IIT-JEE 2008 Paper 1 Offline | JEE Advanced Year Wise Previous Years Questions

IIT-JEE 2008 Paper 1 Offline

MCQ (More than One Correct Answer)

-2

Let $$P\left( {{x_1},{y_1}} \right)$$ and $$Q\left( {{x_2},{y_2}} \right),{y_1} < 0,{y_2} < 0,$$ be the end points of the latus rectum of the ellipse $${x^2} + 4{y^2} = 4.$$ The equations of parabolas with latus rectum $$PQ$$ are :

$${x^2} + 2\sqrt 3y = 3 + \sqrt 3 $$

$${x^2} - 2\sqrt 3y = 3 + \sqrt 3 $$

$${x^2} + 2\sqrt 3y = 3 - \sqrt 3 $$

$${x^2} - 2\sqrt 3 y = 3 - \sqrt 3 $$

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-1

A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation $$\sqrt 3 x\, + \,y\, - \,6 = 0$$ and the point D is $$\left( {{{3\,\sqrt 3 } \over 2},\,{3 \over 2}} \right)$$. Further, it is given that the origin and the centre of C are on the same side of the line PQ.

The equation of circle C is

$${\left( {x\, - 2\sqrt 3 \,} \right)^2} + {(y - 1)^2} = 1$$

$${\left( {x\, - 2\sqrt 3 \,} \right)^2} + {(y + {1 \over 2})^2} = 1$$

$${\left( {x\, - \sqrt 3 \,} \right)^2} + {(y + 1)^2} = 1$$

$${\left( {x\, - \sqrt 3 \,} \right)^2} + {(y - 1)^2} = 1$$

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-1

Let a and b be non-zero real numbers. Then, the equation

$$(a{x^2} + b{y^2} + c)({x^2} - 5xy + 6{y^2}) = 0$$ represents :

four straight lines, when c = 0 and a, b are of the same sign

two straight lines and a circle, when a = b, and c is of sign opposite to that of a

two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a

a circle and an ellipse, when a and b are of the same sign and c is of sign opposite to that of a

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-1

Let $$g(x) = {{{{(x - 1)}^n}} \over {\log {{\cos }^m}(x - 1)}};0 < x < 2,m$$ and $$n$$ are integers, $$m \ne 0,n > 0$$, and let $$p$$ be the left hand derivative of $$|x - 1|$$ at $$x = 1$$. If $$\mathop {\lim }\limits_{x \to {1^ + }} g(x) = p$$, then

$$n = 1,m = 1$$

$$n = 1,m = - 1$$

$$n = 2,m = 2$$

$$n > 2,m = n$$

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