1
IIT-JEE 1980
MCQ (Single Correct Answer)
+1
-0.25
Two circles $${x^2} + {y^2} = 6$$ and $${x^2} + {y^2} - 6x + 8 = 0$$ are given. Then the equation of the circle through their points of intersection and the point (1, 1) is
A
$${x^2} + {y^2} - 6x + 4 = 0$$
B
$${x^2} + {y^2} - 3x + 1 = 0$$
C
$${x^2} + {y^2} - 4y + 2 = 0$$
D
none of these
2
IIT-JEE 1980
Subjective
+4
-0
Given $$y = {{5x} \over {3\sqrt {{{\left( {1 - x} \right)}^2}} }} + {\cos ^2}\left( {2x + 1} \right)$$; Find $${{dy} \over {dx}}$$.
3
IIT-JEE 1980
Fill in the Blanks
+2
-0
$$ABC$$ is a triangle with $$\angle B$$ greater than $$\angle C.\,D$$ and $$E$$ are points on $$BC$$ such that $$AD$$ is perpendicular to $$BC$$ and $$AE$$ is the bisector of angle $$A$$. Complete the relation $$$\angle DAE = {1 \over 2}\left[ {\left( {} \right) - \angle C} \right]$$$
4
IIT-JEE 1980
Fill in the Blanks
+2
-0
$$ABC$$ is a triangle, $$P$$ is a point on $$AB$$, and $$Q$$ is point on $$AC$$ such that $$\angle AQP = \angle ABC$$. Complete the relation $$${{area\,\,of\,\,\Delta APQ} \over {area\,\,of\,\,\Delta ABC}} = {{\left( {...} \right)} \over {A{C^2}}}$$$
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