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Question
If the function f(x) = cos |x| − 2ax + b increases along the entire number scale, then
Options
a = b
\[a = \frac{1}{2}b\]
\[a \leq - \frac{1}{2}\]
\[a > - \frac{3}{2}\]
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Solution
\[a \leq - \frac{1}{2}\]
\[Given: f\left( x \right) = \cos \left| x \right| - 2ax + b\]
\[\text { Now}, \left| x \right| =\begin{cases} x ,& x \geq 0 \\ - x, & x < 0 \end{cases}\]
\[\text { and } \cos \left| x \right| = \begin{cases} \cos\left( x \right) , & x \geq 0 \\cos\left( - x \right) = cos\left( x \right), & x < 0\end{cases}\]
\[ \therefore \cos\left| x \right| = \cos x , \forall x \in R\]
\[ \therefore f\left( x \right) = \cos x - 2ax + b\]
\[ \Rightarrow f'\left( x \right) = - \sin x - 2a\]
\[\text { It is given that f(x) is increasing } . \]
\[ \Rightarrow f'\left( x \right) \geq 0\]
\[ \Rightarrow - \sin x - 2a \geq 0\]
\[ \Rightarrow \sin x + 2a \leq 0\]
\[ \Rightarrow 2a \leq - \sin x\]
\[\text { The least value of -sin x is -1 }.\]
\[ \Rightarrow 2a \leq - 1\]
\[ \Rightarrow a \leq \frac{- 1}{2}\]
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