1
JEE Advanced 2015 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-1
Consider the hyperbola $$H:{x^2} - {y^2} = 1$$ and a circle $$S$$ with center $$N\left( {{x_2},0} \right)$$. Suppose that $$H$$ and $$S$$ touch each other at a point $$P\left( {{x_1},{y_1}} \right)$$ with $${{x_1} > 1}$$ and $${{y_1} > 0}$$. The common tangent to $$H$$ and $$S$$ at $$P$$ intersects the $$x$$-axis at point $$M$$. If $$(l, m)$$ is the centroid of the triangle $$PMN$$, then the correct expressions(s) is(are)
A
$${{dl} \over {d{x_1}}} = 1 - {1 \over {3x_1^2}}$$ for $${x_1} > 1$$
B
$${{dm} \over {d{x_1}}} = {{{x_1}} \over {3\left( {\sqrt {x_1^2 - 1} } \right)}}$$ for $${x_1} > 1$$
C
$${{dl} \over {d{x_1}}} = 1 + {1 \over {3x_1^2}}$$ for $${x_1} > 1$$
D
$${{dm} \over {d{y_1}}} = {1 \over 3}$$ for $${y_1} > 0$$
2
JEE Advanced 2015 Paper 2 Offline
Numerical
+4
-0
Suppose that the foci of the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 5} = 1$$ are $$\left( {{f_1},0} \right)$$ and $$\left( {{f_2},0} \right)$$ where $${{f_1} > 0}$$ and $${{f_2} < 0}$$. Let $${P_1}$$ and $${P_2}$$ be two parabolas with a common vertex at $$(0,0)$$ and with foci at $$\left( {{f_1},0} \right)$$ and $$\left( 2{{f_2},0} \right)$$, respectively. Let $${T_1}$$ be a tangent to $${P_1}$$ which passes through $$\left( 2{{f_2},0} \right)$$ and $${T_2}$$ be a tangent to $${P_2}$$ which passes through $$\left( {{f_1},0} \right)$$. If $${m_1}$$ is the slope of $${T_1}$$ and $${m_2}$$ is the slope of $${T_2}$$, then the value of $$\left( {{1 \over {m_1^2}} + m_2^2} \right)$$ is
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3
JEE Advanced 2015 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-1
If $$\alpha $$ $$ = 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right)$$ and $$\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right),$$ where the inverse trigonimetric functions take only the principal values, then the correct options(s) is (are)
A
$$cos\beta > 0$$
B
$$\sin \beta < 0$$
C
$$\cos \left( {\alpha + \beta } \right) > 0$$
D
$$\cos \alpha < 0$$
4
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 $$(0, 2)$$
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