NIMCET 2019 Question Paper with solutions

Question : 9 of 120

Marks: +1, -0

Let f(x) be a polynomial satisfying, f(0) = 2, f'(0) = 3 and f''(x) = f(x). Then f(4) is equal to

Solution:

Given,

f(0)=2,f'(0)=3

and

f''(x)=f(x)

⇒{d^2y}/{dx^2}-y=

0
Auxiliary equation

=m^2-1=0

m=±1

Solution of given differential equation is

y=ae^x+be^{-x}

⇒f(x)=ae^x+be^{-x}

f(0)=a+b=2

....(i)

f'(x)=ae^x-be^{-x}

f'(0)=a-b=3

...(ii)
Solving Eqs. (i) and (ii), we get

a=5/2\;b==-1/2

∴f(x)=1/2(5e^x-e^{-x})

f(4)=1/2(5e^{4}-e^{-4})

f(4)=1/2({5e^8-1}/e^4)

f(4)={5e^8-1}/{2e^4}

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