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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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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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