Let the point on X-axis be P(x,0,0).
\(Then,\quad { (x-3) }^{ 2 }+{ (0-2) }^{ 2 }+{ (0-2) }^{ 2 }\)
\(={ (x-5) }^{ 2 }+{ (0-5) }^{ 2 }+{ (0-4) }^{ 2 }\)
\(Ans.\left( \frac { 49 }{ 4 } ,0,0 \right) \)

Given a, b, c, d are in GP.
and we know that AM > GM then for the first three terms
 \(\frac { a+c }{ 2 } >b\)   [\(\because \sqrt { ac } =b\)]
\(\Rightarrow \) a + c > 2b...(iii)
Similarly, for the last three terms
\(\frac { b+d }{ 2 } >c\)   [\(\because \sqrt { bd} =c\)]
\(\Rightarrow \) b + d > 2c....(iv)
On adding Eqs. (iii) and (iv), we get
(a + c) + (b + d) > 2b + 2c
\(\Rightarrow \) a + d > b + c

We have , foci of the hyperbola lies on X-axis. So, the equation of hyperbola is
    \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) ---- (i)
Foci \(\equiv \left( \pm c,0 \right) \equiv \left( \pm 2,0 \right) \Rightarrow c=2\)
Eccentricity of the hyperbola, e=3/2
We know that, c=ae
\(\therefore \quad 2=a\left( \frac { 3 }{ 2 } \right) \Rightarrow a=\frac { 4 }{ 3 } \\ Also\quad \quad { c }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }\\ \Rightarrow \quad ({ 2 })^{ 2 }=\left( \frac { 4 }{ 3 } \right) ^{ 2 }+{ b }^{ 2 }\Rightarrow { b }^{ 2 }=\frac { 20 }{ 9 } \\ Thus,\quad { a }^{ 2 }=\frac { 16 }{ 9 } ,{ b }^{ 2 }=\frac { 20 }{ 9 } \)
Hence, equation of hyperbola is
\(\frac { 9{ x }^{ 2 } }{ 16 } -\frac { 9{ y }^{ 2 } }{ 20 } =1\quad \quad \quad [from\quad Eq.(i)]\\ \Rightarrow \quad 45{ x }^{ 2 }-36{ y }^{ 2 }=80\)

Let us first seat  the 5 girls (or 5 boys). This can be done in 5! ways.
Now, for each such arrangement, the 5 boys can be seated only at the cross marked the place as shown below:
(I)  \(\times G_{ 1\times }G_{ 2 }\times { G }_{ 3 }\times { G }_{ 4 }\times { G }_{ 5 }\times \)
(II) \(G_{ 1\times }G_{ 2 }\times { G }_{ 3 }\times { G }_{ 4 }\times { G }_{ 5 }\times \)
In case I, 5 boys can be seated in P = 5! ways
Thus, number of ways of seating = 5! \(\times \) 5! = \({ (5! })^{ 2 }\)
Similarly, in case II, number of ways of seating = \({ (5! })^{ 2 }\) 
Hence, required number of ways = \({ (5! })^{ 2 }\) + \({ (5! })^{ 2 }\) \({ =2.(5! })^{ 2 }=28800\)
For solving this type of problem, we used the following steps
Step I Firstly, decide that from how many digits the required number will be formed.
Step II Fill up the places on which restrictions are present and let the number of ways of filling up these places be k.
Step III Find the number of ways of filling the remaining places with remaining digits by using the formula ⁿP _r  .
Step IV Required number of numbers is k. P . ⁿP _r .

We have 5 boys and 5 girls,
Since no two girls sit together, therefore the possible choices for girls are the places marked as '\(\times\)'.
B1  \(\times\)B2   \(\times\)B3 \(\times\)B4 \(\times\)B5 \(\times\)
Clearly, the girls can be arranged in ⁶P5 ways and the boys can be arranged in 5! ways.
Hence, by fundamental principle multiplication, required number of ways = ⁶P_{5 \(\times \)5!}
 \(=\frac { 6! }{ (6-5)! } = 5!=6!\times 5!=86400\)
For solving this type of problem, we used the following steps
Step I Firstly, decide that from how many digits the required number will be formed.
Step II Fill up the places on which restrictions are present and let the number of ways of filling up these places be k.
Step III Find the number of ways of filling the remaining places with remaining digits by using the formula  ⁿP_r
Step IV Required number of numbers is k. ⁿ P _r .

Ans . 41472

There will be as many ways as there are ways of filling 2 vacant places  in succession by the five given digits. Here, in this case, we start filling in unit’s place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten’s place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is 2 × 5, i.e., 10.

