CBSE Class 12 Sample Paper for 2024 Boards
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CBSE Class 12 Sample Paper for 2024 Boards
Last updated at August 8, 2026 by Teachoo
Transcript
f(š„)=1/(4š„^2 + 2š„ + 1) Finding fā(š) fā(š„)= ((1)^ā² " " (4š„^2 + 2š„ + 1)" ā " (ć4š„^2 + 2š„ + 1)ć^ā² (1))/((ć4š„^2 + 2š„ + 1)ć^2 ) fā(š„)= (0 (4š„^2 + 2š„ + 1)" ā " (8š„ + 2)(1))/((ć4š„^2 + 2š„ + 1)ć^2 ) fā(š„)= ("ā" (8š„ + 2) )/((ć4š„^2 + 2š„ + 1)ć^2 ) Putting fā(š)=š ("ā" (8š„ + 2) )/((ć4š„^2 + 2š„ + 1)ć^2 ) = 0 -(8x + 2) = 0 8x + 2 = 0 8x = -2 ā(8x + 2) = 0 8x + 2 = 0 8x = ā2 x = (ā2)/8 x = (āš)/š Finding fāā(š) fā(š„)=("ā" (8š„ + 2) )/((ć4š„^2 + 2š„ + 1)ć^2 ) " " Differentiating again w.r.t x fāā(x) =ā((8š„ + 2)^ā² (ć4š„^2 + 2š„ + 1)ć^2ā((ć4š„^2+2š„+1)ć^2 )^ā² (8š„ + 2))/(((ć4š„^2 + 2š„ + 1)ć^2 )^2 ) fāā(x) =ā(8(ć4š„^2 + 2š„ + 1)ć^2 ā 2(4š„^2 + 2š„ + 1)(8š„ + 2)(8š„ + 2))/(4š„^2 + 2š„ + 1)^4 fāā(x) =ā(8(ć4š„^2 + 2š„ + 1)ć^2 ā 2(4š„^2 + 2š„ + 1)(8š„ + 2)(8š„ + 2))/(4š„^2 + 2š„ + 1)^4 fāā(x) =ā(8(ć4š„^2 + 2š„ + 1)ć^2 ā 2(4š„^2 + 2š„ + 1) (8š„ + 2)^2)/(4š„^2 + 2š„ + 1)^4 fāā (āš/š) = ā(8(ć4(ā1/4)^2+ 2(ā1/4) + 1)ć^2 ā 2(4(ā1/4)^2+ 2(ā1/4)+ 1) (8(ā1/4)+ 2)^2)/(4(ā1/4)^2+ 2(ā1/4)+ 1)^4 fāā (āš/š) = ā(8(ć4(ā1/4)^2+ 2(ā1/4) + 1)ć^2 ā 2(4(ā1/4)^2+ 2(ā1/4)+ 1) (ā2 + 2)^2)/(4(ā1/4)^2+ 2(ā1/4)+ 1)^4 fāā (āš/š) = ā(8(3/4)^2ā0)/(3/4)^4 = ā8/(3/4)^2 fāā (āš/š) < 0 Since fāā (āš/š) < 0 , š„ = āš/š is point of local maxima Putting š„ = āš/š , we can calculate maximum value f(š„) =1/(4š„^2+2š„+1) f(āš/š)=1/(4(ā1/4)^2+ 2(ā1/4)+ 1) =1/(4(1/16)+ 2(ā1/4)+ 1) =1/(1/4 ā 2/4+ 1) = 4/(1 ā2+ 4) = š/š
