Question 17 - NCERT Exemplar - MCQs - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at Nov. 18, 2021 by Teachoo

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The function f (x) = e
^{
|x|
}
is

(A) continuous everywhere but not differentiable at x = 0

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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₯0@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)

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