Maths

$T_{n+1}={}^n{C}_{r}11^{\frac{1}{6}}\cdot 7^{\frac{824-4}{2}}$

For integral term

6 should divide r

and  $\frac{824-r}{2}$  must be integer

->2 most divide r

->r divisible by 6

->possible values of r Î {0, 1, 2, …824}

->For integer terms

r Î {0, 6, 12, …822} (822 = 0 + (n – 1)6 Þ n = 138)

= 138 terms

If x = 0, y = 6, 7, 8, 9, 10

If x = 1, y = 7, 8, 9, 10

If x = 2, y = 8, 9, 10

If x = 3, y = 9, 10

If x = 4, y = 10

If x = 5, y = no possible value

Total possible ways = (5 + 4 + 3 + 2 + 1) * 2

= 30

Required probability  $=\frac{30}{11*11}=\frac{30}{121}$

Given $\left|\overrightarrow{a}\right|=1$ , $\left|\overrightarrow{b}\right|=4$ , $\overrightarrow{a}\cdot \overrightarrow{b}=2$

$\overrightarrow{c}=2\left(\overrightarrow{a}*\overrightarrow{b}\right)-3\overrightarrow{b}$  

Dot product with  $\overrightarrow{a}$ on both sides

$\overrightarrow{c}\cdot \overrightarrow{a}=-6$ . (1)

Dot product with  $\overrightarrow{b}$  on both sides

$\overrightarrow{b}\cdot \overrightarrow{c}=-48$ . (2)

$\overrightarrow{c}\cdot \overrightarrow{c}=4{\left|\overrightarrow{a}*\overrightarrow{b}\right|}^2+9{\left|\overrightarrow{b}\right|}^2$

${\left|\overrightarrow{c}\right|}^2=4\left[{\left|\overrightarrow{a}\right|}^2\cdot {\left|\overrightarrow{b}\right|}^2-{\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)}^2\right]+9{\left|\overrightarrow{b}\right|}^2$

${\left|\overrightarrow{c}\right|}^2=4\left[\left(1\right){\left(4\right)}^2-\left(4\right)\right]+9\left(16\right)$

${\left|\overrightarrow{c}\right|}^2=4\left[12\right]+144$

${\left|\overrightarrow{c}\right|}^2=48+144$

${\left|\overrightarrow{c}\right|}^2=192$

$cos\theta =\frac{\overrightarrow{b}\cdot \overrightarrow{c}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|}$

$cos\theta =\frac{-48}{\sqrt[]{192}\cdot 4}$

$cos\theta =\frac{-48}{8\sqrt[]{3}\cdot 4}$

$cos\theta =\frac{-3}{2\sqrt[]{3}}$

$cos\theta =\frac{-\sqrt[]{3}}{2}$ $\theta =co{s}^{-1}\left(\frac{-\sqrt[]{3}}{2}\right)$

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

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

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

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

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

|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 -> $e=\frac{2}{\sqrt[]{5}}$