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JEE Main Previous Year Questions (2025): Application of Derivatives
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Page 1 JEE Main Previous Year Questions (2025): Application of Derivatives Q1: If the set of all values of ?? , for which the equation ?? ?? ?? -???? ?? -?? =?? has three distinct real roots, is the interval (?? ,?? ), then ?? -?? ?? is equal to ____ . JEE Main 2025 (Online) 23rd January Morning Shift Ans: 30 Solution: 5?? 3 -15?? -?? =0 ?? (?? )=5?? 3 -15?? ?? ' (?? )=15?? 2 -15=15(?? -1)(?? +1) a?(-10,10) ?? =-10,?? =10 ?? -2?? =10+20=30 Q2: Let ?? (?? ,-?? ),?? (?? ,?? ) and ?? (?? ,-?? ) be the vertices of a triangle ABC . Then the maximum area of the parallelogram ???????? , formed with vertices ?? ,?? and ?? on the sides ???? ,???? and ???? of the triangle ?????? respectively, is ____ JEE Main 2025 (Online) 2nd April Evening Shift Page 2 JEE Main Previous Year Questions (2025): Application of Derivatives Q1: If the set of all values of ?? , for which the equation ?? ?? ?? -???? ?? -?? =?? has three distinct real roots, is the interval (?? ,?? ), then ?? -?? ?? is equal to ____ . JEE Main 2025 (Online) 23rd January Morning Shift Ans: 30 Solution: 5?? 3 -15?? -?? =0 ?? (?? )=5?? 3 -15?? ?? ' (?? )=15?? 2 -15=15(?? -1)(?? +1) a?(-10,10) ?? =-10,?? =10 ?? -2?? =10+20=30 Q2: Let ?? (?? ,-?? ),?? (?? ,?? ) and ?? (?? ,-?? ) be the vertices of a triangle ABC . Then the maximum area of the parallelogram ???????? , formed with vertices ?? ,?? and ?? on the sides ???? ,???? and ???? of the triangle ?????? respectively, is ____ JEE Main 2025 (Online) 2nd April Evening Shift Ans: 3 Solution: The maximum area of such a parallelogram ???????? , with one vertex fixed at ?? and the other three points lying on the sides of triangle ?????? , is half the area of triangle ?????? . Using the determinant formula for area of triangle with vertices ?? (?? 1 ,?? 1 ),?? (?? 2 ,?? 2 ),?? (?? 3 ,?? 3 ) : Area ??????? = 1 2 ?? 1 (?? 2 -?? 3 2 )+?? 2 (?? 3 -?? 1 )+?? 3 (?? 1 -?? 2 )| Substitute the coordinates: = 1 2 |4(1-(-3))+1((-3)-(-2))+9((-2)-1)| = 1 2 |4(4)+1(-1)+9(-3)| = 1 2 |16-1-27|= 1 2 |-12|= 12 2 =6 Maximum area of parallelogram ???????? = 1 2 × area of triangle = 1 2 ×6=3 Q3: Let ?? (?? )=? ?? ?? ?? ? ?? ?? -???? +???? ?? ?? ???? ,?? ??? . Then the numbers of local maximum and local minimum points of ?? , respectively, are : JEE Main 2025 (Online) 22nd January Evening Shift Options: A. 3 and 2 B. 2 and 2 C. 2 and 3 D. 1 and 3 Page 3 JEE Main Previous Year Questions (2025): Application of Derivatives Q1: If the set of all values of ?? , for which the equation ?? ?? ?? -???? ?? -?? =?? has three distinct real roots, is the interval (?? ,?? ), then ?? -?? ?? is equal to ____ . JEE Main 2025 (Online) 23rd January Morning Shift Ans: 30 Solution: 5?? 3 -15?? -?? =0 ?? (?? )=5?? 3 -15?? ?? ' (?? )=15?? 2 -15=15(?? -1)(?? +1) a?