\(\begin{aligned} (p \vee q) & \rightarrow r \end{aligned}\)
நேர்மாறு \(\begin{aligned} \neg r & \rightarrow \neg(p \vee q) \end{aligned}\)
\(\begin{aligned} \neg r & \rightarrow(\neg p \wedge \neg q) \end{aligned}\)

கூற்றுகளின் எண்ணிக்கை =3
மெய்மை அட்டவணையில் நிரைகளின் எண்ணிக்கை = 2³ = 8 நிரைகள்

\(\mathrm{p} \wedge(\neg \mathrm{p} \vee \mathrm{q}) \equiv(\mathrm{p} \wedge \neg \mathrm{p}) \vee(\mathrm{p} \vee \mathrm{q}) \equiv \mathbb{F} \vee(\mathrm{p} \wedge \mathrm{q}) \equiv \mathrm{p} \wedge \mathrm{q}\)

\(\lim _{x \rightarrow 0}\left(\cot x-\frac{1}{x}\right)  =\lim _{x \rightarrow 0}\left(\frac{\cos x}{\sin x}-\frac{1}{x}\right) \quad\left(\frac{0}{0}\right) \) வடிவம் 
\( =\lim _{x \rightarrow 0}\left(\frac{x \cos x-\sin x}{x \sin x}\right) \)
\( =\lim _{x \rightarrow 0}\left(\frac{x(-\sin x)+\cos x-\cos x}{x \cos x+\sin x}\right) \)
\( =\lim _{x \rightarrow 0}\left(\frac{-x \sin x}{x \cos x+\sin x}\right) \quad\left(\frac{0}{0}\right) \) வடிவம் 
\( =\lim _{x \rightarrow 0} \frac{-x \cos x-\sin x}{\cos x-x \sin x+\cos x}=0\)

\(f(x, y) =\mathrm{e}^{x y} \)
\(\frac{\partial f}{\partial y} =\mathrm{e}^{x y}(x)=x \mathrm{e}^{x y} \)
\(\frac{\partial^2 f}{\partial x \partial y}  =x \mathrm{e}^{x y}(\mathrm{y})+\mathrm{e}^{x y}=\mathrm{e}^{x y}(1+x y)\)

\( \vec{a} \times(\vec{b} \times \vec{c})  =(\vec{a} \times \vec{b}) \times \vec{c} \\ (\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}  =(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a} \\ \vec{c}  =\frac{\vec{b} \cdot \vec{c}}{\vec{a} \cdot \vec{b}} \vec{a}=\lambda \vec{a}\)
\( \Rightarrow \vec{a} \) என்பது \(\vec{c}\) க்கு இணையானவை

'ω' என்பது ஒன்றின் முப்படி மூலம் 
\(\omega=\left(\frac{-1+i \sqrt{3}}{2}\right) \) மற்றும் \(\omega^2=\left(\frac{-1-i \sqrt{3}}{2}\right) \) என்க.
கொடுக்கப்பட்டது :  \(\left| \begin{matrix} 1 & 1 & 1 \\ 1 & { -\omega }^{ 2 }-1 & { \omega }^{ 2 } \\ 1 & { \omega }^{ 2 } & { \omega }^{ 7 } \end{matrix} \right| \) = 3k
\(\left| \begin{matrix} 1 & 1 & 1 \\ 1 & { \omega }^{ } & { \omega }^{ 2 } \\ 1 & { \omega }^{ 2 } & { \omega }^{ } \end{matrix} \right| \) = 3k
[ ∵ 1 + ω + ω² = 0 மற்றும் ω³ = 1]
\( 1\left(\omega^2-\omega^4\right)-1\left(\omega-\omega^2\right)+1\left(\omega^2-\omega\right)=3 \mathrm{k} \)
\( \omega^2-\omega-\omega+\omega^2+\omega^2-\omega=3 \mathrm{k} \)
\( 3 \omega^2-3 \omega=3 \mathrm{k} \Rightarrow 3\left(\omega^2-\omega\right)=3 \mathrm{k} \)
\( \therefore \mathrm{k}=\omega^2-\omega=\left(\frac{-1-i \sqrt{3}}{2}\right)-\left(\frac{-1+i \sqrt{3}}{2}\right) \)
\( =\frac{-1-i \sqrt{3}}{2}+\frac{1-i \sqrt{3}}{2}=-i \sqrt{3}=-\sqrt{3} i \)

\((1+\omega)^7  =A+B \omega \Rightarrow\left(-\omega^2\right)^7=A+B \omega \)
\(-\omega^{14}  =A+B \omega \Rightarrow-\omega^2=A+B \omega \)
\(1+\omega  =A+B \omega\)
ஒப்பிடக் கிடைப்பது \(\mathrm{A}=1, \mathrm{~B}=1\).

\(\mathrm{z}  =\left(\sin 40^{\circ}+i \cos 40^{\circ}\right)^{5}=\left(\cos 50^{\circ}+i \sin 50^{\circ}\right)^5 \)
\(=\cos 250^{\circ}+i \sin 250^{\circ} \)
\( =\cos \left(360^{\circ}-110^{\circ}\right)+i \sin \left(360^{\circ}-110^{\circ}\right) \)
\( =\cos 110^{\circ}-i \sin 110^{\circ} \)
\( =\cos \left(-110^{\circ}\right)+i \sin \left(-110^{\circ}\right)\)
முதன்மை வீச்சு = 110^o

a < 0 எனில், ஒரே ஒரு குறிமாற்றம் மட்டுமே கிடைக்கும்.

