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Ex 6.3
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
Ex 6.3, 3 (ii)
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
Ex 6.3,11 Important
Ex 6.3,12 Important
Ex 6.3,13
Ex 6.3,14 Important
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19 Important
Ex 6.3, 20 Important
Ex 6.3,21
Ex 6.3,22 Important
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Last updated at June 12, 2023 by Teachoo
Ex 6.3, 1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (i) f (𝑥) = (2𝑥 – 1)^2 + 3 Square of number cant be negative It can be 0 or greater than 0 𝑓(𝑥)=(2𝑥−1)^2+3 Hence, Minimum value of (2𝑥−1)^2 = 0 Minimum value of (2𝑥−1^2 )+3 = 0 + 3 = 3 Also, there is no maximum value of 𝑥 ∴ There is no maximum value of f(x) Ex 6.3, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (i) f (𝑥) = (2𝑥 – 1)^2+3Finding f’(x) f(𝑥)=(2𝑥−1)^2+3 f’(𝑥)= 2(2𝑥−1) Putting f’(𝒙)=𝟎 2(2𝑥−1)=0 2𝑥 – 1 = 0 2𝑥 = 1 𝒙 = 𝟏/𝟐 Thus, x = 1/2 is the minima Finding minimum value f(𝑥)=(2𝑥−1)^2+3 Putting 𝑥 = 1/2 f(1/2)=(2 × 1/2−1)^2+3= (1−1)^2+3= 3 ∴ Minimum value = 3 There is no maximum value Ex 6.3, 1 (Method 3) Find the maximum and minimum values, if any, of the following functions given by (i) 𝑓 (𝑥)= (2𝑥 – 1)^2 + 3Double Derivative Test f(𝑥)=(2𝑥−1)^2+3 Finding f’(𝒙) f’(𝑥)=2(2𝑥−1)^(2−1) = 2(2𝑥−1) Putting f’(𝒙)=𝟎 2(2𝑥−1)=0 (2𝑥−1)=0 2𝑥 = 0 + 1 𝒙 = 𝟏/𝟐 Finding f’’(𝒙) f’(𝑥)=2(2𝑥−1) f’(𝑥) = 4𝑥 – 2 f’’(𝑥)= 4 f’’ (𝟏/𝟐) = 4 Since f’’ (𝟏/𝟐) > 0 , 𝑥 = 1/2 is point of local minima Putting 𝑥 = 1/2 , we can calculate minimum value f(𝑥) = (2𝑥−1)^2+3 f(1/2)= (2(1/2)−1)^2+3= (1−1)^2+3= 3 Hence, Minimum value = 3 There is no Maximum value