P( not A  and not B) =  \(P(\overline { A } \cap \overline { B } )\)
\(=P(\overline { A\cup B } )=1-P(A\cup B)\)
\(=1-[P(A)+P(B)-P(A\cap B)]\)
\(=1-\left[ \frac { 1 }{ 4 } +\frac { 1 }{ 2 } -\frac { 1 }{ 8 } \right] =1-\left[ \frac { 2+4-1 }{ 8 } \right] \)
\(=1-\frac { 5 }{ 8 } =\frac { 3 }{ 8 } \)

When the digits are not repeated, then first place may have one of 10 digits, the second 9, third 8 and fourth 7. Number of 4-digit numbers, n(S)=10 x 9 x 8 x 7 = 5040. Now lock can be opened only in 1 way.
\(\therefore\)  n(E) = 1
Hence, probability of opening the lock =  \(\frac { n(E) }{ n(S) } =\frac { 1 }{ 5040 } \)

On arranging the given data in ascending order, we have 22,24,25,27,28,29,30,31,41,42
Now, Mean
\(\bar { x } =\frac { 22+24+25+27+28+29+30+31+41+42 }{ 10 } =\frac { 299 }{ 10 } =29.9\)
and mean deviation from the mean, \(MD\left( \bar { x } \right) =\frac { \sum _{ i=1 }^{ n }{ \left| { { x }_{ i }-\bar { x } } \right| } }{ n } \)
\(=\frac { \left[ \left| 22-29.9 \right| +\left| 24-29.9 \right| +\left| 25-29.9 \right| +\left| 27-29.9 \right| +\left| 28-29.9 \right| +\left| 29-29.9 \right| +\left| 30-29.9 \right| +\left| 31-29.9 \right| +\left| 41-29.9 \right| +\left| 42-29.9 \right| \right] }{ 10 } \)
\(=\frac { 7.9+5.9+4.9+2.9+1.9+0.9+0.1+1.1+11.1+12.1 }{ 10 } \)
\(=\frac { 48.8 }{ 10 } =4.88=4.9\left( approx \right) .\)
\(\therefore \ \bar { x } -MD\left( \bar { x } \right) =29.9-4.9=25\)
\(\bar { x } +MD\left( \bar { x } \right) =29.9+4.9=34.8\)
On examining the arranged data, we find that observations between 25 and 34.8 are 27,28,29,30,31.
Hence, there are 5 observations lying between \(\bar { x } -MD\left( \bar { x } \right) \) and \(\bar { x } +MD\left( \bar { x } \right) \)

\(\lim_ { x\rightarrow 0 }{ lim } \frac { x\quad tan4x }{ 1-cos4x } =\lim_ { x\rightarrow 0 }{ lim } \frac { x.\frac { sin\quad 4x }{ cos\quad 4x } }{ 1-cos\quad 2.(2x) } \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { x.\frac { sin\quad 4x }{ cos\quad 4x } }{ 1-\left( 1-2{ sin }^{ 2 }2x \right) } \quad [\because \quad cos2\theta =1-2{ sin }^{ 2 }]\)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { x.sin\quad 4x }{ cos\quad 4x.\quad 2\quad { sin }^{ 2 }2x } \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { x.sin\quad 2(2x) }{ cos\quad 4x(2.sin\quad 2x.sin\quad 2x) } \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { x\quad (2\quad sin\quad 2x.cos\quad 2x) }{ (2\quad sin\quad 2x.sin\quad 2x)cos\quad 4x } \quad [\because \quad sin\ 2\theta =2\quad sin\quad \theta \quad cos\quad \theta ]\)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { x.\quad cos\quad 2x }{ (sin\quad 2x).cos\quad 4x } =\lim_ { x\rightarrow 0 }{ lim } \frac { cos\quad 2x }{ cos\quad 4x } .\left( \frac { x }{ sin\quad 2x } \right) \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { 1 }{ \left( \frac { sin\quad 2x }{ x } \times \frac { 2 }{ 2 } \right) } \times \lim_ { x\rightarrow 0 }{ lim } \frac { cos\quad 2x }{ cos\quad 4x } \)
\(=\frac { 1 }{ 2 } \lim_ { x\rightarrow 0 }{ lim } \frac { 1 }{ \left( \frac { sin\quad 2x }{ 2x } \right) } \times \lim_ { x\rightarrow 0 }{ lim } \frac { cos\quad 2x }{ cos\quad 4x } \)
\(=\frac { 1 }{ 2 } \times \frac { 1 }{ 1 } \times \frac { cos\quad 0 }{ cos\quad 0 } =\frac { 1 }{ 2 } \times \frac { 1 }{ 1 } \times \frac { 1 }{ 1 } =\frac { 1 }{ 2 } \quad \left[ \because \lim_{ x\rightarrow 0 }{ lim } \frac { sin\quad \theta }{ \theta } \quad =1 \right] \)

