Question

If the function f(x) = cos |x| − 2ax + b increases along the entire number scale, then

 

पर्याय

•  a = b

• $$a=\frac{1}{2}b$$

• $$a\le -\frac{1}{2}$$

• $$a>-\frac{3}{2}$$

Answer 1

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

$$Given:f\left(x\right)=\cos \left|x\right|-2ax+b$$

$$\text{ Now},\left|x\right|=\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}$$

$$\text{ and }\cos \left|x\right|=\{\begin{array}{ll}\cos \left(x\right),& x\ge 0\\ cos(-x)=cos\left(x\right),& x<0\end{array}$$

$$\therefore \cos \left|x\right|=\cos x,\mathrm{\forall }x\in R$$

$$\therefore f\left(x\right)=\cos x-2ax+b$$

$$\Rightarrow f^{\prime }\left(x\right)=-\sin x-2a$$

$$\text{ It is given that f(x) is increasing }.$$

$$\Rightarrow f^{\prime }\left(x\right)\ge 0$$

$$\Rightarrow -\sin x-2a\ge 0$$

$$\Rightarrow \sin x+2a\le 0$$

$$\Rightarrow 2a\le -\sin x$$

$$\text{ The least value of -sin x is -1 }.$$

$$\Rightarrow 2a\le -1$$

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