Tag: mathematics and statistics

Questions Related to mathematics and statistics

If $\Delta =\begin{vmatrix} a _{11} & a _{12} & a _{13}\ a _{21} & a _{22} & a _{23}\ a _{31} & a _{32} & a _{33} \end{vmatrix}$ and $c _{ij}=\left ( -1 \right )^{i+j}$ (determinant obtained by deleting ith row and jth column), then $\begin{vmatrix} c _{11} & c _{12} & c _{13}\ c _{21} & c _{22} & c _{23}\ c _{31} & c _{32} & c _{33} \end{vmatrix}=\Delta ^{2}$



If $\begin{vmatrix} 1 & x & x^{ 2 } \ x & x^{ 2 } & 1 \ x^{ 2 } & 1 & x \end{vmatrix}=7$ and $\Delta =\begin{vmatrix}
x^{3}-1 & 0 & x-x^{4}\
0 & x-x^{4} & x^{3}-1\
x-x^{4} & x^{3}-1 & 0
\end{vmatrix}$, then

  1. $\Delta =7$

  2. $\Delta =343$

  3. $\Delta =-49$

  4. $\Delta =49$


Correct Option: D
Explanation:

For $\begin{vmatrix} 1 & x & x^{ 2 } \ x & x^{ 2 } & 1 \ x^{ 2 } & 1 & x \end{vmatrix}=7$
$\begin{vmatrix} c _{ 11 } & c _{ 12 } & c _{ 13 } \ c _{ 21 } & c _{ 22 } & c _{ 23 } \ c _{ 31 } & c _{ 32 } & c _{ 33 } \end{vmatrix}=\begin{vmatrix} x^{ 3 }-1 & 0 & x-x^{ 4 } \ 0 & x-x^{ 4 } & x^{ 3 }-1 \ x-x^{ 4 } & x^{ 3 }-1 & 0 \end{vmatrix}$
$\Delta ={ 7 }^{ 2 }=49$

Let $\Delta _0=\begin{bmatrix}a _{11} & a _{12}  & a _{13}\a _{21}  & a _{22} &a _{23} \ a _{31} & a _{32} & a _{33}\end{bmatrix}$ (where $\Delta _0 \neq  0$) and let $\Delta _1$ denote the determinant formed by the cofactors of elements of $\Delta _0$ and $\Delta _2$ denote the determinant formed by the cofactor at $\Delta _1$ and so on $\Delta _n$ denotes the determinant formed by the cofactors at $\Delta _{n-1}$ then the determinant value of $\Delta _{n}$ is

  1. $\Delta _0^{2n}$

  2. $\Delta _0^{2^n}$

  3. $\Delta _0^{n^2}$

  4. $\Delta _0^{2}$


Correct Option: B
Explanation:

$\Delta _1=\Delta ^2 _0,\Delta _2=\Delta ^2 _1=\Delta ^{2^2} _0$
$\therefore \Delta _n=\Delta ^{2n} _0$

Circle on which the coordinates of any point are $(2+4 \cos \theta,-1+4 \sin \theta)$ where $\theta$ is the parameter is given by $(x-2)^2+(y+1)^2=16$.

  1. True

  2. False


Correct Option: A
Explanation:
From given conditions, we have,
$x=2+4 \cos \theta $
$x-2=4\cos \theta$           .......(1)
$y=-1+4\sin \theta$
$y+1=4 \sin \theta$           .......(2)
Squaring and adding equation 1 and 2, we get,
$(x-2)^2+(y+1)^2=16$.

For any line let $m$ and $b$ represent its slope and $y$ intercept respectively and related by $2m+b=3$. These lines all have a specific common point from where tangents are drawn to $x^{2}+y^{2}=1$.

  1. $2x+3y=1$

  2. $2x-3y=1$

  3. $3x-2y=1$

  4. $3x+2y=1$


Correct Option: A

Slope of $\left{ (x,y)/x=2t+3,y=2t+5,t\epsilon R \right} $ is _______.

  1. 2

  2. 1

  3. $\frac { -1 }{ 1 } $

  4. -1


Correct Option: A

The equation of a line which passes through (2,3) and the product of whose intercepts on the coordinate axis is 27, can be

  1. 5x+4y=22

  2. 3x-y=3

  3. 3x+4y=18

  4. 2x+3y=13


Correct Option: C
Explanation:
Let the equation of the line is $\dfrac{x}{a}+\dfrac{y}{b}=1$
$\dfrac{2}{a}+\dfrac{3}{b}=1$

$2b+3a=ab$
$3a+2b=27$
$ab=27$
$b=\dfrac{27}{a}$
$3a+2\times \dfrac{27}{a}=27$
$3a^2+54=27a$
$3a^2-27a+54=0$
$a^2-9a+18=0$
$(a-6)(a-3)=0$
$a=6,3$
$b=\dfrac{9}{2}$ or $9$
Required equation is
$\dfrac{x}{6}+\dfrac{y}{\dfrac{9}{2}}=1 \implies 3x+4y=18$
Or,
$\dfrac{x}{3}+\dfrac{y}{9}=1 \implies 3x+y=9$

If the straight lines joining the origin and the points of intersection of the curve
$ {5x}^{2} + 12xy-{6y}^{2} +4x -2y+3 =0$ and $x+ky-1=0 $ are equally inclined to the co ordinate axis,then the  value of k-

  1. is equal to 1

  2. is equal to -1

  3. is equal to 2

  4. does not exist in the set of real numbers


Correct Option: B

If the line $AX+BY=1$ passes through point of intersection of $y=x\tan\alpha+p\sec\alpha$,$y\sin(30-\alpha)-x\cos(30^ {o}-\alpha)=p$ and is inclined at $30^ {o}$ with $y=(x\tan\alpha+p\sec\alpha)$ then the value of $a^ {2}+b^ {2}=?$

  1. $\dfrac {1}{p^ {2}}$

  2. $\dfrac {2}{p^ {2}}$

  3. $\dfrac {3}{2p^ {2}}$

  4. $\dfrac {3}{4p^ {2}}$


Correct Option: A

A line $OP$ through origin $O$ is inclined at $30^{o}$ and $45^{o}$ to $OX$ and $OY$ respectively. The angle at which it is inclined to $OZ$ is-

  1. $\cos^{-1} \sqrt{\dfrac{4}{6}}$

  2. $\cos^{-1} \left(\dfrac{2}{6}\right)$

  3. $\cos^{-1} \left(\dfrac{1}{2}\right)$

  4. $Not\ defined$


Correct Option: A
Find the equation of the straight line whose $x$ and $y$-intercepts on the axes are given by $2$ and $3$.
  1. $3x-2y+6=0$

  2. $3x+2y-6=0$

  3. $3x-2y-6=0$

  4. none of these


Correct Option: B
Explanation:

Consider the given points.

$(2, 0)$ and $(0, 3)$

We know that the equation of the line which is passing through the points
$y-y _1=\dfrac{y _2-y _1}{x _2-x _1}(x-x _1)$

So,

$y-0=\dfrac{3-0}{0-2}(x-2)$

$y=\dfrac{3}{-2}(x-2)$

$-2y=3x-6$

$3x+2y-6=0$

Hence, Option $B$ is the answer.