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# Solution for The Function F ( X ) = λ Sin X + 2 Cos X Sin X + Cos X is Increasing, If (A) λ < 1 (B) λ > 1 (C) λ < 2 (D) λ > 2 - CBSE (Science) Class 12 - Mathematics

ConceptIncreasing and Decreasing Functions

#### Question

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

(a) λ < 1
(b) λ > 1
(c) λ < 2
(d) λ > 2

#### Solution

(d) λ > 2

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

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

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

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

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

$\text { Forf(x) to be increasing, we must have }$

$f'\left( x \right) > 0$

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

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

$\Rightarrow \lambda > 2$

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Solution for question: The Function F ( X ) = λ Sin X + 2 Cos X Sin X + Cos X is Increasing, If (A) λ < 1 (B) λ > 1 (C) λ < 2 (D) λ > 2 concept: Increasing and Decreasing Functions. For the courses CBSE (Science), PUC Karnataka Science, CBSE (Arts), CBSE (Commerce)
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