\(P(\overline { A } \cup \overline { B } )\)=\(\left( \overline { A\cap B } \right) \)(By de Morgan's law)
1-P(A∩B)=1-0.24
=0.76​

\(\int x^{11} d x=\frac{x^{11+1}}{11+1}+c=\frac{x^{12}}{12}+c\)

Dividing by x, we get 
\({1-x^3\over 3x+2}={{1\over x}-x^2\over 3+{2\over x}}\rightarrow -\infty \ as x\rightarrow \infty\)
Therefore the limit does not exist.

\(\frac{1}{x-3}\) can be made arbitrarily large by choosing x suficiently close to 1 on the right side but not equal to 1.
\(\therefore \frac{1}{x-3}\) does not approach any value when x approaches 3 from the right.
\(\therefore \lim _{x \rightarrow 3^{+}} \frac{1}{x-3}=x\) and hence the limit does not exist.

\(lim_{x\rightarrow3}(4-x)\)

At x = 3, the value of the curve on y-axis is 1.
\(\therefore lim_{x\rightarrow3}(4-x)=1\)

Let \( f(x)=\frac{x-2}{x^2-4}=\frac{x-2}{(x-2)(x+2)}=\frac{1}{x+2} \)
\( \therefore \lim _{x \rightarrow 2} \frac{x-2}{x^2-4}=\lim _{x \rightarrow 2} \frac{1}{x+2}=\frac{1}{4}=0.25 \)

90°

3x - 8y + 13= 0

\(-{3\over \sqrt{13}}\)
\(cos \theta=-{2\over\sqrt{13}}\)
\(tan\theta={3\over 2}\)
\(cosec \theta=-{\sqrt{13}\over3}\)
\(sec\theta=-{\sqrt{13}\over2}\)
\(cot \theta={2\over3}\)

Let tan^-1 \(({-1\over\sqrt{3}})\) = y, where -\({\pi\over 2}\le y \le {\pi\over 2}\)
\(tan \ y=-{1\over \sqrt{3}} \Rightarrow tan \ y =tan (-{\pi\over 6})\Rightarrow y =-{\pi\over6}\)
Thus the principal value of tan​​^-1 \(({-1\over\sqrt{3}})=-{\pi\over6}\)

⁸P₄ = 8 \(\times\)7 \(\times\) 6 \(\times\) 5 = 1680

2018
Let a_n = 2018
then the first 6 terms are 2018, 2018, 2018, 2018, 2018, 2018
It is not an AP, GP, AGP and HP.

∴ \(\frac { n! }{ r!(n-r)! } =\frac { 10! }{ 3!(10-3)! } =\frac { 10! }{ 3!7! } \)
=   \(\frac { 10\times 9\times 8\times 7! }{ 3\times 2\times 7! } =\frac { 10\times 9\times 8 }{ 3\times 2 } \) = 120

cos 65^o + cos 15^o = \(2\cos { \left( \frac { 65+15 }{ 2 } \right) } .\cos { \left( \frac { 65-15 }{ 2 } \right) } \)
= 2 cos 40^o cos 25^o

The relation R is defined by xRy if x + 2y = 1 for x, y \(\in \)  N.
Reflexivity : Let x, y \(\in \)  N
xRx \(\Rightarrow\)  x + 2x = 1 \(\Rightarrow\) 3x = 1 \(\Rightarrow\) x = \(\frac { 1 }{ 3 } \notin N\)
\(\therefore\)  R is reflexive.
Symmetricity: xRy \(\Rightarrow\) yRx for x, y \(\in \)  N
xRy \(\Rightarrow\)  x + 2y = 1 which is not possible for any values of x, Y \(\in \) N
\(\therefore\) R is not symmetric
Transitivity: xRy and yRz \(\Rightarrow\) xRz.
xRy and yRz are not possible for any values of x, y, z \(\in \) N
\(\therefore\)  R is not transitive.
\(\therefore\) R is neither reflexive, nor symmetric and not transitive.

