For every twice differentiable function f: ℝ → [- 2, 2]with (f(0))...

f\left(x\right)$can't be constant throughout the domain. Hence we can find$x \in \left(r, s\right)$such that$ƒ ( x )$is one-one option$( A )$is true.<br >$( B )\left|{f}^{&prime;}\left(x\right)\right| = \left|\dfrac{f\left(0\right) - f\left(- 4\right)}{4}\right| \le 1\left(LMVT\right)$<br >$( C ) f\left(x\right) = \sin \left(\sqrt{85x}\right)$satisfies given condition<br >But$\lim_{x \to \infty} \sin \left(\sqrt{85}\right)D . N . E$<br >$⇒$Incorrect<br > \left(D\right)$g\left(x\right) = {f}^{2}\left(x\right) + {\left({f}^{′}\left(x\right)\right)}^{2}<br > \left|{f}^{&prime;}\left({x}_{1}\right)\right| \le 1 (by LMVT)
\left|f\left({x}_{1}\right)\right| \le 2$(given)<br >$⇒ g\left({x}_{1}\right) \le 5 ∃ {x}_{1} \in \left(- 4,0\right)$<br >Similarly$g\left({x}_{2}\right) \le 5 ∃ {x}_{2} \in \left(0,4\right)$<br >$g\left(0\right) = 85 ⇒ g\left(x\right)$has maxima in \left({x}_{1}, {x}_{2}\right)sayat\alpha$
{g}^{&prime;}\left(\alpha\right) = 0andg\left(\alpha\right) \ge 85
2{f}^{&prime;}\left(\alpha\right)\left(f\left(\alpha\right) + {f}^{&prime;&prime;}\left(\alpha\right)\right) = 0
If {f}^{&prime;}\left(\alpha\right) = 0 ⇒ g\left(\alpha\right) = {f}^{2}\left(\alpha\right) \ge 85 not possible
⇒ f\left(\alpha\right) + {f}^{&prime;&prime;}\left(\alpha\right) = 0 ∃ \alpha \in \left({x}_{1}, {x}_{2}\right) \in \left(- 4,4\right)
$( D )$ correct