1

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

+3

-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

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

2

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

+3

-1

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

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

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

Which of the following is true?

3

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

+3

-1

Let $$g(x) = {{{{(x - 1)}^n}} \over {\log {{\cos }^m}(x - 1)}};0 < x < 2,m$$ and $$n$$ are integers, $$m \ne 0,n > 0$$, and let $$p$$ be the left hand derivative of $$|x - 1|$$ at $$x = 1$$. If $$\mathop {\lim }\limits_{x \to {1^ + }} g(x) = p$$, then

Questions Asked from Limits, Continuity and Differentiability (MCQ (Single Correct Answer))

Number in Brackets after Paper Indicates No. of Questions

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