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Q.

f(x)={1−sin3x3cos2xifx<π/2Pifx=π/2q(1−sinx)(π−2x)2ifx>π/2 if continuous at x=π/2 then (p, q)

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An Intiative by Sri Chaitanya

a

(4, 12)

b

(−12, 4)

c

(4, −12)

d

(12, 4)

answer is D.

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Detailed Solution

Limx→π2−1−sin3x3cos2x=Limx→π2+q(1−sinx)4(π2−x)2=P 12=Limh→01−coshh2.q4=P 12=q8=P (12, 4)

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