Class 11th

${\left(si{n}^{-1}x\right)}^2-{\left(co{s}^{-1}x\right)}^2=a,0<x<1$           

$\left(si{n}^{-1}x+co{s}^{-1}x\right)\left(si{n}^{-1}x-co{s}^{-1}x\right)=a$

$2co{s}^{-1}x=\frac{\pi }{2}-\frac{2a}{\pi }..........\left(i\right)$    

Let $co{s}^{-1}x=\theta thenx=cos\theta$

So, $2x^2-1=2co{s}^2\theta -1=cos2\theta ..........\left(ii\right)$

Now, $2\theta =2co{s}^{-1}x=\frac{\pi }{2}-\frac{2a}{\pi }from\left(i\right)$

So, cos 2θ = cos $\left(\frac{\pi }{2}-\frac{2a}{\pi }\right)=sin\frac{2a}{\pi }$

S = {1, 2, 3, 4, 5, 6, 9}

Elements of type 3n -> 3, 6, 9

Type 3n + 1 ->1, 4

3n + 2 -> 2, 5

Number of subset of S containing one element which are not divisible by  $3{=}^2{C}_1{+}^2{C}_1=4$ number of subset of S containing two numbers whose sum is not divisible $3{=}^3{C}_1{*}^2{C}_1{+}^3{C}_1{*}^2{C}_1{+}^2{C}_2{+}^2{C}_2=14$ by

Number of subset of S containing 3 elements whose sum is not divisible by

Number of subset containing 4 elements whose sum is not divisible by 3

Number of subset of S containing 6 elements = 4

Hence total subset = 80

sin^4_θ + cos⁴θ - sinθ cosθ = 0

${\left(si{n}^2\theta +co{s}^2\theta \right)}^2-2si{n}^2\theta co{s}^2\theta -sin\theta cos\theta =0$

$\frac{{\left(2sin\theta cos\theta \right)}^2}{2}-\frac{2sin\theta cos\theta }{2}+1=0$               

$si{n}^{2}2\theta +sin2\theta -2=0$              

sin 2θ = 1 .(i)

$as\theta \in \left[0,4\pi \right]so2\theta \in \left[0,8\pi \right]$              

$\left(i\right)2\theta =\frac{\pi }{2},\frac{5\pi }{2},\frac{9\pi }{2},\frac{18\pi }{2}$

Hence $\frac{8s}{\pi }=56$

B. O. D. value < 5 ppm for clean water and B.O.D value of polluted water 17 ppm.

Given let α, β be the roots of the equation x² + bx + c = 0

So, ${\alpha }^2+b\alpha +c=0\&{\beta }^2+b\beta +c=0$

Also x² + bx + c = (x - α) (x - β)

$NowL=\underset{x\to \beta }{lim}\frac{e^{2\left(x^2+bx+c\right)}-1-2\left(x^2+bx+c\right)}{{\left(x-\beta \right)}^2}$           

$\underset{x\to \beta }{lim}\frac{2{\left(x-\alpha \right)}^2{\left(x-\beta \right)}^2+\frac{8}{6}{\left(x-\alpha \right)}^3{\left(x-\beta \right)}^3+........}{{\left(x-\beta \right)}^2}$

=  $2{\left(\beta -\alpha \right)}^2=2\left[{\left(\beta +\alpha \right)}^2-4\alpha \beta \right]=2\left[b^2-4c\right]$

Portland cement contains

Dicalcium silicate = 26%

Tricalcium silicate = 51%

Tricalcium aluminate = 11%

Hence major percentage is of tricalcium silicate

number of elements in $A\cap B=5$ which is

(0, 0) (1, 0) (1, 1) (1, -1) (2, 0)

Similarly number of elements in $A\cap C=5$ which is

(2, 0) (2, 2) (1, 1) (2, 1) (3, 1)

Hence number of relation from

$\left(A\cap B\right)to\left(A\cap C\right)=2^{5*5}=2^{25}$            

->P = 25

$\frac{z-i}{z+2i}\in R$

So, $\frac{z-i}{z+2i}=\left(\frac{\overline{z-i}}{z+2i}\right)$

$\frac{z-i}{z+2i}=\frac{\overline{z}+i}{\overline{z}-2i}orz\overline{z}-i\overline{z}-2iz-2=z\overline{z}+2i\overline{z}+iz-2$       

$\Rightarrow z+\overline{z}=0$    

=> z is purely imaginary

i.e. x = 0 if z = x + iy

so, z = iy

=> S is a straight line in complex plane

$\left(p\wedge \left(p\to q\right)\wedge \left(q\to r\right)\right)\to r$

$\left(\left(p\wedge q\right)\wedge \left(\sim p\vee r\right)\right)\to r$

$=\sim \left(p\wedge q\wedge r\right)\vee r$

$=\sim p\vee \sim q\vee \sim r\vee r$

=> tautology

In $\Delta BCD,tan?=\frac{x}{a+b}........\left(i\right)$  

In $\Delta APC,tan\left(\theta +?\right)=\frac{x}{b}..........\left(ii\right)$  

Now tan θ = tan $\left(\theta +?-?\right)$  

$=\frac{tan\left(\theta +?\right)-tan?}{1+tan\left(\theta +?\right)tan?}$  

given , $tan\theta =\frac{1}{2}so\frac{ax}{b\left(a+b\right)+x^2}=\frac{1}{2}$  

-> x² – 2ax + b (a + b) = 0