JEE PYQ: Trigonometric Equations And Inequations Question 1
Question 1 - 2021 (16 Mar Shift 1)
If for $x \in \left(0, \frac{\pi}{2}\right)$, $\log_{10} \sin x + \log_{10} \cos x = -1$ and $\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - 1)$, $n > 0$ then the value of $n$ is equal to:
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(1) 20
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(2) 12
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(3) 9
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(4) 16
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Answer:
Solution
$\log_{10} (\sin x \cdot \cos x) = -1 \Rightarrow \sin x \cdot \cos x = \frac{1}{10}$
$\sin x + \cos x = \sqrt{\frac{n}{10}}$
Squaring: $1 + 2 \sin x \cdot \cos x = \frac{n}{10} \Rightarrow 1 + \frac{1}{5} = \frac{n}{10} \Rightarrow n = 12$
BETA
