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Question

If fx=1-sin2 x3 cos2 x ,x<π2 a ,x=π2b 1-sin xπ-2x2,x>π2. Then, f (x) is continuous at x=π2, if
(a) a=13, b = 2

(b) a=13, b=83

(c) a=23, b=83

(d) none of these

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Solution

(b) a=13 , b=83


Given: fx=1-sin2x3cos2x, if x<π2a, if x=π2b1-sinxπ-2x2, if x>π2

We have
(LHL at x = π2) = limxπ2-fx=limh0fπ2-h

=limh01-sin2 π2-h3 cos2 π2-h=limh01-cos2 h3 sin2 h=13limh0 sin2 h sin2 h=13

(RHL at x = π2) = limxπ2+fx=limh0fπ2+h

=limh0b1-sin π2+hπ-2π2+h2=limh0b1-cos h-2h2=limh02b sin2h24h2=limh02b sin2h216h24=b8limh0sinh2h22=b8×1=b8

Also, fπ2=a

If f(x) is continuous at x = π2, then

limxπ2-fx =lim xπ2+fx=fπ2

13 =b8 = a

a=13 and b=83


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