JEE Advanced 2013 Paper 2 Offline | Application of Derivatives Question 19 | Mathematics

JEE Advanced 2013 Paper 2 Offline

MCQ (More than One Correct Answer)

-2

The function $$f(x) = 2\left| x \right| + \left| {x + 2} \right| - \left| {\left| {x + 2} \right| - 2\left| x \right|} \right|$$ has a local minimum or a local maximum at x =

$$-$$2

$${{ - 2} \over 3}$$

$${{ 2} \over 3}$$

JEE Advanced 2013 Paper 1 Offline

MCQ (More than One Correct Answer)

-1

A rectangular sheet of fixed perimeter with sides having their lengths in the ratio $$8:15$$ is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is $$100$$, the resulting box has maximum volume. Then the lengths of the vsides of the rectangular sheet are

$$24$$

$$32$$

$$45$$

$$60$$

IIT-JEE 2012 Paper 2 Offline

MCQ (More than One Correct Answer)

-1

If $$f\left( x \right) = \int_0^x {{e^{{t^2}}}} \left( {t - 2} \right)\left( {t - 3} \right)dt$$ for all $$x \in \left( {0,\infty } \right),$$ then

$$f$$ has a local maximum at $$x=2$$

$$f$$ is decreasing on $$(2, 3)$$

there exists some $$c \in \left( {0,\infty } \right),$$ such that $$f'(c)=0$$

$$f$$ has a local minimum at $$x=3$$

IIT-JEE 2009 Paper 2 Offline

MCQ (More than One Correct Answer)

-2

For the function $$$f\left( x \right) = x\cos \,{1 \over x},x \ge 1,$$$

for at least one $$x$$ in the interval $$\left[ {1,\infty } \right)$$, $$f\left( {x + 2} \right) - f\left( x \right) < 2$$

$$\mathop {\lim }\limits_{x \to \infty } f'\left( x \right) = 1$$

for all $$x$$ in the interval $$\left[ {1,\infty } \right)f\left( {x + 2} \right) - f\left( x \right) > 2$$

$$f'(x)$$ is strictly decreasing in the interval $$\left[ {1,\infty } \right)$$

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