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Solve! - Let f: be a function such that,. Then f(2) equals : - Differential equations - JEE Main

Let f:R\rightarrow R be a function such that f(x)= x^3+x^2{f}'(1)+x{f}''(2)+{f}'''(3),x\epsilon R. Then f(2) equals :

  • Option 1)

    -4

  • Option 2)

    30

  • Option 3)

    -2

  • Option 4)

    8

Answers (1)
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A admin

 

Differential Equations -

An equation involving independent variable (x), dependent variable (y) and derivative of dependent variable with respect to independent variable 
\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )

- wherein

eg:

  \frac{d^{2}y}{dx^{2}}- 3\frac{dy}{dx}+5x=0

 

 

Order of a Differential Equation -

The order of a differential equation is order of highest order occuring in differential equation

- wherein

order of

\frac{d^{2}y} {dx^{2}}+5=0  

is 2.

Given f(x)=x^{3}+x^{2}f{}'(1)+xf{}''(2)+f{}''{}'(3)

=>f{}'(x)=3x^{2}+2xf{}'(1)+f{}''(2)................................(1)

=>f{}{}''(x)=6x+2f{}'(1).....................................................(2)

=>f{}{}''{}'(x)=6..........................................................................(3)

put x=1 in equation (1)

=>f{}'(1)=3+2f{}'(1)+f{}'{}'(2).......................................(4)

put x=2 in equation (2)

=>f{}''(2)=12+2f{}'(1).....................................................(5)

from eqn (4) and (5)

-3-f{}'(1)=12+2f{}'(1)

=>3f{}'(1)=-15

=>f{}'(1)=-5     and     f{}'{}'(2)=2

Now,

put x = 3 in eqn (3)

f{}'{}'{}'(3)=6

\therefore f(x)=x^{3}-5x^{2}+2x+6

\therefore f(2)=8-20+4+6=-2

 

 

 


Option 1)

-4

Option 2)

30

Option 3)

-2

Option 4)

8

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