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EXERCISE 8.1 · Q23

Q.Discuss the continuity of the following function at the point indicated against it: f(x)=4x−2x+1+11−cos⁡2xf(x) = \dfrac{4^x - 2^{x+1}+1}{1-\cos 2x}, for x≠0x \ne 0, =(log⁡2)22= \dfrac{(\log 2)^2}{2}, for x=0x = 0, at x=0x = 0.

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f(x)=4x−2x+1+11−cos⁡2xf(x)=\dfrac{4^x-2^{x+1}+1}{1-\cos2x} for x≠0x\ne0, f(0)=(log⁡2)22f(0)=\dfrac{(\log2)^2}{2}.

Let u=2xu=2^x. Then 4x=u24^x=u^2 and 2x+1=2u2^{x+1}=2u, so the numerator is u2−2u+1=(u−1)2=(2x−1)2u^2-2u+1=(u-1)^2=(2^x-1)^2. The denominator is 1−cos⁡2x=2sin⁡2x1-\cos2x=2\sin^2x.

f(x)=(2x−1)22sin⁡2x=(2x−1x)2⋅x22sin⁡2x.f(x)=\frac{(2^x-1)^2}{2\sin^2x}=\left(\frac{2^x-1}{x}\right)^2\cdot\frac{x^2}{2\sin^2x}.

As x→0x\to0: 2x−1x→log⁡2\dfrac{2^x-1}{x}\to\log2, and x2sin⁡2x→1\dfrac{x^2}{\sin^2x}\to1, so …

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