Solutions%20%2001%20-%20Mathematics%20-%20March%202006

Solutions%20%2001%20-%20Mathematics%20-%20March%202006 -...

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Time : 3 hours Solutions 01 - Mathematics - March 2006 Marks : 100 Pg - 1 Instructions : 1. Answer all questions. 2. Write your answers according to the instructions given below with the questions. 3. Begin each section on a new page. S E CT I O N - A Given below are 1 to 15 multiple choice questions. Each carries one mark. Write the serial number ( a or b or c or d ) in your answer book of the alternative which you feel is the correct answer of the question. 1. If the co-ordinates of a point ( 6, -1 ) changes to ( 8, - 4 ), then write the co-ordinates of the point where the origin is shifted. ( a ) ( 2, - 3 ) ( b ) ( -3, 2 ) ( c ) ( 3, - 2 ) ( d ) ( - 2, 3 ) Solution : ( x, y ) = ( 6, - 1 ) are the coordinates of the point before shifting of the origin. ( x’, y’ ) = (8, - 4 ) are the coordinates of the point after shifting of the origin. ( h, k ) are the coordinates where the origin is shifted. => h = x - x’ = 6 - 8 = - 2 and k = y - y’ = - 1 - ( - 4 ) = 3. 2. Write the measure of the angle between the lines x + y = 0 and y = [ π ] ( a ) 4 π ( b ) 3 π ( c ) 2 π ( d ) 0 Solution : Slope of the line x + y = 0 is m - 1. [ π ] is a horizontal line. Angle between the horizontal line and the line of slope m is tan - 1 l m l = tan - 1 l - 1 l = 4 π . 3. How many tangents to the circle x 2 + y 2 = 29 pass through the point ( 5, 2 ) ? ( a ) 0 ( b ) 1 ( c ) 2 ( d ) none of these Solution : The point ( 5, 2 ) satisfies the equation of the circle and hence lies on the circle. Only one tangent can be drawn at a point on the circle. 4. The standard equation of the parabola having vertex at the origin, passing through ( -1, 1 ) and symmetric about Y - axis is ( a ) y 2 = - x ( b ) x 2 = y ( c ) y 2 = x ( d ) x 2 = - y Solution : The standard equation of the parabola having vertex at the origin, and symmetric about Y - axis is x 2 = ky. If it passes through the point ( -1, 1 ), ( -1 ) 2 = k ( 1 ) => k = 1 => the required equation of the parabola is x 2 = y.
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Time : 3 hours Solutions 01 - Mathematics - March 2006 Marks : 100 Pg - 2 5. Write the equation of the auxiliary circle of 1 9 y 4 x 2 2 - = . ( a ) x 2 + y 2 = - 5 ( b ) x 2 + y 2 = 5 ( c ) x 2 + y 2 = 4 ( d ) x 2 + y 2 = 9 Solution : The equation of the auxiliary circle of the hyperbola 1 b y a x 2 2 2 2 = - is x 2 + y 2 = a 2 . Here, a 2 = 4 => the equation of the auxiliary circle of the given hyperbola is x 2 + y 2 = 4. 6. Obtain l cos θ cos α , cos θ sin α , sin θ l ( a ) - 1 ( b ) 0 ( c ) 1 ( d ) none of these Solution : l cos θ cos α , cos θ sin α , sin θ l = θ sin α sin θ cos α cos θ cos 2 2 2 2 2 + + = θ sin ) α sin α cos ( θ cos 2 2 2 2 + + = θ sin θ cos 2 2 + = 1. 7. Find the resultant force of ( 1, 2, - 1 ) and ( 1, - 2, 1 ) ( a ) ( 2, 0, 0 ) ( b ) ( - 1, 4, 2 ) ( c ) ( 2, 4, 2 ) ( d ) ( - 2, 0, 0 ) Solution : The resultant of ( x, y, z ) = ( 1, 2, - 1 ) and ( x’, y’, z’ ) = ( 1, - 2, 1 ) is ( x + x’, y + y’, z + z’ ) = ( 1 + 1, 2 - 2, - 1 + 1 ) = ( 2, 0, 0 ). 8. Find the centre of the sphere x 2 + y 2 + z 2 - 2x - 4y - 6z - 11 = 0. ( a ) ( - 1, - 2, - 3 ) ( b ) ( 3, 2, 1 ) ( c ) ( 1, 2, 3 ) ( d ) ( 1, 2, - 3 ) Solution : Comparing the given equation of the sphere x 2 + y 2 + z 2 - 2x - 4y - 6z - 11 = 0 with the general equation of the sphere x 2 + y 2 + z 2 + 2ux + 2vy + 2wz + c = 0, the centre of the sphere is ( - u, - v, - w ) = ( 1, 2, 3 ). 9. Find x 1 e lim - 3x 0 x ( a ) 3 ( b ) 3 1 ( c ) log e e ( d ) log e 3 Solution : x 1 e lim - 3x 0 x = 3x 1 e lim 3 - 3x 0 x = 3.
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This note was uploaded on 11/28/2011 for the course MATH 300 taught by Professor Jones during the Spring '06 term at ITT Tech Flint.

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Solutions%20%2001%20-%20Mathematics%20-%20March%202006 -...

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