Question 15
The function f (x) = 𝑒^(|𝑥|) 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(𝑥) = 𝑒^(|𝑥|)
We need to check continuity and differentiability of f(𝑥)
Continuity of f(𝒙)
Let 𝒈(𝒙)=𝒆^𝒙 & 𝒉(𝒙)=|𝒙|
Then,
𝒈𝒐𝒉(𝒙)=𝑔(ℎ(𝑥))
=𝑔(|𝑥|)
=𝑒^|𝑥|
=𝒇(𝒙)
∴ 𝑓(𝑥)=𝑔𝑜ℎ(𝑥)
We know that,
𝒉(𝒙)=|𝒙| is continuous as it is modulus function
𝒈(𝒙)=𝑒^𝑥 is continuous as it is an exponential function
Hence, g(𝑥) & h(𝑥) both are continuous
And
If two functions g(𝑥) & h(𝑥) are continuous then their composition 𝑔𝑜ℎ(𝑥) is also continuous
∴ 𝒇(𝒙) is continuous
Differentiability of 𝒇(𝒙)
𝑓(𝑥)=𝑒^(|𝑥|)
𝑓(𝑥)={■8(𝑒^𝑥, 𝑥≥0@𝑒^(−𝑥), 𝑥<0)┤
Now, 𝑓(𝑥) is differentiable at 𝑥=0, if
LHD = RHD
(𝒍𝒊𝒎 )┬(𝐡→𝟎) (𝒇(𝒙) − 𝒇(𝒙 − 𝒉))/𝒉
= (𝑙𝑖𝑚)┬(h→0) (𝑓(0) − 𝑓(0 − ℎ))/ℎ
= (𝑙𝑖𝑚)┬(h→0) (𝑒^(|0|)− 𝑒^(|0 −ℎ|))/ℎ
= (𝑙𝑖𝑚)┬(h→0) (𝑒^(|0|)− 𝑒^(| −ℎ|))/ℎ
= (𝑙𝑖𝑚)┬(h→0) (𝑒^0 − 𝑒^ℎ)/ℎ
= (𝑙𝑖𝑚)┬(h→0) (1 − 𝑒^ℎ)/ℎ
= (𝑙𝑖𝑚)┬(h→0) (−(𝑒^ℎ − 1))/ℎ
Using (𝑙𝑖𝑚)┬(x→0) (𝑒^𝑥 − 1)/𝑥=1
= (𝑙𝑖𝑚)┬(h→0) −1
= −1
(𝒍𝒊𝒎 )┬(𝐡→𝟎) (𝒇(𝒙 + 𝒉) − 𝒇(𝒙 ))/𝒉
= (𝑙𝑖𝑚)┬(h→0) (𝑓(0 + ℎ) − 𝑓(0))/ℎ
= (𝑙𝑖𝑚)┬(h→0) (𝑒^(|0 + ℎ|) −𝑒^(|0|))/ℎ
= (𝑙𝑖𝑚)┬(h→0) (𝑒^(|ℎ|) −𝑒^0)/ℎ
= (𝑙𝑖𝑚)┬(h→0) (𝑒^ℎ − 1)/ℎ
Using (𝑙𝑖𝑚)┬(x→0) (𝑒^𝑥 − 1)/𝑥=1
= (𝑙𝑖𝑚)┬(h→0) 1
= 𝟏
Since,
LHD ≠ RHD
∴ 𝑓(𝑥) is not differentiable at 𝑥=0
Thus,
𝑓(𝑥) 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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