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)
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Davneet Singh
Davneet Singh is an IIT Kanpur graduate and has been teaching for 16+ years. At Teachoo, he breaks down Maths, Science and Computer Science into simple steps so students understand concepts deeply and score with confidence.
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