IIT-JEE 2008 Paper 1 Offline | Application of Integration Question 28 | Mathematics

Consider the functions defined implicitly by the equation $$y^3-3y+x=0$$ on various intervals in the real line. If $$x\in(-\infty,-2)\cup(2,\infty)$$, the equation implicitly defines a unique real valued differentiable function $$y=f(x)$$. If $$x\in(-2,2)$$, the equation implicitly defines a unique real valued differentiable function $$y=g(x)$$ satisfying $$g(0)=0$$

The area of the region bounded by the curve $$y=f(x),$$ the
$$x$$-axis, and the lines $$x=a$$ and $$x=b$$, where $$ - \infty < a < b < - 2,$$ is :

$$\int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx + bf\left( b \right) - af\left( a \right)$$

$$ - \int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx + bf\left( b \right) - af\left( a \right)$$

$$\int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx - bf\left( b \right) + af\left( a \right)$$

$$ - \int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx - bf\left( b \right) + af\left( a \right)$$

IIT-JEE 2005 Screening

MCQ (Single Correct Answer)

-0.75

The area bounded by the parabola $$y = {\left( {x + 1} \right)^2}$$ and
$$y = {\left( {x - 1} \right)^2}$$ and the line $$y=1/4$$ is

$$4$$ sq. units

$$1/6$$ sq. units

$$4/3$$ sq. units

$$1/3$$ sq. units

IIT-JEE 2004 Screening

MCQ (Single Correct Answer)

-0.75

The area enclosed between the curves $$y = a{x^2}$$ and
$$x = a{y^2}\left( {a > 0} \right)$$ is $$1$$ sq. unit, then the value of $$a$$ is

$$1/\sqrt 3 $$

$$1/2$$

$$1$$

$$1/3$$

IIT-JEE 2003 Screening

MCQ (Single Correct Answer)

-0.75

The area bounded by the curves $$y = \sqrt x ,2y + 3 = x$$ and
$$x$$-axis in the 1^st quadrant is

$$9$$

$$27/4$$

$$36$$

$$18$$

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