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Question

If fx=sin cos x-cos xπ-2x2,xπ2 k ,x=π2is continuous at x = π/2, then k is equal to
(a) 0
(b) 12
(c) 1
(d) −1

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Solution

(a) 0

Given: fx=sincos x-cos xπ-2x2, xπ2k, x=π2

If f(x) is continuous at x=π2, then
limxπ2fx=fπ2

limxπ2sincos x-cos xπ-2x2=k


Now,
π2-x=y
π-2x=2y

Also, xπ2, y0

limy0sincosπ2-y-cosπ2-y4y2=k

limy0sinsin y-siny4y2=k

limy02 sinsin y-y2 cossin y+y24y2=k sin C - sin D=2 sinC-D2 cosC+D212limy0sinsin y-y2 ycossin y+y2y=k12limy0sin y-y2 sinsin y-y2 ysin y-y2cossin y+y2y=k12limy0sin y-y2ysinsin y-y2sin y-y2cossin y+y2y=k12limy0sin y-y2ylimy0sinsin y-y2sin y-y2limy0cossin y+y2y=k14limy0sin yy-1limy0sinsin y-y2sin y-y2limy0cossin y+y2y=k14×0×1×limy0cossin y+y2y=k0=k

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