Maths

$\left(p\Delta q\right)\Rightarrow \left(p\Delta \sim q\right)\vee \left(\sim p\Delta q\right)$

Case I

When  $\Delta$  is same as $\vee .$  

Then  $\left(p\Delta \sim q\right)\vee \left(-p\Delta q\right)$ becomes

$\left(p\vee \sim q\right)\vee \left(\sim p\vee q\right)$ which is always true, so x becomes tautology.

Case II

When  $\Delta$  is same as $\vee$  

Then  $\left(p\wedge q\right)\Rightarrow \left(p\wedge \sim q\right)\vee \left(\sim p\wedge q\right)$  becomes $p\wedge q$ is T, then $\left(p\wedge \sim q\right)\vee \left(\sim p\wedge q\right)$ is false, so x cannot be tautology.

Case III

When  $\Delta$  is same as $\Rightarrow$  

Then  $\left(p\Rightarrow \sim q\right)\vee \left(-p\Rightarrow q\right)$ is same as $\left(\sim p\vee \sim q\right)\vee \left(p\vee q\right)$ which is true, so x becomes tautology.

Case IV

When  $\Delta$ is same as $\iff$  

Then   $\left(p\iff q\right)\Rightarrow \left(p\iff \sim q\right)\vee \left(\sim p\iff q\right)$

$p\iff q$ is true when p and q have same truth values $p\iff \sim qand\sim p\iff q$  both are false. Hence x cannot be tautology.

  $f\left(g\left(x\right)\right)=x\forall x\in R$

$\Rightarrow g\left(x\right)=f^{-1}\left(x\right)$      

For y = g (x)

x = y³ + y – 5

$\frac{dx}{dy}=3y^2+1\Rightarrow g\text{'}\left(63\right)=\frac{1}{3\left(16\right)+1}=\frac{1}{49}$

$g\text{'}\left(63\right)=\left(\frac{dy}{dx}\right)$

$\left(\begin{array}{l}x=63\\ y=4\end{array}\right)$                

$A\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right)=\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right)$

$A\left(\begin{array}{l}1\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}-1\\ 0\\ 1\end{array}\right)$

$A\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{l}1\\ 1\\ 2\end{array}\right)$

$\Rightarrow A\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$

$AB=C\Rightarrow A=C{B}^{-1}$

$\left|A-2l\right|=\left(-4\right)\left(-1\right)+3\left(-1\right)+1\left(-1\right)$

4 – 3 – 1 = 0

$\mathcal{l}\left(-1,0,1\right)+m\left(-1,1,0\right)=\left(-4,3,1\right)\Rightarrow -\mathcal{l}-m=-4$

$m=3,\mathcal{l}=1$

$f\left(x+y\right)=2f\left(x\right)f\left(y\right),x,y\in N$

$f\left(1\right)=2$

${\displaystyle \sum _{k=1}^{10}f\left(\alpha +k\right)={\displaystyle \sum _{k=1}^{10}2f\left(\alpha \right)f\left(k\right)}}$

$=2f\left(\alpha \right){\displaystyle \sum _{k=1}^{10}f\left(k\right)=2.2^{2\alpha -1}.\frac{2}{3}\left(4^{10}-1\right)}$

$=\frac{2^{2\alpha +1}}{3}.\left(4^{10}-1\right)$

$\Rightarrow 2\alpha +1=9$

=4

  $\frac{sin\theta sin\theta }{cos\theta }+\frac{sin\theta }{cos\theta }=2sin\theta cos\theta$

$\Rightarrow \frac{sin\theta }{cos\theta }\left[sin\theta +1\right]=2sin\theta cos\theta$            

$\therefore$ sinθ = 0 and 1 + sin θ = 2 cos² θ = 2 – 2 sin² θ ………….(i)

θ = np

θ = -p, p, 0

From (i), 2 sin² θ + sin θ - 1 = 0

(2 sin θ - 1) (sin θ + 1) = 0

sin θ = -1, $\frac{1}{2}$  

$\theta =\frac{3\pi }{2},\theta =\frac{\pi }{6},\frac{5\pi }{6}$           

$\theta =\frac{3\pi }{2}$ is rejected.

T = cos (-2p) + cos 2p + cos θ + cos $\frac{\pi }{3}+cos\frac{5\pi }{3}=4$  

$\therefore$ T + n(s) = 4 + 5 = 9

$\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=\frac{2\left(a+b+c\right)}{24}=\frac{c}{5}=\frac{a}{4}=\frac{b}{3}$

$r=\frac{\Delta }{S}=\frac{6k^2}{6k}=k$

$R=\frac{5k}{2}\Rightarrow \frac{R}{r}=\frac{5}{2}$

  $l={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{e^{cosx}sinxdx}{\left(1+co{s}^{2}x\right)\left(e^{cosx}+e^{-cosx}\right)}}$ ${\displaystyle \underset{}{\overset{}{}}\frac{{}^{}}{{}^{}{}^{}{}^{}}}$

$2l={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{sindx}{1+co{s}^{2}x}=2{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{sinxdx}{1+co{s}^{2}x}}}$

$l={\displaystyle \underset{1}{\overset{0}{\int }}\frac{-dt}{1+t^2}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{1+t^2}=\frac{\pi }{4}}}$

$\frac{x^2-5x+6}{x^2-9}\ge -1\&\frac{x^2-5x+6}{x^2-9}\le 1$

$\frac{2x+1}{x+3}\ge 0,x\ne -3\&\frac{1}{x+3}\ge 0,x\ne -3\Rightarrow x>-3$

$x\in \left[-\frac{1}{2},\infty \right]...........(i)$

$x^2-3x+2>0and{x}^2-3x+1\ne 0$

(x – 2) (x – 1) > 0 and $x\ne \frac{3\pm \sqrt[]{5}}{2}$

$x\in \left(-\infty ,1\right)\cup \left(2,\infty \right)-\left\{\frac{3\pm \sqrt[]{5}}{2}\right\}$

From (i) and (ii)   $x\in \left[-\frac{1}{2},1\right)\cup \left(2,\infty \right)-\left\{\frac{3\pm \sqrt[]{5}}{2}\right\}$

4x – 3y + k₂ = 0

$2r=\frac{40}{5}=8\Rightarrow r=4$

${\left(x-1\right)}^2+{\left(y+2\right)}^2=16$