IIT-JEE 2008 Paper 2 Offline | Definite Integrals and Applications of Integrals Question 68 | Mathematics

1

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

+3

-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

A

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

B

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

C

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

D

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

2

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

+4

-1

Consider the function $$f:\left( { - \infty ,\infty } \right) \to \left( { - \infty ,\infty } \right)$$ defined by
$$f\left( x \right) = {{{x^2} - ax + 1} \over {{x^2} + ax + 1}},0 < a < 2.$$

Let $$g\left( x \right) = \int\limits_0^{{e^x}} {{{f'\left( t \right)} \over {1 + {t^2}}}} \,dt.$$

Which of the following is true?

A

$$g'(x)$$ is positive on $$\left( { - \infty ,0} \right)$$ and negative on $$\left( {0,\infty } \right)$$

B

$$g'(x)$$ is negative on $$\left( { - \infty ,0} \right)$$ and positive on $$\left( {0,\infty } \right)$$

C

$$g'(x)$$ changes sign on both $$\left( { - \infty ,0} \right)$$ and $$\left( {0,\infty } \right)$$

D

$$g'(x)$$ does not change sign on $$\left( { - \infty ,0} \right)$$

3

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

+3

-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 :

A

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

B

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

C

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

D

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

4

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

+3

-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 $$g(0)=0$$

$$\int\limits_{ - 1}^1 {g'\left( x \right)dx = } $$

A

$$2g(-1)$$

B

$$0$$

C

$$-2g(1)$$

D

$$2g(1)$$