MCQ
$\int \frac{x d x}{(x-1)(x-2)}$ equals
- A$\log \left|\frac{(x-1)^{2}}{x-2}\right|+C$
- B$\log |(x-1)(x-2)|+C$
- C$\log \left| {{{\left( {\frac{{x - 1}}{{x - 2}}} \right)}^2}} \right| + C$
- ✓$\log \left|\frac{(x-2)^{2}}{x-1}\right|+C$
$x=A(x-2)+B(x-1)$ .........$(1)$
Equating the coefficients of $x$ and constant, we obtain
$A+B=1$ and $-2 A-B=0$
$A=-1$ and $B=2$
$\therefore \frac{x}{(x-1)(x-2)}=-\frac{1}{(x-1)}+\frac{2}{(x-2)}$
$\Rightarrow \int \frac{x}{(x-1)(x-2)} d x=\int\left\{\frac{-1}{(x-1)}+\frac{2}{(x-2)}\right\} d x$
$=-\log |x-1|+2 \log |x-2|+C$
$=\log \left|\frac{(x-2)^{2}}{x-1}\right|+C$
Hence, the correct Answer is $D$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Statement $-1 :$ $f\left( c \right) = \frac{1}{3}$ for some $c\; \in R$
Statement $-2 :$$0 < f\left( x \right) < \frac{1}{{2\sqrt 2 }}\;,\forall x\; \in R$
[Note: Here $z$ takes the values in the complex plane and $\operatorname{Im} z$ and $\operatorname{Re} z$ denote, respectively, the imaginary part and the real part of $z]$
| column-$I$ | column-$II$ |
| $(A)$ The set of points $z$ satisfying $|z-i| z||=|z+i| z||$ is contained in or equal to | $(p)$ an ellipse with eccentricity $\frac{4}{5}$ |
| $(B)$ The set of points $z$ satisfying $|z+4|+|z-4|=10$ is contained in or equal to | $(q)$ the set of points $z$ satisfying $\operatorname{Im} z=0$ |
| $(C)$ If $|\omega|=2$, then the set of points $z=\omega-1 / \omega$ is contained in or equal to | $(r)$ the set of points $z$ satisfying $|\operatorname{Im} z| \leq 1$ |
| $(D)$ If $|\omega|=1$, then the set of points $z=\omega+1 / \omega$ is contained in or equal to | $(s)$ the set of points $z$ satisfying $|\operatorname{Re} z| \leq 1$ |
| $(t)$ the set of points $z$ satisfying $|z| \leq 3$ |