_ ^ dx 2x - x^2 = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{\sqrt {2x - {x^2}} }} = } $

A
${\cos ^{ - 1}}(x - 1) + c$
✓
${\sin ^{ - 1}}(x - 1) + c$
C
${\cos ^{ - 1}}(1 + x) + c$
D
${\sin ^{ - 1}}(1 - x) + c$

✓

Answer

Correct option: B.

${\sin ^{ - 1}}(x - 1) + c$

b
(b)$\int_{}^{} {\frac{{dx}}{{\sqrt {2x - {x^2}} }}} = \int_{}^{} {\frac{{dx}}{{\sqrt {1 - {{(x - 1)}^2}} }}} = {\sin ^{ - 1}}(x - 1) + c.$

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 - 7.1 inderfinite integral→STD 12 - 7.1 inderfinite integral — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

$\int {\frac{{2\left( {{x^3} - 1} \right)}}{{x\left( {2{x^3} + 1} \right)}}} \,dx$ is equal to (where $C$ is the constant of integration)

The value of $ \cos^{-1}\left (\cot \left (\dfrac {\pi}{2}\right )\right ) + \cos^{-1} \left (\sin \left (\dfrac {2\pi}{3}\right )\right )$ is:

The triangle of maximum area that can be inscribed in a given circle of radius $'r'$ is ...... .

Let $f: R \rightarrow R$ be defined by $f(x)=x+|x|$. Then $f(x)$ is

If the co-ordinates of $A$ and $B$ be $(1, 2, 3)$ and $(7, 8, 7)$, then the projections of the line segment $AB$ on the co-ordinate axes are

The degree of the differntial equation $\left\{5+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}\right\}^{\frac{5}{3}}=\text{x}^{5}\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)$ is:

The probability distribution of a discrete random variable $X$ is given below :

$X$	2	3	4	5
$P(X)$	$\frac{5}{k}$	$\frac{7}{k}$	$\frac{9}{k}$	$\frac{11}{k}$

The value of $k$ is

Find the inverse of each of the matrices (if it exists). $\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos a & \sin a \\ 0 & \sin a & -\cos a\end{array}\right]$

On the set $Q^+$ of all positive rational numbers a binary operation $*$ is defined by $\text{a}^*\text{b}=\frac{\text{ab}}2\forall\text{ a, b}\in \text{Q}^+$. The inverse of 8 is:

The value of the integral $\int_0^{\frac{1}{2}} \frac{1+\sqrt{3}}{\left((x+1)^2(1-x)^6\right)^{\frac{1}{4}}} d x$ is . . . . . . . .