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

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 :

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

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

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

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

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Let the function $$g:\left( { - \infty ,\infty } \right) \to \left( { - {\pi \over 2},{\pi \over 2}} \right)$$ be given by

$$g\left( u \right) = 2{\tan ^{ - 1}}\left( {{e^u}} \right) - {\pi \over 2}.$$ Then, $$g$$ is

even and is strictly increasing in $$\left( {0,\infty } \right)$$

odd and is strictly decreasing in $$\left( { - \infty ,\infty } \right)$$

odd and is strictly increasing in $$\left( { - \infty ,\infty } \right)$$

neither even nor odd, but is strictly increasing in $$\left( { - \infty ,\infty } \right)$$

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Let $$g(x) = \log f(x)$$, where $$f(x)$$ is a twice differentiable positive function on (0, $$\infty$$) such that $$f(x + 1) = xf(x)$$. Then for N = 1, 2, 3, ..., $$g''\left( {N + {1 \over 2}} \right) - g''\left( {{1 \over 2}} \right) = $$

$$ - 4\left\{ {1 + {1 \over 9} + {1 \over {25}} + ....... + {1 \over {{{\left( {2N - 1} \right)}^2}}}} \right\}$$

$$4\left\{ {1 + {1 \over 9} + {1 \over {25}} + ....... + {1 \over {{{\left( {2N - 1} \right)}^2}}}} \right\}$$

$$ - 4\left\{ {1 + {1 \over 9} + {1 \over {25}} + ....... + {1 \over {{{\left( {2N + 1} \right)}^2}}}} \right\}$$

$$4\left\{ {1 + {1 \over 9} + {1 \over {25}} + ....... + {1 \over {{{\left( {2N + 1} \right)}^2}}}} \right\}$$

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