Q&A - Ask Doubts and Get Answers

A random variable X has the following probability distribution: $X: \: \: 1\; \; \: \: 2\; \;\: \: \: \: 3\; \; \: \: \: \: \: 4\: \: \: \; \; \: \: 5$ $P(X): \;K^{2}\; \; 2K\; \; K\; \; 2K\; \; 5K^{2}$ Then $P(X>2)$ is equal to:
Option: 1 $\frac{7}{12}$
Option: 2 $\frac{23}{36}$
Option: 3 $\frac{1}{36}$
Option: 4 $\frac{1}{6}$

7/12

View Full Answer(6)

Posted by

Kamlesh kumar

Let $a_{n}$ be the n^th term of a G.P. of positive terms. If $\sum_{n=1}^{100}a_{2n+1}=200\: \: and\: \: \sum_{n=1}^{100}a_{2n}=100,\: \: then\: \sum_{n=1}^{200}a_n$ is equal to :
Option: 1 $300$
Option: 2 $175$
Option: 3 $225$
Option: 4 $150$

225

View Full Answer(2)

Posted by

Anakha

If $x=\sum_{n=0}^{\infty }(-1)^{n}\tan ^{2n}\theta \: \: and\: \: y=\sum_{n=0}^{\infty }\cos ^{2n}\theta ,$ for $0<\theta < \frac{\pi }{4},$ then
Option: 1 $y(1+x)=1$
Option: 2 $x(1-y)=1$
Option: 3 $y(1-x)=1$
Option: 4 $x(1+y)=1$

$x=\sum_{n=0}^{\infty}(-1)^{n} \tan ^{2 n} \theta=1-\tan^2\theta+\tan^4\theta..........$

$y=\sum_{n=0}^{\infty} \cos ^{2 n} \theta=1+\cos^2\theta+\cos^4\theta......$

Use $\text S_{\infty}=\frac{1}{1-r}$

${x=\frac{1}{1+\tan ^{2} \alpha}=\cos ^{2} \theta} \\ {y=\frac{1}{1-\cos ^{2} \theta}=\frac{1}{\sin ^{2} \theta}}$

$\Rightarrow (1-x)= \sin ^{2} \theta$

$\Rightarrow y(1-x)=1$

Correct Option (3)

View Full Answer(1)

Posted by

avinash.dongre

If $z$ be a complex number satisfying $\left | Re\left ( z \right ) \right |+\left | Im(z) \right |=4,$ then $\left | z \right |$ cannot be :
Option: 1 $\sqrt{7}$

Option: 2 $\sqrt{\frac{17}{2}}$

Option: 3 $\sqrt{10}$

Option: 4 $\sqrt{8}$

Complex number -

A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form as a + bi where a is the real part and b is the imaginary part. For example, 5 + 2i is a complex number. So, too, is 3 + 4i√3.

We write the complex number by C or z = a + ib, a and b are real number (a, b ∈ R).

a is real part of the complex number and denoted by Re(z),
b is the imaginary part of the complex number and denoted by Im(z),

E.g : z = 2 + 3i is a complex number.

With Re(z) = 2 and Im(z) = 3

-

Area of triangle, circle (formula) -

Equation of Circle:

The equation of the circle whose center is at the point $z_0$ and have radius r is given by

$|z-z_0| = r$

If the center is origin then, $z_0=0$ , hence equation reduces to |z| = r

Interior of the circle is represented by $|z-z_0| < r$

The exterior is represented by $|z-z_0| > r$

Here z can be represented as x + iy and $z_0$ is represented by $x_0 + iy_0$

-

z = x + iy

|x| + |y| = 4

Minimum value of |z| = $2\sqrt2$

Maximum value of |z| = 4

$z\in[\sqrt8,\sqrt{16}]$

So |z| can't be $\sqrt7$

Correct Option (1)

View Full Answer(1)

Posted by

avinash.dongre

The following system of linear equations $7x+6y-2z=0$ $3x+4+2z=0$ $x-2y-6z=0,$ has
Option: 1 infinitely many solutions, $(x,y,z)$ satisfying $y=2z.$
Option: 2 infinitely many solutions, $(x,y,z)$ satisfying $x=2z.$
Option: 3 no solution
Option: 4 only the trivial solution.

