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Byju's Answer

Standard XII

Mathematics

Definition of Function

If f'x=x-1x...

Question

If $f^{'} (x) = (x - 1) (x - 2)^{2}$ , then find $f (x) :$

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Solution

$f^{'} (x) = (x - 1) (x - 2)^{2}$
Integrating ; $f (x) = \int (x - 1) (x - 2)^{2} . d x = \int (x - 1) (x^{2} + 4 - 4 x) . d x$
$= \int (x^{3} + 4 x - 4 x^{2} - x^{2} - 4 + 4 x) . d x$
$= \int x^{3} - 5 x^{2} + 8 x - 4. d x$
$∴ f (x) = \frac{x^{4}}{4} - \frac{5 x^{3}}{3} + \frac{8 x^{2}}{2} - 4 x + c$

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Similar questions

Q. If

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

g (x) = f^{'} (x)

F (x) = {(f (\frac{x}{2}))}^{2} + {(g (\frac{x}{2}))}^{2}

and given that

F (5) = 5

, then

F (10)

If $F (x) = {(f (\frac{x}{2}))}^{2} + {(g (\frac{x}{2}))}^{2}$ where $f^{'} (x) = - f (x)$ and $g (x) = f^{'} (x)$ and given that $F (5) = 5$ , then $F (10)$ is equal to

Q. Let

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

and

g (x) = f^{'} (x) .

F (x) = {f (\frac{x}{2})}^{2} + {g (\frac{x}{2})}^{2}

and

F (5) = 5

then

F (10)

is equal to

Q. If

f " (x) = - f (x)

and

g (x) = f^{'} (x)

and

F (x) = {(f (\frac{x}{2}))}^{2} + {(g (\frac{x}{2}))}^{2}

and given that

F (5) = 5

, then

F (10)

is equal to

Q. If

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{| x + 2 |}{t a n^{- 1} (x + 2)} & x \neq - 2 2, & x = - 2 \end{matrix}

then,

f (x)

is:

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If f′(x)=(x−1)(x−2)2, then find f(x):

f′(x)=(x−1)(x−2)2Integrating ; f(x)=∫(x−1)(x−2)2.dx=∫(x−1)(x2+4−4x).dx=∫(x3+4x−4x2−x2−4+4x).dx=∫x3−5x2+8x−4.dx∴f(x)=x44−5x33+8x22−4x+c

If $f^{'} (x) = (x - 1) (x - 2)^{2}$ , then find $f (x) :$