1well Q2 is exjactly a jee2007 q except that it has that area also asked
u just calculate the area enclosed... (without any condition)
Q1. Find the least integral value of a such that
\sqrt{9-a^2+2ax-x^2} > \sqrt{16-x^2} for at least one positive x
(ans given 6 while i got 2)
(integer type)
Q2. matrix match
Let (x,y) be such that sin-1(ax) + cos-1(y) + cos-1(bxy) = \pi /2
\begin{matrix} COLUMN\; \; 1 & COLUMN\;\; 2\\\ \textup{(A) If a=1 and b=0, then (x,y)} & \textup{1. lies on the circle}\; x^2+y^2=1 \\ \textup{(B) If a=1 and b=1, then (x,y)} & \textup{2. lies on}\; (x^2-1)(y^2-1)=0\\ \textup{(C) If a=1 and b=2, then (x,y)} & \textup{3. lies on}\; 4x^2+y^2=1\\ \textup{(D) If a=2 and b=0, then (x,y)} & \textup{4. lies on the curve having bounds area }\; \pi /2 \\ & \textup{5. lies on curve having bounds area}\; \pi \end{matrix}
I have got the equations to the curves as
A-1
B-2
C-1
D-3
But to find bound area do we just calculate the area enclosed... (without any condition) or we use the conditions given also i,e,sin-1(ax) + cos-1(y) + cos-1(bxy) = \pi /2
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6 Answers
4@Asish : If x ε R
Then both values of a r wrong
If a=2
√5 + 4x -x2 > √16-x2
Put x =3
=> √3 > √7 ----> x wrong
If a=6
√-27 +12x - x2 > √16-x2
Put x=3
0 > √7 ---> x wrong
1@ ashish : The answer to the first one is very clearly two. I dont know why you need to be worrying about this test now.
1
106@che: Why without any condition??
yeah i know it was JEE 07 Q
1well conditions were jus to find the equation of the curves....
for ex for first option lies on curve x2+y2=1....area enclosed is pie