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Let $f’(x)=4x^3-3x^2+2x+k,$  $f(0)=1$ and $f(1)=4.$ Then $f(x)$ is equal to

1. $4x^4-3x^3+2x^2+x+1$
2. $x^4-x^3+x^2+2x+1$
3. $x^4-x^3+x^2+2(x+1)$
4. none of these
in Calculus
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Answer $B$?

+1 vote

$\underline{\textbf{Answer: B}}$

$\underline{\textbf{Explanation:}}$

Given:

$\mathrm {f'(x) = 4x^3 -3x^2 + 2x + k,\; f(0) = 1 \;\text{and}\; f(1) = 4}$

Now, we can integrate the $\mathrm {f'(x)}$ to get the $\mathrm {f(x)}$:

$\text{Let} \;\mathrm{I=f(x) = \int(4x^3-3x^2+2x+k)}$

$\Rightarrow \mathrm{f(x)=\frac{4x^4}{4} -\frac{3x^3}{3}+\frac{2x^2}{2}+kx + C, \;where\;c\;is\;a\;constant}$

$\Rightarrow \mathrm{f(x) =x^4 -x^3+x^2+kx + C}\tag{1}$

Now,

$\mathrm f(0) = 0\;\;\;\text{[Given]}$

$\therefore \mathrm {f(0)} = 0-0+0+\mathrm k\times 0 + \mathrm C = 1$

This is possible only when $\mathrm C = 1$

$\therefore \bf{C = 1}$

Now,

$\mathrm{f(1) = 1-1+1+k.1+C = 4 \\ \Rightarrow 1 + k + 1 = 4\\ \Rightarrow \bf{k = 2}}$

Now, substitute $\mathrm C = 1,\;\text{and}\; \mathrm k = 2$, we get:

$\mathrm {x^4-x^3+x^2+2x + 1}$

$\therefore \mathbf B$ is the correct option.

by Boss (19.2k points)
edited by

+1 vote