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Applications of Derivatives: JEE Main Previous Year Questions (2021-2026)
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Page 1 JEE Main Previous Year Questions (2021-2026): Application of Derivatives (January 2026) Q1: Consider the following three statements for the function f : (0, 8) ? R defined by f(x) = |log? x| - |x - 1| : (I) f is differentiable at all x > 0. (II) f is increasing in (0, 1). (III) f is decreasing in (1, 8). Then. A: Only (I) is TRUE. B: Only (I) and (III) are TRUE. C: Only (II) and (III) are TRUE. D: All (I), (II) and (III) are TRUE. Answer: B Explanation: Step 1: Write f(x) without absolute values for different intervals When x = 1, ln x is positive and x - 1 is also positive. So: f(x) = |ln x| - |x - 1| = ln x - (x - 1) = ln x - x + 1 for x = 1. When 0 < x < 1, ln x is negative and x - 1 is negative. So: f(x) = |ln x| - |x - 1| = - ln x - (-(x - 1)) = - ln x + x - 1 for 0 < x < 1. Step 2: Find the derivative f'(x) in both cases For x = 1, f'(x) = d/dx (ln x - x + 1) = 1/x - 1 For 0 < x < 1, f'(x) = d/dx (- ln x + x - 1) = -1/x + 1 Page 2 JEE Main Previous Year Questions (2021-2026): Application of Derivatives (January 2026) Q1: Consider the following three statements for the function f : (0, 8) ? R defined by f(x) = |log? x| - |x - 1| : (I) f is differentiable at all x > 0. (II) f is increasing in (0, 1). (III) f is decreasing in (1, 8). Then. A: Only (I) is TRUE. B: Only (I) and (III) are TRUE. C: Only (II) and (III) are TRUE. D: All (I), (II) and (III) are TRUE. Answer: B Explanation: Step 1: Write f(x) without absolute values for different intervals When x = 1, ln x is positive and x - 1 is also positive. So: f(x) = |ln x| - |x - 1| = ln x - (x - 1) = ln x - x + 1 for x = 1. When 0 < x < 1, ln x is negative and x - 1 is negative. So: f(x) = |ln x| - |x - 1| = - ln x - (-(x - 1)) = - ln x + x - 1 for 0 < x < 1. Step 2: Find the derivative f'(x) in both cases For x = 1, f'(x) = d/dx (ln x - x + 1) = 1/x - 1 For 0 < x < 1, f'(x) = d/dx (- ln x + x - 1) = -1/x + 1 Step 3: Check differentiability at x = 1 From the left of x = 1: f'(1 ?) = -1/1 + 1 = 0 From the right of x = 1: f'(1 ?) = 1/1 - 1 = 0 Both one-sided derivatives are equal. So f(x) is differentiable at x = 1. Since the formulas for f'(x) are valid everywhere for x > 0, f(x) is differentiable for all x > 0. Step 4: Check if f(x) is increasing or decreasing Step 5: Check the statements (I) f is differentiable at all x > 0: True (II) f is increasing in (0, 1): False (III) f is decreasing in (1, 8): True So only statements (I) and (III) are correct. Q2: The least value of (cos²? - 6 sin ? cos ? + 3 sin²? + 2) is A: 4 - v10 B: -1 C: 4 + v10 D: 1 Answer: A Explanation: f(?) = cos²? - 6 sin ? cos ? + 3 sin²? + 2 f(?) = (1 + cos 2?)/2 - 3 sin 2? + 3((1 - cos 2?)/2) + 2 f(?) = 1/2 + cos 2?/2 - 3 sin 2? + 3/2 - 3 cos 2?/2 + 2 f(?) = 4 - cos 2? - 3 sin 2? f(?) = 4 - (cos 2? + 3 sin 2?) Let g(?) = cos 2? + 3 sin 2? Page 3 JEE Main Previous Year Questions (2021-2026): Application of Derivatives (January 2026) Q1: Consider the following three statements for the function f : (0, 8) ? R defined by f(x) = |log? x| - |x - 1| : (I) f is differentiable at all x > 0. (II) f is increasing in (0, 1). (III) f is decreasing in (1, 8). Then. A: Only (I) is TRUE. B: Only (I) and (III) are TRUE. C: Only (II) and (III) are TRUE. D: All (I), (II) and (III) are TRUE. Answer: B Explanation: Step 1: Write f(x) without absolute values for different intervals When x = 1, ln