The sum of the maximum and minimum values of the function f(x)=|5x-7|+[x^2+2x] in the interval [ 5 4,2], where [t] is the greatest integer ≤ t, is .

Given,

$$f(x)=|5x-7|+[{x}^{2}+2x]$$f(x)=∣5x−7∣+[x2+2x]

$$=|5x-7|+[{(x+1)}^{2}-1]$$=∣5x−7∣+[(x+1)2−1]

$$=|5x-7|+[{(x+1)}^{2}]-1$$=∣5x−7∣+[(x+1)2]−1

$$(\text{as }[x-1]\text{=}[x]\text{-1 where }[.]\text{ is greatest integer function})$$(as [x−1]=[x]-1 where [.] is greatest integer function)

Now critical points of

$$f(x)=\frac{7}{5},\sqrt{5}-1,\sqrt{6}-1,\sqrt{7}-1,\sqrt{8}-1,2$$f(x)=57​,5​−1,6​−1,7​−1,8​−1,2

$\therefore$∴ Maximum or minimum value of $f(x)$f(x) occur at critical points or boundary points.

So, $f(\frac{5}{4})=\frac{3}{4}+4=\frac{19}{4}$f(45​)=43​+4=419​ and $f(\frac{7}{5})=0+4=4$f(57​)=0+4=4

As both $|5x-7|$∣5x−7∣ and ${x}^{2}+2x$x2+2x are increasing in nature after $x=\frac{7}{5}$x=57​

So, $f(2)=|10-7|+[4+4]=3+8=11$f(2)=∣10−7∣+[4+4]=3+8=11

$\therefore f{(\frac{7}{5})}_{\mathrm{min}}=4$∴f(57​)min​=4 and $f{(2)}_{\mathrm{max}}=11$f(2)max​=11

So sum of minimum and maximum value is $4+11=15$4+11=15