The area bounded by the curves $$y = \ln x,y = \ln \left| x \right|,y = \left| {\ln {\mkern 1mu} x} \right|$$ and $$y = \left| {\ln \left| x \right|} \right|$$ is
A
$$4$$sq. units
B
$$6$$sq. units
C
$$10$$sq. units
D
none of these
Explanation
First we draw each curve as separate graph
NOTE : Graph of $$y = \left| {f\left( x \right)} \right|$$ can be obtained from the graph of the curve $$y = f\left( x \right)$$ by drawing the mirror image of the portion of the graph below $$x$$-axis, with respect to $$x$$-axis.
Clearly the bounded area is as shown in the following figure.
$$ = - 4\left[ {x\,\ln x + - x_0^1} \right] = 4\,\,$$ sq. units
2
AIEEE 2002
MCQ (Single Correct Answer)
If $$y=f(x)$$ makes +$$ve$$ intercept of $$2$$ and $$0$$ unit on $$x$$ and $$y$$ axes and encloses an area of $$3/4$$ square unit with the axes then $$\int\limits_0^2 {xf'\left( x \right)dx} $$ is
A
$$3/2$$
B
$$1$$
C
$$5/4$$
D
$$-3/4$$
Explanation
We have $$\int\limits_0^2 {f\left( x \right)} dx = {3 \over 4};Now,$$
$$\int\limits_0^2 {xf'\left( x \right)} dx$$
$$ = x\int\limits_0^2 {f'\left( x \right)dx} - \int\limits_0^2 {f\left( x \right)} dx$$