Question 17
If f (x) = {β(ππ₯+1, ππ π₯β€π/2@sinβ‘γπ₯+π, ππ π₯> π/2γ )β€ , is continuous at x = π/2, then
(A) m = 1, n = 0 (B) m = ππ/2 + 1
(C) n = ππ/2 (D) none of these
Given that function is continuous at π₯=π/2
Now,
π is continuous at π₯=π/2
If L.H.L = R.H.L = π(π/2)
i.e. limβ¬(xβγπ/2γ^β ) π(π₯)=limβ¬(xβγπ/2γ^+ ) " " π(π₯)= π(π/2)
LHL at x β π /π
(πππ)β¬(π₯βγπ/2γ^β ) f(x) = (πππ)β¬(ββ0) f ( π/2 β h)
= limβ¬(hβ0) m (Ο/2 β h) + 1
= m (π/2 β 0) + 1
= mπ /π+ 1
RHL at x β π /π
(πππ)β¬(π₯βγπ/2γ^+ ) f(x) = (πππ)β¬(ββ0) f ( π/2 + h)
= sin = limβ¬(hβ0) sin (Ο/2 + h) + n
(Ο/2 + 0) + n
= 1 + n
Since, f is continuous at π₯ =π/2
β΄ L.H.L = R.H.L
"m" π/2 "+ 1"="1 + n"
"m" π/2="n"
π§=ππ /π
So, the correct answer is (C)

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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