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,
πππ(π)=π(β(π₯))
=π(|π₯|)
=π^|π₯|
=π(π)
β΄ π(π₯)=ππβ(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 ππβ(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^(βπ₯), π₯<0)β€
Now, π(π₯) is differentiable at π₯=0, if
LHD = RHD
(πππ )β¬(π‘βπ) (π(π) β π(π β π))/π
= (πππ)β¬(hβ0) (π(0) β π(0 β β))/β
= (πππ)β¬(hβ0) (π^(|0|)β π^(|0 ββ|))/β
= (πππ)β¬(hβ0) (π^(|0|)β π^(| ββ|))/β
= (πππ)β¬(hβ0) (π^0 β π^β)/β
= (πππ)β¬(hβ0) (1 β π^β)/β
= (πππ)β¬(hβ0) (β(π^β β 1))/β
Using (πππ)β¬(xβ0) (π^π₯ β 1)/π₯=1
= (πππ)β¬(hβ0) β1
= β1
(πππ )β¬(π‘βπ) (π(π + π) β π(π ))/π
= (πππ)β¬(hβ0) (π(0 + β) β π(0))/β
= (πππ)β¬(hβ0) (π^(|0 + β|) βπ^(|0|))/β
= (πππ)β¬(hβ0) (π^(|β|) βπ^0)/β
= (πππ)β¬(hβ0) (π^β β 1)/β
Using (πππ)β¬(xβ0) (π^π₯ β 1)/π₯=1
= (πππ)β¬(hβ0) 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)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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