# Ex 6.5,28

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.5,28 For all real values of x, the minimum value of 1 − 𝑥 + 𝑥21 + 𝑥 + 𝑥2 is (A) 0 (B) 1 (C) 3 (D) 13 Let 𝑓 𝑥= 1 − 𝑥 + 𝑥21 + 𝑥 + 𝑥2 Step 1 :- Finding 𝑓′ 𝑥 𝑓 𝑥= 1 − 𝑥 + 𝑥21 + 𝑥 + 𝑥2 𝑓′ 𝑥= 1 − 𝑥 + 𝑥2′ 1 + 𝑥 + 𝑥2 − 1 − 𝑥 + 𝑥2 1 + 𝑥 + 𝑥2′ 1 + 𝑥 + 𝑥22 𝑓′ 𝑥= −1 + 2𝑥 1 + 𝑥 + 𝑥2 − 1 − 𝑥 + 𝑥2 1 + 2𝑥 1 + 𝑥 + 𝑥22 𝑓′ 𝑥= −1 − 𝑥 − 𝑥2 + 2𝑥 + 2 𝑥2+ 2 𝑥3 − 1 − 𝑥 + 𝑥2+ 2𝑥 − 2 𝑥2 + 2 𝑥3 1 + 𝑥 + 𝑥22 𝑓 𝑥= −1 + 𝑥 + 𝑥2 + 2 𝑥3 − 1 + 𝑥 − 𝑥2 + 2 𝑥3 1 + 𝑥 + 𝑥22 𝑓′ 𝑥= −2 + 2 𝑥2 1 + 𝑥 + 𝑥22 Step 2 :- Putting 𝑓′ 𝑥=0 −2 + 2 𝑥2 1 + 𝑥 + 𝑥22=0 2 𝑥2−2=0 2 𝑥2=2 𝑥2=1 𝑥=±1 Hence, x = 1 or x = –1 are the critical points Finding value of 𝑓 𝑥 at critical points Hence, minimum value of f(x) is 13. So, (D) is the correct answer

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Class 12

Important Question 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
- 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

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.