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) $a\le -\frac{1}{2}$

(d) $a>-\frac{3}{2}$

Answer 1

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

$$\begin{gathered} \mathrm{Given}:f\left(x\right)=\cos \left|x\right|-2ax+b \\ \mathrm{Now},\left|x\right|=\left\{\begin{array}{l}x,x\ge 0 \\ \\ -x,x<0\end{array}\right. \\ \mathrm{and}\cos \left|x\right|=\left\{\begin{array}{l}\cos \left(x\right),x\ge 0 \\ \\ \cos \left(-x\right)=\cos \left(x\right),x<0\end{array}\right. \\ \therefore \cos \left|x\right|=\cos x,\forall x\in R \\ \therefore f\left(x\right)=\cos x-2ax+b \\ \Rightarrow f\text{'}\left(x\right)=-\sin x-2a \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}f\left(x\right)\mathrm{is}\mathrm{increasing}. \\ \Rightarrow f\text{'}\left(x\right)\ge 0 \\ \Rightarrow -\sin x-2a\ge 0 \\ \Rightarrow \sin x+2a\le 0 \\ \Rightarrow 2a\le -\sin x \\ \mathrm{The}\mathrm{l}\text{east value of -sin x is -1.} \\ \Rightarrow 2a\le -1 \\ \Rightarrow a\le \frac{-1}{2} \end{gathered}$$