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. Determine the hybridization of the central atom in PCl5.
A.
sp
B.
sp2
C.
sp3
D.
dsp3
Show solution
Solution
Phosphorus in PCl5 is dsp3 hybridized, allowing it to form five bonds.
Correct Answer:
D
— dsp3
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Q. Determine the intervals where f(x) = -x^2 + 4x is concave up. (2023)
A.
(-∞, 0)
B.
(0, 2)
C.
(2, ∞)
D.
(0, 4)
Show solution
Solution
f''(x) = -2, which is always negative, indicating concave down everywhere.
Correct Answer:
C
— (2, ∞)
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Q. Determine the intervals where f(x) = x^3 - 3x is increasing. (2021)
A.
(-∞, -1)
B.
(-1, 1)
C.
(1, ∞)
D.
(-∞, 1)
Show solution
Solution
f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = -1, 1. f'(x) > 0 for x > 1.
Correct Answer:
C
— (1, ∞)
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Q. Determine the intervals where f(x) = x^4 - 4x^3 has increasing behavior. (2023)
A.
(-∞, 0)
B.
(0, 2)
C.
(2, ∞)
D.
(0, 4)
Show solution
Solution
f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3). f'(x) > 0 for x in (0, 3).
Correct Answer:
B
— (0, 2)
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Q. Determine the intervals where f(x) = x^4 - 4x^3 has local minima. (2020)
A.
(0, 2)
B.
(1, 3)
C.
(2, 4)
D.
(0, 1)
Show solution
Solution
f'(x) = 4x^3 - 12x^2. Setting f'(x) = 0 gives x = 0, 3. Testing intervals shows local minima at (0, 2).
Correct Answer:
A
— (0, 2)
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Q. Determine the intervals where the function f(x) = x^3 - 3x is increasing.
A.
(-∞, -1)
B.
(-1, 1)
C.
(1, ∞)
D.
(-∞, 1)
Show solution
Solution
f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = ±1. f'(x) > 0 for x > 1, so f(x) is increasing on (1, ∞).
Correct Answer:
C
— (1, ∞)
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Q. Determine the intervals where the function f(x) = x^4 - 4x^3 has increasing behavior.
A.
(-∞, 0) U (2, ∞)
B.
(0, 2)
C.
(0, ∞)
D.
(2, ∞)
Show solution
Solution
f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3). The function is increasing where f'(x) > 0, which is in the intervals (-∞, 0) and (3, ∞).
Correct Answer:
A
— (-∞, 0) U (2, ∞)
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Q. Determine the length of the latus rectum of the parabola y^2 = 16x.
Show solution
Solution
The length of the latus rectum for the parabola y^2 = 4px is given by 4p. Here, p = 4, so the length is 16.
Correct Answer:
B
— 8
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Q. Determine the limit: lim (x -> 0) (tan(5x)/x) (2022)
A.
0
B.
1
C.
5
D.
Undefined
Show solution
Solution
Using the standard limit lim (x -> 0) (tan(kx)/x) = k, we have k = 5. Thus, lim (x -> 0) (tan(5x)/x) = 5.
Correct Answer:
C
— 5
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Q. Determine the limit: lim (x -> 1) (x^3 - 1)/(x - 1) (2020)
Show solution
Solution
Factoring gives (x - 1)(x^2 + x + 1)/(x - 1). For x ≠ 1, this simplifies to x^2 + x + 1. Evaluating at x = 1 gives 3.
Correct Answer:
C
— 3
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Q. Determine the limit: lim (x -> 1) (x^4 - 1)/(x - 1) (2021)
A.
0
B.
1
C.
4
D.
Undefined
Show solution
Solution
Factoring gives (x - 1)(x^3 + x^2 + x + 1)/(x - 1). For x ≠ 1, this simplifies to x^3 + x^2 + x + 1. Evaluating at x = 1 gives 4.
Correct Answer:
D
— Undefined
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Q. Determine the local maxima and minima of f(x) = x^2 - 4x + 3.
A.
Maxima at x=2
B.
Minima at x=2
C.
Maxima at x=1
D.
Minima at x=1
Show solution
Solution
f'(x) = 2x - 4. Setting f'(x) = 0 gives x = 2. f''(x) = 2 > 0 indicates a local minimum at x = 2.
Correct Answer:
B
— Minima at x=2
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Q. Determine the local maxima and minima of f(x) = x^3 - 3x.
A.
Maxima at (1, -2)
B.
Minima at (0, 0)
C.
Maxima at (0, 0)
D.
Minima at (1, -2)
Show solution
Solution
f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = ±1. f''(1) = 6 > 0 (min), f''(-1) = 6 > 0 (min). Local maxima at (0, 0).
Correct Answer:
A
— Maxima at (1, -2)
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Q. Determine the local maxima and minima of f(x) = x^4 - 8x^2 + 16. (2023)
A.
Maxima at x = 0
B.
Minima at x = 2
C.
Maxima at x = 2
D.