\(\frac { 2\pi }{ 5 } \)
\(\frac { 2\pi }{ 5 } \) \(\times\) \(\frac { 180 }{ \pi } \) = 2 \(\times\) 36⁰ = 72⁰

135⁰
135⁰ = 135\(\times\) \(\frac { \pi }{ 180 } =\frac { 3\pi }{ 4 } \)

x > - 1 and x < 4.
⇒ x ∈ (-1, 4)

Given |x| - 10 < -3
⇒ |x| < -3 + 1
⇒ |x| < 7
This means -7 < x < 7.​​​​​​
\(\therefore\) The Solution set is (-7, 7)

\(\left\{ x:\frac { x-4 }{ x+2 } =3,x\in R-\{ -2\} \right\} \)
Let D = \(\left\{ x:\frac { x-4 }{ x+2 } =3,x\in R-\{ -2\} \right\} \)
\(\Rightarrow\) D = {x: x - 4 = 3x + 6, x \(\in \) R}
\(\Rightarrow\) D = {x:-4-6 = 3x-x, x  ​​​​​​\(\in \) R}
\(\Rightarrow\) D = {x:2x = -10, x \(\in \) R}
\(\Rightarrow\) D = {x:x = -5, x\(\in \)R}
\(\Rightarrow\) D = {-5}

{x \(\in \)​ ​​​​​​N : 4x + 9 < 52}
Let C = {x\(\in \)N:4x + 9 < 52}
\(\Rightarrow\) C = {x\(\in \)N:4x < 52 - 9}
\(\Rightarrow\) C = {x\(\in \)N: 4x < 43}
\(\Rightarrow\) C = \(\left\{ x\in N:x<\frac { 43 }{ 4 } \right\} \) \(\Rightarrow\) C = {x \(\in \) N : x < 10.75}
\(\Rightarrow\) C = {1,2,3,4,5,6,7,8,9,10}.

\((i) \text {Let } A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \text { and } B=\left[\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right]\)
\(A B=\left[\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right]=\left[\begin{array}{cc} 1+0 & -1+4 \\ 3+0 & -3+8 \end{array}\right]=\left[\begin{array}{ll} 1 & 3 \\ 3 & 5 \end{array}\right]\)
\(B A=\left[\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right]\left[\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{cc} 1-3 & 2-4 \\ 0+6 & 0+8 \end{array}\right]=\left[\begin{array}{cc} -2 & -2 \\ 6 & 8 \end{array}\right]\)
\(\therefore A B \neq B A\)
\(\text {(ii)Let } A=\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right] \neq 0, B=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right] \neq 0\)
\(A B=\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]=\left[\begin{array}{ll} 0+0 & 0+0 \\ 0+0 & 0+0 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0\)
\(B A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 0+0 & 0+0 \\ 0+0 & 0+0 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0\)
\(\therefore A B=0=B A\)
\(\text { (iii) } A=\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right], \quad B=\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]\)
\(A B=\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]=\left[\begin{array}{ll} 0+0 & 0+0 \\ 0+0 & 0+0 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0\)
\(B A=\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 0+0 & 0+1 \\ 0+0 & 0+0 \end{array}\right]=\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right] \neq 0\)
\(\therefore A B =0\)
\(B A \neq 0 \)

X = {a, b, c, d} 
R = {(a, a), (b, b), (a, c)} 
(i) To make R reflexive we need to include (c, c) and (d, d) 
(ii) To make R symmetric we need to include (c, a) 
(iii) R is transitive
(iv) To make R reflexive we need to include (c, c) 
To make R symmetric we need to include (c, c) and (c, a) for transitive 
∴ The relation now becomes 
R = {(a, a), (b, b), (a, c), (c, c), (c, a)} 
∴ R is equivalence relation.

The given vector is \(\hat{i}\) - \(\hat{k}\)
The direction ratio are 1, 0, -1
x = \(\sqrt{x^2+y^2+z^2}=\sqrt{1^2+0^2+(-1)^2}=\sqrt{2}\)
Hence, the direction cosines are \({1\over\sqrt{2}},{0\over\sqrt{2}},{-1\over\sqrt{2}}\Rightarrow {1\over\sqrt{2}},0,{-1\over\sqrt{2}}\)

The given vector is 5\(\hat{i}\) - 3\(\hat{j}\) - 48\(\hat{k}\)
The direction ratios are 5, -3, -48.
r = \(\sqrt{x^2+y^2+z^2}=\sqrt{5^2+(-3)^2+(-48)^2}\)
\(=\sqrt{25+9+2304}=\sqrt{2338}\)
Hence, the direction cosines are \({5\over \sqrt{2338}},{-3\over \sqrt{2338}},{-48\over \sqrt{2338}}\)