(-10,10) ?? =-10,?? =10 ?? -2?? =10+20=30 Q2: Let ?? (?? ,-?? ),?? (?? ,?? ) and ?? (?? ,-?? ) be the vertices of a triangle ABC . Then the maximum area of the parallelogram ???????? , formed with vertices ?? ,?? and ?? on the sides ???? ,???? and ???? of the triangle ?????? respectively, is ____ JEE Main 2025 (Online) 2nd April Evening Shift Ans: 3 Solution: The maximum area of such a parallelogram ???????? , with one vertex fixed at ?? and the other three points lying on the sides of triangle ?????? , is half the area of triangle ?????? . Using the determinant formula for area of triangle with vertices ?? (?? 1 ,?? 1 ),?? (?? 2 ,?? 2 ),?? (?? 3 ,?? 3 ) : Area ??????? = 1 2 ?? 1 (?? 2 -?? 3 2 )+?? 2 (?? 3 -?? 1 )+?? 3 (?? 1 -?? 2 )| Substitute the coordinates: = 1 2 |4(1-(-3))+1((-3)-(-2))+9((-2)-1)| = 1 2 |4(4)+1(-1)+9(-3)| = 1 2 |16-1-27|= 1 2 |-12|= 12 2 =6 Maximum area of parallelogram ???????? = 1 2 × area of triangle = 1 2 ×6=3 Q3: Let ?? (?? )=? ?? ?? ?? ? ?? ?? -???? +???? ?? ?? ???? ,?? ??? . Then the numbers of local maximum and local minimum points of ?? , respectively, are : JEE Main 2025 (Online) 22nd January Evening Shift Options: A. 3 and 2 B. 2 and 2 C. 2 and 3 D. 1 and 3 Ans: C Solution: We are given ?? (?? )=? 0 ?? 2 ? ?? 2 -8?? +15 ?? ?? ???? ,?? ?R. To find the local extrema, we first compute the derivative using the Fundamental Theorem of Calculus and the chain rule. Step 1. Rewrite the derivative: Let ?? (?? )=? 0 ?? ? ?? 2 -8?? +15 ?? ?? ???? , so that ?? (?? )=?? (?? 2 ). Then by the chain rule, ?? ' (?? )=?? ' (?? 2 )·2?? . Since ?? ' (?? )= ?? 2 -8?? +15 ?? ?? , we substitute ?? =?? 2 to get ?? ' (?? )= (?? 2 ) 2 -8?? 2 +15 ?? ?? 2 ·2?? = 2?? (?? 4 -8?? 2 +15) ?? ?? 2 . Step 2. Factor and Identify Critical Points: Factor the polynomial ?? 4 -8?? 2 +15 by writing it in terms of ?? 2 . Let ?? =?? 2 , then ?? 2 -8?? +15=(?? -3)(?? -5) so that ?? 4 -8?? 2 +15=(?? 2 -3)(?? 2 -5). Thus, ?? ' (?? )= 2?? (?? 2 -3)(?? 2 -5) ?? ?? 2 . Since ?? ?? 2 >0 for all ?? , the zeros of ?? ' (?? ) are determined by 2?? (?? 2 -3)(?? 2 -5)=0. That gives the critical points: 2?? =0??? =0, ?? 2 -3=0??? =±v3, ?? 2 -5=0??? =±v5. Step 3. Analyzing the Sign of ?? ' (?? ) : We need to determine the nature (maximum or minimum) by looking at the sign changes of ?? ' (?? ) on the intervals determined by the critical points ?? =-v5,?? =-v3,?? =0,?? =v3, and ?? =v5. Notice that the factor 2?? (?? 2 -3)(?? 2 -5) will dictate the sign. Let's define: h(?? )=?? (?? 2 -3)(?? 2 -5). Examine the sign of h(?? ) in each interval: For ?? <-v5 : ?? is negative. ?? 2 >5 so ?? 2 -3>0 and ?? 2 -5>0. Page 4 JEE Main Previous Year Questions (2025): Application of Derivatives Q1: If the set of all values of ?? , for which the equation ?? ?? ?? -???? ?? -?? =?? has three distinct real roots, is the interval (?? ,?? ), then ?? -?? ?? is equal to ____ . JEE Main 2025 (Online) 23rd January Morning Shift Ans: 30 Solution: 5?? 3 -15?? -?? =0 ?? (?? )=5?? 3 -15?? ?? ' (?? )=15?? 2 -15=15(?? -1)(?? +1) a?