Download App

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.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from INTEGRALS→INTEGRALS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The order and degree of the differential equation $\sqrt {\frac{{dy}}{{dx}}} - 4\frac{{dy}}{{dx}} - 7x = 0$ are
→
Let $A = \{p, q, r\}$. Which of the following is an equivalence relation on $A$
→
If $ \displaystyle \begin{vmatrix}\text{a} &\text{amp; }\text{b} &\text{amp; 0}\\ 0 &\text{amp; a} &\text{amp; b}\\\text{b}&\text{amp; a}&\text{amp; 0}\end{vmatrix}=0,$ then the order is:
→
Choose the correct answer from the given four options.
If $|\vec{{\text{a}}}|=10,|\vec{{\text{b}}}|=2$ and $\vec{{\text{a}}}\cdot\vec{{\text{b}}}=12,$ then value of $|\vec{{\text{a}}}\times\vec{\text{b}}|$ is :
→
The order and degree of the differential equation $\frac{{{d^2}y}}{{d{x^2}}} = \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} $ is
→
Find the values of $a, b, c$ and $d$ respectively if $\left[\begin{array}{cc}2 a+b & a-2 b \\ 5 c-d & 4 c+3 d\end{array}\right]=\left[\begin{array}{cc}4 & -3 \\ 11 & 24\end{array}\right]$.
→
Suppose $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is a real matrix with non-zero entries, $\alpha d-b c=0$ and $A^2=A$. Then, $a + d$ equals
→
Let $f(x)$ be a polynomial function such that $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^{5}+64$. Then, the value of $\lim _{x \rightarrow 1} \frac{f(x)}{x-1}$
→
Let $f(x)$ be a differentiable function in $[0, 2], f(0) = 0$ and $f'(x) \le \frac{1}{2}\,\forall x \in \left[ {0,\,2} \right]$ . Then
→
If $y = 2sin^{-1} \sqrt {(1-x)} + sin^{-1} (2\sqrt{x(1-x)})$ for $0 < x < \frac{1}{2}$, then $\frac{dy}{dx}$ equals-
→