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Question 23 (Choice 1) - CBSE Class 12 Sample Paper for 2024 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by

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

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Transcript

f(𝑥)=1/(4𝑥^2 + 2𝑥 + 1) Finding f’(𝒙) f’(𝑥)= ((1)^′ " " (4𝑥^2 + 2𝑥 + 1)" − " (〖4𝑥^2 + 2𝑥 + 1)〗^′ (1))/((〖4𝑥^2 + 2𝑥 + 1)〗^2 ) f’(𝑥)= (0 (4𝑥^2 + 2𝑥 + 1)" − " (8𝑥 + 2)(1))/((〖4𝑥^2 + 2𝑥 + 1)〗^2 ) f’(𝑥)= ("−" (8𝑥 + 2) )/((〖4𝑥^2 + 2𝑥 + 1)〗^2 ) Putting f’(𝒙)=𝟎 ("−" (8𝑥 + 2) )/((〖4𝑥^2 + 2𝑥 + 1)〗^2 ) = 0 -(8x + 2) = 0 8x + 2 = 0 8x = -2 −(8x + 2) = 0 8x + 2 = 0 8x = −2 x = (−2)/8 x = (−𝟏)/𝟒 Finding f’’(𝒙) f’(𝑥)=("−" (8𝑥 + 2) )/((〖4𝑥^2 + 2𝑥 + 1)〗^2 ) " " Differentiating again w.r.t x f’’(x) =−((8𝑥 + 2)^′ (〖4𝑥^2 + 2𝑥 + 1)〗^2−((〖4𝑥^2+2𝑥+1)〗^2 )^′ (8𝑥 + 2))/(((〖4𝑥^2 + 2𝑥 + 1)〗^2 )^2 ) f’’(x) =−(8(〖4𝑥^2 + 2𝑥 + 1)〗^2 − 2(4𝑥^2 + 2𝑥 + 1)(8𝑥 + 2)(8𝑥 + 2))/(4𝑥^2 + 2𝑥 + 1)^4 f’’(x) =−(8(〖4𝑥^2 + 2𝑥 + 1)〗^2 − 2(4𝑥^2 + 2𝑥 + 1)(8𝑥 + 2)(8𝑥 + 2))/(4𝑥^2 + 2𝑥 + 1)^4 f’’(x) =−(8(〖4𝑥^2 + 2𝑥 + 1)〗^2 − 2(4𝑥^2 + 2𝑥 + 1) (8𝑥 + 2)^2)/(4𝑥^2 + 2𝑥 + 1)^4 f’’ (−𝟏/𝟒) = −(8(〖4(−1/4)^2+ 2(−1/4) + 1)〗^2 − 2(4(−1/4)^2+ 2(−1/4)+ 1) (8(−1/4)+ 2)^2)/(4(−1/4)^2+ 2(−1/4)+ 1)^4 f’’ (−𝟏/𝟒) = −(8(〖4(−1/4)^2+ 2(−1/4) + 1)〗^2 − 2(4(−1/4)^2+ 2(−1/4)+ 1) (−2 + 2)^2)/(4(−1/4)^2+ 2(−1/4)+ 1)^4 f’’ (−𝟏/𝟒) = −(8(3/4)^2−0)/(3/4)^4 = −8/(3/4)^2 f’’ (−𝟏/𝟒) < 0 Since f’’ (−𝟏/𝟒) < 0 , 𝑥 = −𝟏/𝟒 is point of local maxima Putting 𝑥 = −𝟏/𝟒 , we can calculate maximum value f(𝑥) =1/(4𝑥^2+2𝑥+1) f(−𝟏/𝟒)=1/(4(−1/4)^2+ 2(−1/4)+ 1) =1/(4(1/16)+ 2(−1/4)+ 1) =1/(1/4 − 2/4+ 1) = 4/(1 −2+ 4) = 𝟒/𝟑

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.