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MISCELLANEOUS EXERCISE-8 · Q51

Q.Select the correct answer from the given alternatives. If f(x)=12x−4x−3x+11−cos⁡2xf(x) = \dfrac{12^x-4^x-3^x+1}{1-\cos 2x}, for x≠0x \ne 0 is continuous at x=0x=0 then the value of f(0)f(0) is (A) log⁡122\dfrac{\log 12}{2} (B) log⁡2.log⁡3\log 2 . \log 3 (C) log⁡2.log⁡32\dfrac{\log 2 . \log 3}{2} (D) None of these

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f(x)=12x−4x−3x+11−cos⁡2xf(x)=\dfrac{12^x-4^x-3^x+1}{1-\cos2x} for x≠0x\ne0, continuous at 00.

Numerator: since 12x=4x⋅3x12^x=4^x\cdot3^x, 12x−4x−3x+1=4x(3x−1)−(3x−1)=(3x−1)(4x−1)12^x-4^x-3^x+1=4^x(3^x-1)-(3^x-1)=(3^x-1)(4^x-1).

Denominator: 1−cos⁡2x=2sin⁡2x1-\cos2x=2\sin^2x.

f(x)=(3x−1)(4x−1)2sin⁡2x=12(3x−1x)(4x−1x)(x2sin⁡2x).f(x)=\frac{(3^x-1)(4^x-1)}{2\sin^2x}=\frac{1}{2}\left(\frac{3^x-1}{x}\right)\left(\frac{4^x-1}{x}\right)\left(\frac{x^2}{\sin^2x}\right). …

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