System of Homogeneous linear equations -

$\\\mathrm{Let,} \\\mathrm{a_1x+b_1y +c_1z=0\;\;\; ...(i)} \\\mathrm{a_2x+b_2y +c_2z=0\;\;\; ...(ii)} \\\mathrm{a_3x+b_3y +c_3z=0\;\;\; ...(iii)} \\\mathrm{be \; three\; homogeneous\; equation} \\\mathrm{and \; let\; \Delta = \begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}}$

If ? ≠ 0, then x= 0, y = 0, z = 0 is the only solution of the above system. This solution is also known as a trivial solution.

If ? = 0, at least one of x, y and z are non-zero. This solution is called a non-trivial solution.

Explanation: using equation (ii) and (iii), we have

$\\\mathrm{\frac{x}{b_2c_3-b_3c_2} = \frac{y}{c_2a_3-c_3a_2}=\frac{z}{a_2b_3-a_3b_2}} \\\\\mathrm{or \;\; \frac{x}{\begin{vmatrix} b_2 &c_2 \\ b_3 & c_3 \end{vmatrix}}=\frac{y}{\begin{vmatrix} c_2 &a_2 \\ c_3 & a_3 \end{vmatrix}} = \frac{z}{\begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix}} = k (say \neq 0)} \\\mathrm{\therefore x = k\begin{vmatrix} b_2 &c_2 \\ b_3 & c_3 \end{vmatrix}, y = k\begin{vmatrix} c_2& a_2\\ c_3 & a_3 \end{vmatrix} \; and \; z = k\begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix}} \\\mathrm{putting\; these\; value\; in \; equation \; (i), we\; have} \\\mathrm{a_1\left \{ k\begin{vmatrix} b_2 & c_2\\ b_3 & c_3 \end{vmatrix} \right \}+b_1\left \{ k\begin{vmatrix} c_2 & a_2\\ c_3 & a_3 \end{vmatrix} \right \}+c_1\left \{ \begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix} \right \}=0}$

$\\\mathrm{\Rightarrow a_1\begin{vmatrix} b_2 &c_2 \\ b_3 & c_2 \end{vmatrix}-b_1\begin{vmatrix} a_2 & c_2\\ a_3 &c_3 \end{vmatrix}+c_1\begin{vmatrix} a_2 &b_2 \\ a_3 &b_3 \end{vmatrix} = 0 \;\;\;[\because \; k \neq 0]} \\\mathrm{or \;\; \begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix} = 0 \; or \; \Delta = 0}$

This is the condition for a system have Non-trivial solution.

-

$\begin{aligned} &(1)\;\;7 x+6 y-2 z=0\\ &(2)\;\;3 x+4 y+2 z=0\\ &(3)\;\;x-2 y-6 z=0 \end{aligned}$

$\left|\begin{array}{ccc}{7} & {6} & {-2} \\ {3} & {4} & {2} \\ {1} & {-2} & {-6}\end{array}\right| =7(-20)-6(-20)-2(-10)=-140+120+20=0$

so infinite non-trivial solution exist

now equation (1) + 3 equation (3)

10x - 20z = 0

x = 2z

Correct Option 2

View Full Answer(1)

Posted by

avinash.dongre

Let $a-2b+c=1.$ If $f\left ( x \right )= \begin{vmatrix} x+a & x+2 &x+1 \\ x+b &x+3 &x+2 \\ x+c &x+4 & x+3 \end{vmatrix},$ then :
Option: 1 $f(-50)=501$
Option: 2 $f(-50)=-1$
Option: 3 $f(50)=1$
Option: 4 $f(50)=-501$

Elementary row operations -

Elementary row operations

Row transformation: Following three types of operation (Transformation) on the rows of a given matrix are known as elementary row operation (transformation).

i) Interchange of i^th row with j^th row, this operation is denoted by

$\\\mathrm{R_i \leftrightarrow R_j \;\;or\; R_{ij}}$

ii) The multiplication of ith row by a constant k (k≠0) is denoted by

$\\\mathrm{R_i \leftrightarrow kR_i \;\; or \; k\cdot R_i}$

iii) The addition of ith row to the elements of jth row multiplied by constant k (k≠0) is denoted by

$\\\mathrm{R_i \leftrightarrow R_i + kR_j \;\; or \; k\cdot R_{ij}}$

In the same way, three-column operations can also be defined too.

