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

Transcript

Ex 6.5,28 For all real values of x, the minimum value of 1 − 𝑥 + 𝑥2﷮1 + 𝑥 + 𝑥2﷯ is (A) 0 (B) 1 (C) 3 (D) 1﷮3﷯ Let 𝑓 𝑥﷯= 1 − 𝑥 + 𝑥2﷮1 + 𝑥 + 𝑥2﷯ Step 1 :- Finding 𝑓′ 𝑥﷯ 𝑓 𝑥﷯= 1 − 𝑥 + 𝑥2﷮1 + 𝑥 + 𝑥2﷯ 𝑓′ 𝑥﷯= 1 − 𝑥 + 𝑥﷮2﷯﷯﷮′﷯ 1 + 𝑥 + 𝑥﷮2﷯﷯ − 1 − 𝑥 + 𝑥﷮2﷯﷯ 1 + 𝑥 + 𝑥﷮2﷯﷯﷮′﷯﷮ 1 + 𝑥 + 𝑥﷮2﷯﷯﷮2﷯﷯ 𝑓′ 𝑥﷯= −1 + 2𝑥﷯ 1 + 𝑥 + 𝑥﷮2﷯﷯ − 1 − 𝑥 + 𝑥﷮2﷯﷯ 1 + 2𝑥﷯﷮ 1 + 𝑥 + 𝑥﷮2﷯﷯﷮2﷯﷯ 𝑓′ 𝑥﷯= −1 − 𝑥 − 𝑥﷮2﷯ + 2𝑥 + 2 𝑥﷮2﷯+ 2 𝑥﷮3﷯ − 1 − 𝑥 + 𝑥﷮2﷯+ 2𝑥 − 2 𝑥﷮2﷯ + 2 𝑥﷮3﷯﷯﷮ 1 + 𝑥 + 𝑥﷮2﷯﷯﷮2﷯﷯ 𝑓 𝑥﷯= −1 + 𝑥 + 𝑥﷮2﷯ + 2 𝑥﷮3﷯ − 1 + 𝑥 − 𝑥﷮2﷯ + 2 𝑥﷮3﷯﷯﷮ 1 + 𝑥 + 𝑥﷮2﷯﷯﷮2﷯﷯ 𝑓′ 𝑥﷯= −2 + 2 𝑥﷮2﷯﷮ 1 + 𝑥 + 𝑥﷮2﷯﷯﷮2﷯﷯ Step 2 :- Putting 𝑓﷮′﷯ 𝑥﷯=0 −2 + 2 𝑥﷮2﷯﷮ 1 + 𝑥 + 𝑥﷮2﷯﷯﷮2﷯﷯=0 2 𝑥﷮2﷯−2=0 2 𝑥﷮2﷯=2 𝑥﷮2﷯=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 1﷮3﷯. So, (D) is the correct answer