1
IIT-JEE 2010 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-1
For $$r = 0,\,1,....,$$ let $${A_r},\,{B_r}$$ and $${C_r}$$ denote, respectively, the coefficient of $${X^r}$$ in the expansions of $${\left( {1 + x} \right)^{10}},$$ $${\left( {1 + x} \right)^{20}}$$ and $${\left( {1 + x} \right)^{30}}.$$
Then $$\sum\limits_{r = 1}^{10} {{A_r}\left( {{B_{10}}{B_r} - {C_{10}}{A_r}} \right)} $$ is equal to
A
$$\left( {{B_{10}} - {C_{10}}} \right)$$
B
$${A_{10}}\left( {{B^2}_{10}{C_{10}}{A_{10}}} \right)$$
C
$$0$$
D
$${{C_{10}} - {B_{10}}}$$
2
IIT-JEE 2010 Paper 2 Offline
Numerical
+4
-0
Let $${a_1},\,{a_{2\,}},\,{a_3}$$......,$${a_{11}}$$ be real numbers satisfying $${a_1} = 15,27 - 2{a_2} > 0\,\,and\,\,{a_k} = 2{a_{k - 1}} - {a_{k - 2}}\,\,for\,k = 3,4,........11$$. if $$\,\,\,{{a_1^2 + a_2^2 + a_{11}^2} \over {11}} = 90$$, then the value of $${{{a_1} + {a_2} + .... + {a_{11}}} \over {11}}$$ is equal to
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3
IIT-JEE 2010 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-1

Tangents are drawn from the point $$P(3, 4)$$ to the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 4} = 1$$ touching the ellipse at points $$A$$ and $$B$$.

The coordinates of $$A$$ and $$B$$ are

A
$$(3,0)$$ and $$(0,2)$$
B
$$\left( { - {8 \over 5},{{2\sqrt {161} } \over {15}}} \right)$$ and $$\left( { - {9 \over 5},{8 \over 5}} \right)$$
C
$$\left( { - {8 \over 5},{{2\sqrt {161} } \over {15}}} \right)$$ and $$(0,2)$$
D
$$(3,0)$$ and $$\left( { - {9 \over 5},{8 \over 5}} \right)$$
4
IIT-JEE 2010 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-1

Tangents are drawn from the point $$P(3, 4)$$ to the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 4} = 1$$ touching the ellipse at points $$A$$ and $$B$$.

The equation of the locus of the point whose distances from the point $$P$$ and the line $$AB$$ are equal, is

A
$$9{x^2} + {y^2} - 6xy - 54x - 62y + 241 = 0$$
B
$${x^2} + 9{y^2} + 6xy - 54x + 62y - 241 = 0$$
C
$$9{x^2} + 9{y^2} - 6xy - 54x - 62y - 241 = 0$$
D
$${x^2} + {y^2} - 2xy + 27x + 31y - 120 = 0$$
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