Use \(|x|>a\Longrightarrow x>a\) or  \(x<-a\)  
\(\left( -\infty ,\frac { 5 }{ 3 } \right) \) \(\cup \) \(\left( 3,\infty \right) \)

Use \(|x|\ge a\Longrightarrow x\ge a\) or  \(x\le -a\)
(-\(\infty \), -13] \(\cup \) [7,\(\infty \))

\(\frac { { \left( 1-i \right) }^{ 3 } }{ 1-{ i }^{ 3 } } =\frac { { 1 }^{ 3 }-{ i }^{ 3 }-3i+3{ i }^{ 2 } }{ 1+i } =\frac { 1+i-3i-3 }{ 1+i } =\frac { -2-2i }{ 1+i } \)
Ans.-2

\(z=\frac { 1+icos\theta }{ 1-2icos\theta } \times \frac { 1+2icos\theta }{ 1+2icos\theta } =\frac { 1-2{ cos }^{ 2 }\theta +3icos\theta }{ 1+4{ cos }^{ 2 }\theta } \)
\(For\quad Re(z),put\quad 3cos\theta =0\Rightarrow cos\theta =0\)
Ans. \(2n\pi \pm \frac { \pi }{ 2 } \)

We have,
\({ \left( 1+i \right) }^{ 6 }={ \left( { \left( 1+i \right) }^{ 2 } \right) }^{ 3 }\)
\(={ \left( 1-{ i }^{ 2 }+2i \right) }^{ 3 }\)
\(\left[ \because { \left( { z }_{ 1 }+{ z }_{ 2 } \right) }^{ 2 }={ z }_{ 1 }^{ 2 }+{ z }_{ 2 }^{ 2 }+{ 2z }_{ 1 }{ z }_{ 2 } \right] \)
\(={ \left( 1-1+2i \right) }^{ 3 }\ \left[ \because { i }^{ 3 }=-1 \right] \)
\(\Rightarrow { \left( 1+i \right) }^{ 6 }={ \left( 2i \right) }^{ 3 }={ 8 }i^{ 3 }=8i \left[ \because { i }^{ 3 }=-1 \right] \left( i \right) \)
\(and\quad { \left( 1-i \right) }^{ 3 }={ i }^{ 3 }-{ i }^{ 3 }-3{ \left( 1 \right) }^{ 2 }i+3\left( 1 \right) { \left( i \right) }^{ 2 }\)
\(\left[ \because { \left( { z }_{ 1 }-{ z }_{ 2 } \right) }^{ 2 }={ z }_{ 1 }^{ 3 }-{ 3z }_{ 1 }^{ 2 }{ z }_{ 2 }+3{ z }_{ 1 }{ z }_{ 2 }^{ 2 }-{ z }_{ 2 }^{ 3 } \right] \)
\(=1-\left( -i \right) -3i-3\ \left[ \because { i }^{ 3 }=-1\ { i }^{ 2 }=-1 \right]\) 
\(\Rightarrow(1-i)^{3}=-2-2 i\)
On adding eqs. (i) and (ii), we get
(1 + i)⁶ + (1-i)³ 
= - 8i - 2 - 2i
= -2 - 10i.

\(z=\frac { 1-i }{ 1+i } x \frac { 1-i }{ 1-i } =\frac { 1-1-2i }{ 1+1 } =-i\)   = i

Consider :\(P(k) : \sum _{ t=1 }^{ k-1 }{ t(t+1) } =\frac { k(k-1)(k+1) }{ 3 } ​​​​k\ge 2\) 
P(k):1.2+2.3+3.4+....+(k-1)k \(=\frac { k(k-1)(k+1) }{ 3 } \)
Now P(k+1):1.2+2.3+3.4+...+(k1)k+k(k+1)
\(=\frac { k(k-1)(k+1) }{ 3 } +k(k+1)\)
\(=\frac { k(k+1)(k+2) }{ 3 } k\ge 2\)

Consider P(k) : 1+2+2²+.... +2^k = 2^k+1-1
Now P(+1):1+2+2² +..+2^k = 2^k+1 
=2^k+1-1+2^k+1
=2^(k+1)+1 -1

