Question 15
The function f (x) = ๐^(|๐ฅ|) 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(๐ฅ) = ๐^(|๐ฅ|)
We need to check continuity and differentiability of f(๐ฅ)
Continuity of f(๐)
Let ๐(๐)=๐^๐ & ๐(๐)=|๐|
Then,
๐๐๐(๐)=๐(โ(๐ฅ))
=๐(|๐ฅ|)
=๐^|๐ฅ|
=๐(๐)
โด ๐(๐ฅ)=๐๐โ(๐ฅ)
We know that,
๐(๐)=|๐| is continuous as it is modulus function
๐(๐)=๐^๐ฅ is continuous as it is an exponential function
Hence, g(๐ฅ) & h(๐ฅ) both are continuous
And
If two functions g(๐ฅ) & h(๐ฅ) are continuous then their composition ๐๐โ(๐ฅ) is also continuous
โด ๐(๐) is continuous
Differentiability of ๐(๐)
๐(๐ฅ)=๐^(|๐ฅ|)
๐(๐ฅ)={โ 8(๐^๐ฅ, ๐ฅโฅ0@๐^(โ๐ฅ), ๐ฅ<0)โค
Now, ๐(๐ฅ) is differentiable at ๐ฅ=0, if
LHD = RHD
(๐๐๐ )โฌ(๐กโ๐) (๐(๐) โ ๐(๐ โ ๐))/๐
= (๐๐๐)โฌ(hโ0) (๐(0) โ ๐(0 โ โ))/โ
= (๐๐๐)โฌ(hโ0) (๐^(|0|)โ ๐^(|0 โโ|))/โ
= (๐๐๐)โฌ(hโ0) (๐^(|0|)โ ๐^(| โโ|))/โ
= (๐๐๐)โฌ(hโ0) (๐^0 โ ๐^โ)/โ
= (๐๐๐)โฌ(hโ0) (1 โ ๐^โ)/โ
= (๐๐๐)โฌ(hโ0) (โ(๐^โ โ 1))/โ
Using (๐๐๐)โฌ(xโ0) (๐^๐ฅ โ 1)/๐ฅ=1
= (๐๐๐)โฌ(hโ0) โ1
= โ1
(๐๐๐ )โฌ(๐กโ๐) (๐(๐ + ๐) โ ๐(๐ ))/๐
= (๐๐๐)โฌ(hโ0) (๐(0 + โ) โ ๐(0))/โ
= (๐๐๐)โฌ(hโ0) (๐^(|0 + โ|) โ๐^(|0|))/โ
= (๐๐๐)โฌ(hโ0) (๐^(|โ|) โ๐^0)/โ
= (๐๐๐)โฌ(hโ0) (๐^โ โ 1)/โ
Using (๐๐๐)โฌ(xโ0) (๐^๐ฅ โ 1)/๐ฅ=1
= (๐๐๐)โฌ(hโ0) 1
= ๐
Since,
LHD โ RHD
โด ๐(๐ฅ) is not differentiable at ๐ฅ=0
Thus,
๐(๐ฅ) 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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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