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Chapter 5 Class 12 Continuity and Differentiability
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## (D) none of these

This question is similar to Ex 5.1, 28 - Chapter 5 Class 12 - Continuity and Differentiability

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### Transcript

Question 19 If f (x) = {β(ππ₯+1, ππ π₯β€π/[email protected]β‘γπ₯+π, ππ π₯> π/2γ )β€ , is continuous at x = π/2, then (A) m = 1, n = 0 (B) m = ππ/2 + 1 (C) n = ππ/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¬(xβγπ/2γ^β ) π(π₯)=limβ¬(xβγπ/2γ^+ ) " " π(π₯)= π(π/2) LHL at x β π/π (πππ)β¬(π₯βγπ/2γ^β ) f(x) = (πππ)β¬(ββ0) f ( π/2 β h) = limβ¬(hβ0) m (Ο/2 β h) + 1 = m (π/2 β 0) + 1 = mπ/π+ 1 RHL at x β π/π (πππ)β¬(π₯βγπ/2γ^+ ) f(x) = (πππ)β¬(ββ0) f ( π/2 + h) = sin = limβ¬(hβ0) 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§=ππ/π So, the correct answer is (C)