JEE Main Differential Calculus: Limits, Continuity and Differentiability – Chapter Test

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Difficulty: 10 medium · 10 hard

Limits, Continuity and Differentiability MCQ questions with answers

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

1. Q1. Let f(x) = e^{1/x}/(1 + e^{1/x}) for x ≠ 0 and f(0) = 1/2. At x = 0 the function f has:

   मान लीजिए x ≠ 0 के लिए f(x) = e^{1/x}/(1 + e^{1/x}) तथा f(0) = 1/2। x = 0 पर फलन f में है:

   • A. a jump discontinuity (discontinuity of the first kind) / उछाल असांतत्यता (प्रथम प्रकार की असांतत्यता)
   • B. no discontinuity (f is continuous at 0) / कोई असांतत्यता नहीं (f, 0 पर संतत है)
   • C. an infinite discontinuity / अनंत असांतत्यता
   • D. a removable discontinuity / निवार्य (removable) असांतत्यता
   Solution

   As x → 0⁺, e^{1/x} → ∞ so f → 1; as x → 0⁻, e^{1/x} → 0 so f → 0. The two one-sided limits are finite but unequal, giving a jump discontinuity (of the first kind).

   जब x → 0⁺, तब e^{1/x} → ∞ अतः f → 1; जब x → 0⁻, तब e^{1/x} → 0 अतः f → 0। दोनों एकपक्षीय सीमाएँ परिमित हैं परंतु असमान हैं, जिससे प्रथम प्रकार की उछाल असांतत्यता (jump discontinuity) प्राप्त होती है।

2. Q2. Applying Rolle's theorem to f(x) = x³ − 3x on the interval [−√3, √3], the value(s) of c in the open interval satisfying f′(c) = 0 are:

   अंतराल [−√3, √3] पर f(x) = x³ − 3x के लिए रोल के प्रमेय को लागू करने पर, खुले अंतराल में f′(c) = 0 को संतुष्ट करने वाले c का/के मान है/हैं:

   • A. ±1
   • B. 0
   • C. ±1/√3
   • D. ±√3
   Solution

   f(−√3) = f(√3) = 0 and f′(x) = 3x² − 3 = 0 gives x = ±1, both lying in (−√3, √3).

   f(−√3) = f(√3) = 0 तथा f′(x) = 3x² − 3 = 0 से x = ±1 प्राप्त होता है, जो दोनों (−√3, √3) में स्थित हैं।

3. Q3. Assertion (A): For f(x) = |x| on [−1, 1], there is no point c in (−1, 1) at which f′(c) = 0. Reason (R): Rolle's theorem requires f to be differentiable throughout the open interval, and |x| is not differentiable at x = 0. Choose the correct option.

   अभिकथन (A): [−1, 1] पर f(x) = |x| के लिए, (−1, 1) में ऐसा कोई बिंदु c नहीं है जिस पर f′(c) = 0 हो। कारण (R): रोल के प्रमेय के लिए आवश्यक है कि f संपूर्ण विवृत अंतराल में अवकलनीय हो, और |x| x = 0 पर अवकलनीय नहीं है। सही विकल्प चुनें।

   • A. A is false but R is true / A असत्य है परंतु R सत्य है
   • B. Both A and R are true but R is not the correct explanation of A / A और R दोनों सत्य हैं परंतु R, A का सही स्पष्टीकरण नहीं है
   • C. Both A and R are true and R is the correct explanation of A / A और R दोनों सत्य हैं और R, A का सही स्पष्टीकरण है
   • D. A is true but R is false / A सत्य है परंतु R असत्य है
   Solution

   f′(x) = ±1 (never 0) wherever it exists, so A is true; R correctly states the failed hypothesis of Rolle's theorem that explains why the conclusion need not hold, so R is the correct explanation.

   f′(x) = ±1 (कभी 0 नहीं) जहाँ भी यह अस्तित्व में है, अतः A सत्य है; R रोल के प्रमेय की उस असफल परिकल्पना को सही ढंग से बताता है जो समझाती है कि निष्कर्ष क्यों आवश्यक नहीं होता, अतः R, A का सही स्पष्टीकरण है।

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