If f (x) = 2x and g (x) = x^2/2+1 , then which of the following can be a discontinuous function

(A) f (x) + g (x) 

(B) f (x) – g (x)

(C) f (x) . g (x) 

(D) g(x)/(f(x))

This question is similar to Ex 5.1, 21 - Chapter 5 Class 12 - Continuity and Differentiability

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  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

Question 13 If f (x) = 2x and g (x) = ๐‘ฅ^2/2+1 , then which of the following can be a discontinuous function (A) ๐‘“ (๐‘ฅ) + ๐‘” (๐‘ฅ) (B) f (x) โ€“ g (x) (C) ๐‘“ (๐‘ฅ) . ๐‘” (๐‘ฅ) (D) ๐‘”(๐‘ฅ)/(๐‘“(๐‘ฅ)) Given functions ๐‘“(๐‘ฅ)=2๐‘ฅ & ๐‘”(๐‘ฅ)=๐‘ฅ^2/2+1 Checking each option one by one Option 1 ๐’‡ (๐’™)+ ๐’ˆ (๐’™)=2๐‘ฅ+๐‘ฅ^2/2+1 =๐Ÿ/๐Ÿ ๐’™^๐Ÿ+๐Ÿ๐’™+๐Ÿ Continuous since it is a polynomial function Option 2 ๐’‡ (๐’™)โˆ’ ๐’ˆ (๐’™)=2๐‘ฅโˆ’(๐‘ฅ^2/2+1) =2๐‘ฅโˆ’๐‘ฅ^2/2โˆ’1 =โˆ’๐Ÿ/๐Ÿ ๐’™^๐Ÿ+๐Ÿ๐’™โˆ’๐Ÿ Continuous since it is a polynomial function Option 3 ๐’‡ (๐’™)โˆ™๐’ˆ (๐’™)=2๐‘ฅโˆ™(๐‘ฅ^2/2+1) =(2๐‘ฅ^3)/2+2๐‘ฅ =๐’™^๐Ÿ‘+๐Ÿ๐’™ Continuous since it is a polynomial function Option 4 (๐’ˆ(๐’™))/(๐’‡(๐’™))=((๐‘ฅ^2/2+ 1))/2๐‘ฅ =(((๐‘ฅ^2 + 2)/2))/2๐‘ฅ =(๐’™^๐Ÿ+๐Ÿ)/๐Ÿ’๐’™ Not defined when 4๐‘ฅ=0 i.e., ๐’™=๐ŸŽ โˆด (๐‘”(๐‘ฅ))/(๐‘“(๐‘ฅ)) is a discontinuous function when ๐‘ฅ=0 So, the correct answer is (D)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.