Maths

Three consecutive integers belong to 98 sets and four consecutive integers belongs to 97 sets.

Þ Number of permutations of b₁ b₂ b₃ b₄ = number of permutations when b₁ b₂ b₃ are consecutive + number of permutations when b₂, b₃, b₄ are consecutive – b₁ b₂ b₃ b₄ are consecutive = 98 * 97 * 98 * 97 – 97 = 18915

2x + y – 3z = 4

${\pi }_2=\left|\begin{array}{ccc}\hfill x-2\hfill & \hfill y+3\hfill & \hfill z-2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -5\hfill \\ \hfill -1\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right|=0$

${\pi }_2:-5x+5y+z+23=0$

now line lying in both the planes have DR.

$\frac{a}{1+15}=\frac{b}{15-2}=\frac{c}{10+5}$

$\frac{a}{16}=\frac{b}{13}=\frac{c}{15}$

So direction ratio's a : b : c = 16 : 13 : 15

$x+f\text{'}\left(x\right)=f\text{'}\left(0\right)$

$f\text{'}\left(0\right)=1-c^{2}fromequation\left(i\right)$

$x+f\text{'}\left(x\right)=1-c^2$

$\frac{x^2}{2}+f\left(x\right)=\left(1-c^2\right)x+d$

f(0) = 0

$f\left(x\right)=\frac{-x^2}{\alpha }+\left(1-c^2\right)x$

c = 3/2

$f"\left(x\right)=-1=2\left(c+1\right)$

$f\left(x\right)=\frac{-x^2}{2}+\left(\frac{-5}{4}\right)x$

now $\left(f\left(1\right)+f\left(2\right)+.......f\left(20\right)\right)$

$=\frac{-1}{2}\left(1^2+2^2+......+20^2\right)-\frac{5}{4}\left(1+2+....+20\right)$

$=\frac{20\cdot 21}{2}\cdot \frac{\left(-82-15\right)}{12}$

now $2\left|f\left(1\right)+f\left(2\right)+....+f\left(20\right)\right|$

$=\frac{20\cdot 21\cdot 97}{12}=3395$

Given 2a + 2b = $4\left(2\sqrt[]{2}+\sqrt[]{14}\right)$

$=4\sqrt[]{2}\left(2+\sqrt[]{7}\right)$

$b^2=a^2\left(\frac{11}{4}-1\right)$

4b2 = 7a2

b = $\frac{\sqrt[]{7}}{2}a$

f (x) = (c + 1)x² + (1 – c²)x + 2k

$f\text{'}\left(x\right)=\left(\left(c+1\right)\right)2x+\left(1-c^2\right).............\left(i\right)$

$f\text{'}\left(x\right)=\underset{y\to 0}{lim}I\frac{f\left(x+y\right)-f\left(x\right)}{x+y-x}$

$=\underset{y\to 0}{lim}\frac{f\left(y\right)-xy}{y}$

$f\text{'}\left(x\right)=\underset{y\to 0}{lim}\frac{f\left(y\right)}{y}-x$

$x+f\text{'}\left(x\right)=\underset{y\to 0}{lim}\frac{f\left(y\right)-f\left(0\right)}{y-0}asf\left(c\right)=0$

 $co{s}^{-1}\frac{1}{\sqrt[]{5}}=co{t}^{-1}\frac{1}{2}$

$tan\left(2\left(ta{n}^{-1}\frac{1}{2}+co{t}^{-1}\frac{1}{2}\right)\right)+ta{n}^{-1}\left(\frac{1}{2}\right)=tan\left(\pi +ta{n}^{-1}\frac{1}{2}\right)=\frac{1}{2}$

$tan2\alpha =2\sqrt[]{2}\Rightarrow \frac{2tan\alpha }{1-ta{n}^2\alpha }=2\sqrt[]{2}$

$\sqrt[]{2}ta{n}^2\alpha +tan\alpha -\sqrt[]{2}=0$

$tan\alpha =\frac{1}{\sqrt[]{2}},\left(-\sqrt[]{2}\right)\to rejected$

$co{s}^{-1}\frac{1}{\sqrt[]{5}}=co{t}^{-1}\frac{1}{2}$

$tan\left(2\left(ta{n}^{-1}\frac{1}{2}+co{t}^{-1}\frac{1}{2}\right)\right)+ta{n}^{-1}\left(\frac{1}{2}\right)=tan\left(\pi +ta{n}^{-1}\frac{1}{2}\right)=\frac{1}{2}$

