Question 7
The value of k which makes the function defined by f (x) = {β 8(π ππ 1/π₯," if " π₯β "0 " @π ", if x " ="0" )β€ , continuous at x = 0 is
8 (B) 1
(C) β1 (D) None of these
At π = 0
f(x) is continuous at π₯ =0
if L.H.L = R.H.L = π(0)
if if limβ¬(xβ0^β ) π(π₯) = limβ¬(xβ0^+ ) π(π₯) = π(0)
LHL at x β 0
limβ¬(xβ0^β ) f(x) = limβ¬(hβ0) f(0 β h)
= limβ¬(hβ0) f(βh)
= limβ¬(hβ0) sinβ‘(1/(ββ))
= (πππ)β¬(π‘βπ) γβπππγβ‘(π/π)
= (πππ)β¬(ββ0) (βm)
= β m
RHL at x β 0
limβ¬(xβ0^+ ) f(x) = limβ¬(hβ0) f(0 + h)
= limβ¬(hβ0) f(h)
= (πππ)β¬(π‘βπ) πππβ‘(π/π)
= (πππ)β¬(ββ0) (m)
= m
β΄ L.H.L and R.H.L can never be equal as one is always negative of another.
Hence, there does not exist any value of k for which f(x) is continuous at π₯=0
So, the correct answer is (D)

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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