मराठी

The Function F ( X ) = λ Sin X + 2 Cos X Sin X + Cos X is Increasing, If

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प्रश्न

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

 

पर्याय

  • λ < 1

  • λ > 1

  • λ < 2

  • λ > 2

MCQ
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उत्तर

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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पाठ 16: Increasing and Decreasing Functions - Exercise 17.4 [पृष्ठ ४१]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 16 Increasing and Decreasing Functions
Exercise 17.4 | Q 23 | पृष्ठ ४१

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