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Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

## (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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### Transcript

Question 13 If f (x) = 2x and g (x) = π₯^2/2+1 , then which of the following can be a discontinuous function (A) π (π₯) + π (π₯) (B) f (x) β g (x) (C) π (π₯) . π (π₯) (D) π(π₯)/(π(π₯)) Given functions π(π₯)=2π₯ & π(π₯)=π₯^2/2+1 Checking each option one by one Option 1 π (π)+ π (π)=2π₯+π₯^2/2+1 =π/π π^π+ππ+π Continuous since it is a polynomial function Option 2 π (π)β π (π)=2π₯β(π₯^2/2+1) =2π₯βπ₯^2/2β1 =βπ/π π^π+ππβπ Continuous since it is a polynomial function Option 3 π (π)βπ (π)=2π₯β(π₯^2/2+1) =(2π₯^3)/2+2π₯ =π^π+ππ Continuous since it is a polynomial function Option 4 (π(π))/(π(π))=((π₯^2/2+ 1))/2π₯ =(((π₯^2 + 2)/2))/2π₯ =(π^π+π)/ππ Not defined when 4π₯=0 i.e., π=π β΄ (π(π₯))/(π(π₯)) is a discontinuous function when π₯=0 So, the correct answer is (D)