(-10,10) ?? =-10,?? =10 ?? -2?? =10+20=30 Q2: Let ?? (?? ,-?? ),?? (?? ,?? ) and ?? (?? ,-?? ) be the vertices of a triangle ABC . Then the maximum area of the parallelogram ???????? , formed with vertices ?? ,?? and ?? on the sides ???? ,???? and ???? of the triangle ?????? respectively, is ____ JEE Main 2025 (Online) 2nd April Evening Shift Ans: 3 Solution: The maximum area of such a parallelogram ???????? , with one vertex fixed at ?? and the other three points lying on the sides of triangle ?????? , is half the area of triangle ?????? . Using the determinant formula for area of triangle with vertices ?? (?? 1 ,?? 1 ),?? (?? 2 ,?? 2 ),?? (?? 3 ,?? 3 ) : Area ??????? = 1 2 ?? 1 (?? 2 -?? 3 2 )+?? 2 (?? 3 -?? 1 )+?? 3 (?? 1 -?? 2 )| Substitute the coordinates: = 1 2 |4(1-(-3))+1((-3)-(-2))+9((-2)-1)| = 1 2 |4(4)+1(-1)+9(-3)| = 1 2 |16-1-27|= 1 2 |-12|= 12 2 =6 Maximum area of parallelogram ???????? = 1 2 × area of triangle = 1 2 ×6=3 Q3: Let ?? (?? )=? ?? ?? ?? ? ?? ?? -???? +???? ?? ?? ???? ,?? ??? . Then the numbers of local maximum and local minimum points of ?? , respectively, are : JEE Main 2025 (Online) 22nd January Evening Shift Options: A. 3 and 2 B. 2 and 2 C. 2 and 3 D. 1 and 3 Ans: C Solution: We are given ?? (?? )=? 0 ?? 2 ? ?? 2 -8?? +15 ?? ?? ???? ,?? ?R. To find the local extrema, we first compute the derivative using the Fundamental Theorem of Calculus and the chain rule. Step 1. Rewrite the derivative: Let ?? (?? )=? 0 ?? ? ?? 2 -8?? +15 ?? ?? ???? , so that ?? (?? )=?? (?? 2 ). Then by the chain rule, ?? ' (?? )=?? ' (?? 2 )·2?? . Since ?? ' (?? )= ?? 2 -8?? +15 ?? ?? , we substitute ?? =?? 2 to get ?? ' (?? )= (?? 2 ) 2 -8?? 2 +15 ?? ?? 2 ·2?? = 2?? (?? 4 -8?? 2 +15) ?? ?? 2 . Step 2. Factor and Identify Critical Points: Factor the polynomial ?? 4 -8?? 2 +15 by writing it in terms of ?? 2 . Let ?? =?? 2 , then ?? 2 -8?? +15=(?? -3)(?? -5) so that ?? 4 -8?? 2 +15=(?? 2 -3)(?? 2 -5). Thus, ?? ' (?? )= 2?? (?? 2 -3)(?? 2 -5) ?? ?? 2 . Since ?? ?? 2 >0 for all ?? , the zeros of ?? ' (?? ) are determined by 2?? (?? 2 -3)(?? 2 -5)=0. That gives the critical points: 2?? =0??? =0, ?? 2 -3=0??? =±v3, ?? 2 -5=0??? =±v5. Step 3. Analyzing the Sign of ?? ' (?? ) : We need to determine the nature (maximum or minimum) by looking at the sign changes of ?? ' (?? ) on the intervals determined by the critical points ?? =-v5,?? =-v3,?? =0,?? =v3, and ?? =v5. Notice that the factor 2?? (?? 2 -3)(?? 2 -5) will dictate the sign. Let's define: h(?? )=?? (?? 2 -3)(?? 2 -5). Examine the sign of h(?? ) in each interval: For ?? <-v5 : ?? is negative. ?? 2 >5 so ?? 2 -3>0 and ?? 2 -5>0. Product: negative × positive × positive = negative . Thus, ?? ' (?? )<0. For -v5<?? <-v3 : ?? is negative. ?? 2 is between 3 and 5 so ?? 2 -3>0 while ?? 2 -5<0. Product: negative × positive × negative = positive . Thus, ?? ' (?? )>0. For -v3<?? <0 : ?? is negative. ?? 2 <3 so both ?? 2 -3<0 and ?? 2 -5<0. Product: negative × negative × negative = negative. Thus, ?? ' (?? )<0. For 0<?? <v3 : ?? is positive. ?? 2 <3 so both ?? 2 -3<0 and ?? 2 -5<0. Product: positive × negative × negative = positive . Thus, ?? ' (?? )>0. For v3<?? <v5 : ?? is positive. ?? 