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Let and   be the roots of equation

is equal to:

• Option 1)

6

• Option 2)

-6

• Option 3)

3

• Option 4)

-3

3

All the pairs ( x, y ) that satisfy the inequality

$2^{\sqrt{sin^{2}x-2sinx+5}\cdot \frac{1}{4^{sin^{2}y}}}\leq 1$  also satisfy the equation :

• Option 1)

$2|sinx|=3siny$

• Option 2)

$2 sinx=siny$

• Option 3)

$sinx=2siny$

• Option 4)

$sinx=|siny|$

=>  &  So, correct option is (4). Option 1) Option 2) Option 3) Option 4)

If $\alpha ,\beta \: \: and\: \: \gamma$ are three consecutive terms of a non-constant

G.P. such that the equations $\alpha x^{2}+2\beta x+\gamma =0$  and

$x^{2}+ x-1=0$ have a common root , then $\alpha (\beta +\gamma )$ is equal to :

• Option 1)

0

• Option 2)

$\alpha \beta$

• Option 3)

$\alpha \gamma$

• Option 4)

$\beta \gamma$

are in G.P. =>  For equation,                  Hence, roots are equal & equals to  Since, given equation have common roots , hence   must be root of                     Option 1) 0 Option 2) Option 3) Option 4)

Let $z\epsilon C$  with $Im(z)=10$  and  it satisfies

$\frac{2z-n}{2z+n}=2i-1$  for some natural number n . Then :

• Option 1)

n = 20 and Re(z) = -10

• Option 2)

n = 40 and Re(z) = 10

• Option 3)

n = 40 and Re(z) = -10

• Option 4)

n = 20 and Re(z) = 10

,  ,  Option 1)  n = 20 and Re(z) = -10 Option 2)  n = 40 and Re(z) = 10 Option 3)  n = 40 and Re(z) = -10 Option 4)  n = 20 and Re(z) = 10

A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal to :

• Option 1)

28

• Option 2)

27

• Option 3)

25

• Option 4)

24

atleast one boy & one girl :  ( 1B & 2G) + ( 2B & 1G)   As, n cannot be -ve so, n = 25 Option 1) 28 Option 2) 27 Option 3) 25 Option 4) 24

The number of ways of choosing 10 objects out of $31$ objects of which $10$ are identical and the remaining $21$ are distinct, is:

• Option 1)

$2^{20}-1$

• Option 2)

$2^{20}+1$

• Option 3)

$2^{21}$

• Option 4)

$2^{20}$

No. of ways Option 1) Option 2) Option 3) Option 4)

If $B=\begin{bmatrix} 5 &2\alpha &1 \\ 0 &2 &1 \\ \alpha &3 &-1 \end{bmatrix}$ is the inverse of a $3\times 3$ matrix A, then the sum of all values of $\alpha$ for which det $\left ( A \right )+1=0,$ is :



• Option 1)

$0$

• Option 2)

$1$

• Option 3)

$2$

• Option 4)

$-1$

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

The equation $\left | z-i \right |=\left | z-1 \right |,i=\sqrt{-1},$ represents :

• Option 1)

a circle of radius $\frac{1}{2}$.

• Option 2)

a circle of radius $1.$

• Option 3)

the line through the origin with slope $1.$

• Option 4)

the line through the origin with slope $-1.$

The line through the origin with slope 1Option 1)a circle of radius .  Option 2)a circle of radius Option 3)the line through the origin with slope Option 4)the line through the origin with slope

if $\alpha \: and\: \beta$ are roots of the equation $375x^{2}-25x-2=0,$ then $\lim_{n\rightarrow \infty }\sum_{r=1}^{n}\alpha ^{r}+\lim_{n\rightarrow \infty }\sum_{r=1}^{n}\beta ^{r}$ is equal to :

• Option 1)

$\frac{1}{12}$

• Option 2)

$\frac{7}{116}$

• Option 3)

$\frac{21}{346}$

• Option 4)

$\frac{29}{358}$

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

iIf three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is :

• Option 1)

$\frac{1}{5}$

• Option 2)

$\frac{3}{20}$

• Option 3)

$\frac{3}{10}$

• Option 4)

$\frac{1}{10}$

Only two equilateral triangle are possible  and    Option 1) Option 2) Option 3) Option 4)

Let a,b and c be in G.P. with common ratio r , where $a\neq 0$ and

$0. If  3a , 7b and 15c are the first three terms of an A.P.,

then the 4th term of this A.P. is :

• Option 1)

$\frac{2}{3}a$

• Option 2)

5 a

• Option 3)

$\frac{7}{3}a$

• Option 4)

a

Since a,b,c are in G.P. with common ratio r then  b = ar ,  Also 3a, 7b and 15c are in A.P. =>  =>  =>  =>  =>  =>  =>  So, terms are                         or                         =>    or    So, 4th term     or    So, option (4) is correct.   Option 1) Option 2) 5 a Option 3) Option 4) a

