# Ex 6.5,28 - Chapter 6 Class 12 Application of Derivatives

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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Chapter 6 Class 12 Application of Derivatives

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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.