1
IIT-JEE 2008 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
Let a solution $$y=y(x)$$ of the differential equation,

$$x\sqrt {{x^2} - 1} \,\,dy - y\sqrt {{y^2} - 1} \,dx = 0$$ satify $$y\left( 2 \right) = {2 \over {\sqrt 3 }}.$$

STATEMENT-1 : $$y\left( x \right) = \sec \left( {{{\sec }^{ - 1}}x - {\pi \over 6}} \right)$$ and

STATEMENT-2 : $$y\left( x \right)$$ given by $${1 \over y} = {{2\sqrt 3 } \over x} - \sqrt {1 - {1 \over {{x^2}}}} $$

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 2008 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
Let two non-collinear unit vectors $$\widehat a$$ and $$\widehat b$$ form an acute angle. A point $$P$$ moves so that at any time $$t$$ the position vector $$\overrightarrow {OP} $$ (where $$O$$ is the origin) is given by $$\widehat a\cos t + \widehat b\sin t.$$ When $$P$$ is farthest from origin $$O,$$ let $$M$$ be the length of $$\overrightarrow {OP} $$ and $$\widehat u$$ be the unit vector along $$\overrightarrow {OP} $$. Then :
A
$$\widehat u = {{\widehat a + \widehat b} \over {\left| {\widehat a + \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + \widehat a.\,\widehat b} \right)^{1/2}}$$
B
$$\widehat u = {{\widehat a - \widehat b} \over {\left| {\widehat a - \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + \widehat a.\,\widehat b} \right)^{1/2}}$$
C
$$\widehat u = {{\widehat a + \widehat b} \over {\left| {\widehat a + \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + 2\widehat a.\,\widehat b} \right)^{1/2}}$$
D
$$\widehat u = {{\widehat a - \widehat b} \over {\left| {\widehat a - \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + 2\widehat a.\,\widehat b} \right)^{1/2}}$$
3
IIT-JEE 2008 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
Consider the lines

$${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y + 2} \over 2} = {{z - 3} \over 3}$$

The unit vector perpendicular to both $${L_1}$$ and $${L_2}$$ is :

A
$${{ - \widehat i + 7\widehat j + 7\widehat k} \over {\sqrt {99} }}$$
B
$${{ - \widehat i - 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}$$
C
$${{ - \widehat i + 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}$$
D
$${{7\widehat i - 7\widehat j - \widehat k} \over {\sqrt {99} }}$$
4
IIT-JEE 2008 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

Consider the lines,

$${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$

The distance of the point $$(1, 1, 1)$$ from the plane passing through the point $$(-1, -2, -1)$$ and whose normal is perpendicular to both the lines $${L_1}$$ and $${L_2}$$ is :
A
$${2 \over {\sqrt {75} }}$$
B
$${7 \over {\sqrt {75} }}$$
C
$${13 \over {\sqrt {75} }}$$
D
$${23 \over {\sqrt {75} }}$$
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