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Question: If the function \(f\left( x \right)=\cos \left| x \right|-2ax+b\) increases along the entire number ...

If the function f(x)=cos⁡∣x∣−2ax+bf\left( x \right)=\cos \left| x \right|-2ax+b increases along the entire number scale. The range of 'a' is given by:

& A.a=b \\\ & B.a=\dfrac{b}{2} \\\ & C.a\le -\dfrac{1}{2} \\\ & D.a\le -\dfrac{3}{2} \\\ \end{aligned}$$
Explanation

Solution

In this question, we are given a function f(x) which is increasing along the entire number scale. We have to calculate the range of 'a' which is a part of the function f(x). As we know, if a function f(x) is increasing, then f′(x)≥0f'\left( x \right)\ge 0 so we will use this property and evaluate the range of 'a' in terms of sinx. Then we will put the minimum value of sinx to find the required range of 'a'. Properties that we will use in the question are:
(i)cos⁡(−x)=cos⁡x (ii)ddxcos⁡x=−sin⁡x \begin{aligned} & \left( i \right)\cos \left( -x \right)=\cos x \\\ & \left( ii \right)\dfrac{d}{dx}\cos x=-\sin x \\\ \end{aligned}
(iii) Value of sinx lies between -1 and 1.

Complete step by step solution:
Here, we are given the function as f(x)=cos⁡∣x∣−2ax+bf\left( x \right)=\cos \left| x \right|-2ax+b.
Now, we know that \left| x \right|=\left\\{ \begin{aligned} & x,x\ge 0 \\\ & -x,x\text{ }<\text{ }0 \\\ \end{aligned} \right..
So, value of \cos \left| x \right|=\left\\{ \begin{aligned} & \cos \left( x \right),x\ge 0 \\\ & \cos \left( -x \right),x\text{ }<\text{ }0 \\\ \end{aligned} \right..
But, we know that cos⁡(−x)=cos⁡x\cos \left( -x \right)=\cos x.
So, value of cos⁡∣x∣\cos \left| x \right| becomes \cos \left| x \right|=\left\\{ \begin{aligned} & \cos \left( x \right),x\ge 0 \\\ & \cos \left( x \right),x\text{ }<\text{ }0 \\\ \end{aligned} \right..
Therefore, cos⁡∣x∣=cos⁡x\cos \left| x \right|=\cos x. Putting this values in the function f(x), hence our function becomes,
f(x)=cos⁡x−2ax+bf\left( x \right)=\cos x-2ax+b.
Now, we are given that f(x) is increasing function.
As we know, if a function f(x) is increasing then f′(x)≥0f'\left( x \right)\ge 0 so taking derivative of f(x) we get:

& \Rightarrow \dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( \cos x-2ax+b \right) \\\ & \Rightarrow f'\left( x \right)=-\sin x-2a \\\ \end{aligned}$$ Now, $f'\left( x \right)\ge 0$ so $-\sin x-2a\ge 0$. Changing the inequality by removing negative signs we get: $\sin x+2a\le 0\Rightarrow 2a\le -\sin x$. Now if we take the least value of -sinx, then we get the exact range of 'a'. As we know, the range of -sinx is (-1,1) so the least value of -sinx will be -1. Hence, putting -1 in place of -sinx, we get: $2a\le -1\Rightarrow a\le -\dfrac{1}{2}$. Hence the required range of a is $a\le -\dfrac{1}{2}$. **Hence option C is the correct answer.** **Note:** Students should note that, when increasing function is given, we take $f'\left( x \right)\ge 0$ but when strictly increasing function is given, we take $f\left( x \right)\ge 0$. Students should know that derivative of a constant is zero which is why we have taken the derivative of b as zero. While calculating range, take care of the signs. While changing signs, inequality signs also change.