JEE Main Differential Calculus: Applications of Derivatives – Chapter Test

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Difficulty: 10 medium · 10 hard

Applications of Derivatives MCQ questions with answers

The first 3 of 20 questions. Unlock the series for the rest, each with a solution.

1. Q1. Match List I (function) with List II (its maximum value). List I: A. 3 sin x + 4 cos x; B. sin x + cos x; C. sin x · cos x; D. 5 sin x + 12 cos x. List II: 1. √2; 2. 13; 3. 5; 4. 1/2. Choose the correct matching.

   सूची I (फलन) का सूची II (उसका अधिकतम मान) से मिलान कीजिए। सूची I: A. 3 sin x + 4 cos x; B. sin x + cos x; C. sin x · cos x; D. 5 sin x + 12 cos x। सूची II: 1. √2; 2. 13; 3. 5; 4. 1/2। सही मिलान चुनिए।

   • A. A-2, B-1, C-4, D-3
   • B. A-3, B-4, C-1, D-2
   • C. A-3, B-1, C-4, D-2
   • D. A-1, B-3, C-4, D-2
   Solution

   The maximum of a sin x + b cos x is √(a² + b²), giving 5 for A and 13 for D; sin x + cos x has maximum √2, and sin x cos x = (1/2)sin 2x has maximum 1/2, so A-3, B-1, C-4, D-2.

   a sin x + b cos x का अधिकतम मान √(a² + b²) होता है, जिससे A के लिए 5 और D के लिए 13 प्राप्त होता है; sin x + cos x का अधिकतम मान √2 है, और sin x cos x = (1/2)sin 2x का अधिकतम मान 1/2 है, अतः A-3, B-1, C-4, D-2।

2. Q2. If f(x) = x³ and x changes from 2 to 2.01, the approximate change in the value of f (using differentials) is:

   यदि f(x) = x³ है और x, 2 से 2.01 तक बदलता है, तो f के मान में सन्निकट परिवर्तन (अवकलज का उपयोग करते हुए) है:

   • A. 0.06
   • B. 1.2
   • C. 0.012
   • D. 0.12
   Solution

   df = 3x²·dx = 3(4)(0.01) = 0.12.

   df = 3x²·dx = 3(4)(0.01) = 0.12।

3. Q3. For the function f(x) = xˣ defined on (0, ∞), consider the following statements: 1. f has a local minimum at x = 1/e. 2. The minimum value of f is e^(−1/e). 3. f is decreasing on (0, 1/e). Which of these statements is/are correct?

   (0, ∞) पर परिभाषित फलन f(x) = xˣ के लिए, निम्नलिखित कथनों पर विचार कीजिए: 1. f का x = 1/e पर एक स्थानीय निम्निष्ठ (local minimum) है। 2. f का न्यूनतम मान e^(−1/e) है। 3. f, (0, 1/e) पर ह्रासमान है। इनमें से कौन-सा/से कथन सही है/हैं?

   • A. 1, 2 and 3 / 1, 2 और 3
   • B. 1 and 2 only / केवल 1 और 2
   • C. 1 and 3 only / केवल 1 और 3
   • D. 2 and 3 only / केवल 2 और 3
   Solution

   Since ln f = x ln x, f′(x) = xˣ(1 + ln x), which is zero at x = 1/e and negative on (0, 1/e), confirming a minimum there with value (1/e)^(1/e) = e^(−1/e); all three statements are correct.

   चूँकि ln f = x ln x, अतः f′(x) = xˣ(1 + ln x), जो x = 1/e पर शून्य है और (0, 1/e) पर ऋणात्मक है, जिससे वहाँ निम्निष्ठ की पुष्टि होती है जिसका मान (1/e)^(1/e) = e^(−1/e) है; तीनों कथन सही हैं।

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