If f (x) = {(mx+1, if x≤π/2 sin⁡γ€–x+n, if x> π/2γ€—)┤ , is continuous at x = π/2, then

(A) m = 1, n = 0     

(B) m = nπ/2 + 1

(C) n = mπ/2     

(D) none of these

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

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Question 17 If f (x) = {β–ˆ(π‘šπ‘₯+1, 𝑖𝑓 π‘₯β‰€πœ‹/2@sin⁑〖π‘₯+𝑛, 𝑖𝑓 π‘₯> πœ‹/2γ€— )─ , is continuous at x = πœ‹/2, then (A) m = 1, n = 0 (B) m = π‘›πœ‹/2 + 1 (C) n = π‘šπœ‹/2 (D) none of these Given that function is continuous at π‘₯=πœ‹/2 Now, 𝒇 is continuous at π‘₯=πœ‹/2 If L.H.L = R.H.L = 𝑓(πœ‹/2) i.e. lim┬(xβ†’γ€–πœ‹/2γ€—^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’γ€–πœ‹/2γ€—^+ ) " " 𝑓(π‘₯)= 𝑓(πœ‹/2) LHL at x β†’ 𝝅/𝟐 (π‘™π‘–π‘š)┬(π‘₯β†’γ€–πœ‹/2γ€—^βˆ’ ) f(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) f ( πœ‹/2 βˆ’ h) = lim┬(hβ†’0) m (Ο€/2 – h) + 1 = m (πœ‹/2 βˆ’ 0) + 1 = m𝝅/𝟐+ 1 RHL at x β†’ 𝝅/𝟐 (π‘™π‘–π‘š)┬(π‘₯β†’γ€–πœ‹/2γ€—^+ ) f(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) f ( πœ‹/2 + h) = sin = lim┬(hβ†’0) sin (Ο€/2 + h) + n (Ο€/2 + 0) + n = 1 + n Since, f is continuous at π‘₯ =πœ‹/2 ∴ L.H.L = R.H.L "m" πœ‹/2 "+ 1"="1 + n" "m" πœ‹/2="n" 𝐧=π’Žπ…/𝟐 So, the correct answer is (C)

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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.