Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

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.

Complex Numbers and Quadratic Equations

Multiple Choice Questions

261.

$If Z_{R} = \cos (\frac{π}{2^{r}}) + isin (\frac{π}{2^{r}}) for r = 1, 2, 3, . . ., then Z_{1} Z_{2} Z_{3} . . . \infty = ?$

262.

If x₁ and x₂are the real roots of the equation x² - kx + c = 0, then the distance between the points A(x₁, 0) and B(x₂, 0) is

$\sqrt{k^{2} + 4 c}$
$\sqrt{k^{2} - c}$
$\sqrt{c - k^{2}}$
$\sqrt{k^{2} - 4 c}$

263.

If p and q are distinct prime numbers and if the equation x² - px + q = 0 has positive integers as its roots, then the roots of the equation are

1, - 1
2, 3
1, 2
3, 1

264.

The cubic equation whose roots are the squares of the roots of x³ - 2x² + 10x - 8 = 0, is

$x^{3} + 16 x^{2} + 68 x - 64 = 0$
$x^{3} + 8 x^{2} + 68 x - 64 = 0$
$x^{3} + 16 x^{2} - 68 x - 64 = 0$

265.

$If ω is a complex cube root of unity, then ω^{(\frac{1}{3} + \frac{2}{9} + \frac{4}{27} + \infty)} + ω (\frac{1}{2} + \frac{3}{8} + \frac{9}{32} + \infty) = ?$

1
- 1
$ω$
i

266.
The common roots of the equations $z^{3} + 2 z^{2} + 2 z + 1 = 0, z^{2014} + z^{2015} + 1 = 0 are$

$ω, ω^{2}$

$1, ω, ω^{2}$

$- 1, ω, ω^{2}$

$- ω, - ω^{2}$

$ω, ω^{2}$

$The given equation z^{3} + 2 z^{2} + 2 z + 1 = 0, z^{2014} + z^{2015} + 1 = 0 can be written as (z + 1) (z^{2} + z + 1) = 0 Since, its roots are - 1, ω, ω^{2} Let f (z) = z^{2014} + z^{2015} + 1 = 0 Put z = - 1, ω, ω^{2} respectively, we get f (- 1) = (- 1^{2014}) + {(- 1)}^{2015} + 1 = 0 = 1 \neq 0 Therefore, - 1 is not a root of the equation f (z) = 0 Again, f (ω) = ω^{2014} + ω^{2015} + 1 = 0 = {(ω^{3})}^{671} ω + {(ω^{3})}^{671} ω^{2} + 1 = 0 \Rightarrow ω + ω^{2} + 1 \Rightarrow ω^{2} + ω + 1 = 0 \Rightarrow 0 = 0 Therefore, ω is a root of the equation f (z) = 0 Similarly, f (ω^{3}) = {(ω^{2})}^{2014} + {(ω^{3})}^{2015} + 1 = 0 \Rightarrow {(ω^{3})}^{1342} . ω^{2} + {(ω^{3})}^{1343} ω 1 = 0 \Rightarrow ω^{2} + ω + 1 = 0 \Rightarrow 0 = 0 Hence, ω and ω^{2} are the common roots$