The number of points in $(-\infty, \infty)$, for which $x^2 - x \sin x - \cos x = 0$, is

The number of points in (-∞, ∞), for which x² - x sin x - cos x = 0, is $f(x) = x^2 - x\sin x - \cos x$ $\text{To find: number of real solutions of } f(x) = 0$ $f(-x) = x^2 - (-x)\sin(-x) - \cos(-x)$ $= x^2 - x\sin x - \cos x = f(x)\quad (f \text{ is even})$ $f'(x) = 2x - (\sin x + x\cos x) + \sin x$ $= 2x - x\cos x = x(2 - \cos x)$ $2 - \cos x \geq 1 > 0 \Rightarrow \operatorname{sgn} f'(x) = \operatorname{sgn} x$ $f \searrow \text{ on } (-\infty, 0),\qquad f \nearrow \text{ on } (0, \infty)$ $f(0) = -1 \quad (\text{global minimum})$ $f(x) \to +\infty \text{ as } x \to \pm\infty$ $\text{on } (0, \infty):\ f \text{ rises from } -1 \text{ to } +\infty \Rightarrow \text{exactly one root}$ $\text{even} \Rightarrow \text{exactly one root on } (-\infty, 0)$ $\text{2 real roots} \Rightarrow \text{option (3)}$