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Question
The function \[f\left( x \right) = \frac{\lambda \sin x + 2 \cos x}{\sin x + \cos x}\] is increasing, if
Options
λ < 1
λ > 1
λ < 2
λ > 2
Solution
λ > 2
\[f\left( x \right) = \frac{\lambda \sin x + 2 \cos x}{\sin x + \cos x}\]
\[f'\left( x \right) = \frac{\left( \sin x + \cos x \right)\left( \lambda \cos x - 2 \sin x \right) + \left( \lambda \sin x + 2 \cos x \right)\left( \cos x - \sin x \right)}{\left( \sin x + \cos x \right)^2}\]
\[ = \frac{\lambda\cos x \sin x + \lambda \cos^2 x - 2 \sin^2 x - 2 \sin x \cos x - \lambda\sin x \cos x - 2 \cos^2 x + \lambda \sin^2 x + 2 \cos x \sin x}{\left( \sin x + \cos x \right)^2}\]
\[ = \frac{- 2 \left( \sin^2 x + \cos^2 x \right) + \lambda \left( \sin^2 x + \cos^2 x \right)}{\left( \sin x + \cos x \right)^2}\]
\[ = \frac{- 2 + \lambda}{\left( \sin x + \cos x \right)^2}\]
\[\text { For f(x) to be increasing, we must have }\]
\[f'\left( x \right) > 0\]
\[ \Rightarrow \frac{- 2 + \lambda}{\left( \sin x + \cos x \right)^2} > 0 \]
\[ \Rightarrow \lambda - 2 > 0 \left[ \because \left( \sin x + \cos x \right)^2 > 0 \right]\]
\[ \Rightarrow \lambda > 2\]
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