Maths

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

Factors of 36 = 2².3².1

Five-digit combinations can be

(1, 2, 3, 3), (1, 4, 3, 1), (1, 9, 2, 1), (1, 4, 9, 11), (1, 2, 3, 6, 1), (1, 6, 1, 1)

i.e., total numbers $\frac{5!5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{3!}+\frac{5!}{2!}+\frac{5!}{3!2!}=\left(30*3\right)+20+60+10=180.$

$d_1=\frac{199-100}{2}\cancel{\in }l$

$d_2=\frac{199-100}{3}=33$

$d_3=\frac{199-100}{4}\cancel{\in }l$

$d_n=\frac{199-100}{i+1}\in l$

$\Rightarrow di=33+11or9$

$\therefore$ sum of common differences = 33 + 11 + 9 = 53

Let and are the roots of $\left(p^2+q^2\right)x^2-2q\left(p+r\right)x+q^2+r^2=0$

$\therefore \alpha +\beta >0and\alpha \beta >0$ Also, it has a common root with x² + 2x – 8 = 0

$\therefore$ The common root between above two equations is 4.

$\Rightarrow 16\left(p^2+q^2\right)-8q\left(p+r\right)+q^2+r^2=0\Rightarrow \left(16p^2-8pq+q^2\right)+\left(16q^2-8qr+r^2\right)=0$

$\Rightarrow {\left(4p-q\right)}^2+{\left(4q-r\right)}^2=0\Rightarrow q=4pandr=16p\therefore \frac{q^2+r^2}{p^2}=\frac{16p^2+256p^2}{p^2}=272$