Let f be a differential function such that fx=f2-x and gx=f1+x then
g(x) is an odd function
g(x) is an even function
Graph f(x) is symmetrical about the line x = 1
f'1=0
We have fx=f2-xReplacing x by f1+x f1+x=f1-xHence graph of fx is symmetrical about the line x=1Also gx=f1+x=f1-x=g-x∴ gx is an even functionFurther f'1+x=-f'1-x∴ f'1=-f'1∴ f'1=0