IIT-JEE 2012 Paper 1 Offline | Limits, Continuity and Differentiability Question 39 | Mathematics

IIT-JEE 2012 Paper 1 Offline

MCQ (Single Correct Answer)

-1

Let $$f(x) = \left\{ {\matrix{ {{x^2}\left| {\cos {\pi \over x}} \right|,} & {x \ne 0} \cr {0,} & {x = 0} \cr } } \right.$$

x$$\in$$R, then f is

differentiable both at x = 0 and at x = 2.

differentiable at x = 0 but not differentiable at x = 2.

not differentiable at x = 0 but differentiable at x = 2.

differentiable neither at x = 0 nor at x = 2.

IIT-JEE 2011 Paper 2 Offline

MCQ (Single Correct Answer)

-1

If $$\mathop {\lim }\limits_{x \to 0} {[1 + x\ln (1 + {b^2})]^{1/x}} = 2b{\sin ^2}\theta $$, $$b > 0$$ and $$\theta \in ( - \pi ,\pi ]$$, then the value of $$\theta$$ is

$$ \pm {\pi \over 4}$$

$$ \pm {\pi \over 3}$$

$$ \pm {\pi \over 6}$$

$$ \pm {\pi \over 2}$$

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 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 $$\left( { - \infty ,\infty } \right)$$

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