Maths

$f\left(x\right)=[\begin{array}{ccc}\hfill 2n\hfill & \hfill \hfill & \hfill n=2,4,6....\hfill \\ \hfill n-1\hfill & \hfill \hfill & \hfill n=3,7,11,15....\hfill \\ \hfill \frac{n+1}{2}\hfill & \hfill \hfill & \hfill n=1,5,9,13....\hfill \end{array}$

for n = 2, 4, 6 ……

f (n) = 4, 8, 12, ….4 (n) form        

for n = 3, 7, 11, 15, ….

f (n) = 1, 3, 5, 7, …. (4n + 1) or, (4n + 3) from

f is one and onto.

$f\left(x\right)=a{x}^2+bx+c$

g (x) = px + q

$f\left(g\left(x\right)\right)=a{\left(px+q\right)}^2+b\left(p\right)\left(+q\right)+c$  

$\Rightarrow 8x^2-2x=a{\left(px+q\right)}^2+b\left(px+q\right)+c$  

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

$9\left(f\left(x\right)\right)=p\left(a{x}^2+bx+c\right)+q$  

->4x² + 6x + 1 = apx² + bpx + cp + q

->ap = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

->b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

$N=M^2+M^4+.....+M^{98}$

$=\left(-{\alpha }^{2}I\right)+{\left(-{\alpha }^{2}I\right)}^2+....+{\left(-{\alpha }^{2}I\right)}^{49}$

$=I\left(-{\alpha }^2+{\alpha }^4-{\alpha }^6+....-{\alpha }^{98}\right)$

N =  $-I\left({\alpha }^2-{\alpha }^4+{\alpha }^6.......+{\alpha }^{98}\right)$

$=-I\frac{{\alpha }^2\left(1-{\left(-{\alpha }^2\right)}^{49}\right)}{1-\left(-{\alpha }^2\right)}$

N =  $\frac{-I{\alpha }^2\left(1+{\alpha }^{98}\right)}{1+{\alpha }^2}$  

Now  $\left(I-m^2\right)N=-2I$

$\left(I+{\alpha }^{2}I\right)\frac{\left(-I{\alpha }^2\cdot (1+{\alpha }^{98}\right)}{1+{\alpha }^2}=-2I$  

-> a¹⁰⁰ + a² = 2

->a = $\pm 1$

Fix the unit place, find the chances for the first three digits

unit digit as 1, total ways = 9.10²

unit digit as 2, total ways = 4.5²

unit digit as 3 total ways = 3.4²

unit digit as 4 total ways = 2.3²

unit digit as 5 total ways = 1.2²

unit digit as 6 total ways = 1.2²

unit digit as 7 total ways = 1.2²

unit digit as 8 total ways = 1.2²

unit digit as 9 total ways = 1.2²

$T_{r+1}{=}^{15}\text{?}{C}_r\cdot {\left(2x^{1/5}\right)}^{15-r}{\left(-x^{-1/5}\right)}^r$  

= ¹⁵C_{r $\cdot 2^{15-r}\cdot x^{\frac{15-2r}{5}}\cdot {\left(-1\right)}^r$}

Coefficient of x^-1 -> r = 10 ->m = ¹⁵C_{10 $\cdot 2^5$}

$x^{-3}\Rightarrow r=15\Rightarrow n{=}^{15}\text{?}{C}_{15}\cdot 2^0=1$  

now mn² = ¹⁵C_{r $\cdot 2^r$}  

r = 5

f (x) is an even function

$f\left(-\frac{1}{4}\right)=f\left(-\frac{1}{2}\right)=f\left(\frac{1}{2}\right)=f\left(\frac{1}{4}\right)=0$  

So, f (x) has at least four roots in (-2, 2)

$g\left(\frac{-3}{4}\right)=g\left(\frac{3}{4}\right)=0$  

So, g (x) has at least two roots in (-2, 2)

now number of roots of f (x) $\cdot g"\left(x\right)=f\text{'}\left(x\right)\cdot g\text{'}\left(x\right)=0$  

It is same as number of roots of  $\frac{d}{dx}\left(f\left(x\right)\cdot g\text{'}\left(x\right)\right)=0$ will have atleast 4 roots in (-2, 2)

Given a > b

Area common to x² + y² $\le a^{2}and\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1$  

is  $\pi a^2-\pi ab=30\pi ..............\left(i\right)$  

Similarly  $\pi ab-\pi b^2=18\pi .................\left(ii\right)$  

Equation (i) and equation (ii)  $\frac{a}{b}=\frac{5}{3}$  

Equation (i) + equation (ii)  $a^2-b^2=48$  

a² = 75, b² = 27

sin x = 1 – sin² x

sin x =  $\frac{-1+\sqrt[]{5}}{2},\frac{-1-\sqrt[]{5}}{2}\left(rejected\right)$  

draw y = sin x

y =  $\frac{\sqrt[]{5}-1}{2},$ find their pt. of intersection.