IIT-JEE 2011 Paper 2 Offline | Limits, Continuity and Differentiability Question 42 | Mathematics

IIT-JEE 2011 Paper 2 Offline

MCQ (More than One Correct Answer)

-1

If $$f(x) = \left\{ {\matrix{ { - x - {\pi \over 2},} & {x \le - {\pi \over 2}} \cr { - \cos x} & { - {\pi \over 2} < x \le 0} \cr {x - 1} & {0 < x \le 1} \cr {\ln x} & {x > 1} \cr } } \right.$$, then

f(x) is continuous at x = $$-$$ $$\pi$$/2.

f(x) is not differentiable at x = 0.

f(x) is differentiable at x = 1.

f(x) is differentiable at x = $$-$$3/2.

IIT-JEE 2009 Paper 1 Offline

MCQ (More than One Correct Answer)

-2

Let $$L = \mathop {\lim }\limits_{x \to 0} {{a - \sqrt {{a^2} - {x^2}} - {{{x^2}} \over 4}} \over {{x^4}}},a > 0$$. If L is finite, then

$$a = 2$$

$$a = 1$$

$$L = {1 \over {64}}$$

$$L = {1 \over {32}}$$

IIT-JEE 2008 Paper 1 Offline

MCQ (More than One Correct Answer)

-2

Let $$f(x)$$ be a non-constant twice differentiable function defined on $$\left( { - \infty ,\infty } \right)$$

such that $$f\left( x \right) = f\left( {1 - x} \right)$$ and $$f'\left( {{1 \over 4}} \right) = 0.$$ Then,

$$f''\left( x \right)$$ vanishes at least twice on $$\left[ {0,1} \right]$$

$$f'\left( {{1 \over 2}} \right) = 0$$

$$\int\limits_{ - 1/2}^{1/2} {f\left( {x + {1 \over 2}} \right)\sin x\,dx} = 0$$

$$\int\limits_0^{1/2} {f\left( t \right){e^{\sin \,\pi t}}dt = } \int\limits_{1/2}^1 {f\left( {1 - t} \right){e^{\sin \,\pi t}}dt} $$

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