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Question

If fx=1-sin xπ-2x2, when x ≠ π/2 and f (π/2) = λ, then f (x) will be continuous function at x = π/2, where λ =
(a) 1/8
(b) 1/4
(c) 1/2
(d) none of these

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Solution

(a) 18

If fx is continuous at x=π2, then
limxπ2fx=fπ2

limxπ21-sin xπ-2x2=fπ2 ...(1)

Suppose π2-x=t, then

limt01-sin π2-t2t2=fπ2 From eq. (1)limt01-cos t4t2=fπ214limt02 sin2 t2t2=fπ214limt024 sin2 t2t24=fπ218limt0sin2 t2t24=fπ218limt0sin t2t22=fπ2fπ2=λ=18

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