2 is between 3 and 5 so ?? 2 -3>0 while ?? 2 -5<0. Product: positive × positive × negative = negative . Thus, ?? ' (?? )<0. For ?? >v5 : ?? is positive. ?? 2 >5 so both ?? 2 -3>0 and ?? 2 -5>0. Product: positive × positive × positive = positive . Thus, ?? ' (?? )>0. Step 4. Classify the Critical Points: At ?? =-v5:?? ' (?? ) changes from negative to positive ? local minimum. At ?? =-v3:?? ' (?? ) changes from positive to negative ? local maximum. At ?? =0 : ?? ' (?? ) changes from negative to positive ? local minimum. At ?? =v3:?? ' (?? ) changes from positive to negative ? local maximum. At ?? =v5:?? ' (?? ) changes from negative to positive ? local minimum. Step 5. Conclusion: There are local maximum points at ?? =-v3 and ?? =v3 (2 points in total), and local minimum points at ?? =-v5,?? =0, and ?? =v5 ( 3 points in total). Thus, the numbers of local maximum and local minimum points of ?? are 2 and 3 , respectively. The correct option is Option C. Q4: A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm , the ice-cream melts at the rate of ???? ???? ?? /?????? and the thickness of the ice-cream layer decreases at the rate of Page 5 JEE Main Previous Year Questions (2025): Application of Derivatives Q1: If the set of all values of ?? , for which the equation ?? ?? ?? -???? ?? -?? =?? has three distinct real roots, is the interval (?? ,?? ), then ?? -?? ?? is equal to ____ . JEE Main 2025 (Online) 23rd January Morning Shift Ans: 30 Solution: 5?? 3 -15?? -?? =0 ?? (?? )=5?? 3 -15?? ?? ' (?? )=15?? 2 -15=15(?? -1)(?? +1) a?(-10,10) ?? =-10,?? =10 ?? -2?? =10+20=30 Q2: Let ?? (?? ,-?? ),?? (?? ,?? ) and ?? (?? ,-?? ) be the vertices of a triangle ABC . Then the maximum area of the parallelogram ???????? , formed with vertices ?? ,?? and ?? on the sides ???? ,???? and ???? of the triangle ?????? respectively, is ____ JEE Main 2025 (Online) 2nd April Evening Shift Ans: 3 Solution: The maximum area of such a parallelogram ???????? , with one vertex fixed at ?? and the other three points lying on the sides of triangle ?????? , is half the area of triangle ?????? . Using the determinant formula for area of triangle with vertices ?? (?? 1 ,?? 1 ),?? (?? 2 ,?? 2 ),?? (?? 3 ,?? 3 ) : Area ??????? = 1 2 ?? 1 (?? 2 -?? 3 2 )+?? 2 (?? 3 -?? 1 )+?? 3 (?? 1 -?? 2 )| Substitute the coordinates: = 1 2 |4(1-(-3))+1((-3)-(-2))+9((-2)-1)| = 1 2 |4(4)+1(-1)+9(-3)| = 1 2 |16-1-27|= 1 2 |-12|= 12 2 =6 Maximum area of parallelogram ???????? = 1 2 × area of triangle = 1 2 ×6=3 Q3: Let ?? (?? )=? ?? ?? ?? ? ?? ?? -???? +???? ?? ?? ???? ,?? ??? . Then the numbers of local maximum and local minimum points of ?? , respectively, are : JEE Main 2025 (Online) 22nd January Evening Shift Options: A. 3 and 2 B. 2 and 2 C. 2 and 3 D. 1 and 3 Ans: C Solution: We are given ?? (?? )=? 0 ?? 2 ? ?? 2 -8?? +15 ?? ?? ???? ,?? ?R. To find the local extrema, we first compute the derivative using the Fundamental Theorem of Calculus and the chain rule. Step 1. Rewrite the derivative: Let ?? (?? )=? 0 ?? ? ?? 