$\frac{\left|{\mathrm{x}}^2-10\mathrm{x}+9\right|}{\mathrm{x}}=\mathrm{k}$ 
 $\mathrm{f}\left(\mathrm{x}\right)=\frac{\left|{\mathrm{x}}^2-10\mathrm{x}+9\right|}{\mathrm{x}}=\left[\begin{array}{l}\frac{(\mathrm{x}-1)(\mathrm{x}-9)}{\mathrm{x}}\text{ if }\mathrm{x}\in (-\mathrm{\infty },1)\cup (9,\mathrm{\infty })\\ \frac{-(\mathrm{x}-1)(\mathrm{x}-9)}{}\text{ if }\mathrm{x}\in (1,9)\end{array}\right.$
${\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\frac{\mathrm{x}(2\mathrm{x}-10)-\left({\mathrm{x}}^2-10\mathrm{x}+9\right)}{{\mathrm{x}}^2}=\frac{2{\mathrm{x}}^2-10\mathrm{x}-{\mathrm{x}}^2+10\mathrm{x}-9}{{\mathrm{x}}^2}$
$=\frac{{\mathrm{x}}^2-9}{{\mathrm{x}}^2}=\frac{(\mathrm{x}-3)(\mathrm{x}+3)}{{\mathrm{x}}^2}$
 $\Rightarrow {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\left[\begin{array}{cc}\frac{{\mathrm{x}}^2-9}{{\mathrm{x}}^2}& \text{ for }\mathrm{x}\in (-\mathrm{\infty }\\ \uparrow \text{ in }\left|\mathbf{x}\right|>3\text{ and }& \end{array}\right.$           
From the graph  (– ∞, – 16] ⋃ [4, ∞) Ans.

Aliter-1:   For the given condition to be satisfied, the value of  k must lie between the slopes of line  OA  and  OB.
For slope of line  OA,
– (x² – 10x + 9) = kx  must have equal roots.
⇒    (k – 10)² – 36 = 0   ⇒     k = 4, 16 ⇒ k = 4
For slope of the line  OB,
 x² – 10x + 9 = kx  must have equal roots
⇒  (k + 10)² – 36 = 0  ⇒  k = – 4, – 16
∴   k Î [4, ∞) ⋃ (– ∞, – 16]   Ans.

 
Aliter-2:    x² – 10x + 9 = ± kx
x² – (10 ± k) x + 9 = 0
D = (10 ± k)² – 36 = 0
10 + k = ± 6  ⇒  k = – 16 or – 4
 and
10 – k = ± 6  ⇒  k = 16 or 4
Now, draw the graph of y = | x² – 10x + 9 |    
$\frac{\mathrm{dy}}{\mathrm{dx}}$= 0, when x = 5
y(5) = – 16
For y = kx to intersect y = | x² – 10x + 9 |, 
k ≤ 4  or  k ≤ – 16 
Hence, k ∈ (– ∞, – 16] ⋃ [4, ∞).