-

$\\\text { Apply } \mathrm{R}_{1}=\mathrm{R}_{1}+\mathrm{R}_{3}-2 \mathrm{R}_{2}\\\Rightarrow f(x)=\left|\begin{array}{ccc}{1} & {0} & {0} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right| \quad \Rightarrow f(x)=1 \\\quad \Rightarrow f(50)=1$

Correct Option (3)

View Full Answer(1)

Posted by

avinash.dongre

Let $\left [ t \right ]$ denote the greatest integer $\leq t$ and $\lim_{x\rightarrow 0}x\left [ \frac{4}{x} \right ]=A.$ Then the function, $f\left ( x \right )=\left [ x^{2} \right ]\sin \left ( \pi x \right )$ is discontinuous, when x is equal to:
Option: 1 $\sqrt{A+1}$

Option: 2 $\sqrt{A}$

Option: 3 $\sqrt{A+5}$

Option: 4 $\sqrt{A+21}$

$\\\lim _{x \rightarrow 0} x\left(\frac{4}{x}-\left\{\frac{4}{x}\right\}\right)=A \quad \Rightarrow \lim _{x \rightarrow 0} 4-x\left\{\frac{4}{x}\right\}=A\\\Rightarrow 4-0=A$

$\\\text { (1) } x=\sqrt{A+1} \Rightarrow x=\sqrt{5} \Rightarrow \text { discontinuous }\\\text { (2) } x=\sqrt{A+21} \Rightarrow x=5 \Rightarrow \text { continuous }\\\text { (3) } x=\sqrt{A} \Rightarrow x=2 \Rightarrow \text { continuous }\\\text { (4) } x=\sqrt{A+5} \Rightarrow x=3 \Rightarrow \text { continuous }$

Correct Option (1)

View Full Answer(1)

Posted by

avinash.dongre

Let $a,b\epsilon \textbf{R},a\neq 0$ be such that the equation, $ax^{2}-2bx+5=0$ has a repeated root $\alpha$ , which is also a root of the equation, $x^{2}-2bx-10=0$ . If $\beta$ is the other root of this equation, then $\alpha ^{2}+\beta ^{2}$ is equal to:
Option: 1 $24$
Option: 2 $25$
Option: 3 $26$
Option: 4 $28$

Nature of Roots -

Let the quadratic equation is ax² + bx + c = 0

D is the discriminant of the equation.

iii) if roots D = 0, then roots will be real and equal, then

$\\\mathrm{x_1=x_2 = \frac{-b}{2a} } \\\mathrm{Then, \;\; ax^2+bx +c =a(x-x_1)^2 }$

-

ax² – 2bx + 5 = 0 having equal roots or $D=0$ and $\alpha=\frac{b}{a}$

$(2b)^2=4\times5\times a\;\;\Rightarrow \;\;b^2=5a$

Put $\alpha=\frac{b}{a}$ in the second equation

${x^{2}-2 b x-10=0} \\ {\Rightarrow b^{2}-2 a b^{2}-10 a^{2}=0}$

$\\\Rightarrow 5 a-10 a^{2}-10 a^{2}=0 \\ \Rightarrow 20 a^{2}=5 a \\ \Rightarrow a=\frac{1}{4} \text { and } \mathrm{b}^{2}=\frac{5}{4} \\ \alpha^{2}= 20 \text { and } \beta^{2}=5 \\ \alpha^{2}+\beta^{2} \\ =5+20 \\ =25$

Correct Option 2

View Full Answer(1)

