Question

solve the attachment

Answer 1

Let us suppose that there exists a positive root of the function.

If so then there are 3 possibilities

Case 1: The root is greater than 1 or x>1

Is so then x32>x25x32>x25

orx32−x25>0orx32−x25>0

Similarly x18>x11x18>x11

orx18−x11>0orx18−x11>0

x4>x3x4>x3

orx4−x3>0orx4−x3>0

Adding all these we get

x32−x25+x18−x11+x4−x3>0x32−x25+x18−x11+x4−x3>0

which is not equal to 0

So the root can not be greater than 1

Case 2: The root is equal to 1 or x=1

Substituting the Left Hand Side becomes 1–1+1–1+1–1+1=1 which is not equal to 0

So 1 is also not a root

Case 3: The root is less than 1 or x<1

x3<1or1−x3>0x3<1or1−x3>0

Similarly x4−x11>0x4−x11>0

x18−x25>0x18−x25>0

x32>0x32>0

Adding all these inequalities we get the Left hand side greater than 0

So the root can not be less than 1 also

So there is not positive root of the function

( ist option is right now )