Maths

  $f\left(x\right)+f\left(x+k\right)=n\forall x\in R,\&k>0............\left(i\right)$

Replace x by x + k.          

$f\left(x+k\right)+f\left(x+2k\right)=n...............\left(ii\right)$

From (i) & (ii), f(x + 2k) = f(x).

$\therefore f\left(x\right)$  is periodic with period = 2k.

$I_1={\displaystyle \underset{0}{\overset{4nk}{\int }}f\left(x\right)}dx=2x{\displaystyle \underset{0}{\overset{2k}{\int }}f\left(x\right)dx}...................\left(ii\right)$

$I_2={\displaystyle \underset{-k}{\overset{3k}{\int }}f\left(x\right)}dx$ put x = t + k

$={\displaystyle \underset{-2k}{\overset{2k}{\int }}t\left(t+k\right)dt}=2{\displaystyle \underset{0}{\overset{2k}{\int }}f\left(t+k\right)dt}$

$=2n{\displaystyle \underset{0}{\overset{2k}{\int }}ndx}=4n^{2}k.$

  $f\left(\frac{r}{4}\right)=\sqrt[]{2},f\left(\frac{r}{2}\right)=0\&f\text{'}\left(\frac{r}{2}\right)=1$

g(x)   ${\displaystyle \underset{}{\overset{}{\int }}\left(f\text{'}\left(t\right)sect+tantsectf\left(t\right)\right)dt}$

$={\left[f\left(t\right)sect\right]}_x^{\pi /4}$

=   $\sqrt[]{2}*\sqrt[]{2}-f\left(x\right)secx$

$=\frac{2cosx-f\left(x\right)}{cosx}$

=   $\frac{2sinx+1}{sinx}=3$

$f\left(x\right)=\{\begin{array}{cc}\hfill \left[x\right],x<0\hfill & \hfill \left|1-x\right|,x\ge 0\hfill \end{array}$ $g\left(x\right)=\{\begin{array}{cc}\hfill c^x-x,x<0\hfill & \hfill {\left(x-1\right)}^2-1,x\ge 0\hfill \end{array}$

fog

$fog(\begin{array}{}\end{array}\left[e^x-x\right],\left(e^x-x<0\right)\left(x<0\right)\left[{\left(x-1\right)}^2-1\right]{\left(x-1\right)}^2-1<0x\ge 0\left|1-e^x+x\right|,e^x-x\ge 0x<0\left|1-{\left(x+1\right)}^2+1\right|,{\left(x-1\right)}^2-1\ge 0x\ge 0,x\in \left(2,\infty \right)($

 Not possible as of inequalities give $?.$  

e^x – x < 0, x < 0                 (x 1)² – 1 < 0

Not possible                     (x – 1 + 1)(x – 1 – 1) < 0

(x)(x – 2) < 0

x   $\in \left(0,2\right)$

continuous  $\forall$ x < 0

$fog\{\begin{array}{ccc}\hfill \left|1-e^x+x\right|,x<0\hfill & \hfill \left|1-{\left(x-1\right)}^2+1\right|,x=0\hfill & \hfill \left|1-{\left(x-1\right)}^2+1\right|,x\ge 2\hfill \end{array}$ Discontinuous at 0

continuous   $\forall x$

  $\therefore$  fog is discontinuous at 0

  a + a₁……a_n, 100               n arithmetic mean 100

a + n = 33 ……. (i)

$\left(\frac{a_1}{a_n}=\frac{1}{7}\right)$

$\frac{\left(a+d\right)}{\left(100-d\right)}=\frac{7}{7}$

7a + 8d = 100…………… (ii)

a + (n + 1)d = 100………………. (iiI)

Solving these equations (i), (ii) & (iii), we get

n = 23 & d =    $\frac{15}{4}$

a = 10

  $\left(1-x^2+3x^3\right){\left(\frac{5}{2}{x}^3-\frac{1}{5x^2}\right)}^{11},x\ne 0$

(r + 1)^th term for expansion of   ${\left(\frac{5}{2}{x}^3-\frac{1}{5x^2}\right)}^{11}$

  n – r = x₃

r = y   $0\le x,y\le 11shouldbenaturalnumber.$

= ¹¹C_y   $*{\left(\frac{5}{2}\right)}^x*{\left(-\frac{1}{5}\right)}^y*x^{\left(3x-2y\right)}$

for term to be independent of x, power of x should be zero.