2 -8?? +15 ?? ?? ???? , so that ?? (?? )=?? (?? 2 ). Then by the chain rule, ?? ' (?? )=?? ' (?? 2 )·2?? . Since ?? ' (?? )= ?? 2 -8?? +15 ?? ?? , we substitute ?? =?? 2 to get ?? ' (?? )= (?? 2 ) 2 -8?? 2 +15 ?? ?? 2 ·2?? = 2?? (?? 4 -8?? 2 +15) ?? ?? 2 . Step 2. Factor and Identify Critical Points: Factor the polynomial ?? 4 -8?? 2 +15 by writing it in terms of ?? 2 . Let ?? =?? 2 , then ?? 2 -8?? +15=(?? -3)(?? -5) so that ?? 4 -8?? 2 +15=(?? 2 -3)(?? 2 -5). Thus, ?? ' (?? )= 2?? (?? 2 -3)(?? 2 -5) ?? ?? 2 . Since ?? ?? 2 >0 for all ?? , the zeros of ?? ' (?? ) are determined by 2?? (?? 2 -3)(?? 2 -5)=0. That gives the critical points: 2?? =0??? =0, ?? 2 -3=0??? =±v3, ?? 2 -5=0??? =±v5. Step 3. Analyzing the Sign of ?? ' (?? ) : We need to determine the nature (maximum or minimum) by looking at the sign changes of ?? ' (?? ) on the intervals determined by the critical points ?? =-v5,?? =-v3,?? =0,?? =v3, and ?? =v5. Notice that the factor 2?? (?? 2 -3)(?? 2 -5) will dictate the sign. Let's define: h(?? )=?? (?? 2 -3)(?? 2 -5). Examine the sign of h(?? ) in each interval: For ?? <-v5 : ?? is negative. ?? 2 >5 so ?? 2 -3>0 and ?? 2 -5>0. Product: negative × positive × positive = negative . Thus, ?? ' (?? )<0. For -v5<?? <-v3 : ?? is negative. ?? 2 is between 3 and 5 so ?? 2 -3>0 while ?? 2 -5<0. Product: negative × positive × negative = positive . Thus, ?? ' (?? )>0. For -v3<?? <0 : ?? is negative. ?? 2 <3 so both ?? 2 -3<0 and ?? 2 -5<0. Product: negative × negative × negative = negative. Thus, ?? ' (?? )<0. For 0<?? <v3 : ?? is positive. ?? 2 <3 so both ?? 2 -3<0 and ?? 2 -5<0. Product: positive × negative × negative = positive . Thus, ?? ' (?? )>0. For v3<?? <v5 : ?? is positive. ?? 2 is between 3 and 5 so ?? 2 -3>0 while ?? 2 -5<0. Product: positive × positive × negative = negative . Thus, ?? ' (?? )<0. For ?? >v5 : ?? is positive. ?? 2 >5 so both ?? 2 -3>0 and ?? 2 -5>0. Product: positive × positive × positive = positive . Thus, ?? ' (?? )>0. Step 4. Classify the Critical Points: At ?? =-v5:?? ' (?? ) changes from negative to positive ? local minimum. At ?? =-v3:?? ' (?? ) changes from positive to negative ? local maximum. At ?? =0 : ?? ' (?? ) changes from negative to positive ? local minimum. At ?? =v3:?? ' (?? ) changes from positive to negative ? local maximum. At ?? =v5:?? ' (?? ) changes from negative to positive ? local minimum. Step 5. Conclusion: There are local maximum points at ?? =-v3 and ?? =v3 (2 points in total), and local minimum points at ?? =-v5,?? =0, and ?? =v5 ( 3 points in total). Thus, the numbers of local maximum and local minimum points of ?? are 2 and 3 , respectively. The correct option is Option C. Q4: A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm , the ice-cream melts at the rate of ???? ???? ?? /?????? and the thickness of the ice-cream layer decreases at the rate of ?? ?? ?? ???? /?????? . The surface area (in ???? ?? ) of the chocolate ball (without the ice-cream layer) is : JEE Main 2025 (Online) 23rd January Evening Shift Options: A. 128?? B. 196?? C. 225?? D. 256?? Ans: D Solution: v= 4 3 ?? r 3 dv dt =4?? r 2 dr dt 81=4?? r 2 × 1 4?? r 2 =81 r=9 surface area of chocolate =4?? (?? -1) 2 =256??Read More