Minima at x = 0
Show solution
Solution
f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. f''(x) = 12x^2 - 16. Minima at x = 0.
Correct Answer:
D
— Minima at x = 0
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Q. Determine the local maxima and minima of the function f(x) = x^3 - 6x^2 + 9x.
A.
(0, 0)
B.
(2, 0)
C.
(3, 0)
D.
(1, 0)
Show solution
Solution
f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives x = 1, 3. f''(1) > 0 (min), f''(3) < 0 (max).
Correct Answer:
C
— (3, 0)
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Q. Determine the local maxima and minima of the function f(x) = x^4 - 4x^3 + 4x.
A.
Maxima at (0, 0)
B.
Minima at (2, 0)
C.
Maxima at (2, 0)
D.
Minima at (0, 0)
Show solution
Solution
f'(x) = 4x^3 - 12x^2 + 4. Setting f'(x) = 0 gives x = 0 and x = 2. f''(0) = 4 > 0 (min), f''(2) = -8 < 0 (max).
Correct Answer:
B
— Minima at (2, 0)
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Q. Determine the local maxima of f(x) = -x^2 + 4x. (2022)
A.
(2, 4)
B.
(0, 0)
C.
(4, 0)
D.
(1, 1)
Show solution
Solution
f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. f(2) = -2^2 + 4(2) = 4.
Correct Answer:
A
— (2, 4)
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Q. Determine the local maxima of f(x) = -x^3 + 3x^2 + 1. (2021)
A.
(0, 1)
B.
(1, 3)
C.
(2, 5)
D.
(3, 4)
Show solution
Solution
f'(x) = -3x^2 + 6x. Setting f'(x) = 0 gives x = 0 or x = 2. f(2) = 5 is a local maximum.
Correct Answer:
B
— (1, 3)
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Q. Determine the local maxima of f(x) = x^4 - 8x^2 + 16. (2021)
A.
(0, 16)
B.
(2, 12)
C.
(4, 0)
D.
(1, 9)
Show solution
Solution
Find f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. f(2) = 12 is a local maximum.
Correct Answer:
B
— (2, 12)
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Q. Determine the local maxima or minima of f(x) = -x^2 + 4x. (2019)
A.
Maxima at x=2
B.
Minima at x=2
C.
Maxima at x=4
D.
Minima at x=4
Show solution
Solution
f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. Since f''(x) = -2 < 0, it is a maxima.
Correct Answer:
A
— Maxima at x=2
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Q. Determine the local minima of f(x) = x^3 - 3x + 2. (2021)
Show solution
Solution
f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = 1. f(1) = 0.
Correct Answer:
B
— 0
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Q. Determine the local minima of f(x) = x^4 - 4x^2. (2021)
Show solution
Solution
f'(x) = 4x^3 - 8x. Setting f'(x) = 0 gives x = 0, ±2. f(0) = 0.
Correct Answer:
B
— 0
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Q. Determine the maximum area of a triangle with a base of 10 units and height as a function of x. (2020)
Show solution
Solution
Area = 1/2 * base * height = 5h. Max area occurs when h is maximized, thus Area = 50 when h = 10.
Correct Answer:
B
— 50
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Q. Determine the maximum height of the function f(x) = -x^2 + 6x + 5. (2020) 2020
Show solution
Solution
The vertex occurs at x = 3. f(3) = -3^2 + 6*3 + 5 = 8.
Correct Answer:
A
— 8
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Q. Determine the maximum height of the projectile given by h(t) = -16t^2 + 64t + 80. (2023)
Show solution
Solution
The maximum height occurs at t = -b/(2a) = -64/(2*-16) = 2. h(2) = -16(2^2) + 64(2) + 80 = 80.
Correct Answer:
A
— 80
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Q. Determine the maximum height of the projectile modeled by h(t) = -16t^2 + 64t + 80. (2020)
Show solution
Solution
The maximum height occurs at t = -b/(2a) = 64/(2*16) = 2. h(2) = -16(2^2) + 64(2) + 80 = 80.
Correct Answer:
A
— 80
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Q. Determine the maximum value of f(x) = -2x^2 + 4x + 1. (2023)
Show solution
Solution
The vertex is at x = -4/(2*(-2)) = 1. The maximum value is f(1) = -2(1)^2 + 4(1) + 1 = 3.
Correct Answer:
C
— 3
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Q. Determine the maximum value of f(x) = -x^2 + 4x + 1.
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Solution
The vertex occurs at x = 2. f(2) = -2^2 + 4(2) + 1 = 5, which is the maximum value.
Correct Answer:
B
— 5
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Q. Determine the maximum value of f(x) = -x^2 + 4x. (2020)
Show solution
Solution
f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. f(2) = -2^2 + 4(2) = 8.
Correct Answer:
A
— 4
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Q. Determine the maximum value of f(x) = -x^2 + 6x - 8. (2022)
Show solution
Solution
The maximum occurs at x = 3. f(3) = -3^2 + 6(3) - 8 = 6.
Correct Answer:
C
— 6
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