Maths

 $lo{g}_4\left(x-1\right)=lo{g}_2\left(x-3\right)$

$\Rightarrow \frac{1}{2}lo{g}_2\left(x-1\right)=lo{g}_2\left(x-3\right)$

$\Rightarrow lo{g}_2\left(x-1\right)=lo{g}_2{\left(x-3\right)}^2$

$\Rightarrow x-1=x^2-6x+9$

$\Rightarrow x=2,5$

also x – 1 > 0 and x – 3 > 0

x > 1 & x > 3

Hence, x = 5 possible. Only one solution.

$\sqrt[]{3}co{s}^{2}x=\left(\sqrt[]{3}-1\right)cosx+1$

$\left(\sqrt[]{3}cosx+1\right)\left(cosx-1\right)=0$

$\therefore cosx=-\frac{1}{\sqrt[]{3}}\left(rejected\right)$

Hence, cos x = 1 x = 0 one solution

Information missing. The question was droppedby NTA.

 $y=\left|\left|x-1\right|-2\right|$

Area bounded region = $\frac{1}{2}*4*2=4$

Let a, ar, ar², . an increasing G.P then r > 1 & a > 0

given : ar + ar5 = $\frac{25}{2}$ .(i)

and $a{r}^{2}a{r}^4=25\Rightarrow a^2{r}^6=25\Rightarrow {\left(a{r}^3\right)}^2=25$

$\Rightarrow a{r}^3=5.......\left(ii\right)$

From (i) and (ii), $\frac{ar\left(1+r^4\right)}{a{r}^3}=\frac{25}{2*5}$

$\Rightarrow 2+2r^4=5r^2\Rightarrow 2r^4-5r^2+2=0$

$\Rightarrow \left(2r^2-1\right)\left(r^2-2\right)=0$

$\Rightarrow r^2=\frac{1}{2},2\therefore r^2=2\Rightarrow r=\sqrt[]{2}$

$\therefore t_4+t_6+t_8=5+10+20=35$

R = { (P, Q) |P and Q are at the same distance from the origin}.

at $\left(1,-1\right),x^2+y^2={\left(1\right)}^2+{\left(-1\right)}^2=1+1=2$

$P_1:3x+15+21z=9$           

P₂ : x – 3y – z = 5

P₃ : 2x + 10y + 14z = 5

Ratio of the direction cosines of P₁ and P₂

$\frac{3}{2}=\frac{15}{10}=\frac{21}{14}$

Hence, P₁ and P₃ are parallel.

 $\overrightarrow{a}*\left(\overrightarrow{a}*\overrightarrow{b}\right)=\left(\overrightarrow{a}.\overrightarrow{b}\right)\overrightarrow{a}-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}=0-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}\left(as\overrightarrow{a}.\overrightarrow{b}=0given\right)$

$\overrightarrow{a}*\left(\overrightarrow{a}*\left(\overrightarrow{a}*\overrightarrow{b}\right)\right)=-{\left|\overrightarrow{a}\right|}^2\overrightarrow{a}*\overrightarrow{b}$

$\overrightarrow{a}*\left(\overrightarrow{a}*\left(\overrightarrow{a}*\left(\overrightarrow{a}*\overrightarrow{b}\right)\right)\right)=-{\left|\overrightarrow{a}\right|}^2\overrightarrow{a}*\left(\overrightarrow{a}*\overrightarrow{b}\right)=-{\left|\overrightarrow{a}\right|}^2\left(-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}\right)={\left|\overrightarrow{a}\right|}^4\overrightarrow{b}$

$T_{r+1}{=}^{10}{C}_r{\left(t{x}^{\frac{1}{5}}\right)}^{10-r}{\left(\frac{{\left(1-x\right)}^{\frac{1}{10}}}{t}\right)}^r$

According to question, 10 – 2r = 0 ⇒ r = 5

$\therefore T_6{=}^{10}{C}_5*{\left(1-x\right)}^{\frac{1}{2}}$

T₆ is maximum, when f(x) = $x{\left(1-x\right)}^{\frac{1}{2}}$ is maximum.

$f\text{'}\left(x\right)={\left(1-x\right)}^{\frac{1}{2}}-\frac{x}{2\sqrt[]{1-x}}=\frac{2\left(1-x\right)-x}{2\sqrt[]{1-x}}$

For maximum, $f\text{'}\left(x\right)=0\Rightarrow x=\frac{2}{3}$

$\therefore T_6{=}^{10}{C}_5\frac{2}{3}.{\left(\frac{1}{3}\right)}^{\frac{1}{2}}=\frac{2\left(10!\right)}{3\sqrt[]{3}{\left(5!\right)}^2}$

Let n be the number of times.

$p=\frac{1}{2},q=\frac{1}{2}$

According to question,

${\Rightarrow }^n{C}_7{=}^n{C}_9\Rightarrow n-7=9\Rightarrow n=16$

$p\left(x=2\right){=}^{16}{C}_2{\left(\frac{1}{2}\right)}^2.{\left(\frac{1}{2}\right)}^{14}=\frac{15}{2^{13}}$

Put x = $\frac{1}{3}$

S = 1 + 2x + 7x² + 12x³ + 17x⁴ + 22x⁵ + .

$\underset{\_}{xS=x+2x^2+7x^3+12x^4+17x^5+.........}$

Subtracting,

$\left(1-x\right)S=1+x+5x^2+5x^3+5x^4+5x^5+......$

= $1-4x+5x+5x^2+5x^3+.........$

$=\frac{1-x-4x+4x^2+5x}{1-x}=\frac{4x^2+1}{1-x}$

$S=\frac{4x^2+1}{{\left(1-x\right)}^2}putx=\frac{1}{3}wegets=\frac{\frac{4}{9}+1}{\frac{4}{9}}=\frac{13}{4}$