Major Competitive Exams MCQ & Objective Questions
Major Competitive Exams play a crucial role in shaping the academic and professional futures of students in India. These exams not only assess knowledge but also test problem-solving skills and time management. Practicing MCQs and objective questions is essential for scoring better, as they help in familiarizing students with the exam format and identifying important questions that frequently appear in tests.
What You Will Practise Here
Key concepts and theories related to major subjects
Important formulas and their applications
Definitions of critical terms and terminologies
Diagrams and illustrations to enhance understanding
Practice questions that mirror actual exam patterns
Strategies for solving objective questions efficiently
Time management techniques for competitive exams
Exam Relevance
The topics covered under Major Competitive Exams are integral to various examinations such as CBSE, State Boards, NEET, and JEE. Students can expect to encounter a mix of conceptual and application-based questions that require a solid understanding of the subjects. Common question patterns include multiple-choice questions that test both knowledge and analytical skills, making it essential to be well-prepared with practice MCQs.
Common Mistakes Students Make
Rushing through questions without reading them carefully
Overlooking the negative marking scheme in MCQs
Confusing similar concepts or terms
Neglecting to review previous years’ question papers
Failing to manage time effectively during the exam
FAQs
Question: How can I improve my performance in Major Competitive Exams?Answer: Regular practice of MCQs and understanding key concepts will significantly enhance your performance.
Question: What types of questions should I focus on for these exams?Answer: Concentrate on important Major Competitive Exams questions that frequently appear in past papers and mock tests.
Question: Are there specific strategies for tackling objective questions?Answer: Yes, practicing under timed conditions and reviewing mistakes can help develop effective strategies.
Start your journey towards success by solving practice MCQs today! Test your understanding and build confidence for your upcoming exams. Remember, consistent practice is the key to mastering Major Competitive Exams!
Q. Find the value of cos^(-1)(0).
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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.
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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.
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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 equation x² + 4x + k = 0 has no real roots. (2020)
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Solution
The discriminant must be negative: 4² - 4*1*k < 0, which gives k > 4, so the minimum value is -6.
Correct Answer:
B
— -6
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Q. Find the value of k for which the equation x² + kx + 16 = 0 has equal roots. (2022)
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Solution
For equal roots, the discriminant must be zero: k² - 4*1*16 = 0, thus k² = 64, k = ±8. The value of k can be -8.
Correct Answer:
A
— -8
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Q. Find the value of k for which the equation x² + kx + 9 = 0 has no real roots. (2023)
Show solution
Solution
For no real roots, the discriminant must be negative: k² - 4*1*9 < 0, thus k² < 36, hence k < -6 or k > 6.
Correct Answer:
A
— -6
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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
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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 quadratic equation x^2 + kx + 16 = 0 has no real roots. (2020)
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Solution
The discriminant must be less than zero: k^2 - 4*1*16 < 0 leads to k < -8.
Correct Answer:
A
— -8
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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
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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 coefficient of x^2 in the expansion of (x + k)^4 is 6.
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Solution
The coefficient of x^2 in (x + k)^4 is C(4, 2) * k^2 = 6. Thus, 6k^2 = 6, giving k^2 = 1, so k = 1 or -1.
Correct Answer:
B
— 2
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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 equation x² + kx + 16 = 0 has no real roots. (2022)
A.
k < 8
B.
k > 8
C.
k < 0
D.
k > 0
Show solution
Solution
For no real roots, the discriminant must be less than zero: k² - 4*1*16 < 0, which gives k > 8.
Correct Answer:
B
— k > 8
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Q. Find the value of k if the vectors A = (1, k, 2) and B = (2, 3, 4) are perpendicular.
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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 in the expansion of (x + 2)^6 such that the term containing x^4 is 240.
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Solution
The term containing x^4 is C(6,4) * (2)^2 * x^4 = 15 * 4 * x^4 = 60x^4. Setting 60 = 240 gives k = 4.
Correct Answer:
A
— 4
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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).
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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.
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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.
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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.
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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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