Filters

Sort by :
Clear All
Q

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 $[x]$ denotes the greatest integer $\leq x$ , then the system

of linear equations $[sin\theta]x+[-cos\theta]y=0$

$[cot\theta]x+y=0$

• Option 1)

Have infinitely many solutions if $\theta\epsilon (\frac{\pi}{2},\frac{2\pi}{3})$ and has a unique solution if $\theta\epsilon (\pi,\frac{7\pi}{6})$.

• Option 2)

has a unique solution if $\theta\epsilon(\frac{\pi}{2},\frac{2\pi}{3})\cup (\pi,\frac{7\pi}{6})$

• Option 3)

has a unique solution if $\theta\epsilon(\frac{\pi}{2},\frac{2\pi}{3})$ and have infinitely many solutions if $\theta\epsilon (\pi,\frac{7\pi}{6})$

• Option 4)

infinitely many solutions if $\theta\epsilon(\frac{\pi}{2},\frac{2\pi}{3})\cup$$(\pi,\frac{7\pi}{6})$

linear equations  &                              For infinite many solution, i.e. ...................(1) * when                         so,     * when                         so,     *when   *when                          Option 1) Have infinitely many solutions if  and has a unique solution if . Option 2) has a unique solution if  Option 3) has a unique solution if  and have infinitely many...

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 value of $\theta \epsilon (0,\pi /3)$, for which

$\begin{vmatrix} 1+cos^{2}\theta &sin^{2}\theta &4cos6\theta \\ cos^{2}\theta& 1+sin^{2}\theta &4cos6\theta \\ cos^{2}\theta& sin^{2}\theta & 1+4cos6\theta \end{vmatrix}=0$, is:

• Option 1)

$\frac{\pi }{9}$

• Option 2)

$\frac{\pi }{18}$

• Option 3)

$\frac{7\pi }{24}$

• Option 4)

$\frac{7\pi }{36}$

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

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)

If A is a syymmetric matrix and B is a skew-symmetrix matrix such that $A+B=\begin{bmatrix} 2 &3 \\5 &-1 \end{bmatrix}$, then AB is equal to :

• Option 1)

$\begin{bmatrix} -4 &-2 \\ -1 &4 \end{bmatrix}$

• Option 2)

$\begin{bmatrix} 4 &-2 \\ -1 & -4 \end{bmatrix}$

• Option 3)

$\begin{bmatrix} 4 &-2 \\ 1 &-4 \end{bmatrix}$

• Option 4)

$\begin{bmatrix} -4 &2 \\ 1 & 4 \end{bmatrix}$

A is symmetric matrix  B is skew-symmetrix (1) + (2)  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

Let $\lambda$ be a real number for which the system of linear equations

$x+y+z=6$

$4x+\lambda y-\lambda z=\lambda -2$

$3x+2 y-4 z= -5$

has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation :

• Option 1)

$\lambda^{2}+3\lambda-4=0$

• Option 2)

$\lambda^{2}-3\lambda-4=0$

• Option 3)

$\lambda^{2}+\lambda-6=0$

• Option 4)

$\lambda^{2}-\lambda-6=0$

linear equations             Now, using cramers law for infinite solution   all will be zero      Now put  in options and check for the correct one  (1) (2) (3) (4) So, option (4) is correct.     Option 1) Option 2) Option 3) Option 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)

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 the system of linear equations

x + y + z = 5

x + 2y + 2z = 6

x + 3y + $\lambda$ z = $\mu$ , ($\lambda$,$\mu$ $\epsilon R$ ) , has infinitely

many solutions, then the value of $\lambda$ + $\mu$ is :

• Option 1)

12

• Option 2)

9

• Option 3)

7

• Option 4)

10

x + 3y +  z -  = p ( x + y + z - 5) + q ( x + 2y + 2z - 6 )  On comparing the coefficients  p + q = 1   and    p + 2q = 3 => ( p , q ) = ( -1 , 2 ) Hence, x + 3y +  z -  = x + 3y + 3z - 7  =>  =>  So, option (4) is correct. Option 1) 12 Option 2) 9 Option 3) 7 Option 4) 10

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 $\bigtriangleup _{1}=\begin{vmatrix} x &\sin \theta &\cos \theta \\ -\sin \theta &-x &1 \\ \cos \theta & 1 & x \end{vmatrix}$  and

$\bigtriangleup _{2}=\begin{vmatrix} x &\sin 2\theta &\cos 2\theta \\ -\sin 2\theta &-x &1 \\ \cos2 \theta & 1 & x \end{vmatrix}$ , $x\neq0$ ; then

for all $\theta \epsilon (0,\frac{\pi}{2}) :$

• Option 1)

$\triangle_{1}-\triangle_{2}=-2x^{3}$

• Option 2)

$\triangle_{1}-\triangle_{2}=x(cos2\theta-cos4\theta)$

• Option 3)

$\triangle_{1}+\triangle_{2}=-2(x^{3}+x-1)$

• Option 4)

$\triangle_{1}+\triangle_{2}=-2x^{3}$

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

If the system of equations  $2x+3y-z=0,x+ky-2z=0\:\:and\:\:2x-y+z=0$ has a non-trivial solution $\left ( x,y,z \right )$  , then $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k$   is equal to :

• Option 1)

$\frac{3}{4}$

• Option 2)

$\frac{1}{2}$

• Option 3)

$-\frac{1}{4}$

• Option 4)

$-4$

for non-trivial solution  A=0   so       Option 1) Option 2) Option 3) Option 4)
Exams
Articles
Questions