, k must be positive and k^2 > 36, thus k > 6.
Correct Answer:
B
— -4
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Q. Find the value of k for which the function f(x) = kx^2 + 2x + 1 is differentiable at x = 0.
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Solution
f'(x) = 2kx + 2. At x = 0, f'(0) = 2. The function is differentiable for any k, but k = 0 gives a constant function.
Correct Answer:
A
— 0
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Q. Find the value of k for which the function f(x) = kx^2 + 3x + 2 is differentiable everywhere.
A.
k = 0
B.
k = -3
C.
k = 1
D.
k = 2
Show solution
Solution
The function is a polynomial and is differentiable for all k, hence k can be any real number.
Correct Answer:
A
— k = 0
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Q. Find the value of k for which the function f(x) = x^3 - 3kx^2 + 3k^2x - k^3 is differentiable at x = k.
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Solution
For f(x) to be differentiable at x = k, f'(k) must exist. Setting k = 1 makes f'(k) continuous.
Correct Answer:
B
— 1
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Q. Find the value of k for which the roots of the equation x^2 - kx + 9 = 0 are real and distinct.
A.
k < 6
B.
k > 6
C.
k = 6
D.
k ≤ 6
Show solution
Solution
The discriminant must be positive: k^2 - 4*1*9 > 0, which gives k < 6 or k > -6.
Correct Answer:
A
— k < 6
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Q. Find the value of k if the equation x^2 + kx + 16 = 0 has no real roots.
Show solution
Solution
For no real roots, the discriminant must be less than zero: k^2 - 4*1*16 < 0 => k^2 < 64 => |k| < 8.
Correct Answer:
B
— -4
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Q. Find the value of k if the equation x^2 + kx + 9 = 0 has no real roots.
Show solution
Solution
For no real roots, the discriminant must be less than zero: k^2 - 36 < 0, hence k < -6.
Correct Answer:
A
— -6
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Q. Find the value of k if the vectors A = (1, k, 2) and B = (2, 3, 4) are perpendicular.
Show solution
Solution
A · B = 1*2 + k*3 + 2*4 = 0. Thus, 2 + 3k + 8 = 0, so 3k = -10, k = -10/3.
Correct Answer:
A
— 1
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Q. Find the value of k such that the coefficient of x^4 in the expansion of (x + k)^6 is 240.
Show solution
Solution
The coefficient of x^4 is C(6,4) * k^2 = 15k^2. Setting 15k^2 = 240 gives k^2 = 16, so k = 4 or -4.
Correct Answer:
B
— 5
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Q. Find the value of k such that the function f(x) = x^2 + kx has a maximum at x = -2.
Show solution
Solution
For a maximum, f'(x) = 2x + k = 0 at x = -2. Thus, k = 4.
Correct Answer:
A
— -4
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Q. Find the value of k such that the function f(x) = { kx + 1, x < 1; 2x - 1, x >= 1 } is continuous at x = 1.
Show solution
Solution
Setting k(1) + 1 = 2(1) - 1 gives k + 1 = 1, so k = 0.
Correct Answer:
B
— 1
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Q. Find the value of k such that the function f(x) = { kx + 1, x < 1; 3, x = 1; x^2 + 1, x > 1 is continuous at x = 1.
Show solution
Solution
Setting k(1) + 1 = 3 gives k = 2 for continuity.
Correct Answer:
C
— 3
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Q. Find the value of k such that the function f(x) = { kx + 1, x < 2; x^2 - 3, x >= 2 } is continuous at x = 2.
Show solution
Solution
Setting k(2) + 1 = 2^2 - 3 gives 2k + 1 = 1, leading to k = 0.
Correct Answer:
B
— 2
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Q. Find the value of k such that the function f(x) = { kx + 2, x < 1; 3, x = 1; 2x + 1, x > 1 } is continuous at x = 1.
Show solution
Solution
Setting k(1) + 2 = 3 gives k = 1.
Correct Answer:
A
— 1
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Q. Find the value of k such that the function f(x) = { kx, x < 0; 0, x = 0; x^2 + k, x > 0 is continuous at x = 0.
Show solution
Solution
Setting k = 0 for continuity at x = 0 gives f(0) = 0.
Correct Answer:
B
— 0
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Q. Find the value of k such that the function f(x) = { kx, x < 0; x^2 + 1, x >= 0 is continuous at x = 0.
Show solution
Solution
Setting k(0) = 0^2 + 1 gives k = 1.
Correct Answer:
A
— 0
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Q. Find the value of log2(8).
Show solution
Solution
log2(8) = log2(2^3) = 3.
Correct Answer:
B
— 3
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Q. Find the value of m for which the function f(x) = { 2x + m, x < 1; mx + 3, x >= 1 is continuous at x = 1.
Show solution
Solution
Setting 2(1) + m = m(1) + 3 gives m = 1.
Correct Answer:
B
— 2
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Q. Find the value of m for which the function f(x) = { 2x + m, x < 1; x^2 + 1, x >= 1 is continuous at x = 1.
Show solution
Solution
Setting 2(1) + m = 1^2 + 1 gives m = 0.
Correct Answer:
A
— 0
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Q. Find the value of m for which the function f(x) = { 2x + m, x < 3; x^2 - 3, x >= 3 } is continuous at x = 3.
Show solution
Solution
Setting the two pieces equal at x = 3 gives us 6 + m = 6. Thus, m = 0.
Correct Answer:
C
— 2
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Q. Find the value of m for which the function f(x) = { 3x + m, x < 1; 2x^2, x >= 1 is continuous at x = 1.
Show solution
Solution
Setting 3(1) + m = 2(1)^2 gives m = -1.
Correct Answer:
B
— 0
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Q. Find the value of m such that the function f(x) = { x^2 + m, x < 1; 4 - x, x >= 1 } is continuous at x = 1.
Show solution
Solution
Setting the two pieces equal at x = 1: 1^2 + m = 4 - 1. Solving gives m = 2.
Correct Answer:
B
— 1
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Q. Find the value of m such that the function f(x) = { x^2 + m, x < 1; mx + 1, x >= 1 is continuous at x = 1.
Show solution
Solution
Setting x^2 + m = mx + 1 at x = 1 gives m = 1 for continuity.
Correct Answer:
A
— 0
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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!
