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If $z=\frac{\sqrt{3}}{2}+\frac{i}{2}\left ( i=\sqrt{-1} \right ),$ then $\left ( 1+iz+z^{5}+iz^{8} \right )^{9}$ is equal to :

• Option 1)

$0$

• Option 2)

$1$

• Option 3)

$\left ( -1+2i \right )^{9}$

• Option 4)

$-1$

(cube root of unity) Option 1) Option 2) Option 3)   Option 4)

If the fourth term of the binomial expansion $\left ( \sqrt{\frac{1}{x^{1+\log_{10}x}}}+x^{\frac{1}{12}} \right )^{6}$ is equal to $200,$ and $x>1$, then the value of x is :

• Option 1)

$100$

• Option 2)

$10$

• Option 3)

$10^{3}$

• Option 4)

$10^{4}$

Fourth term is given  So,  take log both side.   put  Option 1) Option 2) Option 3) Option 4)

If $\alpha \; and\; \beta$ be the roots of the equation $x^{2}-2x+2=0$, then the least value of n for which $\left ( \frac{\alpha }{\beta } \right )^{n}=1$ is :

• Option 1)

$2$

• Option 2)

$5$

• Option 3)

$4$

• Option 4)

$3$

So   Option 1) Option 2) Option 3) Option 4)

The sum of the solutions of the equation $\left | \sqrt{x}-2 \right |+\sqrt{x}\left (\sqrt{x}-4 \right )+2=0,(x>0)$ is equal to :

• Option 1)

$12$

• Option 2)

$10$

• Option 3)

$9$

• Option 4)

$4$

Sum       Option 1) Option 2) Option 3) Option 4)
Let Z = x +iy Hence z will lie on imaginary axis for every x
15519652398521476122459.jpg Solve it

The number of complex number z which satisfy z^2+2|z|^2=2

@mannika If you look at problem, , it is clear that  must be real. if you solve you get so, if we represent
15519582136261468787462.jpg The number of complex number z such that |z-i|=|z+i|=|z+1| is
@mannika We are given that, |z-i| = |z- (-i)| = |z- (-1)| we know that in a complex plane, |z-a| represents distance of z from complex number a. Now we are given, 3 points (0,1);(0,-1);(-1,0) from which distance of z is equal.But we know in a plane there exist only 1 point which are equidistant from these 3 points which is centroid of triangle made by these 3 points.
i have not understood anything in cube root of unity topic video

z3 = 1

$\\z^3-1=0\\(z-1)(z^2+z+1)=0\\in\:the\:above\:z-1=0\:\:or\:\:z^2+z+1=0\\so,z=1,z=-\frac{1}{2}\pm\frac{i\sqrt3}{2}\\cube\:root\:of\:unity\:are\:1,-\frac{1}{2}+\frac{i\sqrt3}{2}\:\:and\:\:-\frac{1}{2}-\frac{i\sqrt3}{2}\\$

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If the roots of the quadratic equation$x^2 + px + q = 0$ are tan30° and tan15°, respectively then the value of 2 + q − p is

• Option 1)

2

• Option 2)

3

• Option 3)

0

• Option 4)

1

Option 1) 2 Option 2) 3 Option 3) 0 Option 4) 1

All the values of m for which both roots of the equations x2 − 2mx + m2 − 1 = 0 are greater than −2 but less than 4, lie in the interval

• Option 1)

−2 < m < 0

• Option 2)

m > 3

• Option 3)

−1 < m < 3

• Option 4)

1 < m < 4

Option 1) −2 < m < 0 Option 2) m > 3 Option 3) −1 < m < 3 Option 4) 1 < m < 4
IMG_20190109_35438.jpg Sum of roots of the equation (x + â€” 41x + 31+ 3 = O is -
@Arvind 1.
15470031512802109424855.jpg Number of negative integral value of x satisfying
@Vijay Q42
IMG_20190115_201103.jpg

The solution of the inequation |x2-2x-3|<|x2-x+5| is-

15475677420921088009151.jpg |f(x)|=|x2-2x-3| |g(x)|=|x2-x+5| |f(x)|<|g(x)| plot graph of f(x) and g(x)
Sir in |z-2+3i| how we get Centre as (2,3)
@Arvind
how do you get y is less than or equal to 2 from that (y-2)^2 =y (y-2)^2 is less than or equal to 2
@santhosh we are taking different cases (1) when y is less than or equal 2 (2) when 2<y less than or equal to 3 (3) y is greater than 3
Screenshot_718.png
Screenshot_719.png
Screenshot_716.png
santosh2.jpg for atleast one negative value of x (x-x_1)(x-x_2) is less than 0 and Discriminant is greater than 0
Screenshot_715.png
santosh.jpg put x-1=y^2 simplify use the concept that value under any square root is always greater then or equal to 0
rps20190122_174705.jpg

so roots are (x,y ) = $(\frac{1}{\sqrt{2}},-\:\frac{1}{\sqrt{2}}\:)$ or $(-\:\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\:)$

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