1
IIT-JEE 2007
MCQ (Single Correct Answer)
+3
-0.75
STATEMENT-1: The curve $$y = {{ - {x^2}} \over 2} + x + 1$$ is symmetric with respect to the line $$x=1$$. because

STATEMENT-2: A parabola is symmetric about its axis.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
C
Statement-1 is True, Statement-2 is False
D
Statement-1 is False, Statement-2 is True.
2
IIT-JEE 2007
MCQ (Single Correct Answer)
+3
-0.75
$${{{d^2}x} \over {d{y^2}}}$$ equals
A
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}$$
B
$$ - {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
C
$$\left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 2}}$$
D
$$ - \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
3
IIT-JEE 2007
MCQ (Single Correct Answer)
+3
-0.75
Let $$\,\,\,$$$$f\left( x \right) = 2 + \cos x$$ for all real $$X$$.

STATEMENT - 1: for eachreal $$t$$, there exists a point $$c$$ in $$\left[ {t,t + \pi } \right]$$ such that $$f'\left( c \right) = 0$$ because
STATEMENT - 2: $$f\left( t \right) = f\left( {t + 2\pi } \right)$$ for each real $$t$$.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
C
Statement-1 is True, Statement-2 is False
D
Statement-1 is False, Statement-2 is True.
4
IIT-JEE 2007
MCQ (Single Correct Answer)
+3
-0.75
Let $$ABCD$$ be a quadrilateral with area $$18$$, with side $$AB$$ parallel to the side $$CD$$ and $$2AB=CD$$. Let $$AD$$ be perpendicular to $$AB$$ and $$CD$$. If a circle is drawn inside the quadrilateral $$ABCD$$ touching all the sides, then its radius is
A
$$3$$
B
$$2$$
C
$${3 \over 2}$$
D
$$1$$
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