Question

The function \[f\left( x \right) = \frac{\lambda \sin x + 2 \cos x}{\sin x + \cos x}\] is increasing, if

 

विकल्प

• λ < 1

• λ > 1

• λ < 2

• λ > 2

Answer 1

λ > 2

\[f\left( x \right) = \frac{\lambda \sin x + 2 \cos x}{\sin x + \cos x}\]

\[f'\left( x \right) = \frac{\left( \sin x + \cos x \right)\left( \lambda \cos x - 2 \sin x \right) + \left( \lambda \sin x + 2 \cos x \right)\left( \cos x - \sin x \right)}{\left( \sin x + \cos x \right)^2}\]

\[ = \frac{\lambda\cos x \sin x + \lambda \cos^2 x - 2 \sin^2 x - 2 \sin x \cos x - \lambda\sin x \cos x - 2 \cos^2 x + \lambda \sin^2 x + 2 \cos x \sin x}{\left( \sin x + \cos x \right)^2}\]

\[ = \frac{- 2 \left( \sin^2 x + \cos^2 x \right) + \lambda \left( \sin^2 x + \cos^2 x \right)}{\left( \sin x + \cos x \right)^2}\]

\[ = \frac{- 2 + \lambda}{\left( \sin x + \cos x \right)^2}\]

\[\text { For f(x) to be increasing, we must have }\]

\[f'\left( x \right) > 0\]

\[ \Rightarrow \frac{- 2 + \lambda}{\left( \sin x + \cos x \right)^2} > 0 \]

\[ \Rightarrow \lambda - 2 > 0 \left[ \because \left( \sin x + \cos x \right)^2 > 0 \right]\]

\[ \Rightarrow \lambda > 2\]