IIT-JEE 2009 Paper 1 Offline | Application of Integration Question 24 | Mathematics

IIT-JEE 2009 Paper 1 Offline

MCQ (Single Correct Answer)

-1

Let $$f$$ be a non-negative function defined on the interval $$[0,1]$$.

If $$\int\limits_0^x {\sqrt {1 - {{(f'(t))}^2}dt} = \int\limits_0^x {f(t)dt,0 \le x \le 1} } $$, and $$f(0) = 0$$, then

$$f\left( {{1 \over 2}} \right) < {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) > {1 \over 3}$$

$$f\left( {{1 \over 2}} \right) > {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) > {1 \over 3}$$

$$f\left( {{1 \over 2}} \right) < {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) < {1 \over 3}$$

$$f\left( {{1 \over 2}} \right) > {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) < {1 \over 3}$$

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

The area of the region between the curves $$y = \sqrt {{{1 + \sin x} \over {\cos x}}} $$
and $$y = \sqrt {{{1 - \sin x} \over {\cos x}}} $$ bounded by the lines $$x=0$$ and $$x = {\pi \over 4}$$ is

$$\int\limits_0^{\sqrt 2 - 1} {{t \over {\left( {1 + {t^2}} \right)\sqrt {1 - {t^2}} }}dt} $$

$$\int\limits_0^{\sqrt 2 - 1} {{4t \over {\left( {1 + {t^2}} \right)\sqrt {1 - {t^2}} }}dt} $$

$$\int\limits_0^{\sqrt 2 + 1} {{4t \over {\left( {1 + {t^2}} \right)\sqrt {1 - {t^2}} }}dt} $$

$$\int\limits_0^{\sqrt 2 + 1} {{t \over {\left( {1 + {t^2}} \right)\sqrt {1 - {t^2}} }}dt} $$

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-1

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

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