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⎪⎩1−sin2x3cos2x,x<π2a,x=π2b(1−sinx)(π−2x)2,x>π2 then f(x) is continuous at x=π2, if
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Solution
f(x) is continuous at x=π2
So, f(π2−)=f(π2)=f(π2+) ⇒limx→π2−1−sin2x3cos2x=a=limx→π2+b(1−sinx)(π−2x)2 ⇒13=a=limx→π2+b(−cosx)2(π−2x)(−2) a=13=limx→π2+b(sinx)−4(−2)=b8
So, a=13 and b=83