Posted by

avinash.dongre

In the expansion of $\left ( \frac{x}{cos\theta }+\frac{1}{x\sin \theta } \right )^{16}$ , if $L_{1}$ is the least value of the term independent of $x$ when $\frac{\pi }{8}\leq \theta \leq \frac{\pi }{4}$ and $L_{2}$ is the least value of the term independent of $x$ when $\frac{\pi }{16}\leq \theta \leq \frac{\pi }{8}$ , then the ratio $L_{2}:L_{1}$ is equal to :
Option: 1 $16:1$
Option: 2

Exams

Colleges

Predictors

Resources

Quick links

Quick Links

Exams

Colleges

Resources

Colleges

Resources

Exams

Exams

Colleges

Resources

Diploma Colleges

Other Exams

Exams

Resources

Upcoming Events

Exams

Ranking

Products & Resources

NCERT Solutions

Top Countries

Student Visas

Exams

Colleges

Exams

Colleges

Upcoming Events/Predictors

Resources

Exams

Colleges & Courses

Predictors

Resources

Online Courses

Products

Engineering Preparation

Medical Preparation

Top Streams

Specializations

Resources

Top Providers

Exams

Colleges

Predictors

Resources

Resources

Exams

Colleges

Predictors & E-Books

Exams

Animation Courses

Colleges

Resources

Exams

Colleges

Resources

Top Courses & Careers

Colleges

Exams

Resources

Get Answers to all your Questions

All Questions

A random variable X has the following probability distribution: Then is equal to: Option: 1 Option: 2 Option: 3 Option: 4

Posted by

Kamlesh kumar

Let be the nth term of a G.P. of positive terms. If is equal to : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

Anakha

Crack CUET with india's "Best Teachers"

If for then Option: 1 Option: 2 Option: 3 Option: 4

Posted by

avinash.dongre

If be a complex number satisfying then cannot be : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

avinash.dongre

Crack NEET with "AI Coach"

The following system of linear equations has Option: 1 infinitely many solutions, satisfying Option: 2 infinitely many solutions, satisfying Option: 3 no solution Option: 4 only the trivial solution.

Let $a_{n}$ be the n^th term of a G.P. of positive terms. If $\sum_{n=1}^{100}a_{2n+1}=200\: \: and\: \: \sum_{n=1}^{100}a_{2n}=100,\: \: then\: \sum_{n=1}^{200}a_n$ is equal to :
Option: 1 $300$
Option: 2 $175$
Option: 3 $225$
Option: 4 $150$

If $x=\sum_{n=0}^{\infty }(-1)^{n}\tan ^{2n}\theta \: \: and\: \: y=\sum_{n=0}^{\infty }\cos ^{2n}\theta ,$ for $0<\theta < \frac{\pi }{4},$ then
Option: 1 $y(1+x)=1$
Option: 2 $x(1-y)=1$
Option: 3 $y(1-x)=1$
Option: 4 $x(1+y)=1$

If $z$ be a complex number satisfying $\left | Re\left ( z \right ) \right |+\left | Im(z) \right |=4,$ then $\left | z \right |$ cannot be :
Option: 1 $\sqrt{7}$

Option: 2 $\sqrt{\frac{17}{2}}$

Option: 3 $\sqrt{10}$

Option: 4 $\sqrt{8}$

The following system of linear equations $7x+6y-2z=0$ $3x+4+2z=0$ $x-2y-6z=0,$ has
Option: 1 infinitely many solutions, $(x,y,z)$ satisfying $y=2z.$
Option: 2 infinitely many solutions, $(x,y,z)$ satisfying $x=2z.$
Option: 3 no solution
Option: 4 only the trivial solution.

Let $a-2b+c=1.$ If $f\left ( x \right )= \begin{vmatrix} x+a & x+2 &x+1 \\ x+b &x+3 &x+2 \\ x+c &x+4 & x+3 \end{vmatrix},$ then :
Option: 1 $f(-50)=501$
Option: 2 $f(-50)=-1$
Option: 3 $f(50)=1$
Option: 4 $f(50)=-501$