Ex 6.3,7 - Chapter 6 Class 12 Application of Derivatives (Important Question)
Last updated at July 20, 2026 by Teachoo
Chapter 6 Class 12 Application of Derivatives
Question 5
Important
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Class 12
Important Questions for exams Class 12
- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
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Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability
Transcript
Ex 6.3, 7 Find both the maximum value and the minimum value of 3š„4 ā 8š„3 + 12š„2 ā 48š„ + 25 on the interval [0, 3].Let f(x) = 3š„4 ā 8š„3 + 12š„2 ā 48š„ + 25, where š„ ā [0, 3] Finding fā(š) fā(š„)=š(3š„^4 ā 8š„^3 + 12š„^2 ā 48š„ + 25)/šš„ fā(š„)=3 Ć4š„^3ā8 Ć3š„^2+12 Ć2š„ā48+0 fā(š„)=12š„^3ā24š„^2+24š„ā48 fā(š„)=12(š„^3ā2š„^2+2š„ā4) Putting fā(š)=š 12(š„^3ā2š„^2+2š„ā4)=0 š„^3ā2š„^2+2š„ā4=0 š„^2 (š„ā2)+2(š„ā2)=0 (š„^2+2)(š„ā2)=0 Since š„^2=ā2 is not possible Thus š„=2 is only critical point Since are given interval š„ ā [0 , 3] Hence , calculating f(š„) at š„ = 0 , 2 & 3 Hence, Minimum value of f(š„) is ā39 at š = 2 Maximum value of f(š„) is 25 at š = 0
