IIT-JEE 2008 Paper 2 Offline | Limits, Continuity and Differentiability Question 40 | Mathematics

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

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?

$$f(x)$$ is decreasing on $$(-1,1)$$ and has a local minimum at $$x=1$$

$$f(x)$$ is increasing on $$(-1,1)$$ and has a local minimum at $$x=1$$

$$f(x)$$ is increasing on $$(-1,1)$$ but has neither a local maximum nor a local minimum at $$x=1$$

$$f(x)$$ is decreasing on $$(-1,1)$$ but has neither a local maximum nor a local minimum at $$x=1$$

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-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

$$n = 1,m = 1$$

$$n = 1,m = - 1$$

$$n = 2,m = 2$$

$$n > 2,m = n$$