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QJEEM21Ma6ae1: JEE Main 2021 Mathematics - Class 11 JEEM21Ma6ae1 Solution: If for x , log 10 sinx + log 10 cosx = 1 and log 10 (sinx +…

If for x ∈\in (0,π2)\left( {0,{\pi \over 2}} \right) , log10sinx + log10cosx = −- 1 and log10(sinx + cosx) = 12{1 \over 2} (log10 n −- 1), n > 0, then the value of n is equal to :

Multiple Choice Options

  • A 16
  • B 9
  • C 12
  • D 20

Step-by-Step Verified Solution

amp;log⁡10sin⁡x+log⁡10cos⁡x=−1,x∈(0,π/2)amp;log⁡10(sin⁡xcos⁡x)=−1amp;⇒sin⁡xcos⁡x=10−1=110amp;log⁡10(sin⁡x+cos⁡x)=12(log⁡10n−1),ngt;0amp;2log⁡10(sin⁡x+cos⁡x)=(log⁡10n−log⁡1010)amp;⇒log⁡10(sin⁡x+cos⁡x)2=log⁡10(n10)amp;⇒(sin⁡x+cos⁡x)2=n10amp;⇒sin⁡2x+cos⁡2x+2sin⁡xcos⁡x=n10amp;⇒1+2(110)=n10⇒1210=n10amp;∴n=12\begin{aligned}& \log _{10} \sin x+\log _{10} \cos x=-1, x \in(0, \pi / 2) \\\\& \log _{10}(\sin x \cos x)=-1 \\\\& \Rightarrow \sin x \cos x=10^{-1}= {1 \over {10}} \\\\& \log _{10}(\sin x+\cos x)={1 \over {2}}\left(\log _{10} n-1\right), n>0 \\\\& 2 \log _{10}(\sin x+\cos x)=\left(\log _{10} n-\log _{10} 10\right) \\\\& \Rightarrow \log _{10}(\sin x+\cos x)^2=\log _{10}({n \over {10}}) \\\\& \Rightarrow (\sin x+\cos x)^2={n \over {10}} \\\\& \Rightarrow \sin ^2 x+\cos ^2 x+2 \sin x \cos x=\frac{n}{10} \\\\& \Rightarrow 1+2({1 \over {10}})={n \over {10}} \Rightarrow {12 \over {10}}={n \over {10}} \\\\& \therefore n=12\end{aligned}

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