IIT-JEE 2008 Paper 2 Offline | JEE Advanced Year Wise Previous Years Questions

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-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?

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

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

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

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

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Consider the function $$f:\left( { - \infty ,\infty } \right) \to \left( { - \infty ,\infty } \right)$$ defined by

$$f\left( x \right) = {{{x^2} - ax + 1} \over {{x^2} + ax + 1}},0 < a < 2.$$

Which of the following is true?

$${\left( {2 + a} \right)^2}f''\left( 1 \right) + {\left( {2 - a} \right)^2}f''\left( { - 1} \right) = 0$$

$${\left( {2 - a} \right)^2}f''\left( 1 \right) - {\left( {2 + a} \right)^2}f''\left( { - 1} \right) = 0$$

$$f'\left( 1 \right)f'\left( { - 1} \right) = {\left( {2 - a} \right)^2}$$

$$f'\left( 1 \right)f'\left( { - 1} \right) = -{\left( {2 + a} \right)^2}$$

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 2 Offline

MCQ (Single Correct Answer)

-1

Let $$I = \int {{{{e^x}} \over {{e^{4x}} + {e^{2x}} + 1}}dx,\,\,J = \int {{{{e^{ - x}}} \over {{e^{ - 4x}} + {e^{ - 2x}} + 1}}dx.} } $$ Then

for an arbitrary constant $$C$$, the value of $$J -I$$ equals :

$${1 \over 2}\log \left( {{{{e^{4x}} - {e^{2x}} + 1} \over {{e^{4x}} + {e^{2x}} + 1}}} \right) + C$$

$${1 \over 2}\log \left( {{{{e^{2x}} + {e^x} + 1} \over {{e^{2x}} - {e^x} + 1}}} \right) + C$$

$${1 \over 2}\log \left( {{{{e^{2x}} - {e^x} + 1} \over {{e^{2x}} + {e^x} + 1}}} \right) + C$$

$${1 \over 2}\log \left( {{{{e^{4x}} + {e^{2x}} + 1} \over {{e^{4x}} - {e^{2x}} + 1}}} \right) + C$$

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