Download App

STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
$\mathop {Limit}\limits_{x\,\, \to \,\,{x_1}} \,\,\frac{x}{{x\,\, - \,\,{x_1}}}\,\,\,\int\limits_{{x_1}}^x {\,f(t)} \, dt$ is equal to :
  • A
    $\frac{{f\,\,\left( {{x_1}} \right)}}{{{x_1}}}$
  • ✓
    $x_1 f (x_1)$
  • C
    $f (x_1)$
  • D
    does not exist
✓

Answer

Correct option: B.
$x_1 f (x_1)$
b
$\mathop {Limit}\limits_{x \to {x_1}} \,\,\,\frac{{\int\limits_{{x_1}}^x {f(t)} \,\,dt}}{{\left( {\frac{{x - {x_1}}}{x}} \right)}}$ $= \mathop {Limit}\limits_{x \to {x_1}} \,\,\,\frac{{f(x)\,.\,{x^2}}}{{{x_1}}}\,$   (using Lopital’s rule)

$= x_1\, f (x_1) \Rightarrow B$

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 STD 12 - 7.2 definite integral→STD 12 - 7.2 definite integral — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Two cards are drawn from a well shuffled deck of $52$ playing cards with replacement. The probability that both cards are queen is
→
Let $\vec{a}=6 \hat{i}+9 \hat{j}+12 \hat{k}, \vec{b}=\alpha \hat{i}+11 \hat{j}-2 \hat{k}$ and $\vec{c}$ be vectors such that $\vec{a} \times \vec{c}=\vec{a} \times \vec{b}$. If $\vec{a} \cdot \vec{c}=-12$, $\vec{c} .(\hat{i}-2 \hat{j}+\hat{k})=5$, then $\vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})$ is equal to $.............$.
→
$\int_{}^{} {\frac{{dx}}{{{e^{ - 2x}}{{({e^{2x}} + 1)}^2}}} = } $
→
A disease affects two-thirds of the population of a country. A test for the disease gives the correct outcome with probability $\frac{2}{3}$. A person $X$ tested positive for the disease. The probability that $X$ has disease is
→
If $R = \{ (x,\,y)|x,\,y \in Z,\,{x^2} + {y^2} \le 4\} $ is a relation in $Z$, then domain of $R$ is
→
Let $A$ be a $3 \times 3$ real matrix such that $\mathrm{A}\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)=2\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right), \mathrm{A}\left(\begin{array}{l}-1 \\ 0 \\ 1\end{array}\right)=4\left(\begin{array}{l}-1 \\ 0 \\ 1\end{array}\right), \mathrm{A}\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)=2\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)$.  Then, the system $(A-3 I)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$ has
→
The length and foot of the perpendicular from the point $(2, -1, 5)$ to the line $\frac{{x - 11}}{{10}} = \frac{{y + 2}}{{ - 4}} = \frac{{z + 8}}{{ - 11}}$ are
→
Area bounded by the curve $\text{y}=\cos\text{x}$ between $\text{x}=0$ and $\text{x}=3\frac{\pi}{2}$ is :
→
The value of aa for which the area between the curves $y^2=4 a x$ and $x^2=4 a y$ is $1\ sq$. unit, is:
→
Let $f : R \rightarrow R$ be a differentiable function such that $f \left(\frac{\pi}{4}\right)=\sqrt{2}, f \left(\frac{\pi}{2}\right)=0$ and $f ^{\prime}\left(\frac{\pi}{2}\right)=1$ and let $g(x)=\int\limits_{x}^{\pi / 4}\left(f^{\prime}(t) \sec t+\tan t \operatorname{sectf}(t)\right) d t$ for $x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right)$. Then $\lim\limits _{ x \rightarrow\left(\frac{\pi}{2}\right)^{-}} g ( x )$ is equal to
→