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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
Choose the correct answer in Exercises:If $\frac{\text{d}}{\text{dx}}\text{f}\text{(x)}=4\text{x}^3-\frac{3}{\text{x}^4}$such that $\text{f}(2)=0.$Then $\text{f}\text{(x)}$ is
  • ✓
    $\text{x}^4+\frac{1}{\text{x}^3}-\frac{129}{8}$
  • B
    $\text{x}^3+\frac{1}{\text{x}^4}+\frac{129}{8}$
  • C
    $\text{x}^4+\frac{1}{\text{x}^3}+\frac{129}{8}$
  • D
    $\text{x}^3+\frac{1}{\text{x}^4}-\frac{129}{8}$
✓

Answer

Correct option: A.
$\text{x}^4+\frac{1}{\text{x}^3}-\frac{129}{8}$
$\text{f(x)}=\int\bigg(4\text{x}^3-\frac{3}{\text{x}^4}\bigg)\text{ dx}$
$=4\int\text{x}^3\text{ dx}-3\int\frac{1}{\text{x}^4}\text{ dx} $
$=4.\frac{\text{x}^4}{4}-3\int\text{x}^{-4}\text{ dx} =\text{x}^4-3\frac{\text{x}^-3}{-3}+\text{c} $
$\Rightarrow\ \ \ \ \ \ \ \text{f(x)}=\text{x}^4+\frac{1}{\text{x}^3}+\text{c} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ....\text{(i)} $
$\Rightarrow\ \ \ \ \ \ \ \text{f(2)}=16+\frac{1}{8}+\text{c} \ \ \ \ \Rightarrow\ \ \ \ \ \ 0=\frac{128+1}{8}+\text{c} $
$\Rightarrow\ \ \ \ \ \ \ \text{c }+\frac{129}{8}=0 \ \ \ \ \Rightarrow\ \ \ \ \ \ \text{c }=\frac{-129}{8} $
Putting $\text{c}=\frac{-129}{8}$ in eq. (i),
$\text{f(x)}=\text{x}^4+\frac{1}{\text{x}^3}-\frac{129}{8} $
Therefore, option (A) is correct.

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