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f(x)={1−sin3(x)3cos2(x)for x<π2afor x=π2b(1−sin(x))(π−2x)2for x>π2\large{ f(x) = \begin{cases} \dfrac{1-\sin^3(x)}{3\cos^2(x)} & \text{for } x<\dfrac\pi2 \\ a & \text{for } x=\dfrac\pi2 \\ \dfrac{b(1-\sin(x))}{(\pi-2x)^2} & \text{for } x>\dfrac\pi2\\ \end{cases}} f(x)=⎩⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎧3cos2(x)1−sin3(x)a(π−2x)2b(1−sin(x))for x<2πfor x=2πfor x>2π
Find the sum of values of aaa and bbb such that the function f(x)f(x)f(x) above is continuous at x=π2x=\dfrac { \pi }{ 2 } x=2π.
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