f(x)= ∫0 to x g(t)dt
f(-x)= ∫0 to – x g(t)dt
let u= – t
f(-x)= – ∫0 to x g(-u)du= – ∫0 to x g(u)du (since g(u)= g(-u) for it is even)
or f(–x)= –f(x)
g(x)= f(x+5)
g(-x)= f(5 – x)
so f(x+5)= f(5 – x)
replace x by 5 – x
f(10 – x)= f(x)
replace x by 20+x
or f(10 – (20+x))= f(x+20)
f(x+20)= f( – (x+10))= – f(10 – ( – x))= – f( – x)= f(x)
hence f(x)= f(x+20)
now differentiating f(x)= ∫0 to x g(t)dt we get f’(x)= f(x+5).......(1)
diff again f”(x)= f’(x+5)= f(x+10)
again diff f’’’(x)= f’(x+10)= f(x+15)
again f’’’’(x)= f’(x+15)= f(x+20)= f(x)
hence solving f’’’’(x)= f(x) we have
f(x)= ae^x + be^-x + csinx + ecosx. also, f’(x)= ae^x – be^-x + ccosx – esinx. substitute in (1)
ae^x – be^-x + ccosx – esinx = ae^5e^x + be^-5e^-x + csin(x+5) + ecos(x+5) identically
hence a= ae^5 and – b= be^-5 so a=b=0
so f(x)= csinx + ecosx but f(0)=0 so e=0 hence f(x)= csinx
and ccosx= csin(x+5), which can be identically true iff c= 0 hence f(x)= 0 or g(x)= f(x+5)= 0.
since it is given in the ques that g should be non zero, but we have proved it to be 0, hence we deduce that the ques is wrong.
KINDLY APPROVE :))
