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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.