$tan2\alpha =2\sqrt[]{2}\Rightarrow \frac{2tan\alpha }{1-ta{n}^2\alpha }=2\sqrt[]{2}$

$\sqrt[]{2}ta{n}^2\alpha +tan\alpha -\sqrt[]{2}=0$

$tan\alpha =\frac{1}{\sqrt[]{2}},\left(-\sqrt[]{2}\right)\to rejected$

Draw y = cos²x and y = 2x + $\text{exception 25:}$

3cos²2θ + 6cos2θ -   $10\frac{\left(1+cos2\theta \right)}{2}+5=0$

$cos2\theta \left(3cos2\theta +1\right)=0$

 Þ cos2θ = 0,     $\frac{-1}{3}$

Draw y = cos2θ, y = 0 and y =  $\frac{-1}{3},$  find the pt. of intersection.

Let $\pi \equiv -x+y+z-1=0$

1 2 > 0 as both pt.lies on same side

now $P_1=\left|\frac{-1+2-1-1}{\sqrt[]{3}}\right|=\frac{1}{\sqrt[]{3}}$

P2 = $\left|\frac{-2+\left(-1\right)+3-1}{\sqrt[]{3}}\right|=\frac{1}{\sqrt[]{3}}$

as P₁ = P₂ so distance between foot of perpendicular will be same as distance between the points

d = $\sqrt[]{{\left(1-2\right)}^2+{\left(2+1\right)}^2+{\left(-1-3\right)}^2}$

= $\sqrt[]{1+9+16}$$=$

2cos⁴ x – cos²x = $2{\left(\frac{1+cos2x}{2}\right)}^2-cos2x$

$=\frac{1+co{s}^{2}2x}{2}$

$\sqrt[]{2}{\displaystyle \int \frac{2dx}{1+co{s}^{2}2x}=\sqrt[]{2}{\displaystyle \int 2}\cdot \frac{se{c}^{2}2xdx}{2+ta{n}^{2}2x}}$

$=\sqrt[]{2}\cdot \frac{1}{\sqrt[]{2}}\cdot ta{n}^{-1}\left(\frac{tanx}{\sqrt[]{2}}\right)$

now IF = $e^{{\displaystyle \int Pdx}}$

$e^{\sqrt[]{2}{\displaystyle \int 2\frac{se{c}^{2}2xdx}{2+ta{n}^{2}2x}}}=e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}$

Solution: $y\cdot e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}={\displaystyle \int x.e^{ta{n}^{-1}\left(\sqrt[]{2}\cdot cot2x\right)}\cdot e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}dx}$

$y\cdot e^{ta{n}^{-1}\left(\frac{tan2x}{\sqrt[]{2}}\right)}=\frac{x^2}{2}\cdot e^{\pi /2}+C$

at $x=\frac{\pi }{4},y=\frac{{\pi }^2}{32},C=0$

at $x=\frac{\pi }{3},y=\frac{{\pi }^2}{18}{e}^{ta{n}^{-1}\alpha }$

$y\cdot e^{ta{n}^{-1}}\frac{\left(tan2\frac{\pi }{3}\right)}{\sqrt[]{2}}=\frac{{\pi }^2}{18}\cdot e^{\pi /2}$

$\frac{{\pi }^2}{18}\cdot e^{ta{n}^{-1}\alpha }\cdot e^{ta{n}^{-1}}\left(-\frac{\sqrt[]{3}}{2}\right)=\frac{{\pi }^2}{18}\cdot e^{\pi /2}$

tan^-1 a + tan^-1 $\left(-\sqrt[]{3}/2\right)=\frac{\pi }{2}$

cot^-1a = tan^-1 $\left(-\frac{\sqrt[]{3}}{2}\right)$

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

$\frac{1}{\alpha }=-\sqrt[]{3}/2$

2 = $\frac{2}{3}$