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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

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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

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Anakha

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If one end of a focal chord AB of the parabola $y^{2}=8x$ is at $A\left ( \frac{1}{2},-2 \right ),$ then the equation of the tangent to it at B is :
Option: 1 $x+2y+8=0$
Option: 2 $2x-y-24=0$
Option: 3 $x-2y+8=0$
Option: 4 $2x+y-24=0$

Length of the Latus rectum and parametric form -

Parametric Equation:

From the equation of the parabola, we can write
$\\\frac{y}{2a}=\frac{2x}{y}=t\text{ here, t is a parametewr}\\\\\text{Then, }x=at^2 \text{ and }y=2at\text{ are called the parametric equations }\\\text{and the point } (at^2,2at)\text{ lies on the parabola.}$

Tangents of Parabola in Point Form -

Tangents of Parabola in Point Form

$\\ {\text { Equation of the tangent to the parabola } \mathrm{y}^{2}=4 \mathrm{ax} \text { at the point } \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { is }} \\ {\mathrm{y} \mathrm{y}_{1}=2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_{1}\right)}$

$y^{2}=8 x \text { then } A\left(2 t_{1}^{2}, 4 t_1\right) \\ {\text { given } A\left(\frac{1}{2},-2\right) \Rightarrow t_{1}=-1 / 2}$

$t_1\cdot t_2=-1$

$\text { then } t_{2}=2 \Rightarrow B(8,8) \\ \text { Equation of tangent } 8 y=4(x+8) \\ {2 y=x+8}$

Correct Option 3

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avinash.dongre

If $\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{xy}{x^{2}+y^{2}};y(1)=1;$ then a value of $x$ satisfying $y(x)=e$ is :
Option: 1 $\sqrt{3}\: e$

Option: 2 $\frac{1}{2}\sqrt{3}\: e$

Option: 3 $\sqrt{2}\: e$

Option: 4 $\frac{e}{\sqrt{2}}$

Option: 3

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Ayesha Sabeela

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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)

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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)

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The length of the minor axis (along y-axis) of an ellipse in the standard form is $\frac{4}{\sqrt{3}}.$ If this ellipse touches the line, $x+6y=8;$ then its eccentricity is :
Option: 1 $\frac{1}{2}\sqrt{\frac{5}{3}}$

Option: 2 $\frac{1}{2}\sqrt{\frac{11}{3}}$

Option: 3 $\sqrt{\frac{5}{6}}$

Option: 4 $\frac{1}{3}\sqrt{\frac{11}{3}}$

What is Ellipse? -

Ellipse

Standard Equation of Ellipse:

The standard form of the equation of an ellipse with center (0, 0) and major axis on the x-axis is

$\mathbf{\frac{\mathbf{x}^{2}}{\mathbf{a}^{2}}+\frac{\mathbf{y}^{2}}{\mathbf{b}^{2}}=1} \quad \text { where }, \mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)_{(\mathrm{a}>\mathrm{b})}$

a > b
the length of the major axis is 2a
the coordinates of the vertices are (±a, 0)
the length of the minor axis is 2b
the coordinates of the co-vertices are (0, ±b)

Equation of Tangent of Ellipse in Parametric Form and Slope Form -

Slope Form:

$\\ {\text { The equation of tangent of slope m to the ellipse, } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { are }} \\ {y=m x \pm \sqrt{a^{2} m^{2}+b^{2}} \text { and coordinate of point of contact is }} \\ {\left(\mp \frac{a^{2} m}{\sqrt{a^{2} m^{2}+b^{2}}}, \pm \frac{b^{2}}{\sqrt{a^{2} m^{2}+b^{2}}}\right)}$

$\\\begin{array}{l}{2 \mathrm{b}=\frac{4}{\sqrt{3}} \quad \Rightarrow \quad \mathrm{b}=\frac{2}{\sqrt{3}}} \\ {\text { Equation of tangent } \equiv \mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^{2} \mathrm{m}^{2}+\mathrm{b}^{2}}}\end{array}\\\text { comparing with } \equiv y=\frac{-x}{6}+\frac{4}{3}\\$

$\\ {\mathrm{m}=\frac{-1}{6} \text { and } \mathrm{a}^{2} \mathrm{m}^{2}+\mathrm{b}^{2}=\frac{16}{9}} \\ {\Rightarrow \quad \frac{\mathrm{a}^{2}}{36}+\frac{4}{3}=\frac{16}{9}} \\ {\Rightarrow \quad \frac{\mathrm{a}^{2}}{36}=\frac{16}{9}-\frac{4}{3}=\frac{4}{9}} \\ {\Rightarrow \quad a^{2}=16} \\ {\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}}\\e=\sqrt{\frac{11}{12}}$

Correct Option (2)

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avinash.dongre

If $p\rightarrow \left ( p\: \wedge \sim q \right )$ is false, then the truth values of p and q are respectively :
Option: 1 F, T
Option: 2 T, F
Option: 3 F, F
Option: 4 T, T

Relation Between Set Notation and Truth Table -

Sets can be used to identify basic logical structures of statements. Statements have two fundamental roles either it is true or false.

Let us understand with an example of two sets p{1,2} and q{2,3}.

