jee-main 2021 Q61

jee-main · India · session2_16mar_shift1 Laws of Logarithms Solve a Logarithmic Equation

If for $x \in \left( 0 , \frac { \pi } { 2 } \right) , \log _ { 10 } \sin x + \log _ { 10 } \cos x = - 1$ and $\log _ { 10 } ( \sin x + \cos x ) = \frac { 1 } { 2 } \left( \log _ { 10 } n - 1 \right) , n > 0$, then the value of $n$ is equal to :
(1) 20
(2) 12
(3) 9
(4) 16

If for $x \in \left( 0 , \frac { \pi } { 2 } \right) , \log _ { 10 } \sin x + \log _ { 10 } \cos x = - 1$ and $\log _ { 10 } ( \sin x + \cos x ) = \frac { 1 } { 2 } \left( \log _ { 10 } n - 1 \right) , n > 0$, then the value of $n$ is equal to :\\
(1) 20\\
(2) 12\\
(3) 9\\
(4) 16

Paper Questions

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