Guest

Pls help Pls help Pls help Pls help Pls help Pls help Pls help Pls help Pls help Pls help Pls help Pls help Pls help Pls help Pls help Pls help Pls help Pls help Pls help Pls help Pls help

 Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help  Pls help 

Question Image
Grade:11

1 Answers

Aditya Gupta
2086 Points
4 years ago
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 :))
 

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free