JEE Main Differential Calculus – Topic Test

Topics in this test

Difficulty: 15 medium · 15 hard

Differential Calculus MCQ questions with answers

The first 3 of 30 questions. Unlock the series for the rest, each with a solution.

1. Q1. For the greatest integer function f(x) = [x], consider the following statements: 1. f is right-continuous at every integer. 2. f is left-continuous at every integer. 3. The jump of f at each integer equals 1. Which of these statements is/are correct?

   महत्तम पूर्णांक फलन f(x) = [x] के लिए, निम्नलिखित कथनों पर विचार करें: 1. f प्रत्येक पूर्णांक पर दायें-संतत है। 2. f प्रत्येक पूर्णांक पर बायें-संतत है। 3. प्रत्येक पूर्णांक पर f की छलांग 1 के बराबर है। इनमें से कौन-सा/से कथन सही है/हैं?

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

   At an integer n, lim_{x→n⁺}[x] = n = f(n) (right-continuous), but lim_{x→n⁻}[x] = n−1 ≠ n (not left-continuous), and the jump is n − (n−1) = 1. Hence 1 and 3 are correct.

   किसी पूर्णांक n पर, lim_{x→n⁺}[x] = n = f(n) (दायें-संतत), परंतु lim_{x→n⁻}[x] = n−1 ≠ n (बायें-संतत नहीं), और छलांग n − (n−1) = 1 है। अतः 1 और 3 सही हैं।

2. Q2. If y = sin⁻¹(2x/(1 + x²)) for −1 < x < 1, then dy/dx equals:

   यदि −1 < x < 1 के लिए y = sin⁻¹(2x/(1 + x²)), तो dy/dx बराबर है:

   • A. 2/√(1 − x²)
   • B. −2/(1 + x²)
   • C. 2/(1 + x²)
   • D. 1/(1 + x²)
   Solution

   With x = tan θ, 2x/(1 + x²) = sin 2θ, so for −1 < x < 1 we have y = 2 tan⁻¹ x, giving dy/dx = 2/(1 + x²).

   x = tan θ लेने पर, 2x/(1 + x²) = sin 2θ, अतः −1 < x < 1 के लिए y = 2 tan⁻¹ x होता है, जिससे dy/dx = 2/(1 + x²) प्राप्त होता है।

3. Q3. If the line y = 4x − 5 is a tangent to the curve y² = ax³ + b at the point (2, 3), then the value of a + b is:

   यदि रेखा y = 4x − 5 वक्र y² = ax³ + b के बिंदु (2, 3) पर स्पर्शरेखा है, तो a + b का मान है:

   • A. −3
   • B. −5
   • C. 2
   • D. 5
   Solution

   On the curve, 9 = 8a + b. Differentiating, 2y·y′ = 3ax², so at (2,3) the slope is 12a/6 = 2a = 4, giving a = 2 and b = 9 − 16 = −7; hence a + b = −5.

   वक्र पर, 9 = 8a + b। अवकलन करने पर, 2y·y′ = 3ax², अतः (2,3) पर ढाल 12a/6 = 2a = 4 है, जिससे a = 2 और b = 9 − 16 = −7 प्राप्त होता है; अतः a + b = −5।

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