IIT-JEE 2008 Paper 2 Offline | JEE Advanced Year Wise Previous Years Questions

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IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Let $$a,\,b,c$$, $$p,q$$ be real numbers. Suppose $$\alpha ,\,\beta $$ are the roots of the equation $${x^2} + 2px + q = 0$$ and $$\alpha ,{1 \over \beta }$$ are the roots of the equation $$a{x^2} + 2bx + c = 0,$$ where $${\beta ^2} \in \left\{ { - 1,\,0,\,1} \right\}$$

STATEMENT - 1 : $$\left( {{p^2} - q} \right)\left( {{b^2} - ac} \right) \ge 0$$

and STATEMENT - 2 : $$b \ne pa$$ or $$c \ne qa$$

STATEMENT - 1 is True, STATEMENT - 2 is True;
STATEMENT - 2 is a correct explanation for
STATEMENT - 1

STATEMENT - 1 is True, STATEMENT - 2 is True;
STATEMENT - 2 is NOT a correct explanation for
STATEMENT - 1

STATEMENT - 1 is True, STATEMENT - 2 is False

STATEMENT - 1 is False, STATEMENT - 2 is True

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

A particle P stats from the point $${z_0}$$ = 1 +2i, where $$i = \sqrt { - 1} $$. It moves horizontally away from origin by 5 unit and then vertically away from origin by 3 units to reach a point $${z_1}$$. From $${z_1}$$ the particle moves $$\sqrt 2 $$ units in the direction of the vector $$\hat i + \hat j$$ and then it moves through an angle $${\pi \over 2}$$ in anticlockwise direction on a circle with centre at origin, to reach a point $${z_2}$$. The point $${z_2}$$ is given by

6 + 7i

-7 + 6i

7 + 6i

- 6 + 7i

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-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 :

$${2 \over {\sqrt {75} }}$$

$${7 \over {\sqrt {75} }}$$

$${13 \over {\sqrt {75} }}$$

$${23 \over {\sqrt {75} }}$$

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-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 shortest distance between $${L_1}$$ and $${L_2}$$ is :

$$0$$

$${17 \over {\sqrt 3 }}$$

$${41 \over {5\sqrt 3 }}$$

$${17 \over {5\sqrt 3 }}$$

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