CBSE Class 12 Sample Paper for 2024 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## If f(x)=1/(4x^2  + 2x + 1);x∈R, then find the maximum value of f(x).

### Transcript

f(đĽ)=1/(4đĽ^2 + 2đĽ + 1) Finding fâ(đ) fâ(đĽ)= ((1)^â˛ " " (4đĽ^2 + 2đĽ + 1)" â " (ă4đĽ^2 + 2đĽ + 1)ă^â˛ (1))/((ă4đĽ^2 + 2đĽ + 1)ă^2 ) fâ(đĽ)= (0 (4đĽ^2 + 2đĽ + 1)" â " (8đĽ + 2)(1))/((ă4đĽ^2 + 2đĽ + 1)ă^2 ) fâ(đĽ)= ("â" (8đĽ + 2) )/((ă4đĽ^2 + 2đĽ + 1)ă^2 ) Putting fâ(đ)=đ ("â" (8đĽ + 2) )/((ă4đĽ^2 + 2đĽ + 1)ă^2 ) = 0 -(8x + 2) = 0 8x + 2 = 0 8x = -2 â(8x + 2) = 0 8x + 2 = 0 8x = â2 x = (â2)/8 x = (âđ)/đ Finding fââ(đ) fâ(đĽ)=("â" (8đĽ + 2) )/((ă4đĽ^2 + 2đĽ + 1)ă^2 ) " " Differentiating again w.r.t x fââ(x) =â((8đĽ + 2)^â˛ (ă4đĽ^2 + 2đĽ + 1)ă^2â((ă4đĽ^2+2đĽ+1)ă^2 )^â˛ (8đĽ + 2))/(((ă4đĽ^2 + 2đĽ + 1)ă^2 )^2 ) fââ(x) =â(8(ă4đĽ^2 + 2đĽ + 1)ă^2 â 2(4đĽ^2 + 2đĽ + 1)(8đĽ + 2)(8đĽ + 2))/(4đĽ^2 + 2đĽ + 1)^4 fââ(x) =â(8(ă4đĽ^2 + 2đĽ + 1)ă^2 â 2(4đĽ^2 + 2đĽ + 1)(8đĽ + 2)(8đĽ + 2))/(4đĽ^2 + 2đĽ + 1)^4 fââ(x) =â(8(ă4đĽ^2 + 2đĽ + 1)ă^2 â 2(4đĽ^2 + 2đĽ + 1) (8đĽ + 2)^2)/(4đĽ^2 + 2đĽ + 1)^4 fââ (âđ/đ) = â(8(ă4(â1/4)^2+ 2(â1/4) + 1)ă^2 â 2(4(â1/4)^2+ 2(â1/4)+ 1) (8(â1/4)+ 2)^2)/(4(â1/4)^2+ 2(â1/4)+ 1)^4 fââ (âđ/đ) = â(8(ă4(â1/4)^2+ 2(â1/4) + 1)ă^2 â 2(4(â1/4)^2+ 2(â1/4)+ 1) (â2 + 2)^2)/(4(â1/4)^2+ 2(â1/4)+ 1)^4 fââ (âđ/đ) = â(8(3/4)^2â0)/(3/4)^4 = â8/(3/4)^2 fââ (âđ/đ) < 0 Since fââ (âđ/đ) < 0 , đĽ = âđ/đ is point of local maxima Putting đĽ = âđ/đ , we can calculate maximum value f(đĽ) =1/(4đĽ^2+2đĽ+1) f(âđ/đ)=1/(4(â1/4)^2+ 2(â1/4)+ 1) =1/(4(1/16)+ 2(â1/4)+ 1) =1/(1/4 â 2/4+ 1) = 4/(1 â2+ 4) = đ/đ

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.