MCQ
If the function $\text{f}(\text{x})=\cos|\text{x}|-2\text{ax}+\text{b}$ increases along entire number scale, then:
  • A
    $\text{a}=\text{b}$
  • B
    $\text{a}=\frac{1}{2}\text{b}$
  • $\text{a}\leq-\frac{1}{2}$
  • D
    $\text{a}>-\frac{3}{2}$

Answer

Correct option: C.
$\text{a}\leq-\frac{1}{2}$
Given:
$\text{f}(\text{x})=\cos|\text{x}|-2\text{ax}+\text{b}$
Now, $|\text{x}|=\begin{cases}\text{x},&\text{x}\geq0\\-\text{x},&\text{x}<0\end{cases}$
And $\cos|\text{x}|=\begin{cases}\cos(\text{x}),&\text{x}\geq0\\\cos(-\text{x})=\cos(\text{x}),&\text{x}<0\end{cases}$
$\therefore\ \cos|\text{x}|=\cos\text{x},\forall\ \text{x}\in\text{R}$
$\therefore\ \text{f}(\text{x})=\cos\text{x}-2\text{ax}+\text{b}$
$\Rightarrow\text{f}'(\text{x})=-\sin\text{x}-2\text{a}$
It is given that $f(x)$ is increasing.
$\Rightarrow\text{f}'(\text{x})\geq0$
$\Rightarrow-\sin\text{x}-2\text{a}\geq0$
$\Rightarrow\sin\text{x}+2\text{a}\leq0$
$\Rightarrow2\text{a}\leq-\sin\text{x}$
The least value of $-\sin\text{x}$ is $-1.$
$\Rightarrow2\text{a}\leq-1$
$\Rightarrow\text{a}\leq\frac{-1}{2}$

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