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# Solution for If the Function F(X) = Cos |X| − 2ax + B Increases Along the Entire Number Scale, Then (A) a = B (B) a = 1 2 B (C) a ≤ − 1 2 (D) a > − 3 2 - CBSE (Science) Class 12 - Mathematics

ConceptIncreasing and Decreasing Functions

#### Question

If the function f(x) = cos |x| − 2ax + b increases along the entire number scale, then

(a) a = b
(b) $a = \frac{1}{2}b$

(c) $a \leq - \frac{1}{2}$

(d)  $a > - \frac{3}{2}$

#### Solution

(c) $a \leq - \frac{1}{2}$

$Given: f\left( x \right) = \cos \left| x \right| - 2ax + b$

$\text { Now}, \left| x \right| =\begin{cases} x ,& x \geq 0 \\ - x, & x < 0 \end{cases}$

$\text { and } \cos \left| x \right| = \begin{cases} \cos\left( x \right) , & x \geq 0 \\cos\left( - x \right) = cos\left( x \right), & x < 0\end{cases}$

$\therefore \cos\left| x \right| = \cos x , \forall x \in R$

$\therefore f\left( x \right) = \cos x - 2ax + b$

$\Rightarrow f'\left( x \right) = - \sin x - 2a$

$\text { It is given that f(x) is increasing } .$

$\Rightarrow f'\left( x \right) \geq 0$

$\Rightarrow - \sin x - 2a \geq 0$

$\Rightarrow \sin x + 2a \leq 0$

$\Rightarrow 2a \leq - \sin x$

$\text { The least value of -sin x is -1 }.$

$\Rightarrow 2a \leq - 1$

$\Rightarrow a \leq \frac{- 1}{2}$

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [3]

Solution If the Function F(X) = Cos |X| − 2ax + B Increases Along the Entire Number Scale, Then (A) a = B (B) a = 1 2 B (C) a ≤ − 1 2 (D) a > − 3 2 Concept: Increasing and Decreasing Functions.
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