1
JEE Advanced 2015 Paper 2 Offline
Numerical
+4
-0
The coefficient of $${x^9}$$ in the expansion of (1 + x) (1 + $${x^2)}$$ (1 + $${x^3}$$) ....$$(1 + {x^{100}})$$ is
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2
JEE Advanced 2015 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-1
Let $$S$$ be the set of all non-zero real numbers $$\alpha $$ such that the quadratic equation $$\alpha {x^2} - x + \alpha = 0$$ has two distinct real roots $${x_1}$$ and $${x_2}$$ satisfying the inequality $$\left| {{x_1} - {x_2}} \right| < 1.$$ Which of the following intervals is (are) $$a$$ subset(s) os $$S$$?
A
$$\left( { - {1 \over 2} - {1 \over {\sqrt 5 }}} \right)$$
B
$$\left( { - {1 \over {\sqrt 5 }},0} \right)$$
C
$$\left( {0,{1 \over {\sqrt 5 }}} \right)$$
D
$$\left( {{1 \over {\sqrt 5 }},{1 \over 2}} \right)$$
3
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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4
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 $${y_1} > 0$$

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