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Ex 6.5, 28 - Minimum value of 1 - x + x2 / 1 + x + x2  - AOD

Ex 6.5,28 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,28 - Chapter 6 Class 12 Application of Derivatives - Part 3

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Ex 6.5, 28 For all real values of x, the minimum value of (1 βˆ’ π‘₯ + π‘₯2)/(1 + π‘₯ + π‘₯2) is (A) 0 (B) 1 (C) 3 (D) 1/3Let 𝑓(π‘₯)=(1 βˆ’ π‘₯ + π‘₯2)/(1 + π‘₯ + π‘₯2) Finding 𝒇′(𝒙) 𝑓(π‘₯)=(1 βˆ’ π‘₯ + π‘₯2)/(1 + π‘₯ + π‘₯2) 𝑓^β€²(π‘₯) =((1 βˆ’ π‘₯ + π‘₯^2 )^β€² (1 + π‘₯ + π‘₯^2 ) βˆ’ (1 βˆ’ π‘₯ + π‘₯^2 ) (1 + π‘₯ + π‘₯^2 )^β€²)/(1 + π‘₯ + π‘₯^2 )^2 Using quotient Rule As (𝑒/𝑣)^β€²=(𝑒^β€² π‘£βˆ’π‘£^β€² 𝑒)/𝑣^2 𝑓′(π‘₯)=(βˆ’1 βˆ’ π‘₯ βˆ’ π‘₯^2 + 2π‘₯ + 2π‘₯^2+ 2π‘₯^3 βˆ’ (1 βˆ’ π‘₯ + π‘₯^2+ 2π‘₯ βˆ’ 2π‘₯^2 + 2π‘₯^3 ))/(1 + π‘₯ + π‘₯^2 )^2 𝑓(π‘₯)=(βˆ’1 + π‘₯ + π‘₯^2 + 2π‘₯^3 βˆ’ (1 + π‘₯ βˆ’ π‘₯^2 + 2π‘₯^3 ))/(1 + π‘₯ + π‘₯^2 )^2 𝑓′(π‘₯)=(βˆ’2 + 2π‘₯^2)/(1 + π‘₯ + π‘₯^2 )^2 Putting 𝒇^β€² (𝒙)=𝟎 (βˆ’2 + 2π‘₯^2)/(1 + π‘₯ + π‘₯^2 )^2 =0 2π‘₯^2βˆ’2=0 2π‘₯^2=2 π‘₯^2=1 π‘₯=Β±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

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.