Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

$f\left(x\right)$can't be constant throughout the domain. Hence we can find$x\in \left(r,s\right)$such that$ƒ\left(x\right)$is one-one option$\left(A\right)$is true.
$\left(B\right)$$\left|f^{\prime }\left(x\right)\right|=\left|\frac{f\left(0\right)-f\left(-4\right)}{4}\right|\le 1 \left(\mathrm{L}\mathrm{M}\mathrm{V}\mathrm{T}\right)$
$\left(C\right)$$f\left(x\right)=\sin \left(\sqrt{85x}\right)$satisfies given condition
But$\underset{x\to \infty }{\lim }\sin \left(\sqrt{85}\right) D.N.E$
$\Rightarrow$Incorrect
$\left(D\right)$$g\left(x\right)=f^2\left(x\right)+{\left(f^{\prime }\left(x\right)\right)}^2$
$\left|f^{\prime }\left(x_1\right)\right|\le 1$(by LMVT)
$\left|f\left(x_1\right)\right|\le 2$(given)
$\Rightarrow g\left(x_1\right)\le 5 \exists x_1\in \left(-\mathrm{4,0}\right)$
Similarly$g\left(x_2\right)\le 5 \exists x_2\in \left(\mathrm{0,4}\right)$
$g\left(0\right)=85 \Rightarrow g\left(x\right)$has maxima in$\left(x_1,x_2\right) \mathrm{s}\mathrm{a}\mathrm{y} \mathrm{a}\mathrm{t} \alpha$
$g^{\prime }\left(\alpha \right)=0 \mathrm{a}\mathrm{n}\mathrm{d} g\left(\alpha \right)\ge 85$
$2f^{\prime }\left(\alpha \right)\left(f\left(\alpha \right)+f^{\prime \prime }\left(\alpha \right)\right)=0$
If$f^{\prime }\left(\alpha \right)=0\Rightarrow g\left(\alpha \right)=f^2\left(\alpha \right)\ge 85$not possible
$\Rightarrow f\left(\alpha \right)+f^{\prime \prime }\left(\alpha \right)=0 \exists \alpha \in \left(x_1,x_2\right)\in \left(-\mathrm{4,4}\right)$
$\left(D\right)$correct