Maths

$z^2=-\overline{iz}$

$\left|z^2\right|=\left|-\overline{iz}\right|$

$\left|z^2\right|=\left|z\right|$

$\left|z^2\right|-\left|z\right|=0$

$\left|z\right|\left(\left|z\right|-1\right)=0$

|z| = 0 (not acceptable)

|z| = 1

|z|² = 1

Eqⁿ : y – 0 = tan45° (x – 9) Þ y = (x – 9)

Option (B) is correct

Given : x² – 70x + l = 0

->Let roots be a and b

->b = 70 – a

->= a (70 – a)

l is not divisible by 2 and 3

->a = 5, b = 65

-> $\frac{\sqrt[]{5-1}+\sqrt[]{65-1}}{\left|60\right|}=\left|\frac{4+8}{60}\right|=\frac{1}{5}$

$\frac{dy}{dx}=\frac{\left(x+1\right)\left(x^2-x+1\right)+\sqrt[]{\left(1-x\right)\left(1+x\right)}}{\left(x-1\right)\left(x+1\right)}$

$\frac{dy}{dx}=\frac{x\left(x-1\right)+1}{\left(x-1\right)}+\sqrt[]{\frac{\left(1-x\right)\left(1+x\right)}{{\left(x-1\right)}^2{\left(x+1\right)}^2}}$

$\frac{dy}{dx}=x+\frac{1}{x-1}+\frac{1}{\sqrt[]{\left(1-x\right)\left(1+x\right)}}$

$dy=xd+\frac{1}{\left(x-1\right)}dx+\frac{dx}{\sqrt[]{1-x^2}}$

$y=\frac{x^2}{2}+ln\left|x-1\right|+si{n}^{-1}x+c$

at x = 0, y = 2 2 = c

$y=\frac{x^2}{2}+ln\left|x-1\right|+si{n}^{-1}x+2$

$y\left(\frac{1}{2}\right)=\frac{17}{8}+\frac{\pi }{6}-ln2$

-> $-1\le \left|\begin{array}{cc}\hfill \overline{2-\left|x\right|}\hfill & \hfill 4\hfill \end{array}\right|\le 1$

-> $\left|\frac{2-\left|x\right|}{4}\right|\le 1$

$-1\le \frac{2-\left|x\right|}{4}\le 1$

–4 £ 2 – |x| £ 4

–6 £ – |x| £ 2

–2 £ |x| £ 6

|x| £ 6

->x Î [–6, 6]              …(1)

Now, 3 – x ¹ 1

And x ¹ 2                    …(2)

and 3 – x > 0

x < 3                            (3)

From (1), (2) and (3)

->x Î [–6, 3] – {2}

a = 6

b = 3

g = 2

a + b + g = 11

$f\text{'}\left(x\right)=\frac{g\text{'}\left(x\right)-g\text{'}\left(2-x\right)}{2},f\text{'}\left(\frac{3}{2}\right)=\frac{g\text{'}\left(\frac{3}{2}\right)-g\text{'}\left(\frac{1}{2}\right)}{2}=0$  

Also  $f\text{'}\left(\frac{1}{2}\right)=\frac{g\text{'}\left(\frac{1}{2}\right)-g\text{'}\left(\frac{3}{2}\right)}{2}=0$ , f' (1) = 0

-> $f\text{'}\left(\frac{3}{2}\right)=f\text{'}\left(\frac{1}{2}\right)=0$  

->roots in $\left(\frac{1}{2},1\right)$ and $\left(1,\frac{3}{2}\right)$  

->f" (x) is zero at least twice in $\left(\frac{1}{2},\frac{3}{2}\right)$

$?$ ae = 2b

$\frac{4b^2}{a^2}=e^2$  

Or 4 (1 – e²) = e²

4 = 5e² -> $e=\frac{2}{\sqrt[]{5}}$

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

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

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

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

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

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

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