If $m$ is the least value of $f(x)$=$\sqrt{x^2-2x+2}+\sqrt{x^2-4x+29}$, occur at $x = α$ , then $[m]+[α]$ is equal to (where [.] denotes greatest integer function)
(A)6
(B)7
(C)5
(D)4
My approach is as follow
$f(x)=\sqrt{(x-1)^2+1}+\sqrt{(x-2)^2+5^2}$
$\alpha$ lies between 1&2 and $m$ lies between 6 and 7 hence the answer is 7. I checked it and ur us correct. But solution is required
On
In the figure, let $X(x,0)$ be any point on the $x$-axis.
Note that $AX=\sqrt{(x-1)^2+1}$ and $BX=\sqrt{(x-2)^2+5^2}$.
Therefore our job is to minimize $AX+BX$ where $X$ is an arbitrary point on the $x$-axis
Let $A'(1,-1)$ be the mirror image of $A(1,1)$ with respect to the $x$-axis.
From perpendicular bisector theorem,
$AX+XB=A'X+XB \ge A'B=AC+CB$
Thus the least value of $AX+XB$ is $A'B$. It is attained when $X=C$.
Since $A'B=\sqrt{(2-1)^2+(5+1)^2}=\sqrt{37}$
The least value of $f(x)$ is $\sqrt{37}$
Finally let $C=(\alpha, 0)$, then $\frac{0+1}{\alpha-1}=\frac{5+1}{2-1} \implies \alpha = \frac{7}{6}$
By using the Cauchy Schwarz inequality, you can easily obtain: $$\sqrt {a^2+b^2}+\sqrt{c^2+d^2}\ge \sqrt {(a+c)^2+(b+d)^2};$$
and equality holds if $\frac{a}{b}=\frac {c}{d}$. Now, take $a=x-1$, $c=2-x$, $b=1$, $d=5$. We get:
$$\sqrt {(x-1)^2+1^2}+\sqrt{(2-x)^2+5^2}\ge \sqrt {1^2+6^2}=\sqrt {37}\implies m=\sqrt {37}.$$
And,
$$\frac {x-1}{1}=\frac {2-x}{5}\implies a=\frac {7}{6}.$$