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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)

If $a_1,a_2,a_3,.................$ are in A.P. such that $a_1+a_7+a_{16}=40$,

then the sum of the first 15 terms of this A.P. is :

• Option 1)

200

• Option 2)

280

• Option 3)

120

• Option 4)

150

Given ,                 ..................(1) We have to find out , ..............(2) Substituting the value of (1) in (2), Option 1)200Option 2)280Option 3)120Option 4)150

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

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)

Let Sn denote the sum of the first terms of an A.P.. If $S_{4}=16$ and $S_{6}=-48$ then $S_{10}$ is equal to :

• Option 1)

$-380$

• Option 2)

$-320$

• Option 3)

$-260$

• Option 4)

$-410$

So (2) - (1)                                    Option 1) Option 2) Option 3) Option 4)

Let $a_1,a_2,a_3,.........$ be an A.P. with $a_6=2$ . Then the

common difference of this A.P., which maximises the product

$a_1a_4a_5$ , is :

• Option 1)

$\frac{3}{2}$

• Option 2)

• Option 3)

$\frac{6}{5}$

• Option 4)

$\frac{2}{3}$

Assuming the first term of A.P. is a and difference is d. Then, Let  =>  So,  will be maximum at  So, option (2) is correct. 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)

The sum   $\mathrm{1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+..........+\frac{1^3+2^3+3^3+......+15^3}{1+2+3+.......+15}\:-\:\frac{1}{2}\left(1+2+3+.......+15\right)}$

is equal to :

• Option 1)

620

• Option 2)

1240

• Option 3)

1860

• Option 4)

660

So, option (1) is correct.   Option 1) 620 Option 2) 1240 Option 3) 1860 Option 4) 660

The sum

$\frac{3\times 1^{3}}{1^{2}}+\frac{5\times (1^{3}+2^{3})}{1^{2}+2^{2}}+\frac{7\times (1^{3}+2^{3}+3^{3})}{1^{2}+2^{2}+3^{2}}+..........$

upto 10th term , is :

• Option 1)

680

• Option 2)

600

• Option 3)

660

• Option 4)

620

Given, general term will be                                                               So, correct option is (3).Option 1)680Option 2)600Option 3)660Option 4)620

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)

If $a_{1},a_{2},a_{3},..........a_{n}$ are in A.P. and $a_{1}+a_{4}+a_{7}..........+a_{16}=114$,

then $a_{1}+a_{6}+a_{11}+a_{16}$ is equal to :

• Option 1)

98

• Option 2)

76

• Option 3)

38

• Option 4)

64

correct option is (2). Option 1) 98 Option 2) 76 Option 3) 38 Option 4) 64

The sum of the series  $1+2\times3+3\times 5+4\times7+.......upto\:\:\:11^{th}$ term is :

• Option 1)

$915$

• Option 2)

$946$

• Option 3)

$945$

• Option 4)

$916$

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