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Q: ƒ’(x)=ƒ(x)+ -1010 ƒ(x) dx ,where ƒ(x) is a strictly increasing function then

A. ƒ’(x)=ƒ(x) if limx→ -∞ ƒ(x)=0

B.ƒ’(x) - ƒ(x)=0

C.ƒ’’(x) - ƒ(x)=0

D.ƒ’(x)=ƒ(x) if limx→ ∞ ƒ(x)=0

7 years ago

147 Points

Dear Tapacranjan

ƒ’(x)=ƒ(x)+ -1010 ƒ(x) dx

let C= -1010 ƒ(x) dx     because it is a constant

so ƒ’(x)=ƒ(x)+ C

dy/dx -y =C

solve this differential equation

IF =e-∫dx =e-x

so ye-x =Ce-xdx

ye-x =-Ce-x  +C1

y = -C +C1ex

f(x) =-C +C1ex   ..............1

now

C= -1010 ƒ(x) dx

C= -1010 {-C +C1ex} dx

C= -20C +C1 (e10 -e-10)

C = C1 (e10 -e-10)/21

so

f(x)=-C  +21Cex/(e10 -e-10)

let limx→ -∞ ƒ(x)=0

0= -C +0

C=0

so f(x)=f'(x)

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the answer and detailed  solution very  quickly.

All the best.

Regards,

7 years ago
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