CBSE Class 12 Sample Paper for 2024 Boards

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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 Feb. 10, 2025 by

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

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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 15 years. He provides courses for Maths, Science and Computer Science at Teachoo