1
JEE Advanced 2015 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-1
Let $$F:R \to R$$ be a thrice differentiable function. Suppose that
$$F\left( 1 \right) = 0,F\left( 3 \right) = - 4$$ and $$F'\left( x \right) < 0$$ for all $$x \in \left( {{1 \over 2},3} \right).$$ Let $$f\left( x \right) = xF\left( x \right)$$ for all $$x \in R.$$

The correct statement(s) is (are)

A
$$f'\left( 1 \right) < 0$$
B
$$f\left( 2 \right) < 0$$
C
$$f'\left( x \right) \ne 0$$ for any $$x \in \left( {1,3} \right)$$
D
$$f'\left( x \right) = 0$$ for some $$x \in \left( {1,3} \right)$$
2
JEE Advanced 2015 Paper 2 Offline
Numerical
+4
-0
Let $$f:R \to R$$ be a continuous odd function, which vanishes exactly at one point and $$f\left( 1 \right) = {1 \over {2.}}$$ Suppose that $$F\left( x \right) = \int\limits_{ - 1}^x {f\left( t \right)dt} $$ for all $$x \in \,\,\left[ { - 1,2} \right]$$ and $$G(x)=$$ $$\int\limits_{ - 1}^x {t\left| {f\left( {f\left( t \right)} \right)} \right|} dt$$ for all $$x \in \,\,\left[ { - 1,2} \right].$$ If $$\mathop {\lim }\limits_{x \to 1} {{F\left( x \right)} \over {G\left( x \right)}} = {1 \over {14}},$$ then the value of $$f\left( {{1 \over 2}} \right)$$ is
Your input ____
3
JEE Advanced 2015 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-1
Let $$f, g :$$ $$\left[ { - 1,2} \right] \to R$$ be continuous functions which are twice differentiable on the interval $$(-1, 2)$$. Let the values of f and g at the points $$-1, 0$$ and $$2$$ be as given in the following table:
X = -1 X = 0 X = 2
f(x) 3 6 0
g(x) 0 1 -1

In each of the intervals $$(-1, 0)$$ and $$(0, 2)$$ the function $$(f-3g)''$$ never vanishes. Then the correct statement(s) is (are)

A
$$f'\left( x \right) - 3g'\left( x \right) = 0$$ has exactly three solutions in $$\left( { - 1,0} \right) \cup \left( {0,2} \right)$$
B
$$f'\left( x \right) - 3g'\left( x \right) = 0$$ has exactly one solution in $$(-1, 0)$$
C
$$f'\left( x \right) - 3g'\left( x \right) = 0$$ has exactly one solution in $$(0, 2)$$
D
$$f'\left( x \right) - 3g'\left( x \right) = 0$$ has exactly two solutions in $$(-1, 0)$$ and exactly two solutions in $$(0, 2)$$
4
JEE Advanced 2015 Paper 2 Offline
Numerical
+3
-1
Let m and n be two positive integers greater than 1. If $$$\mathop {\lim }\limits_{\alpha \to 0} \left( {{{{e^{\cos \left( {{\alpha ^n}} \right)}} - e} \over {{\alpha ^m}}}} \right) = - \left( {{e \over 2}} \right)$$$ then the value of $${m \over n}$$ is _________.
Your input ____
JEE Advanced Papers
EXAM MAP