Maths

$\overrightarrow{O}P=\left(2,-1,1\right)$

Normal vector to the plane

$=\overrightarrow{A}B*A\overrightarrow{C}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left(3,-1,1\right)=\overrightarrow{n}$

Projection of $\overrightarrow{O}Pon\overrightarrow{n}is\left|\frac{\overrightarrow{O}P.\overrightarrow{n}}{\left|\overrightarrow{n}\right|}\right|$

PN = $\frac{6+1+1}{\sqrt[]{11}}=\frac{8}{\text{exception 25:}}$

$\frac{\text{exception 25:}}{}$Projection of OP on plane = ON =  $\sqrt[]{O{P}^2-P{N}^2}=\sqrt[]{6-\frac{64}{11}}=\sqrt[]{\frac{2}{11}}$

  $\frac{x^2}{a}+\frac{y^2}{b}=1,\frac{x^2}{c}-\frac{y^2}{\left(-d\right)}=1$

Ellipse and Hyperbola are orthogonal so these will be confocal.

$\sqrt[]{a-b}=\sqrt[]{c+\left(-d\right)}$            

a – b = c – d

$l={\displaystyle \int \frac{8x^3+20x^2}{x^4+5x^3-7x^2}}dx=4{\displaystyle \int \frac{2x+5}{x^2+5x-7}dx=4ln\left|x^2+5x-7\right|+c}$

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

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