Question

The greatest positive integral value of x for which 200-x(10+x)

Answer 1

$sry dont know$

Answer 2

Answer: x = 9

Step-by-step explanation:

200−x(10+x)

=200−10x−x

2

=200−20x+10x−x

2

=20(10−x)+x(10−x)

=(20+x)(10−x)

⇒f(x)>0∀xϵ(−20,10)

∴ greatest positive integral value for which f(x)>0 is x=9 (∵ at x=10,f(x)=0)