Ex 6.3,9 - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
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 You are here
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 April 16, 2024 by Teachoo
Ex 6.3, 9 What is the maximum value of the function sin𝑥+cos𝑥? Let f(𝑥)=sin𝑥+cos𝑥 Consider the interval 𝑥 ∈ [0 , 2𝜋] Finding f’(𝒙) f’(𝑥)=𝑑(sin𝑥 + cos𝑥 )/𝑑𝑥 f’(𝑥)=cos𝑥−sin𝑥 Putting f’(𝒙)=𝟎 cos𝑥−sin𝑥=0 cos𝑥=sin𝑥 1 =sin𝑥/cos𝑥 1= tan𝑥 tan 𝑥=1 Since 𝑥 ∈ [0 , 2𝜋] tan 𝑥=1 at 𝑥=𝜋/4 , 𝑥=5𝜋/4 in the interval [0 , 2𝜋] We have given the interval 𝑥 ∈ [0 , 2𝜋] Hence Calculating f(𝑥) at 𝑥=0 ,𝜋/4 , 5𝜋/4 & 2𝜋 Hence Maximum Value of f(𝑥) is √𝟐 at 𝒙 = 𝝅/𝟒