Class 11th

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 $\left(0,3-\frac{t^2}{3}\right)$

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

$2k=6-\frac{1}{3}*4h^2$          

$2x^2+3y-9=0$    

(2E) 2-Bromo hex – 2-en - 4-yne

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

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 = $\frac{-2b^2}{b+1}$  

Synthetic gas is a syn-gas which is also called water gas i.e. mixture of (CO + H₂)

$\underset{\left(Redhot\right)}{C+H_{2}O}\to \underset{\left(1:1\right)mixture}{CO+H_2}$

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

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

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

Locus of point of intersection of perpendicular tangent will be its director circle & director circle of parabola be its directrix.

Given parabola y² = 16 (x – 3) so equation of directrix be x – 3 = - 4 i.e., x + 1 = 0

$y\left(x\right)=co{t}^{-1}\left(\frac{\sqrt[]{1+sinx}+\sqrt[]{1-sinx}}{\sqrt[]{1+sinx}-\sqrt[]{1-sinx}}\right),x\in \left(\frac{\pi }{2},\pi \right)$

$=co{t}^{-1}\left(\frac{sin\frac{x}{2}+cos\frac{x}{2}+sin\frac{x}{2}-cos\frac{x}{2}}{sin\frac{x}{2}+cos\frac{x}{2}-sin\frac{x}{2}+cos\frac{x}{2}}\right)=co{t}^{-1}tan\frac{x}{2}=co{t}^{-1}cot\left(\frac{\pi }{2}-\frac{\pi }{2}\right)=\frac{\pi }{2}-\frac{x}{2}$

$\frac{dy}{dx}=\frac{-1}{2}$

Assertion (A) is correct & Reason (R) is incorrect

→character decreases from left to right & non metallic character increases from left to night

→I.E. increases & electron gain enthalpy also increases from left to right.