Solve all your doubts with Teachoo Black (new monthly pack available now!)
Ex 5.1, 32 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at April 13, 2021 by Teachoo
Ex 5.1
Ex 5.1 ,1
Ex 5.1 ,2
Ex 5.1, 3 (a)
Ex 5.1, 3 (b)
Ex 5.1, 3 (c) Important
Ex 5.1, 3 (d) Important
Ex 5.1 ,4
Ex 5.1 ,5 Important
Ex 5.1 ,6
Ex 5.1 ,7 Important
Ex 5.1 ,8
Ex 5.1, 9 Important
Ex 5.1, 10
Ex 5.1, 11
Ex 5.1, 12 Important
Ex 5.1, 13
Ex 5.1, 14
Ex 5.1, 15 Important
Ex 5.1, 16
Ex 5.1, 17 Important
Ex 5.1, 18 Important
Ex 5.1, 19 Important
Ex 5.1, 20
Ex 5.1, 21
Ex 5.1, 22 (i) Important
Ex 5.1, 22 (ii)
Ex 5.1, 22 (iii)
Ex 5.1, 22 (iv) Important
Ex 5.1, 23
Ex 5.1, 24 Important
Ex 5.1, 25
Ex 5.1, 26 Important
Ex 5.1, 27
Ex 5.1, 28 Important
Ex 5.1, 29
Ex 5.1, 30 Important
Ex 5.1, 31
Ex 5.1, 32 You are here
Ex 5.1, 33
Ex 5.1, 34 Important
Transcript
Ex 5.1, 32 Show that the function defined by 𝑓 (𝑥)= |cos𝑥 | is a continuous function.𝑓(𝑥) = |cos𝑥 | Let 𝒈(𝒙) = |𝑥| & 𝒉(𝒙) = cos𝑥 Now, 𝒈𝒐𝒉(𝒙) = g(ℎ(𝑥)) = 𝑔(cos𝑥 ) = |cos𝑥 | = 𝒇(𝒙) Hence, 𝑓(𝑥) = 𝑔𝑜ℎ(𝑥) We know that, 𝒉(𝒙) = cos𝑥 is continuous as cos is continuous & 𝒈(𝒙) = |𝑥| is continuous as it is a modulus function Hence, 𝑔(𝑥) & ℎ(𝑥) are both continuous . We know that If two function of 𝑔(𝑥) & ℎ(𝑥) both continuous, then their composition 𝒈𝒐𝒉(𝒙) is also continuous Hence, 𝒇(𝒙) is continuous . .