The sum of the real roots of the equation

$\begin{vmatrix} x & -6 &-1 \\ 2 &-3x &x-3 \\ -3& 2x &x+2 \end{vmatrix}=0$, is equal to :

• Option 1)

6

• Option 2)

0

• Option 3)

1

• Option 4)

-4

=>  =>  Root of equation (-3,1,2) So, Sum of real root of equation = -3+1+2=0 So, option (2) is correct.Option 1)6Option 2)0Option 3)1Option 4)-4

If $z$ and $w$ are two complex numbers such that $|zw|=1$

and $arg(z)-arg(w)=\frac{\pi}{2},$  then :

• Option 1)

• Option 2)

$z\bar w=\frac{-1+i}{\sqrt2}$

• Option 3)

$\bar z w=-i$

• Option 4)

$z\bar w=\frac{1-i}{\sqrt2}$

and  Let    =>                                                                  Option 1)Option 2)Option 3)Option 4)

Suppose that 20 pillars of the same height have been erected along the boundary

of a circular stadium. If the top of each pillar has been connected by beams with the

top of all its non-adjacent pillars, then the total number of beams is :

• Option 1)

170

• Option 2)

180

• Option 3)

210

• Option 4)

190

Any two non-adjacent pillars are joined by beams.  no. of beams = no. of diagonals                                                                                                             Option (1) is correct. Option 1) 170 Option 2) 180 Option 3) 210 Option 4) 190

Assume that each born child is equally likely to be a boy or a girl. If two

families have two children each, then the conditional probability that

all children are girls given that at least two are girls is:

• Option 1)

$\frac{1}{11}$

• Option 2)

$\frac{1}{10}$

• Option 3)

$\frac{1}{12}$

• Option 4)

$\frac{1}{17}$

There are 4 children  total number of ways in whcih atleast 2 girls are there Required probabilty =  Option (1) is correct. Option 1) Option 2) Option 3) Option 4)

The number of 6 digit numbers that can be formed using the

digits 0,1,2,5,7 and 9 which are divisible by 11 and no digit is repeated, is:

• Option 1)

72

• Option 2)

60

• Option 3)

48

• Option 4)

36

Sum of the given digit is 0+1+2+5+7+9 = 24 6 digit number let abcdef  abcdef is divisible by 11 if   is a multiple of 11  Only one possible is there a+c+e=b+d+f=12 Case 1: {a,c,e}={7,5,0}              {b,d,f}={9,2,1}  So, 2 x 2! x 3! = 24 Case 2: {a,c,e}={9 , 2, 1}              {b,d,f}={7, 5, 0} So, 3! x 3! = 36 Total = 24 + 36 = 60 So, correct option is (2). Option 1) 72 Option 2) 60 Option...

If $\alpha$ and $\beta$ are the roots of the quadratic equation,

$x^{2}+xsin\theta -2sin\theta =0,\theta \epsilon (0,\frac{\pi}{2})$, then

$\frac{\alpha ^{12}+\beta ^{12}}{(\alpha ^{-12}+\beta ^{-12})\cdot (\alpha -\beta )^{24}}$ is equal to :

• Option 1)

$\frac{2^{12}}{(sin\theta-4)^{12}}$

• Option 2)

$\frac{2^{12}}{(sin\theta+8)^{12}}$

• Option 3)

$\frac{2^{12}}{(sin\theta-8)^{6}}$

• Option 4)

$\frac{2^{6}}{(sin\theta+8)^{12}}$

and  are the roots of the equation  Now,                                                                                                                                                                   correct option (2) Option 1) Option 2) Option 3) Option 4)

If a>0 and $z=\frac{(1+i)^{2}}{a-i}$ , has magnitude $\sqrt{\frac{2}{5}}$ ,

then $\bar{z}$ is equal to :

• Option 1)

$-\frac{1}{5}-\frac{3}{5}i$

• Option 2)

$-\frac{3}{5}-\frac{1}{5}i$

• Option 3)

$\frac{1}{5}-\frac{3}{5}i$

• Option 4)

$-\frac{1}{5}+\frac{3}{5}i$

given that  Option (1) is correct. Option 1) Option 2) Option 3) Option 4)

Let $z \:\varepsilon \:C$ be such that $\left | z \right |<1$. If $\omega =\frac{5+3z}{5(1-z)}$ ,

then :

• Option 1)

$5\:Re (\omega )>4$

• Option 2)

$5\:Im (\omega )>5$

• Option 3)

$5\:Re (\omega )>1$

• Option 4)

$5\:Im (\omega )<1$

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

If m is chosen in the quadratic equation $\left ( m^{2}+1 \right )x^{2}-3x+\left ( m^{2}+1 \right )^{2}=0$ such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is  :

• Option 1)

$10\sqrt{5}$

• Option 2)

$8\sqrt{3}$

• Option 3)

$8\sqrt{5}$

• Option 4)

$4\sqrt{3}$

for sum of root to be greatest     should be minimum  now equation         Option 1) Option 2) Option 3) Option 4)
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