Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12


Last updated at Jan. 7, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Transcript
Ex 6.5, 28 For all real values of x, the minimum value of (1 โ ๐ฅ + ๐ฅ2)/(1 + ๐ฅ + ๐ฅ2) is (A) 0 (B) 1 (C) 3 (D) 1/3Ex 6.5, 28 For all real values of x, the minimum value of (1 โ ๐ฅ + ๐ฅ2)/(1 + ๐ฅ + ๐ฅ2) is (A) 0 (B) 1 (C) 3 (D) 1/3 Let ๐(๐ฅ)=(1 โ ๐ฅ + ๐ฅ2)/(1 + ๐ฅ + ๐ฅ2) Finding ๐โฒ(๐) ๐(๐ฅ)=(1 โ ๐ฅ + ๐ฅ2)/(1 + ๐ฅ + ๐ฅ2) ๐^โฒ(๐ฅ) =((1 โ ๐ฅ + ๐ฅ^2 )^โฒ (1 + ๐ฅ + ๐ฅ^2 ) โ (1 โ ๐ฅ + ๐ฅ^2 ) (1 + ๐ฅ + ๐ฅ^2 )^โฒ)/(1 + ๐ฅ + ๐ฅ^2 )^2 Using quotient Rule As (๐ข/๐ฃ)^โฒ=(๐ข^โฒ ๐ฃโ๐ฃ^โฒ ๐ข)/๐ฃ^2 ๐โฒ(๐ฅ)=(โ1 โ ๐ฅ โ ๐ฅ^2 + 2๐ฅ + 2๐ฅ^2+ 2๐ฅ^3 โ (1 โ ๐ฅ + ๐ฅ^2+ 2๐ฅ โ 2๐ฅ^2 + 2๐ฅ^3 ))/(1 + ๐ฅ + ๐ฅ^2 )^2 ๐(๐ฅ)=(โ1 + ๐ฅ + ๐ฅ^2 + 2๐ฅ^3 โ (1 + ๐ฅ โ ๐ฅ^2 + 2๐ฅ^3 ))/(1 + ๐ฅ + ๐ฅ^2 )^2 ๐โฒ(๐ฅ)=(โ2 + 2๐ฅ^2)/(1 + ๐ฅ + ๐ฅ^2 )^2 Putting ๐^โฒ (๐)=๐ (โ2 + 2๐ฅ^2)/(1 + ๐ฅ + ๐ฅ^2 )^2 =0 2๐ฅ^2โ2=0 2๐ฅ^2=2 ๐ฅ^2=1 ๐ฅ=ยฑ1 Hence, x = 1 or x = โ1 are the critical points Finding value of ๐(๐) at critical points Hence, minimum value of f(x) is 1/3. So, (D) is the correct answer
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