\(\lim_{ h\rightarrow 0 } \frac { (a+h)^{ 2 }sin(a+h)-a^{ 2 }sin\quad a }{ h } \)
\(=\lim_ { lim }{ h\rightarrow 0 } \frac { (a^{ 2 }+h^{ 2 }+2ah)[sin \ a\ cos \ h+cos\ a\ sin \ h)-a^{ 2 }sin\quad a }{ h } \)
\(\left[ \because \ sin(C+D)=sinCcosD+CosCsinD \right] \)
\(=\lim_{ h\rightarrow 0 } \left[ \frac { a^{ 2 }sin \ a(cos \ h-1) }{ h } +\frac { a^{ 2 }cos \ a \ sin \ h) }{ h } +(h+2a)(sin \ a \ cos \ h+cos\ a \ sin\ h) \right] \)
\(=\lim_{ h\rightarrow 0 } \left[ \frac { a^{ 2 }sina(-2sin^{ 2 }\frac { h }{ 2 } ) }{ \frac { h^{ 2 } }{ 2 } } .\frac { h }{ 2 } \right] +\overset { lim }{ h\rightarrow 0 } \frac { a^{ 2 }cosasinh }{ h } +\lim_{ h\rightarrow 0 } (h+2a)sin(a+h)\)
\(\left[ \because cosm \ h-1=-2sin^{ 2 } h/2\quad and\quad sinacos\quad h+cosasin\quad h=sin(a+h) \right] \)
\(=a^{ 2 }sin\quad a\times 0+a^{ 2 }\quad cosa(1)+2asina \quad \left[ \because \lim_{ x\rightarrow 0 } \frac { sin\quad x }{ x } =1 \right]\)
\( =a^{ 2 }cos \ a+2asin \ a\)

Given nth term T_n = (2n - 1)² \(\Rightarrow \) T_n = 4n² + 1 - 4n
Now,\(S=\Sigma { T }_{ n\quad }=\Sigma ({ 4n }^{ 2 }+1-4n)\)
\(=4\Sigma { n }^{ 2 }+\Sigma 1-4\Sigma n\)
\(=\frac { 4n(n+1)(2n+1) }{ 6 } +n-\frac { 4n(n+1) }{ 2 } \quad [\because \Sigma 1=n,\Sigma n=\frac { n(n+1) }{ 2 } and\quad \Sigma { n }^{ 2 }=\frac { n(n+1)(2n+1) }{ 6 } ]\)
\(=n\left[ \frac { 2(n+1)(2n+1) }{ 3 } +\frac { 1 }{ 1 } -\frac { 2(n+1) }{ 1\\ } \right] \)
\(=n\left[ \frac { 2({ 2n }^{ 2 }+n+2n+1)+3-6(n+1) }{ 3 } \right] \)
\( =\frac { n[{ 4n }^{ 2 }+6n+2+3-6n-6] }{ 3 } \)
\(=\frac { n({ 4n }^{ 2 }-1) }{ 3 } =\frac { n }{ 3 } (2n+1)(2n-1) \quad [\because ({ a }^{ 2 }-{ b }^{ 2 })=(a-b)(a+b)]\)

