Mathematics

Mathematics

Problem

Let g(x)=ln⁡x cos⁡xg(x)=\ln x\,\cos x for x>0x>0. On the interval (π/2,π)(\pi/2,\pi), which of the following best explains why g(x)g(x) is decreasing? a) Both ln⁡x\ln x and cos⁡x\cos x decrease on (π/2,π)(\pi/2,\pi), so their product g(x)g(x) must also decrease. b) The derivative g′(x)=1xcos⁡x+ln⁡x sin⁡xg'(x)=\dfrac{1}{x}\cos x+\ln x\,\sin x is negative because cos⁡x<0\cos x<0 and sin⁡x>0\sin x>0 throughout the interval. c) The derivative g′(x)=1xcos⁡x−ln⁡x sin⁡xg'(x)=\dfrac{1}{x}\cos x-\ln x\,\sin x is negative because cos⁡x<0\cos x<0 and sin⁡x>0\sin x>0 on (π/2,π)(\pi/2,\pi), making both terms negative. d) Since ln⁡x\ln x is positive while cos⁡x\cos x is negative, their product is negative and therefore decreasing on (π/2,π)(\pi/2,\pi).

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