1
IIT-JEE 2012 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-1
If $$f\left( x \right) = \int_0^x {{e^{{t^2}}}} \left( {t - 2} \right)\left( {t - 3} \right)dt$$ for all $$x \in \left( {0,\infty } \right),$$ then
A
$$f$$ has a local maximum at $$x=2$$
B
$$f$$ is decreasing on $$(2, 3)$$
C
there exists some $$c \in \left( {0,\infty } \right),$$ such that $$f'(c)=0$$
D
$$f$$ has a local minimum at $$x=3$$
2
IIT-JEE 2012 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-1
Let $$PQR$$ be a triangle of area $$\Delta $$ with $$a=2$$, $$b = {7 \over 2}$$ and $$c = {5 \over 2}$$; where $$a, b,$$ and $$c$$ are the lengths of the sides of the triangle opposite to the angles at $$P.Q$$ and $$R$$ respectively. Then $${{2\sin P - \sin 2P} \over {2\sin P + \sin 2P}}$$ equals.
A
$${3 \over {4\Delta }}$$
B
$${45 \over {4\Delta }}$$
C
$${\left( {{3 \over {4\Delta }}} \right)^2}$$
D
$${\left( {{45 \over {4\Delta }}} \right)^2}$$
3
IIT-JEE 2012 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-1
A tangent PT is drawn to the circle $${x^2}\, + {y^2} = 4$$ at the point P $$\left( {\sqrt 3 ,1} \right)$$. A straight line L, perpendicular to PT is a tangent to the circle $${(x - 3)^2}$$ + $${y^2}$$ = 1

A common tangent of the two circles is

A
x = 4
B
y = 2
C
$${x + \sqrt 3 \,y = 4}$$
D
$${x +2 \sqrt 2 \,y = 6}$$
4
IIT-JEE 2012 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-1
A tangent PT is drawn to the circle $${x^2}\, + {y^2} = 4$$ at the point P $$\left( {\sqrt 3 ,1} \right)$$. A straight line L, perpendicular to PT is a tangent to the circle $${(x - 3)^2}$$ + $${y^2}$$ = 1.

A possible equation of L is

A
$${x - \sqrt 3 \,y = 1}$$
B
$${x + \sqrt 3 \,y = 1}$$
C
$${x - \sqrt 3 \,y = -1}$$
D
$${x + \sqrt 3 \,y = 5}$$
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