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 Multiple Choice QuestionsMultiple Choice Questions

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11.

A line is drawn through the point (1, 2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is

  • -1/4

  • -4

  • -2

  • -2


C.

-2

The slope of line PQ 


Let 'm' be the slope of the line PQ, then the equation of PQ is
y -2 = m (x-1)

Now, PQ meets X-axis at P open parentheses 1 minus 2 over straight m comma space 0 close parentheses and y-axis at Q (0,2-m)
OP space equals space 1 minus 2 over straight m space and space OQ space equals space 2 minus straight m
Also comma space area space space of space increment OPQ space equals space 1 half space left parenthesis OP right parenthesis left parenthesis OQ right parenthesis
space equals space 1 half open vertical bar open parentheses 1 minus 2 over straight m close parentheses left parenthesis 2 minus straight m right parenthesis close vertical bar
equals space 1 half open vertical bar 2 minus straight m minus 4 over straight m plus 2 close vertical bar
equals space 1 half open vertical bar 4 minus open parentheses straight m plus 4 over straight m close parentheses close vertical bar
Let space straight f left parenthesis straight m right parenthesis space equals space 4 minus open parentheses straight m plus 4 over straight m close parentheses
rightwards double arrow space straight f apostrophe left parenthesis straight m right parenthesis space equals space minus space 1 plus 4 over straight m squared
Now, f'(m) = 0 
m = ± 2
f(2) =0
f(-2) = 8
Since, the area cannot be zero, hence the required value of m is -2

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12.

Three distinct points A, B and C are given in the 2 – dimensional coordinate plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point ( - 1, 0) is equal to 1/3 . Then the circumcentre of the triangle ABC is at the point

  • (0,0)

  • (5/4, 0)

  • (5/2, 0)

  • (5/2, 0)

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13.

The point diametrically opposite to the point P (1, 0) on the circle x2+ y2 + 2x + 4y − 3 = 0 is 

  • (-3,4)

  • (-4,3)

  • (-3,-4)

  • (-3,-4)

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14.

The perpendicular bisector of the line segment joining P (1, 4) and Q (k, 3) has y−intercept − 4. Then a possible value of k is

  • 1

  • 2

  • -2

  • -2

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15.

If |z + 4| ≤ 3, then the maximum value of |z + 1| is

  • 4

  • 10

  • 6

  • 6

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16.

A body falling from rest under gravity passes a certain point P. It was at a distance of 400 m from P, 4s prior to passing through P. If g = 10 m/s2 , then the height above the point P from where the body began to fall is

  • 720 m

  • 900 m

  • 320 m

  • 320 m

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17.

A straight line through the point A(3, 4) is such that its intercept between the axes is bisected at A. Its equation is

  • x + y = 7

  • 3x − 4y + 7 = 0

  • 4x + 3y = 24

  • 4x + 3y = 24

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18.

The two lines x = ay + b, z = cy + d; and x = a′y + b′, z = c′y + d′ are perpendicular to each other if

  • aa′ + cc′ = −1

  • aa′ + cc′ = 1

  • fraction numerator straight a over denominator straight a apostrophe end fraction space plus fraction numerator straight c over denominator straight c apostrophe end fraction space equals 1
  • fraction numerator straight a over denominator straight a apostrophe end fraction space plus fraction numerator straight c over denominator straight c apostrophe end fraction space equals 1
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19.

The system of equations 
αx + y + z = α - 1, 
x + αy + z = α - 1, 
x + y + αz = α - 1 

has no solution, if α is

  • -2

  • either-2 or 1

  • not -2

  • not -2

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20.

Two points A and B move from rest along a straight line with constant acceleration f and f′ respectively. If A takes m sec. more than B and describes ‘n’ units more than B in acquiring the same speed then

  • (f - f′)m2 = ff′n

  • (f + f′)m2 = ff′n

  • 1/2(f - f′)m = ff′n2

  • 1/2(f - f′)m = ff′n2

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