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Let \alpha and \beta   be the roots of equation

x^{2}-6x-2=0.\; if a_{n}=\alpha ^{n}-\beta ^{n},\; for\: n\geq 1,\; then\; the \: value\; of\; \frac{a_{10}-2a_{8}}{2a_{9}}

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<r\leq \frac{1}{2}. 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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