Maths

Set of first 10 prime numbers

= {2,3,5,7,11,13,17,19,23,29,31}

So sample space = 104.

Favourable cases

So required probability

$=\frac{10+4{*}^{10}\text{?}{C}_2}{10^4}=\frac{10+\frac{4*10.9}{2}}{10^4}=\frac{19}{1000}$

Let PT perpendicular to QR

$\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}=\lambda \Rightarrow T\left(2\lambda -1,3\lambda -2,2\lambda +1\right)$ therefore

$2\left(2\lambda -5\right)+3\left(3\lambda -4\right)+2\left(2\lambda -6\right)=0\Rightarrow \lambda =2$

$T\left(3,4,5\right)\therefore PT=\sqrt[]{1+4+4}=3\therefore QT=\sqrt[]{26-9}=\sqrt[]{17}$

$\therefore \Delta PQR=\frac{1}{2}*2\sqrt[]{17}*3=3\sqrt[]{17}$

Therefore square of $ar\left(\Delta PQR\right)$ = 153.

 $f\text{'}\left(x\right)=\frac{4x^2?1}{x}$ so f (x) is decreasing in $\left(0,\frac{1}{2}\right)and\text{?}\text{?}\left(\frac{1}{2},?\right)$ $?a=\frac{1}{2}$

Tangent at y² = 2x is y = mx + $\frac{1}{2m}$ it is passing through (4, 3) therefore we get m = $\frac{1}{2}or\text{?}\text{?}\frac{1}{4}$

So tangent may be $y=\frac{1}{2}x+1\text{?}\text{?}or\text{?}\text{?}y=\frac{1}{4}x+2\text{?}\text{?}\text{?}but\text{?}\text{?}y=\frac{1}{2}x+1$ passes through (-2, 0) so rejected.

Equation of normal $\frac{x}{9}+\frac{y}{36}=1$

Slope of AH = $\frac{a+2}{1}$ slope of BC = $-\frac{1}{p}\therefore p=a+2$ $\therefore C\left(\frac{18p-30}{p+1},\frac{15p-33}{p+1}\right)$

slope of HC = $\frac{16p-p^2-31}{16p-32}$

slope of BC * slope of HC = -1 p = 3 or 5

hence p = 3 is only possible value.

 ${\displaystyle \underset{3}{\overset{x}{\int }}f\left(x\right)dx}={\left(\frac{f\left(x\right)}{x}\right)}^3\Rightarrow x^3{\displaystyle \underset{3}{\overset{x}{\int }}f\left(x\right)dx}=f^3\left(x\right),$ differentiating w.r.to x

$x^{3}f\left(x\right)+3x^2\frac{f^3\left(x\right)}{x^3}=3f^2\left(x\right)f\text{'}\left(x\right)\Rightarrow 3y^2\frac{dy}{dx}=x^{3}y=\frac{3y^3}{x}\Rightarrow 3xy\frac{dy}{dx}=x^4+3y^2$

After solving we get $y^2=\frac{x^4}{3}+c{x}^2$ also curve passes through (3, 3) c = -2

$\therefore y^2=\frac{x^4}{3}-2x^2$ which passes through $\left(\alpha ,6\sqrt[]{10}\right)$ $\therefore \frac{{\alpha }^4-6{\alpha }^2}{3}=360\Rightarrow \alpha =6$

 ${\displaystyle \underset{0}{\overset{1}{\int }}1.{\left(1-x^n\right)}^{2n+1}dx}$ using by parts we get,

$\left(2n^2+n+1\right){\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-x^n\right)}^{2n+1}dx=1177{\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-x^n\right)}^{2n+1}dx}}$

$\Rightarrow 2n^2+n+1=1177\Rightarrow n=24or-\frac{49}{2}\therefore n=24$

Coefficient of x in ${\left(1+x\right)}^p{\left(1-x\right)}^q={-}^p{C_0}^q{C}_1{+}^p{C_1}^q{C}_0=-3\Rightarrow p-q=3$

Coefficient of x² in ${\left(1+x\right)}^p{\left(1-x\right)}^q{=}^p{C_0}^q{C}_2{-}^p{C_1}^q{C}_1{-}^p{C_2}^q{C}_0=-5$

$\Rightarrow \frac{q\left(q-1\right)}{2}-pq+\frac{p\left(p-1\right)}{2}=-5\Rightarrow \frac{q\left(q-1\right)}{2}-\left(q-3\right)q+\frac{\left(q-3\right)\left(q-4\right)}{2}=-5\Rightarrow q=11,p=8$

Coefficient of x³ in ${\left(1+x\right)}^8{\left(1-x\right)}^{11}={-}^{11}{C}_3{+}^8{C_1}^{11}{C}_2{-}^8{C_2}^{11}{C}_1{+}^8{C}_3=23$

 $x^8-x^7-x^6+x^5+3x^4-4x^3-2x^2+4x-1=0$

$\Rightarrow x^7\left(x-1\right)-x^5\left(x-1\right)+3x^3\left(x-1\right)-x\left(x^2-1\right)+2x\left(1-x\right)+\left(x-1\right)=0$

$\Rightarrow \left(x-1\right)\left(x^2-1\right)\left(x^5+3x-1\right)=0\therefore x=\pm 1$ are roots of above equation and x5 + 3x – 1 is a monotonic term hence vanishes at exactly one value of x other then 1 or 1.

$\therefore$ 3 real roots.

Given series $\left\{3*1\right\},\left\{3*2,3*3,3*4\right\},\left\{3*5,3*6,3*7,3*8,3*9\right\}.........$

$\therefore$ 11th set will have 1 + (10)2 = 21 terms

Also up to 10th set total 3 * k type terms will be 1 + 3 + 5 + ……… +19 = 100 terms

$\therefore Set11=\left\{3*101,3*102,......3*121\right\}$ $\therefore$ Sum of elements = 3 * (101 + 102 + ….+121)

$=\frac{3*222*21}{2}=6993.$

A = $\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)$

$A^2=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill a^2+bc\hfill & \hfill ab+bd\hfill \\ \hfill ac+dc\hfill & \hfill ac+d^2\hfill \end{array}\right)$

a2 + bc = bc + d2 = 1 ………. (i)

and b (a + d) = c (a + d) = 0 ……… (ii)

Case 1

b = c = 0

then possible ordered pair of

(a, d) $\equiv$  (1, 1) (-1, -1) (-1, 1) (1, -1) total 4 possible case

Case 2

a = -d

then (a, d) $\equiv$  (-1, 1) (1, -1)

then bc = 0

now if b = 0

then possible choice for {-1, 0, 1, 2, …….10} = 12

Similarly if c = 0 then possible choice for $b\in \left\{-1,0,1,2,......10\right\}$ is = 12

but (0, 0) counted twice

$\therefore$ bc = 0 in (12 + 12 – 1) = 23 ways

$\therefore$ total number of ways = 2 * 23 = 46

$\therefore$ total number of required matrices = 46 + 4 = 50