x is positive and x - 1 is also positive. So: f(x) = |ln x| - |x - 1| = ln x - (x - 1) = ln x - x + 1 for x = 1. When 0 < x < 1, ln x is negative and x - 1 is negative. So: f(x) = |ln x| - |x - 1| = - ln x - (-(x - 1)) = - ln x + x - 1 for 0 < x < 1. Step 2: Find the derivative f'(x) in both cases For x = 1, f'(x) = d/dx (ln x - x + 1) = 1/x - 1 For 0 < x < 1, f'(x) = d/dx (- ln x + x - 1) = -1/x + 1 Step 3: Check differentiability at x = 1 From the left of x = 1: f'(1 ?) = -1/1 + 1 = 0 From the right of x = 1: f'(1 ?) = 1/1 - 1 = 0 Both one-sided derivatives are equal. So f(x) is differentiable at x = 1. Since the formulas for f'(x) are valid everywhere for x > 0, f(x) is differentiable for all x > 0. Step 4: Check if f(x) is increasing or decreasing Step 5: Check the statements (I) f is differentiable at all x > 0: True (II) f is increasing in (0, 1): False (III) f is decreasing in (1, 8): True So only statements (I) and (III) are correct. Q2: The least value of (cos²? - 6 sin ? cos ? + 3 sin²? + 2) is A: 4 - v10 B: -1 C: 4 + v10 D: 1 Answer: A Explanation: f(?) = cos²? - 6 sin ? cos ? + 3 sin²? + 2 f(?) = (1 + cos 2?)/2 - 3 sin 2? + 3((1 - cos 2?)/2) + 2 f(?) = 1/2 + cos 2?/2 - 3 sin 2? + 3/2 - 3 cos 2?/2 + 2 f(?) = 4 - cos 2? - 3 sin 2? f(?) = 4 - (cos 2? + 3 sin 2?) Let g(?) = cos 2? + 3 sin 2? Q3: Let a and ß respectively be the maximum and the minimum values of the function f(?) = 4(sin 4(7p/2 - ?) + sin 4(11p + ?)) - 2(sin 6(3p/2 - ?) + sin 6(9p - ?)), ? ? R. Then a + 2ß is equal to : A: 6 B: 5 C: 4 D: 3 Answer: B Explanation: f(?) max = 2 = a f(?) min = 2 - 1/2 = 3/2 = ß a + 2ß = 2 + 3 = 5 Page 4 JEE Main Previous Year Questions (2021-2026): Application of Derivatives (January 2026) Q1: Consider the following three statements for the function f : (0, 8) ? R defined by f(x) = |log? x| - |x - 1| : (I) f is differentiable at all x > 0. (II) f is increasing in (0, 1). (III) f is decreasing in (1, 8). Then. A: Only (I) is TRUE. B: Only (I) and (III) are TRUE. C: Only (II) and (III) are TRUE. D: All (I), (II) and (III) are TRUE. Answer: B Explanation: Step 1: Write f(x) without absolute values for different intervals When x = 1, ln x is positive and x - 1 is also positive. So: f(x) = |ln x| - |x - 1| = ln x - (x - 1) = ln x - x + 1 for x = 1. When 0 < x < 1, ln x is negative and x - 1 is negative. So: f(x) = |ln x| - |x - 1| = - ln x - (-(x - 1)) = - ln x + x - 1 for 0 < x < 1. Step 2: Find the derivative f'(x) in both cases For x = 1, f'(x) = d/dx (ln x - x + 1) = 1/x - 1 For 0 < x < 1, f'(x) = d/dx (- ln x + x - 1) = -1/x + 1 Step 3: Check differentiability at x = 1 From the left of x = 1: f'(1 ?) = -1/1 + 1 = 0 From the right of x = 1: f'(1 ?) = 1/1 - 1 = 0 Both one-sided derivatives are equal. So f(x) is differentiable at x = 1. Since the formulas for f'(x) are valid everywhere for x > 0, f(x) is differentiable for all x > 0. Step 4: Check if f(x) is increasing or decreasing Step 5: Check the statements (I) f is differentiable at all x > 0: True (II) f is increasing in (0, 1): False (III) f is decreasing in (1, 8): True So only statements (I) and (III) are correct. Q2: The least value of (cos²? - 6 sin ? cos ? + 3 sin²? + 2) is A: 4 - v10 B: -1 C: 4 + v10 D: 1 Answer: A Explanation: f(?) = cos²? - 6 sin ? cos ? + 3 sin²? + 2 f(?) = (1 + cos 2?)/2 - 3 sin 2? + 3((1 - cos 2?)/2) + 2 f(?) = 1/2 + cos 2?/2 - 3 sin 2? + 3/2 - 3 cos 2?