Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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### Transcript

Ex 6.3, 28 For all real values of x, the minimum value of (1 β π₯ + π₯2)/(1 + π₯ + π₯2) is (A) 0 (B) 1 (C) 3 (D) 1/3Let π(π₯)=(1 β π₯ + π₯2)/(1 + π₯ + π₯2) Finding πβ²(π) π(π₯)=(1 β π₯ + π₯2)/(1 + π₯ + π₯2) π^β²(π₯) =((1 β π₯ + π₯^2 )^β² (1 + π₯ + π₯^2 ) β (1 β π₯ + π₯^2 ) (1 + π₯ + π₯^2 )^β²)/(1 + π₯ + π₯^2 )^2 Using quotient Rule As (π’/π£)^β²=(π’^β² π£βπ£^β² π’)/π£^2 πβ²(π₯)=(β1 β π₯ β π₯^2 + 2π₯ + 2π₯^2+ 2π₯^3 β (1 β π₯ + π₯^2+ 2π₯ β 2π₯^2 + 2π₯^3 ))/(1 + π₯ + π₯^2 )^2 π(π₯)=(β1 + π₯ + π₯^2 + 2π₯^3 β (1 + π₯ β π₯^2 + 2π₯^3 ))/(1 + π₯ + π₯^2 )^2 πβ²(π₯)=(β2 + 2π₯^2)/(1 + π₯ + π₯^2 )^2 Putting π^β² (π)=π (β2 + 2π₯^2)/(1 + π₯ + π₯^2 )^2 =0 2π₯^2β2=0 2π₯^2=2 π₯^2=1 π₯=Β±1 Hence, x = 1 or x = β1 are the critical points Finding value of π(π) at critical points Hence, minimum value of f(x) is 1/3. So, (D) is the correct answer