\(\Delta \geq 0 \)
\(\mathbf{k}^2-4(1)(9) \geq 0 \)
\(\mathbf{k}^2 \geq 36 \Rightarrow|k| \geq 6\)

\(\alpha+\beta+\gamma=-p, \quad \alpha \beta+\beta \gamma+\gamma \alpha=q, \alpha \beta \gamma=-r \)
\(\sum \frac{1}{\alpha}=\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\beta \gamma+\alpha \gamma+\alpha \beta}{\alpha \beta \gamma}=\frac{q}{-r}\)

வட்டத்தின் சமன்பாடு\((x - 3) ^ 2 + (y + r) ^ 2 = r ^ 2 \) ....(1)
இது (1, -2) என்ற புள்ளி வழிச் செல்கிறது
\( (1-3)^2+(-2+r)^2 =r^2 \\ 4+4-4 r+r^2 =r^2 \Rightarrow r=2 \\ (x-3)^2+(y+2)^2 =4 \)
\((5-3)^2+(-2+2)^2=4 \Rightarrow 4=4\)

\(a^2=9, \quad b^2=4 \\  2 x-y=1 \Rightarrow y=2 x-1 \\  m=2 \\  c=\sqrt{a^2 m^2-b^2}=\sqrt{9(4)-4} \\  =\sqrt{36-4}=\sqrt{32} \\  c= \pm 4 \sqrt{2} \\ =\left(\frac{-a^2 m}{c}, \frac{-b^2}{c}\right) \\  =\left(\frac{-9 \times 2}{-4 \sqrt{2}}, \frac{-4}{-4 \sqrt{2}}\right) \text { (or) }\left(\frac{-9 \times 2}{4 \sqrt{2}}, \frac{-4}{4 \sqrt{2}}\right) \\  =\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right) \text { (or) }\left(\frac{-9}{2 \sqrt{2}}, \frac{-1}{\sqrt{2}}\right) \\ \)

\( y^2=12 x \quad \Rightarrow 4 a =12 \Rightarrow a=3 \\ x+y =k \Rightarrow y=-x+k \\ \therefore m =-1, c=k \\ c=-2 a m-a m^2 \Rightarrow k =-2 a(-1)-a(-1)^3 \\ k =-6(-1)-3(-1)=6+3=9 \)

\( C A =C B \Rightarrow C A^2=C B^2 \\ (1-1)^2+(0-h)^2 =(2-1)^2+(3-h)^2 \\ h^2 =1+h^2+9-6 h \\ 6 h =10 \Rightarrow h=\frac{10}{6}=\frac{5}{3} \\ =2 h =2\left(\frac{5}{3}\right)=\frac{10}{3} \)

செவ்வகல நீளம்
\( \frac{2 b^2}{a} =8 \\ b^2 =4 a ........ (1) \) 
துணையச்சின் நீளம் 
\( 2 b =\frac{1}{2}(2 a e) \\ b =\frac{a e}{2} \\ b^2 =\frac{a^2 e^2}{4} \)
\( \frac{a^2 e^2}{4} =a^2\left(e^2-1\right) \\ \mathrm{e}^2\left(e^2-1\right) =4 \mathrm{e}^2-4 \\ 3 \mathrm{e}^2 =4 \Rightarrow \mathrm{e}^2=\frac{4}{3} \\ \therefore \mathrm{e} =\frac{2}{\sqrt{3}}\)

\(\mathrm{i}^n+\mathrm{i}^{n+1}+\mathrm{i}^{n+2}+\mathrm{i}^{n+3}  =\mathrm{i}^n\left[1+i+i^2+i^3\right] \)
\( =i^n[1+i-1-i] =i^n(0)=0\)

\(\mathrm{AB} =\left[\begin{array}{ll} 2 & 0 \\ 1 & 5 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 0 \end{array}\right]=\left[\begin{array}{cc} 2 & 8 \\ 11 & 4 \end{array}\right] \)
\(\operatorname{adj} \mathrm{AB} =\left[\begin{array}{cc} 4 & -8 \\ -11 & 2 \end{array}\right] \)
\(\therefore|\operatorname{adj} \mathrm{AB}| =8-88=-80\)

AA^T = A^TA மற்றும் B=A^-1A^T
B^T = (A^-1A^T)^T = (A^T)^T (A^-1)^T=A (A^-1)^T
BB^T = (A^-1A^T) (A(A^-1)^T) = A^-1(A^TA)(A^-1)^T
= A^-1(AA^T) (A^-1)^T = (A^-1A) (A^T(A^T)^-1) = I₃

\( \sin ^{-1}\left(2 \cos ^2 x-1\right)+\cos ^{-1}\left(1-2 \sin ^2 x\right) \\ =\sin ^{-1}(\cos 2 x)+\cos ^{-1}(\cos 2 x) \\ =\frac{\pi}{2} \quad\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right] \)

 \(\cot ^{-1} 2+\cot ^{-1} 3 =\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3} \\   =\tan ^{-1}\left[\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{b}{6}}\right]=\tan ^{-1}(1)=\frac{\pi}{4}\)
மூன்றாவது கோணம் = \(\pi-\frac{\pi}{4}=\frac{3 \pi}{4} \)

\( \sin ^{-1} x+\cos ^{-1} x+\cos ^{-1} y+\sin ^{-1} y =\frac{\pi}{2}+\frac{\pi}{2}=\pi \\ \frac{2 \pi}{3}+\cos ^{-1} x+\cos ^{-1} y =\pi \\ \cos ^{-1} x+\cos ^{-1} y =\pi-\frac{2 \pi}{3} \\ =\frac{3 \pi-2 \pi}{3}=\frac{\pi}{3} \)