Maths

 $cosec\left(2co{t}^{-1}\left(5\right)+co{s}^{-1}\left(\frac{4}{5}\right)\right)$

Let cot^-1 (5) = and cos^-1 $\left(\frac{4}{5}\right)=\alpha$

cot = 5 cos = $\frac{4}{5}$

$=cosec\left(2\theta +\alpha \right)=\frac{1}{sin\left(2\theta +\alpha \right)}$

$as\{\begin{array}{l}sin2\theta =\frac{2tan\theta }{1+ta{n}^2\theta }=\frac{2\left(\frac{1}{5}\right)}{1+\frac{1}{25}}=\frac{5}{13}\\ cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }=\frac{1-\frac{1}{25}}{1+\frac{1}{25}}=\frac{12}{13}\end{array}$

Let A = {a, b, c}, B = {1, 2, 3, 4, 5} n (A * B) = 15

x = number of one-one functions from A to B.

${=}^5{C}_3.3!=60$

y = number of one-one functions for A to (A * B)

${=}^{15}{C}_3.3!=15*14*13=2730$

$\frac{Y}{x}=\frac{2730}{60}\Rightarrow 2y=91x$

$f\left(x\right)=x^3-a{x}^2+bx-4$

f (1) = f (2)

->1 – a + b – 4 = 8 – 4a + 2b – 4

->3a – b = 7 . (i)

8a = 16 + 3b . (ii)

(i) and (ii) -> (a, b) = (5, 8)

Statement : “If you will work, you will earn money”

contrapositive : If you will not earn money, you will not work.

${\displaystyle \underset{-1}{\overset{1}{\int }}{x}^2{e}^{\left[x^3\right]}dx=}{\displaystyle \underset{-1}{\overset{0}{\int }}{x}^2.e^{-1}dx+}{\displaystyle \underset{0}{\overset{1}{\int }}{x}^2.e^{0}dx=\frac{1}{e}}{\displaystyle \underset{-1}{\overset{0}{\int }}{x}^{2}dx+{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{2}dx}}$

$=\frac{1}{e}{\left[\frac{x^3}{3}\right]}_{-1}^0+{\left[\frac{x^3}{3}\right]}_0^1=\frac{1}{e}\left[0-\left(\frac{-1}{3}\right)\right]+\left[\frac{1}{3}-0\right]=\frac{1}{3e}+\frac{1}{3}=\frac{e+1}{3e}$

$\frac{x^2}{25}+\frac{y^2}{16}=1$

16 = 25 (1 – e²)

$\Rightarrow e=\frac{3}{5}$

OF₁ = 5 $\left(\frac{3}{5}\right)=3$

For Hyperbola : e' $=\frac{5}{3}$

a = 3

$Hyperbola,\frac{x^2}{9}-\frac{y^2}{16}=1$

f : N ->N

f (n + 1) = f (n) + f (1)

Let f (1) = a, a  $\in$ N

f (2) = f (1) + f (1) = 2a

f (3) = f (2) + f (1) = 3a

and so on

->f (m) = ma, m, a  $\in$ N

->f is one – one, Þ option (2) is true.

Suppose f (g (x) is one-one

then f (g (x₁) $\ne$ f (g (x₂) for x₁ $\ne$  x₂

->g (x₁) $\ne$ g (x₂) (as f is one-one)

->g is one – one

 $x,y\in \left(0,\pi \right)$

cos x + cos y – cos (x + y) = $\frac{3}{2}$

$2cos\frac{x+y}{2}.cos\frac{x-y}{2}-\left(2co{s}^2\frac{x+y}{2}-1\right)=\frac{3}{2}$

$\Rightarrow 2co{s}^2\frac{x+y}{2}-2cos\frac{x-y}{2}.cos\frac{x+y}{2}+\frac{1}{2}=0$

As,

$x,y\in \left(0,\pi \right)$

$\frac{-\pi }{2}<\frac{x-y}{2}<\frac{\pi }{2}$

$sin\frac{x-y}{2}=0\Rightarrow \frac{x-y}{2}=0\Rightarrow x=y$

then equation : cos x + cos y – cos (x + y) = $\frac{3}{2}$

$cosx=\frac{2\pm \sqrt[]{4-4}}{4}=\frac{1}{2}$

$x=y=\frac{\pi }{3}\Rightarrow sinx+cosy=\frac{\sqrt[]{3}}{2}+\frac{1}{2}=\frac{\sqrt[]{3}+1}{2}$

$z^2+\alpha z+\beta =0,\alpha ,\beta \in R$

roots : 1 – 2i, 1 + 2i

Sum of roots = 2 = -α and product of roots = 5 = β

α - β = -2 - 5 = -7

As per questions

$\frac{dy}{dx}=\frac{x^2-4x+y+8}{x-2}$           

$\frac{dy}{dx}=\frac{{\left(x-2\right)}^2+\left(y+4\right)}{\left(x-2\right)}$           

$\frac{dy}{dx}=\left(x-2\right)+\frac{y+4}{x-2}$                       .(i)

Let $\frac{y+4}{x-2}=t$  

(y + 4) = t(x – 2)

Putting in equation (i)

$\left(x-2\right)\frac{dt}{dx}+t=\left(x-2\right)+t$        

    $\frac{dt}{dx}=1$        

dt = dx

Integrating on both the sides t = x + c

$\frac{y+4}{x-2}=x+c$  

Passing through origin C = -2

  $\therefore$         equation of curve $\frac{y+4}{x-2}=x-2$