\(LHS=cos\frac { 2\pi }{ 15 } cos\frac { 4\pi }{ 15 } cos\frac { 8\pi }{ 15 } cos\frac { 16\pi }{ 15 }\)
\(=cos24^{ \circ }cos48^{ \circ }cos96^{ \circ }cos192^{ \circ }\)
\(=\frac { 1 }{ 16sin24^{ \circ } } \left[ (2sin24^{ \circ }cos24^{ \circ })(2cos48^{ \circ })(2cos96^{ \circ })(2cos192^{ \circ }) \right]\)
\(=\frac { 1 }{ 16sin24^{ \circ } } \left[ 2sin48^{ \circ }cos48^{ \circ }(2cos96^{ \circ })(2cos192^{ \circ }) \right]\)
\(=\frac { 1 }{ 16sin24^{ \circ } } \left[ (2sin96^{ \circ }cos96^{ \circ })(2cos192^{ \circ }) \right]\)
\(=\frac { 1 }{ 16sin24^{ \circ } } sin384^{ \circ }=\frac { sin(360^{ \circ }+24^{ \circ }) }{ 16sin24^{ \circ } } \)
\(=\frac { 1 }{ 16 } =RHS\)
Hence proved.

\(LHS=\frac { 1 }{ { cos }^{ 2 }\theta } +\frac { 1 }{ { sin }^{ 2 }\theta } =\frac { { sin }^{ 2 }\theta +{ cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta { cos }^{ 2 }\theta }\)
\(=\frac { 4 }{ \left( sin2\theta \right) ^{ 2 } } =4{ cosec }^{ 2 }2\theta .\)
\(\because \quad cosec^{ 2 }\quad \phi \ge 1,\quad so\quad 4\quad cosec^{ 2 }\quad 2\theta \ge 4\)

given, \(\frac { sin(x+y) }{ sin(x-y) } =\frac { a+b }{ a-b } \)
using componendo and dividendo rule,we get
\(\frac { sin(x+y)+sin(x-y) }{ sin(x+y)-sin(x-y) } =\frac { a+b+a-b }{ a+b-a-b }\)
\( \\ \Rightarrow \frac { 2sin\left( \frac { x+y+x-y }{ 2 } \right) .cos\left( \frac { x+y-x+y }{ 2 } \right) }{ 2cos\left( \frac { x+y+x-y }{ 2 } \right) .sin\left( \frac { x+y-x+y }{ 2 } \right) } =\frac { 2a }{ 2b } \)
\(\left[ sinA+sinB=2sin\frac { A+B }{ 2 } .cos\frac { A-B }{ 2 } and\quad sinA-sinB=2cos\frac { A+B }{ 2 } .sin\frac { A-b }{ 2 } \right] \)
\(\Rightarrow \frac { sin\quad x.cos\quad y }{ cos\quad x.sin\quad y } =\frac { a }{ b } \Rightarrow \quad \frac { tan\quad x }{ tan\quad y } =\frac { a }{ b }\)
Hence proved.

We have, \(f(x)=\frac { 1 }{ \sqrt { x+\left| x \right| } } \)
We know that \(\left| x \right| =\begin{cases} x,\quad \quad x\ge 0 \\ -x,\quad x<0 \end{cases}\) 
\(\Rightarrow x+\left| x \right| =\begin{cases} x+x\quad if\quad x\ge 0 \\ x-x\quad if\quad x<0 \end{cases}\)
\(\Rightarrow x+\left| x \right| =\begin{cases} 2x\quad if\quad x\ge 0 \\ 0\quad if\quad x<0 \end{cases}\)
since,f(x)=\(\frac { 1 }{ \sqrt { x-\left| x \right| } } \) assume real values
When \(x+\left| x \right| >0\) then
x > 0
\(\Rightarrow \ x\in (0,\infty )\)
Hence, domain of (f)=\((0,\infty )\)

Given,   \(t(c)=\frac { 9C }{ 5 } +32\)
On putting C = 28 in Eq.(i), we get
\(t(28)=\frac { 9\times 28 }{ 5 } +32=\frac { 252 }{ 5 } +\frac { 32 }{ 1 }\)
\(=\frac { 252+160 }{ 5 } =\frac { 412 }{ 5 } \)

\(Step\quad I\quad Let\quad P(n)\quad be\quad the\quad given\quad statement. i.e.P(n):{ 7 }^{ n }-{ 2 }^{ n }is\quad divisible\quad by\quad 5.\)
\(Step\quad II\quad For\quad n=1,\quad we\quad have\quad P(1):{ 7 }^{ 1 }-{ 2 }^{ 1 }=5\)
Which is divisible by 5.
Thus,P(1) is true.
\(Step\quad III\quad Let\quad us\quad assume\quad that\quad P(n)\quad is\quad true\quad for\quad n=k. i.e.P(k):{ 7 }^{ k }-{ 2 }^{ k }is\quad divisible\quad by\quad 5.\)
\(Then\quad { 7 }^{ k }-{ 2 }^{ k }\quad =5d\quad for\quad some\quad d\in N.\)
\(\Rightarrow { 7 }^{ k }=5d+{ 2 }^{ k },for\quad some\quad d\in N.\quad ...(i)\)
Step IV Now, we shall prove the statement for n=k+1.
For this, we have to show that N.
\({ 7 }^{ k+1 }-{ 2 }^{ k+1 }is\quad divisible\quad by\quad 5.\)
\(Then,\quad { 7 }^{ k+1 }-{ 2 }^{ k+1 }={ 7 }^{ k }.7-{ 2 }^{ k+1 }\)
\(=7(5d+{ 2 }^{ k })-{ 2.2 }^{ k }\quad \quad [from\quad Eq.(i)]\)
\(=35d+7.2k-{ 2.2 }^{ k }=35d+{ 5.2 }^{ k }\)
\(=5(7d+{ 2 }^{ k }),which\quad s\quad divisible\quad by\quad 5.\)