$\begin{array}{|c|c|c|}\hline\quad p\vee q\quad & \quad p\cup q\quad&\quad 1,2,3\quad \\ \hline p\wedge q& p\cap q&2 \\ \hline p^c& \sim p & 3,4 \\ \hline q^c& \sim q&1,4 \\ \hline\end{array}$

Using this relation we get

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}\hline \text{Element } & \mathrm{\;\;\;\;\;}p\mathrm{\;\;\;\;\;}&\mathrm{\;\;\;}q\mathrm{\;\;\;}&\mathrm{\;\;\;\;\;}\sim p\mathrm{\;\;\;\;\;}&\mathrm{\;\;\;}\sim q\mathrm{\;\;\;} &\mathrm{\;\;\;}p\wedge q\mathrm{\;\;}&\mathrm{\;\;}p\vee q\mathrm{\;\;}&\sim\left (p\wedge q \right )\mathrm{\;\;}&\sim p\wedge\sim q\mathrm{\;\;} \\ \hline \hline 1& \mathrm{T}&\mathrm{F} & \mathrm{F} &\mathrm{T}&\mathrm{F}&\mathrm{T}&\mathrm{T}&\mathrm{F} \\ \hline2& \mathrm{T}&\mathrm{T} & \mathrm{F} &\mathrm{F}&\mathrm{T}&\mathrm{T}&\mathrm{F}&\mathrm{T} \\ \hline 3& \mathrm{F}&\mathrm{T} & \mathrm{T} &\mathrm{F}&\mathrm{F}&\mathrm{T}&\mathrm{T}&\mathrm{F} \\ \hline4& \mathrm{F}&\mathrm{F} & \mathrm{T} &\mathrm{T}&\mathrm{F}&\mathrm{F}&\mathrm{T}&\mathrm{F} \\ \hline\end{array}$

Practise Session - 2 -

Q1. Write the truth table for the following statement pattern:
$\\ \mathrm{1.\;\;\;\sim(p \vee((\sim q \Rightarrow q) \wedge q))}\\\mathrm{2.\;\;\;(p\wedge q)\vee(\sim r)}\\$

$\begin{array}{c|c|c}{\mathbf{p}} & {\mathbf{q}} & {(\mathbf{p} \rightarrow(\mathbf{p} \mathbf{\Lambda}-\mathbf{q}))} \\ \hline F & {F} & {\mathbf{T}} \\ \hline F & {T} & {\mathbf{T}} \\ \hline \mathbf{T} & {F} & {\mathbf{T}} \\ \hline \mathbf{T} & {T} & {F}\end{array}$

Correct Option (4)

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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

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avinash.dongre

Given : $f(x)=\left\{\begin{matrix} x, &0\leq x< \frac{1}{2} \\ \frac{1}{2}, &x=\frac{1}{2} \\ 1-x, &\frac{1}{2} < x\leq 1 \end{matrix}\right.$ and $g(x)=\left ( x-\frac{1}{2} \right )^{2},\: x\epsilon \textbf{R}.$ Then the area (in sq. units) of the region bounded b the curves, $y=f(x)$ and $y=g(x)$ between the lines, $2x=1\: \: \: and\: \: \: 2x=\sqrt{3},$ is :
Option: 1 $\frac{\sqrt{3}}{4}-\frac{1}{3}$

Option: 2 $\frac{1}{3}+\frac{\sqrt{3}}{4}$

Option: 3 $\frac{1}{2}+\frac{\sqrt{3}}{4}$

Option: 4 $\frac{1}{2}-\frac{\sqrt{3}}{4}$

Area Bounded by Curves When Intersects at More Than One Point -

Area bounded by the curves y = f(x), y = g(x) and intersect each other in the interval [a, b]

First find the point of intersection of these curves y = f(x) and y = g(x) , let the point of intersection be x = c

Area of the shaded region

$=\int_{a}^{c}\{f(x)-g(x)\} d x+\int_{c}^{b}\{g(x)-f(x)\} d x$

When two curves intersects more than one point

rea bounded by the curves y=f(x), y=g(x) and intersect each other at three points at x = a, x = b amd x = c.

To find the point of intersection, solve f(x) = g(x).

For x ∈ (a, c), f(x) > g(x) and for x ∈ (c, b), g(x) > f(x).

Area bounded by curves,

$\\\mathrm{A=} \int_{a}^{b}\left |f(x)-g(x) \right |dx\\\\\mathrm{\;\;\;\;=} \int_{a}^{c}\left ( f(x)-g(x) \right )dx+\int_{c}^{b}\left ( g(x)-f(x) \right )dx$

Required area = Area of trapezium ABCD - $\int_{1/2}^{\sqrt3/2}\left ( x-\frac{1}{2} \right )^2dx$

$\\=\frac{1}{2}\left(\frac{\sqrt{3}-1}{2}\right)\left(\frac{1}{2}+1-\frac{\sqrt{3}}{2}\right)-\frac{1}{3}\left(\left(x-\frac{1}{2}\right)^{3}\right)_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\\=\frac{\sqrt3}{4}-\frac{1}{3}$

Correct Option (1)

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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

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If one end of a focal chord AB of the parabola is at then the equation of the tangent to it at B is : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

avinash.dongre

If then a value of satisfying is : Option: 1 Option: 2 Option: 3 Option: 4

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Ayesha Sabeela

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

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 one end of a focal chord AB of the parabola $y^{2}=8x$ is at $A\left ( \frac{1}{2},-2 \right ),$ then the equation of the tangent to it at B is :
Option: 1 $x+2y+8=0$
Option: 2 $2x-y-24=0$
Option: 3 $x-2y+8=0$
Option: 4 $2x+y-24=0$

If $\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{xy}{x^{2}+y^{2}};y(1)=1;$ then a value of $x$ satisfying $y(x)=e$ is :
Option: 1 $\sqrt{3}\: e$

Option: 2 $\frac{1}{2}\sqrt{3}\: e$

Option: 3 $\sqrt{2}\: e$

Option: 4 $\frac{e}{\sqrt{2}}$

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 $\sqrt{8}$

If $p\rightarrow \left ( p\: \wedge \sim q \right )$ is false, then the truth values of p and q are respectively :
Option: 1 F, T
Option: 2 T, F
Option: 3 F, F
Option: 4 T, T

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.