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Question 19 - NCERT Exemplar - MCQs - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 29, 2023 by

If f (x) = {(mx+1, if x≀π/2 sin⁑〖x+n, if x> Ο€/2γ€—)─ , is continuous at x = Ο€/2, then

(A) m = 1, n = 0Β  Β  Β 

(B) m = nΟ€/2 + 1

(C) n = mΟ€/2Β  Β  Β 

(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 19 If f (x) = {β–ˆ(π‘šπ‘₯+1, 𝑖𝑓 π‘₯β‰€πœ‹/[email protected]⁑〖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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