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JEE Mathematics Solved Question Paper 2013

Multiple Choice Questions

Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A × B having 3 or more elements is

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The real number k for which the equation, 2x³ +3x +k = 0 has two distinct real roots in [0,1]

lies between 1 and 2
lies between 2 and 3
lies between -1 and 0
lies between -1 and 0

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The sum of first 20 terms of the sequence 0.7,0.77,0.777...... is

$7 over 81 space left parenthesis 179 minus 10 to the power of negative 20 end exponent right parenthesis$
$7 over 9 left parenthesis 99 minus 10 to the power of negative 20 end exponent right parenthesis$
$7 over 81 left parenthesis 179 plus 10 to the power of negative 20 end exponent right parenthesis$
$7 over 81 left parenthesis 179 plus 10 to the power of negative 20 end exponent right parenthesis$

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A ray of light along $straight x space plus square root of 3 straight y end root space equals space square root of 3$ get reflected upon reaching X -axis, the equation of the reflected ray is

$straight y equals space straight x plus square root of 3$
$square root of 3 straight y end root space equals space straight x minus square root of 3$
$straight y space equals space square root of 3 straight x end root minus square root of 3$
$straight y space equals space square root of 3 straight x end root minus square root of 3$

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The number of values of k, for which the system of equations
(k+1) x + 8y = 4k
kx + (k+3)y = 3k -1
has no solution, is

infinite
1
2
2

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6.
If the equations x² + 2x + 3 = 0 and ax² + bx + c = 0, a, b, c ∈ R, have a common root, then a : b : c is

1:2:3

3:2:1

1:3:2

1:3:2

1:2:3

Given equations are
x² +2x+3 =0 ... (i)
ax² +bx +c =0 .. (ii)
since, Eq(i) has imaginary roots, So eq (ii) will also have both roots same as eq (i)
thus,

$straight a over 1 space equals space straight b over 2 space equals space straight c over 3$
Hence, a:b:c is 1:2:3

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The circle passing through (1,-2) and touching the axis of x at (3,0) also passes through the point

(-5,2)
(2,-5)
(5,-2)
(5,-2)

181 Views

If x, y, z are in A.P. and tan⁻¹ x, tan⁻¹ y and tan⁻¹ z are also in A.P., then

x= y= z
2x =3y = 6z
6x = 3y= 2z
6x = 3y= 2z

207 Views

Consider :
Statement − I : (p ∧ ~ q) ∧ (~ p ∧ q) is a fallacy.
Statement − II : (p → q) ↔ (~ q → ~ p) is a tautology.

Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I
Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I
Statement -I is True; Statement -II is False.
Statement -I is True; Statement -II is False.

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