Maths

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

Area of ?

$=\frac{1}{2}\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill x\hfill & \hfill y\hfill & \hfill 1\hfill \\ \hfill -x\hfill & \hfill y\hfill & \hfill 1\hfill \end{array}\right|$

-> $\left|\frac{1}{2}\left(xy+xy\right)\right|=\left|xy\right|$

->Area (D) = |xy| = |x (– 2x² + 54x)|

$\frac{d\left(\Delta \right)}{dx}=\left|\left(-6x^2+108x\right)\right|\Rightarrow \frac{d\Delta }{dx}=0$  at x = 0 and 18

->at x = 0, minima

and at x = 18 maxima

Area (D) = |18 (– 2 (18)² + 54 * 18)| = 5832

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

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

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

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

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

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

(a – 1) * 2 + (b – 2) * 5 + (g – 3) * 1 = 0

2a + 5b + g – 15 = 0

Also, P lie on line

a + 1 = 2λ

b – 2 = 5λ

g – 4 = λ

2 (2λ – 1) + 5 (5λ + 2) + λ + 4 – 15 = 0

4λ + 25λ + λ – 2 + 10 + 4 – 15 = 0

30λ – 3 = 0

$\lambda =\frac{1}{10}$  

a + b + g = (2λ – 1) + (5λ + 2) + (λ + 4)

$=8\lambda +5=\frac{8}{10}+5=5.8$

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

x + 2y + 3z = 42

0    x + 2y = 42 ->22 cases

1    x + 2y = 39 ->19 cases

2    x + 2y = 36 ->19 cases

3    x + 2y = 33 ->17 cases

4    x + 2y = 30 ->16 cases

5    x + 2y = 27 ->14 cases

6    x + 2y = 24 ->13 cases

7    x + 2y = 21 ->11 cases

8    x + 2y = 18 ->10 cases

9    x + 2y = 15 ->8 cases

10  x + 2y =12 -> 7 cases

11  x + 2y = 9 -> 5 cases

12  x + 2y = 6 -> 4 cases

13  x + 2y = 3 -> 2 cases

14  x + 2y = 0 -> 1 cases.

R₁ = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
(4, 4), (4, 8), . (4, 20), (5, 5), (5, 10), (5, 15), (5, 20), (6, 6), (6, 12), (6, 18), (7. 7),
(7, 14), (8, 8), (8, 16), (9, 9), (9, 18), (10, 10), (10, 20), (11, 11), (12, 12), . (20, 20)}

n (R₁) = 66

R₂ = {a is integral multiple of b}

So n (R₁ – R₂) = 66 – 20 = 46

as R₁ Ç R₂ = { (a, a) : a Î s} = { (1, 1), (2, 2), ., (20, 20)}

RHL $\underset{x\to 0^{+}}{lim}\frac{co{s}^{-1}\left(1-x^2\right)si{n}^{-1}\left(1-x\right)}{x-x^3}$

$\underset{x\to 0^{+}}{lim}\frac{\pi }{2}\cdot \frac{co{s}^{-1}\left(1-x^2\right)}{x}$

$\frac{\pi }{2}\underset{x\to 0^{+}}{lim}\frac{-1}{\sqrt[]{{\left(1-\left(1-x^2\right)\right)}^2}}\left(-2x\right)$

$=\frac{\pi }{2}\underset{x\to 2^{+}}{lim}\frac{2x}{\sqrt[]{2x^2-x^4}}=\pi \underset{x\to 0^{+}}{lim}\frac{x}{x\sqrt[]{2-x^2}}$

$=\frac{\pi }{\sqrt[]{2}}$

LHL $\underset{x\to 0^{+}}{lim}\frac{co{s}^{-1}\left(1-{\left(1+x\right)}^2\right)si{n}^{-1}\left(1-\left(1+x\right)\right)}{1\cdot \left(1-{\left(1+x\right)}^2\right)}$

$=\underset{x\to 0^{-}}{lim}\frac{co{s}^{-1}\left(-x^2-2x\right),si{n}^{-1}\left(-x\right)}{-x^2-2x}$            

$=\frac{\pi }{2}\underset{x\to 0^{-}}{lim}\frac{-si{n}^{-1}x}{-x\left(x+2\right)}=\frac{\pi }{2}*\frac{1}{2}=\frac{\pi }{2}$            

$tanB*tanC=\frac{\sqrt[]{x}}{\sqrt[]{x^2+x+1}}*\frac{1}{\sqrt[]{x\left(x^2+x+1\right)}}$

$=\frac{1}{x^2+x+1}=ta{n}^{2}A$            

tan² A = tan B tan C

It is only possible when A = B = C at x = 1

A = 30°, B = 30°, C = 30° $\left[tanA=tanB=tanC=\frac{1}{\sqrt[]{3}}\right]$                                 

$A+B=\frac{\pi }{2}-C$

$gof\left(x\right)=\{\begin{array}{cc}\hfill f\left(x\right),\hfill & \hfill f\left(x\right)<0\hfill \\ \hfill f\left(x\right),\hfill & \hfill f\left(x\right)>0\hfill \end{array}$

$=\{\begin{array}{lll}{e}^{lnx}=x\hfill & \left(0,1\right)\hfill & e^{-x}\hfill \\ \left(-\infty ,0\right)\hfill & lnx\hfill & \left(1,\infty \right)\hfill \end{array}$

Therefore, gof (x) is many one and into