Ex 5.1
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Last updated at Dec. 16, 2024 by Teachoo
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Ex 5.1, 26 Find the values of k so that the function f is continuous at the indicated point π(π₯)={β((π cosβ‘π₯)/(π β 2π₯ ) , ππ π₯β π/2@& 3, ππ π₯=π/2)β€ at π₯ = π/2 Given that function is continuous at π₯ =π/2 π is continuous at =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 ) π(π₯) = (πππ)β¬(ββ0) π(π/2ββ) = limβ¬(hβ0) (π cosβ‘(π/2 β β))/(π β 2(π/2 β β) ) = limβ¬(hβ0) (π sinβ‘β)/(π β π + 2β ) = limβ¬(hβ0) (π sinβ‘β)/(2β ) = k/2 (πππ)β¬(π‘βπ) π¬π’π§β‘π/(π ) = π/2 Γ 1 = π/π RHL at x β π /π (πππ)β¬(π₯βγπ/2γ^+ ) π(π₯) = (πππ)β¬(ββ0) π(π/2+β) = limβ¬(hβ0) (π cosβ‘(π/2 + β))/(π β 2(π/2 + β) ) = limβ¬(hβ0) (π γ(βsinγβ‘β))/(π β π β 2β ) = limβ¬(hβ0) (βπ sinβ‘β)/(β2β ) = k/2 (πππ)β¬(π‘βπ) πππβ‘π/(π ) = π/2 Γ 1 = π/2 And π(π/2) = 3 Now, L.H.L = R.H.L = π(π/2) π/2 = π/2 = 3 Hence, π/2 = 3 k = 3 Γ 2 k = 6 Hence, k = 6