(3x – 2y) = 0 (when 1 is multiplied).

3x + 3y = 33

y = $\frac{33}{5}\left(Nosolution\right)$  

3x – 2y + 2 = 0 where (-x²) is multiplied).

3x +  3y = 33

y = 7, x = 4

$\begin{array}{cc}\hfill 3x-2y+3=0\hfill & \hfill 3x+3y=33\hfill \end{array}\}$

$\therefore$   coefficient of term independent of x.

$\left(-1\right){*}^{11}\text{?}{C}_7*{\left(\frac{5}{2}\right)}^4*{\left(\frac{-1}{5}\right)}^7$

$=\frac{33}{200}$

Let the number of chocolates given to C₁, C₂, C₃ & C₄ be a, b, c, d respectively.

Given 4   $\le b\le 7$

$2\le c\le 6$

Now using these the maximum number of chocolates that can be given to C₁ or C₄ is 24 (where b & c are given 2 & 4 chocolates).

$\therefore 0\le a\le 24$

$0\le d\le 24$

& a + b + c + d = 30

So, total possible solution to the above equation.

Coefficient of x³⁰ in.

$\left(x^0+x+x^2+....+x^{24}\right)\left(x^4+x^5+....+x^7\right)\left(x^2+x^3+....+x^6\right)\left(x^0+x+x^2+....+x^{24}\right)$

=  ${\left(1+....+x^{24}\right)}^2\left(x^4\right)\left(1+x+....+x^3\right)*x^2\left(1+x+....+x^4\right)$

$=\left(x^{56}-2x^{31}+x^6\right)*\left(x^9-x^4-x^5+1\right)*{\left(x-1\right)}^{-4}$

x⁵⁶ & x³¹ can never give x³⁰ so we discard them.

$x^6*x^9*{\left(x-1\right)}^{-4}{\to }^{15+4-1}\text{?}{C}_{4-1}{=}^{18}\text{?}{C}_3$

Coefficient x³⁰ ® ¹⁸C₃ – ²³C₃ – ²²C₃ + ²⁷C₃

=   $\frac{18*17*16}{6}-\frac{23*22*21}{6}-\frac{22*21*20}{6}+\frac{27*26*25}{6}$

= 430

$f\left(-2\right)+f\left(3\right)=0.$ One root of f (x) = 0 is (-1)

Let's assume other root to be a

$\therefore f\left(x\right)=a\left(x+1\right)\left(x-\alpha \right)$

Given that f (-2) + f (3) = 0

a (-2 + 1) (-2 -a) + a (3 + 1) (3 - a) = 0

Þ 14 -3a = 0 Þ a =    $\frac{14}{3}$

$\therefore$ sum of roots =   $\frac{14}{3}-1=\frac{11}{3}$

R₁ =  $\left\{\left(a,1\right)\in N*N:\left|a-b\right|\le 13\right\}$

 $\left(a,a\right)\in R_{1}as\left|a-a\right|\le 13\left(Reflexive\right)$

$\left(a,b\right)\&\left(b,a\right)\in R_{1}as\left|a-b\right|=\left|b-a\right|\to \left(symmetric\right).$

But it is not necessary that if (a,b) & (b, c)  $\in R,then\left(a,c\right)\in R$

Eg   $-\left(21,10\right)\in R\&\left(10,1\right)\in Rbut\left(21,1\right)\notin R_1$

R₂ =   $\left\{\left(a,b\right)\in N*N:\left|a-b\right|\ne 13\therefore R_2\to Notequivalencesolutoin.\right\}$

$\left(a,b\right)\&\left(b,a\right)\in R_{2}as\left|a-b\right|=\left|b-a\right|$

But it is not necessary that if (a, b) & (b, c)  $\in R_2$  then (a, c) also $\in R_2$ .

Eg – (21, 1)   $\in R_2\&\left(1,8\right)\in R_{2}but\left(21,8\right)\notin R_2$

$f\left(x\right)=a{x}^2+bx+c$  

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

? Andhra Pradesh = 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 $\left(I-m^2\right)N=-2I$

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

? α¹⁰⁰ + α² = 2

? α = $\pm 1$