Ncert Solutions Maths 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 $\left(\frac{\pi }{2}-\frac{2a}{\pi }\right)=sin\frac{2a}{\pi }$

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

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)

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

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

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

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

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

Equation of ${\perp }^r$ bisector of

$AB:y-3=\frac{t}{3}\left(x-t\right)$

For C put x = 0 so C $\left(0,3-\frac{t^2}{3}\right)$

$h=\frac{t}{2}\&K=\frac{6-\frac{t^2}{3}}{2}$

$2k=6-\frac{1}{3}*4h^2$          

$2x^2+3y-9=0$    

$ar\left(\Delta ABC\right)=\frac{1}{2}\left|\begin{array}{ccc}\hfill a\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill b\hfill & \hfill 2b+1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill b\hfill & \hfill 1\hfill \end{array}\right|=\frac{1}{2}\left[a\left(2b+1-b\right)+\left(b^2\right)\right]=\frac{1}{2}\left[ab+a+b^2\right]$  

Given ar (  $\Delta ABC)$ = 1 so |ab + a + b²| = 2

  $\Rightarrow a\left(b+1\right)+b^2=\pm 2$             

a = $\frac{-b^2+2}{b+1},\frac{-b^2-2}{b+1}$  

So sum = $\frac{-2b^2}{b+1}$  

Equation of tangent at P (2, -4)

y (-4) = 4 (x + 2)

x + y + 2 = 0

So, A (-2, 0)

Equation of normal at P:

y + 4 = 1 (x – 2)

x – y = 6

So, B (-2, -8)

For square mid-point of AB = mid-point of PQ

$\Rightarrow \frac{a+2}{2}=-2\Rightarrow a=-6$        

$\frac{b-4}{2}=\frac{-8}{2}\Rightarrow b=-4$   

So, 2a + b = -16

$\frac{sinA}{sinB}=\frac{sin\left(A-C\right)}{sin\left(C-B\right)}$

sin A sin C cos B – sin A cos C sin B = sin B sin A cos C – sin B cos A sin C

2 sin A sin B cos C = sin A sin C cos B + sin B sin C cos A

By sine rule $\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=k$

$2.ak.bk.\frac{\left(a^2+b^2-c^2\right)}{2ab}=ak.ck.\frac{\left(c^2+a^2-b^2\right)}{2ac}+bk.ck.\frac{\left(b^2+c^2-a^2\right)}{2bc}$

$\Rightarrow b^2,c^2,a^{2}areinA.P.$