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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

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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?

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Solution for question: 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. For the courses CBSE (Science), CBSE (Commerce), PUC Karnataka Science, CBSE (Arts)
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