/2 + 2 f(?) = 4 - cos 2? - 3 sin 2? f(?) = 4 - (cos 2? + 3 sin 2?) Let g(?) = cos 2? + 3 sin 2? Q3: Let a and ß respectively be the maximum and the minimum values of the function f(?) = 4(sin 4(7p/2 - ?) + sin 4(11p + ?)) - 2(sin 6(3p/2 - ?) + sin 6(9p - ?)), ? ? R. Then a + 2ß is equal to : A: 6 B: 5 C: 4 D: 3 Answer: B Explanation: f(?) max = 2 = a f(?) min = 2 - 1/2 = 3/2 = ß a + 2ß = 2 + 3 = 5 Q4: Let f(x) = x² °² 5 - x² ° ° °, x ? [0, 1] and the minimum value of the function f(x) in the interval [0, 1] be (80) 8 °(n) ? 8¹. Then n is equal to A: -40 B: -41 C: -80 D: -81 Answer: D Explanation: f(x) = x² °² 5 - x² ° ° ° ; x ? [0, 1] f'(x) = 2025x² °² 4 - 2000x¹ ? ? ? = x¹ ? ? ?(2025x 5 - 2000) f'(x) = 0 => x 5 = 80/81 f(x) = (x 5) 8¹ - (x 5) 8 ° = (80/81) 8¹ - (80/81) 8 ° = (80/81) 8 ° (-1/81) = (80) 8 °(-81) ? 8¹ Q5: Here is the text form of both images: Let f : R ? R be a twice differentiable function such that f?(x) > 0 for all x ? R and f'(a - 1) = 0, where a is a real number. Consider the following two statements: (I) g is increasing in (0, p/4) (II) g is decreasing in (p/4, p/2) Then, A Both (I) and (II) are True B Neither (I) nor (II) is True C Only (I) is True D Only (II) is True Answer: B Explanation: g(x) = f((tan x - 1)² + a - 1) g'(x) = f'((tan x - 1)² + a - 1) · 2(tan x - 1) · sec²x. ...(i) Given f?(x) > 0 ?x ? R Page 5 JEE Main Previous Year Questions (2021-2026): Application of Derivatives (January 2026) Q1: Consider the following three statements for the function f : (0, 8) ? R defined by f(x) = |log? x| - |x - 1| : (I) f is differentiable at all x > 0. (II) f is increasing in (0, 1). (III) f is decreasing in (1, 8). Then. A: Only (I) is TRUE. B: Only (I) and (III) are TRUE. C: Only (II) and (III) are TRUE. D: All (I), (II) and (III) are TRUE. Answer: B Explanation: Step 1: Write f(x) without absolute values for different intervals When x = 1, ln x is positive and x - 1 is also positive. So: f(x) = |ln x| - |x - 1| = ln x - (x - 1) = ln x - x + 1 for x = 1. When 0 < x < 1, ln x is negative and x - 1 is negative. So: f(x) = |ln x| - |x - 1| = - ln x - (-(x - 1)) = - ln x + x - 1 for 0 < x < 1. Step 2: Find the derivative f'(x) in both cases For x = 1, f'(x) = d/dx (ln x - x + 1) = 1/x - 1 For 0 < x < 1, f'(x) = d/dx (- ln x + x - 1) = -1/x + 1 Step 3: Check differentiability at x = 1 From the left of x = 1: f'(1 ?) = -1/1 + 1 = 0 From the right of x = 1: f'(1 ?) = 1/1 - 1 = 0 Both one-sided derivatives are equal. So f(x) is differentiable at x = 1. Since the formulas for f'(x) are valid everywhere for x > 0, f(x) is differentiable for all x > 0. Step 4: Check if f(x) is increasing or decreasing Step 5: Check the statements (I) f is differentiable at all x > 0: True (II) f is increasing in (0, 1): False (III) f is decreasing in (1, 8): True So only statements (I) and (III) are correct. Q2: The least value of (cos²? - 6 sin ? cos ? + 3 sin²? + 2) is A: 4 - v10 B: -1 C: 4 + v10 D: 1 Answer: A Explanation: f(?) = cos²? - 6 sin ? cos ? + 3 sin²? + 2 f(?) = (1 + cos 2?)/2 - 3 sin 2? + 3((1 - cos 2?)/2) + 2 f(?) = 1/2 + cos 2?/2 - 3 sin 2? + 3/2 - 3 cos 2?/2 + 2 f(?) = 4 - cos 2? - 3 sin 2? f(?) = 4 - (cos 2? + 3 sin 2?) Let g(?) = cos 2? + 3 sin 2? Q3: Let a and ß respectively be the maximum and the minimum values of the function f(?) = 4(sin 4(7p/2 - ?) + sin 4(11p + ?)) - 2(sin 6(3p/2 - ?) + sin 6(9p - ?)), ? ? R. Then a + 2ß is equal to : A: 6 B: 5 C: 4 D: 3 Answer: B Explanation: f(?) max = 2 = a f(?) min = 2 - 1/2 = 3/2 = ß a + 2ß = 2 + 3 = 5 Q4: Let f(x) = x² °² 5 - x² ° ° °, x ? [0, 1] and the minimum value of the function f(x) in the interval [0, 1] be (80) 8 °(n) ? 