Class 11th

If two circles intersect at two distinct points

->|r₁ – r₂| < C₁C₂ < r₁ + r₂

| r – 2| <  $\sqrt[]{9+16}$ < r + 2

|r – 2| < 5                     and r + 2 > 5

–5 < r 2 < 5                    r > 3                      … (2)

–3 < r < 7                                                        (1)

From (1) and (2)

3 < r < 7

$\underset{n\to \infty }{lim}{\displaystyle \sum _{k=1}^n\frac{n^3}{n^4\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3k^2}{n^2}\right)}}$

$=\underset{n\to \infty }{lim}\frac{1}{n}{\displaystyle \sum _{k-1}^n\frac{1}{\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3k^2}{n^2}\right)}}$

$={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dx}{3\left(1+x^2\right)\left(\frac{1}{3}+x^2\right)}}$

$={\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{3}*\frac{3}{2}\frac{\left(x^2+1\right)-\left(x^2+\frac{1}{3}\right)}{\left(1+x^2\right)\left(x^2+\frac{1}{3}\right)}dx}$

$=\frac{1}{2}{\left[\sqrt[]{3}ta{n}^{-1}\left(\sqrt[]{3}x\right)\right]}_0^1-\frac{1}{2}{\left(ta{n}^{-1}x\right)}_0^1$

$=\frac{\sqrt[]{3}}{2}\left(\frac{\pi }{3}\right)-\frac{1}{2}\left(\frac{\pi }{4}\right)=\frac{\pi }{2\sqrt[]{3}}-\frac{\pi }{8}$

S₂₀ = $\frac{20}{2}$ [2a + 19d] = 790

2a + 19d = 79              . (1)

$S_{10}\frac{10}{2}\left[2a+9d\right]=145$

2a + 9d = 29                . (2)

from (1) and (2) a = –8, d = 5

$S_{15}-S_5=\frac{15}{2}\left[2a+14d\right]-\frac{5}{2}\left[2a+4d\right]$

$=\frac{15}{2}\left[-16+70\right]-\frac{5}{2}\left[-16+20\right]$  

= 405 – 10

= 395

Release of toxic/undesirable materials in the environment.

Krypton (Kr)  belongs to the zero group element as per Mendeleev's periodic table.

Cu : 4s¹3d¹⁰ Cu⁺ : 4s⁰3d¹⁰

Zn : 4s²3d¹⁰ Zn²⁺ : 4s⁰3d¹⁰

Most stable due to aromatic character. It has 2pe^– and follow (4n + 2)pe^– Huckel rule.

Delocalisation of  electrons in benzene  leads to saturation

BCl₃ is having bond angles of