Class 11th

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

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.

${\left(3x^2+4x+3\right)}^2-\left(k+1\right)\left(3x^2+4x+3\right)\left(3x^2+4x+2\right)+k{\left(3x^2+4x+2\right)}^2=0$

 Let $3x^2+4x+3=a\&3x^2+4x+2=b\Rightarrow a-1$  

So, (i) becomes a² – (k + 1)ab + kb² = 0

(a – kb) (a – b) = 0 Þ a = kb or a = b ® not possible

->3x² + 4x + 3 = k (3x² + 4x + 2)

For real roots $D\ge 0$  

$16{\left(k-1\right)}^2-12\left(k-1\right)\left(2k-3\right)\ge 0$     

So, $k\in \left(1,\frac{5}{2}\right]$  

Ge (Z = 32)

^{$=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^2$}         

$m_{\mathcal{l}}=0$       (for 4s, 3s, 2s, 1s) 4 orbital

$m_{\mathcal{l}}$       = 0 (one p orbital of 2p and 3p)

$m_{\mathcal{l}}$ = 0 (one d orbital)

Total orbitals = 7

Ans. = 7

Variance = $\frac{\sum {\left(x_i-\overline{x}\right)}^2}{n}=\frac{\sum \left({x_i}^2+{\overline{x}}^2-2\overline{x}xi\right)}{n}$  

$=\frac{\sum {x_i}^2+{\overline{x}}^2\sum 1-2\overline{x}\sum x_i}{n}$

$=\frac{\frac{n\left(n+1\right)\left(2n+1\right)}{6}+{\left(\frac{n\left(n+1\right)}{2n}\right)}^2.n-\frac{2\left(n+1\right)}{2n}.\frac{n\left(n+1\right)}{2}}{n}$ $=\frac{n^2-1}{12}$

Now, $\frac{n^2-1}{12}=14son=13$