8¹. Then n is equal to A: -40 B: -41 C: -80 D: -81 Answer: D Explanation: f(x) = x² °² 5 - x² ° ° ° ; x ? [0, 1] f'(x) = 2025x² °² 4 - 2000x¹ ? ? ? = x¹ ? ? ?(2025x 5 - 2000) f'(x) = 0 => x 5 = 80/81 f(x) = (x 5) 8¹ - (x 5) 8 ° = (80/81) 8¹ - (80/81) 8 ° = (80/81) 8 ° (-1/81) = (80) 8 °(-81) ? 8¹ Q5: Here is the text form of both images: Let f : R ? R be a twice differentiable function such that f?(x) > 0 for all x ? R and f'(a - 1) = 0, where a is a real number. Consider the following two statements: (I) g is increasing in (0, p/4) (II) g is decreasing in (p/4, p/2) Then, A Both (I) and (II) are True B Neither (I) nor (II) is True C Only (I) is True D Only (II) is True Answer: B Explanation: g(x) = f((tan x - 1)² + a - 1) g'(x) = f'((tan x - 1)² + a - 1) · 2(tan x - 1) · sec²x. ...(i) Given f?(x) > 0 ?x ? R ? f'(x) is increasing function Case (i) $0 0 < tan x > 1 -1 < tan x - 1 < 0 0 < (tan x - 1)² < 1 a - 1 < (tan x - 1)² + (a - 1) < a ? f'((tan x - 1)² + a - 1) > f'(a - 1) = 0 From (i) g'(x) = (+)(-)(+) = -ve g(x) is decreasing in (0, p/4) Similarly we can prove that g(x) is increasing in (p/4, p/2) Q6: Let (2a, a) be the largest interval in which the function f(t) = |t+1|/t², t < 0, is strictly decreasing. Then the local maximum value of the function g(x) = 2 log?(x - 2) + ax² + 4x - a, x > 2, is ____ Answer: 4 Explanation: If t < -1, f'(t) = (t+2)/t³ roots is t = -2 for t ? (-2, -1), f'(t) < 0, decreasing interval t ? (-8, -2), f'(t) > 0, Increasing interval If t ? (-1, 0), f'(t) = -(t+2)/t³ f'(t) > 0 so t ? (-1, 0) is increasing interval. It is given that (2a, a) is the largest interval in which f(t) is decreasing matches with t ? (-2, -1). so, a = -1. now, g(x) = 2 log?(x - 2) + ax² + 4x - a, x > 2 substitute a = -1 g(x) = 2 log?(x - 2) - x² + 4x + 1, x > 2 ? g'(x) = 2 · 1/(x-2) - 2x + 4Read More
FAQs on Applications of Derivatives: JEE Main Previous Year Questions (2021-2026)
| 1. What are some common applications of derivatives in real life? |  |
Ans. Some common applications of derivatives in real life include determining maximum and minimum values, analyzing rates of change, optimizing functions, and modeling physical systems such as motion and growth.
| 2. How are derivatives used in economics and finance? | |
Ans. In economics and finance, derivatives are used to analyze and predict changes in variables like stock prices, interest rates, and currency exchange rates. They help in risk management, portfolio optimization, and pricing financial instruments.
| 3. Can derivatives be used to solve optimization problems? | |
Ans. Yes, derivatives can be used to solve optimization problems by finding the critical points of a function and determining whether they correspond to a maximum, minimum, or saddle point. This is useful in maximizing profits, minimizing costs, and optimizing processes.
| 4. What is the relationship between derivatives and tangents in calculus? | |
Ans. Derivatives represent the slope of a function at a given point, which is equivalent to the slope of the tangent line to the function at that point. This relationship allows us to approximate the behavior of a function near a specific point.
| 5. How do derivatives help in analyzing motion and velocity? | |
Ans. Derivatives help in analyzing motion and velocity by providing information about the rate at which an object's position changes with respect to time. The derivative of the position function gives the velocity function, which describes the object's speed and direction at any given moment.
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