MCQ
$\int_{-1}^1 \frac{|x-2|}{x-2} d x, x \neq 2$ is equal to
  • A
    1
  • B
    -1
  • C
    2
  • D
    -2

Answer

Let $I=\int_{-1}^1 \frac{|x-2|}{x-2} d x$
$
=\int_{-1}^1 \frac{-(x-2)}{x-2} d x=\int_{-1}^1-1 \cdot d x=[-x]_{-1}^1=-[1-(-1)]=-2
$

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

Similar questions

Let $'a'$ $(a < 0, a \notin  I)$ be a constant and $'t'$ be a parameter, then set of values of $'t'$ for which the function $f(x)=\left( \frac{\left| \left[ t \right]+1 \right|+a}{\left| \left[ t \right]+1 \right|+1-a} \right)x$ ,is a decreasing function of $x$ ([.] denote greatest integer function) is
If $A =\left[\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right]$ then $A ^4$ equals to :
$\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{ dx}=$
  1. $\log(3+4\cos^2\text{x})+\text{C}$
  2. $\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
  3. $-\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
  4. $\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
A function $y = f (x)$ satisfies the condition $f '(x) sin x + f (x) cos x = 1, \, f (x)$ being bounded when $x \rightarrow  0.$ If $I = \int\limits_0^{\frac{\pi }{2}} {{\rm{f}}\,(x)\,dx} $ then
If $f$ is a real-valued differentiable function satisfying $|f(x)-f(y)| \leq(x-y)^2, x, y \in R$ and $f(0)=0$, then $f(1)$ equals
If $a = 2i + 4j + 2k$ and $b = 8i - 3j + \lambda k$ and $a\, \bot \,b,$ then value of $\lambda $ will be
The solution of the differential equation $ydx - \left( {x + 2{y^2}} \right)dy = 0$ is $x\, = f(y)$. If $f(-1)\, = 1$, then $f(1)$ is equal to
The area bounded by the curve $y=\cos x$ between $x=0$ and $x=\frac{3 \pi}{2}$ is _________ sq. unit.
Let $f:[-1,1] \rightarrow R$ be defined as $f(x)=a x^{2}+b x+c$ for all $x \in[-1,1],$ where $a , b , c \in R$ such that $f (-1)=2, f ^{\prime}(-1)=1$ and for $x \in(-1,1)$ the maximum value of $f ^{\prime \prime}( x )$ is $\frac{1}{2} .$ If $f ( x ) \leq \alpha$ , $x \in[-1,1],$ then the least value of $\alpha$ is equal to ...... .
Which of the following differentials equation has $\text{y}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}$ as the general solution?
  1. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
  2. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{y}=0$
  3. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{1}=0$
  4. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{1}=0$