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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


Transcript

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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.