Given, co s\(\alpha \)+cos\(\beta\) = 0 and sin \(\alpha \)+sin\(\beta\) = 0
on squaring both equations, we get
(cos\(\alpha \)+cos\(\beta\) )² =0 ...(i)
and  ( sin \(\alpha \)+sin\(\beta\))² = 0 ....(ii)
On subtracting Eq.(ii) from Eq.(i), we get
(cos\(\alpha \)+cos\(\beta\) )² +( sin \(\alpha \)+sin\(\beta\))² =0
\(\Rightarrow \)(cos² \(\alpha \)+cos² \(\beta\)+2cos\(\alpha \) cos \(\beta\) - (sin²\(\alpha \)+ sin² \(\beta\)+2sin\(\alpha \) sin\(\beta\))=0  [(a+b)²= a²+b²+2ab]
\(\Rightarrow \)cos² \(\alpha \)+cos² \(\beta\)+2cos\(\alpha \) cos \(\beta\)-sin²\(\alpha \) - sin²\(\beta\)-2sin \(\alpha \)+sin\(\beta\)= 0 
\(\Rightarrow \)(cos² \(\alpha \) - sin² \(\alpha \)) + (cos² \(\beta\) - sin² \(\beta\)) + 2[cos² \(\alpha \)+cos² \(\beta\) - 2sin \(\alpha \) sin\(\beta\)]= 0   
\(\left[\begin{array}{l} \because \cos 2 x=\cos ^{2} x-\sin ^{2} x \text { and } \\ \cos A \cos B-\sin A \sin B=\cos (A+B) \end{array}\right]\)
\(\therefore \quad \cos 2 \alpha+\cos 2 \beta=-2 \cos (\alpha+\beta)\)

Given, a cos A=b cos B
\(\Rightarrow \) k sin A cos A=k sin B cos B
\(\Rightarrow \) sin 2A=sin 2B
\(\Rightarrow \)  sin 2A - sin 2B=0
\(\Rightarrow \)  2cos(A+B)sin(A-B)=0
\(\Rightarrow \)  cos(A+B)=0 or sin(A-B)=0
\(\Rightarrow \)  A+B=\(\frac { \pi }{ 2 } \) or A-B=0
Ans. \(\angle \)C=\(\frac { \pi }{ 2 } \) or A=B

We have, T =  { \(x:\frac { x+5 }{ x-7 } -5=\frac { 4x-40 }{ 13-x }\) }
\(\therefore \ \frac { x+5 }{ x-7 } -5=\frac { 4x-40 }{ 13-x } \Rightarrow \frac { x+5 }{ x-7 } -\frac { 5 }{ 1 } =\frac { 4x-40 }{ 13-x }\)
\(\Rightarrow \ \frac { x+5-5x+35 }{ x-7 } =\frac { 4x-40 }{ 13-x } \Rightarrow \frac { 40-4x }{ x-7 } =\frac { 4x-40 }{ 13-x }\)
\(\Rightarrow \ \frac { -(4x-40) }{ x-7 } -\frac { 4x-40 }{ 13-x } =0\)
\(\Rightarrow \ -(4x-40)[\frac { 1 }{ (13-x) } +\frac { 1 }{ (x-7) } ]=0\)
\(\Rightarrow \ (4x-40)[\frac { 6 }{ (13-x)(x-7) } ]=0\quad\)
\( \Rightarrow \ 4x-40=0\)
x = 10
Hence, T is a singleton set.

We have, P ( \(\phi\) )={\(\phi\) }
\(\therefore\)  P (P(\(\phi\)))={\(\phi\) ,{\(\phi\) }}
\(\Rightarrow\) P[P(P\(\phi\) ))]={\(\phi\) },{\(\phi\) },{{\(\phi\)}},{\(\phi\) ,{\(\phi\) }}}
Hence, number of elements in P [P(P (\(\phi\) ))] is 4
i.e. n {P [P(P (\(\phi\)))]}=4