Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

JEE Mathematics Solved Question Paper 2014

Multiple Choice Questions

11.

Let the equation of an ellipse be $\frac{x^{2}}{144} + \frac{y^{2}}{25} = 1$ . Then, the radius of the circle with centre (0, $\sqrt{2}$ ) and passing through the foci of the ellipse is

12.

The straight lines x + y = 0, 5x + y = 4 and x + 5y = 4 form

an isosceles triangle
an equilateral triangle
a scalene triangle
a right angled triangle

13.

The value of $λ$ for which the curve (7x + 5)² + (7y + 3)² = $λ^{2}$ (4x + 3y - 24)²represents a parabola is

$\pm \frac{6}{5}$
$\pm \frac{7}{5}$
$\pm \frac{1}{5}$
$\pm \frac{2}{5}$

14.

Let f(x) = x + 1/2. Then, the number of real values of x for which the three unequal terms f(x), f(2x), f(4x) are in HP is

15.

Let f(x) = 2x²+ 5x + 1. If we write f(x) as f(x) = a(x + 1)(x - 2) + b(x - 2)(x - 1) + c(x - 1)(x + 1) for real numbers a, b, c then

there are infinite number of choices for a, b, c
only one choice for a but infinite number of choices for b and c
exactly one choice for each of a, b, c
more than one but finite number of choices for a, b, c

16.

If $α, β$ are the roots of ax² + bx + c = 0 ( $a \neq 0$ ) and $α + h, β + h$ are the roots of px² + qx + r = 0 (p $\neq$ 0), then the ratio of the squares of their discriminants is

a² : p²
a : p²
a² : p
a : 2p

17.
The equation of the common tangent with positive slope to the parabola y² = 8 $\sqrt{3}$ x and the hyperbola 4x² - y² = 4 is

$y = \sqrt{6} x + \sqrt{2}$

$y = \sqrt{6} x - \sqrt{2}$

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

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

$y = \sqrt{6} x + \sqrt{2}$

Equation of tangent in slope form of parabola $y^{2} = 8 \sqrt{3} x$ is,

$y = mx + c . . . (i) where, c = \frac{a}{m} ∴ c = \frac{2 \sqrt{3}}{m} . . . (ii)$

Also, tangent to the hyperbola 4x² - y² = 4 is,

c² = a²m² - b²

c² = 1m² - 4 $[∵ from Eq . (ii)]$

$\Rightarrow {(\frac{2 \sqrt{3}}{m})}^{2} = m^{2} - 4 [∵ from Eq . (ii)] \Rightarrow \frac{12}{m^{2}} = m^{2} - 4$

$\Rightarrow m^{4} - 4 m^{2} - 12 = 0 \Rightarrow m^{2} (m^{2} - 6) + 2 (m^{2} - 6) = 0 \Rightarrow (m^{2} + 2) (m^{2} - 6) = 0 \Rightarrow m^{2} - 6 = 0 and m^{2} + 2 \neq 0 \Rightarrow m^{2} = 6 \Rightarrow m = \pm \sqrt{6}$

i.e. m = $\sqrt{6}$ is positive slope.

$∴$ From Eq. (i),

$y = \sqrt{6} x + \frac{2 \sqrt{3}}{\sqrt{6}} \Rightarrow y = \sqrt{6} x + \sqrt{2}$

18.

The point on the parabola y² = 64x which is nearest to the line 4x + 3y + 35 = 0 has coordinates

(9, - 24)
(1, 81)
(4, - 16)
(- 9, - 24)

19.

Let z₁, z₂ be two fixed complex numbers in the argand plane and z be an arbitrary point satisfying $|z - z_{1}| + |z - z_{2}| = 2 |z_{1} - z_{2}|$ Then, the locus of z will be

an ellipse
a straight line joining z₁ and z₂
a parabola
a bisector of the line segment joining z₁ and z₂

20.

the coefficient of x⁸ in ${({ax}^{2} + \frac{1}{bx})}^{13}$ is equal to the coefficient of x^{- 8} in ${(ax - \frac{1}{{bx}^{2}})}^{13}$ then a and b will satisfy the relation