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EXERCISE 8.1 · Q31

Q.If f(x)=4x−π+4π−x−2(x−π)2f(x) = \dfrac{4^{x-\pi} + 4^{\pi-x} - 2}{(x-\pi)^2} for x≠πx \ne \pi, is continuous at x=πx = \pi, then find f(π)f(\pi).

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f(x)=4x−π+4π−x−2(x−π)2f(x)=\dfrac{4^{x-\pi}+4^{\pi-x}-2}{(x-\pi)^2} for x≠πx\ne\pi, continuous at π\pi.

Put t=x−πt=x-\pi, t→0t\to0. Let u=4tu=4^t, so 4−t=1/u4^{-t}=1/u; the numerator becomes u+1u−2=u2−2u+1u=(u−1)2uu+\dfrac1u-2=\dfrac{u^2-2u+1}{u}=\dfrac{(u-1)^2}{u}.

f=(u−1)2u t2=(4t−1t)2⋅1u.f=\frac{(u-1)^2}{u\,t^2}=\left(\frac{4^t-1}{t}\right)^2\cdot\frac1u. …

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