NCERT Exemplar - MCQs

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

## (D) none of these.

This question is similar to Ex 5.1, 32 - Chapter 5 Class 12 and Ex 5.2, 9 - Chapter 5 Class 12 - Continuity and Differentiability

Get live Maths 1-on-1 Classs - Class 6 to 12

### Transcript

Question 17 The function f (x) = π^(|π₯|) is (A) continuous everywhere but not differentiable at x = 0 (B) continuous and differentiable everywhere (C) not continuous at x = 0 (D) none of these. f(π₯) = π^(|π₯|) We need to check continuity and differentiability of f(π₯) Continuity of f(π) Let π(π)=π^π & π(π)=|π| Then, πππ(π)=π(β(π₯)) =π(|π₯|) =π^|π₯| =π(π) β΄ π(π₯)=ππβ(π₯) We know that, π(π)=|π| is continuous as it is modulus function π(π)=π^π₯ is continuous as it is an exponential function Hence, g(π₯) & h(π₯) both are continuous And If two functions g(π₯) & h(π₯) are continuous then their composition ππβ(π₯) is also continuous β΄ π(π) is continuous Differentiability of π(π) π(π₯)=π^(|π₯|) π(π₯)={β 8(π^π₯, π₯β₯[email protected]π^(βπ₯), π₯<0)β€ Now, π(π₯) is differentiable at π₯=0, if LHD = RHD (πππ )β¬(π‘βπ) (π(π) β π(π β π))/π = (πππ)β¬(hβ0) (π(0) β π(0 β β))/β = (πππ)β¬(hβ0) (π^(|0|)β π^(|0 ββ|))/β = (πππ)β¬(hβ0) (π^(|0|)β π^(| ββ|))/β = (πππ)β¬(hβ0) (π^0 β π^β)/β = (πππ)β¬(hβ0) (1 β π^β)/β = (πππ)β¬(hβ0) (β(π^β β 1))/β Using (πππ)β¬(xβ0) (π^π₯ β 1)/π₯=1 = (πππ)β¬(hβ0) β1 = β1 (πππ )β¬(π‘βπ) (π(π + π) β π(π ))/π = (πππ)β¬(hβ0) (π(0 + β) β π(0))/β = (πππ)β¬(hβ0) (π^(|0 + β|) βπ^(|0|))/β = (πππ)β¬(hβ0) (π^(|β|) βπ^0)/β = (πππ)β¬(hβ0) (π^β β 1)/β Using (πππ)β¬(xβ0) (π^π₯ β 1)/π₯=1 = (πππ)β¬(hβ0) 1 = π Since, LHD β  RHD β΄ π(π₯) is not differentiable at π₯=0 Thus, π(π₯) continuous everywhere but not differentiable at x = 0 So, the correct answer is (A)