Thus,P(k+1) is true, when ever P(k) is true.Hence,by principle of mathematical induction,P(n) is true for all n∈N.

Step I Let P(n) be the given statement.i.e., P(n) :\(cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ n-1 }\theta =\frac { { sin2 }^{ n }\theta }{ { 2 }^{ n }sin\theta } \)
Step II For n = 1, we have, LHS = \(cos\theta\) and \(RHS=\frac { sin2\theta }{ { 2 }sin\theta } =\frac { { 2 }sin\theta cos\theta }{ { 2 }sin\theta } =cos\theta \)
\(\therefore\) LHS = RHS  P (1) is true.
Step III Let us assume that P(k) is true. i.e., P(k) : \(cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ k-1 }\theta =\frac { { sin2 }^{ k }\theta }{ { 2 }^{ k }sin\theta } \ ...(i)\)
Step IV Now, we shall prove the statement for n = k + 1
For this, we have to show that
\({ cos2 }^{ 2 }\theta ...{ cos2 }^{ (k+1)-1 }\theta \quad \frac { { sin2 }^{ k+1 }\theta }{ { 2 }^{ k+1 }sin\theta }\)
\(Then \ LHS=cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ k }\theta \)
\(=cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ k-1 }\theta \ { cos2 }^{ k }\theta\)
\(=\frac { { sin2 }^{ k }\theta }{ { 2 }^{ k }sin\theta } .{ cos2 }^{ k }\theta\)
\([Multiplying \ numerator \ and \ denominator \ by \ 2]\)
\(=\frac { { sin2.(2 }^{ k }\theta ) }{ { 2 }^{ k+1 }sin\theta } \ \ [\because 2sin\theta \ cos\theta =sin2\theta ]\)
\(=\frac { { sin2 }^{ k+1 }\theta }{ { 2 }^{ k+1 }sin\theta } =RHS\)
\(Thus, \ P(k+1) \ is \ true, \ whenever \ P(k) \ is \ true.\)
\(Hence, \ by \ principle \ of \ mathematical \ induction, \ P(n) \ is \ true \ for \ all \ n\in N\)

If elements are not repeated, then number of elements in 
A_{1 \(\cup\)} A_2\(\cup\)   A_3\(\cup\)  ..._\(\cup\)  A₃₀ is 30 \(\times\)5
But each element is used 10 times, so
n(S) = \(\cfrac{30\times5}{10}\) =15
If elements in B₁ ,B₂ ,.....B_n are not r, then total number of elements is 3n but each element is repeated 9 times, so
n(S) = \(\cfrac{3n}{9} \)
\(\Rightarrow 15=\cfrac{3n}{9}\)
n    = 45

We have, g= {(1,1), (2, 3), (3, 5), (4, 7)}
Since, every element has unique image under g. So, g is a function.
Now, g(x)=\(\alpha x+\beta \)
When x = 1, then  g(1)=\(\alpha (1)+\beta \)     ...(i)
\(\Rightarrow 1=\alpha +\beta \)
\(When\quad x=2,\quad then\quad g(2)=\alpha (2)+\beta\)
\(\Rightarrow 3=2\alpha +\beta \)
On solving Eqs.(i) and (ii),we get
\(\alpha =2,\beta =-1\)

\(LHS=(A\cap B')\cup (B\cup C')\)
\(=\{ A'\cup (B')'\} U(B\cup C)\) [By De Morgan's lae]
\(=(A'\cup B)\cup (B\cup C)\)   \([\quad \because (B')'=B]\)
\(=(A'\cup B)\cup B)\cap (A'\cup B)\cup C\)
\(=(A'\cup (B\cup B))\cap (A'\cup (B\cup C))\)
\(=(A'\cup B)\cap (A'\cup B\cup C)\)
\(=(A'\cup B)\quad =RHS\)