Given    f(x) = x² + 10 sin x $\forall \mathrm{x}\in$ R
Also,    f(0) = 0  and  f(x)  is continuous on R.
$\underset{\mathrm{x}\to \pm \mathrm{\infty }}{\mathrm{Lim}}\mathrm{f}\left(\mathrm{x}\right)\to \mathrm{\infty }\Rightarrow$ there exists some real number c such that    f(x) = 1000     
    (Using intermediate value theorem). Ans.
(B)    Let  f(x) = $|\text{\hspace{0.17em}}{\mathrm{x}}^2+\mathrm{x}|=\left|\text{\hspace{0.17em}}\mathrm{x}(\mathrm{x}+1)\right|=\{\begin{array}{l}\mathrm{x}(\mathrm{x}+1),(-\mathrm{\infty },-1)\cup (0,\mathrm{\infty })\\ -\mathrm{x}(\mathrm{x}+1),[-1,0]\end{array}$

      
So,  f(x) is non-derivable at two points viz   x = – 1, 0. Ans.  

     
(C)    If  y = f(x)  and  x = g(y)  where g = f ^–1, then 

        $\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dy}}^2}=-\frac{\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}}{{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^3},\frac{\mathrm{dy}}{\mathrm{dx}}\ne 0$.

(D)    As  f(x)  is continuous on (0, 5) and  f(x)  takes only irrational values such that  f(2) = π,  
so    f (π) = π ( As  f(x) must be only constant function).