FAQs on JEE Main Previous Year Questions (2025): Application of Derivatives
| 1. What is the application of derivatives in finding the maximum and minimum values of a function? |  |
Ans.Derivatives are used to determine the critical points of a function, where the slope is zero or undefined. By calculating the first derivative of a function and setting it to zero, one can find the points where the function may achieve local maxima or minima. To confirm whether these points are indeed maxima or minima, the second derivative test can be applied. If the second derivative is positive at a critical point, the function has a local minimum; if negative, it has a local maximum.
| 2. How can derivatives help in determining the rate of change of a function? | |
Ans.Derivatives measure the rate of change of a function with respect to its independent variable. For a function f(x), the derivative f'(x) represents how f changes as x changes. For example, if f(x) represents distance traveled over time, then f'(x) represents speed, indicating how quickly distance changes with time. This concept is crucial in physics and engineering for analyzing motion, growth rates, and other dynamic systems.
| 3. What role do derivatives play in curve sketching? | |
Ans.Derivatives are essential for curve sketching as they provide information about the function's behavior. By analyzing the first derivative, one can identify intervals of increase and decrease, while the second derivative indicates concavity. Points of inflection, where the concavity changes, can also be found using the second derivative. By combining this information, one can create a detailed sketch of the function, highlighting key features such as peaks, valleys, and points of inflection.
| 4. How can the concept of derivatives be applied to real-world problems? | |
Ans.Derivatives have numerous real-world applications, including optimization problems in business, engineering, and economics. For instance, companies use derivatives to maximize profit or minimize cost by determining the optimal level of production. In physics, derivatives help in calculating velocity and acceleration. In biology, they can model population growth rates. These applications demonstrate the practicality of derivatives in analyzing and solving various problems across different fields.
| 5. What is the significance of the Mean Value Theorem in calculus? | |
Ans.The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) where the derivative of the function equals the average rate of change over that interval. This theorem is significant as it provides a formal framework for understanding how derivatives relate to function behavior over intervals, and it serves as a foundational concept for many results in calculus, including the Fundamental Theorem of Calculus.
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