Maths

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 $\frac{y+4}{x-2}=x-2$

Combined equation of pair of lines OP and OQ is

$x^2+2y^2=2{\left(x+y\right)}^2$

$\Rightarrow x^2+4xy=0\Rightarrow x\left(x+4y\right)=0\Rightarrow \{\begin{array}{l}x=0\left(lineOP\right)\\ y=-\frac{x}{4}\left(lineOQ\right)\end{array}$

$tan\left(90°+\theta \right)=-\frac{1}{4}$

$-cot\theta =-\frac{1}{4}\Rightarrow tan\theta =4$

$\theta =ta{n}^{-1}4=co{t}^{-1}\frac{1}{4}=\frac{\pi }{2}-ta{n}^{-1}\frac{1}{4}$

$\angle POQ=\pi -\theta =\frac{\pi }{2}+ta{n}^{-1}\frac{1}{4}$

$l_n={\displaystyle \underset{\pi /4}{\overset{\pi /2}{\int }}co{t}^{n}xdx=}{\displaystyle \underset{\pi /4}{\overset{\pi /2}{\int }}co{t}^{n-2}x\left(cose{c}^{2}x-1\right)dx}$

$={\left[\frac{-co{t}^{n-1}x}{n-1}\right]}_{\pi /4}^{\pi /2}-l_{n-2}=\frac{1}{n-1}-l_{n-2}$

$l_n+l_{n-2}=\frac{1}{n-1}$

$\begin{array}{l}n=4\Rightarrow l_4+l_2=\frac{1}{3}\\ n=5\Rightarrow l_5+l_3=\frac{1}{4}\\ n=6\Rightarrow l_6+l_4=\frac{1}{5}\end{array}\}\Rightarrow \frac{1}{l_2+l_4},\frac{1}{l_3+l_5},\frac{1}{l_4+l_6}$ are in A.P.

2x + 3y + 2z = 9 . (i)

3x + 2y + 2z = 9     . (ii)

x – y + 4z = 8          . (iii)

(ii) – (i) ⇒ x = y

Then (iii) ⇒ z = 2

(i) ⇒ 5x + 4 = 9 ⇒ (x = 1, y = 1, z = 2) unique solution