\(\lim _{x \rightarrow \pi / 6} \frac{2 \sin ^{2} x+\sin x-1}{2 \sin ^{2} x-3 \sin x+1} \)
\(=\lim _{x \rightarrow \pi / 6} \frac{(2 \sin x-1)(\sin x+1)}{(2 \sin x-1)(\sin x-1)} \)
\(=\lim _{x \rightarrow \pi / 6} \frac{\sin x+1}{\sin x-1} \)
\(=\frac{1+\sin \frac{\pi}{6}}{\sin \frac{\pi}{6}-1}=\frac{1+\frac{1}{2}}{\frac{1}{2}-1}=-3\)

\(\lim _{x \rightarrow \pi / 6} \frac{\sqrt{3} \sin x-\cos x}{x-\frac{\pi}{6}}\)
\(=\lim _{x \rightarrow \pi / 6} \frac{2\left(\sin x \cos \frac{\pi}{6}-\cos x \sin \frac{\pi}{6}\right)}{\left(x-\frac{\pi}{6}\right)}\)
\(=2 \lim _{x \rightarrow \pi / 6} \frac{\sin \left(x-\frac{\pi}{6}\right)}{\left(x-\frac{\pi}{6}\right)} \)
\([\because \sin A \cos B-\cos A \sin B=\sin (A-B)] \)
\(=2 \left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1 \text { and } x \rightarrow \frac{\pi}{6} \Rightarrow\left(x-\frac{\pi}{6}\right) \rightarrow 0\right]\)

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

Equation of the line passing through (3,2) having slope \(\frac { 3 }{ 4 } \)is given by
\(y-2=\frac{3}{4}(x-3) \quad\left[\because y-y_{0}=m\left(x-x_{0}\right)\right]\)
\(\Rightarrow 4 y-3 x+1=0\)  ..(i)
Let (h,k) be the required point on the line such that distance between (h,k) and (3,2) is 5
\(\Rightarrow (h-3)^{2}+(k-2)^{2}=25\) ...(ii)  [by distance formula]
since, point(h,k) lies on the line (i)
\(4 k-3 h+1 =0\)   ...(iii)
\(\Rightarrow k =\frac{3 h-1}{4}\)   ...(iv)
On putting the value of k in Eq.(ii) and simplifying, we get
\(25 h^{2}-150 h-175=0 \Rightarrow h^{2}-6 h-7=0 \)
\(\Rightarrow (h+1)(h-7)=0 \Rightarrow h=-1 \text { or } h=7\)
On putting these values of h in Eq.(iv), we get
k = -1 or k = 5
therefore, the coordinates of the required points are either(-1,-1) or (7,5)

Given, depreciation in value of machine = 20%
After each year the value of the machine is 80 %
(100-20)% of its value of the previous year, so at the end of 5 yrs, the machine will depreciate as many times as 5.
Hence, we have to find the 6th terms of the G P whise first term a₁ is 1250 and common ratio r is 8.
Hence, value at the end 5 y6rs.
=t₆=a_rr⁵
=1250(0.8)⁵=409.6

9T₉ = 13T₁₃  \(\Longrightarrow \) 9 ( a + 8d ) = 13 ( a + 12d ) \(\Longrightarrow \) a = -21d = 0

Given that, sum of n terms of an AP,
\(Sn=2n+ { 3n }^{ 2 }\)
On replacing (n-1) by n, we get \({ s }_{ n-1 }=2(n-1)+3({ n-1 })^{ 2 }\)
\(Then,\quad { T }_{ n }={ S }_{ n }{ S }_{ n-1 }\)
\( =(2n+{ 3n }^{ 2 })-[2n-2+3({ n }^{ 2 }+1-2n)]\)
\(=(2n+{ 3n }^{ 2 })-[2n-2+3{ n }^{ 2 }+3-6n)]\)
\(=2n+{ 3n }^{ 2 }-2n+2-{ 3n }^{ 2 }-3+6n\)
\(=6n-1\)
\( \therefore \quad rth\quad term,{ T }_{ r }=6r-1\)

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 E = \(\left( { x }^{ 2 }-\sqrt { 1-{ x }^{ 2 } } \right) ^{ 4 }+\left( { x }^{ 2 }+\sqrt { 1-{ x }^{ 2 } } \right) ^{ 4 }\)
Put \(\sqrt { 1-{ x }^{ 2 } } =y,\quad \) , we get
\(E=\left( { x }^{ 2 }-y \right) ^{ 4 }+\left( { x }^{ 2 }+y \right) ^{ 4 }\)
\(=2[^{ 4 }{ C }_{ o }\left( { x }^{ 2 } \right) ^{ 4 }{ y }^{ o }+^{ 4 }{ C }_{ 2 }\left( { x }^{ 2 } \right) ^{ 2 }{ y }^{ 2 }+^{ 4 }{ C }_{ 4 }\left( { x }^{ 2 } \right) ^{ o }{ y }^{ 4 }]\)
\(\left[ if\quad n\quad is \quad even,then(x-a)^{ n }+(x+a)^{ n }=2\left\{ ^{ n }{ C }_{ o }{ x }^{ n }{ a }^{ o }+^{ n }{ C }_{ 2 }{ x }^{ n-2 }{ a }^{ 2 }+^{ n }{ C }_{ 4 }{ x }^{ n-4 }{ a }^{ 4 }+... \right\} \right] \)
\(=2[1\times { x }^{ 8 }\times 1+6\times { x }^{ 4 }{ y }^{ 2 }+1\times 1\times { y }^{ 4 }]\)
\(=2[{ x }^{ 8 }+6{ x }^{ 4 }(1-{ x }^{ 2 })+(1-{ x }^{ 2 })^{ 2 }]\quad [put\quad y=\sqrt { 1-{ x }^{ 2 } } \)
\(=2({ x }^{ 8 }+6{ x }^{ 4 }-6{ x }^{ 6 }+1+{ x }^{ 4 }-2{ x }^{ 2 })\)
\(=2{ x }^{ 8 }-12{ x }^{ 6 }+14{ x }^{ 4 }-4{ x }^{ 2 }+2\)

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 .

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 \(|x|\ge a\Longrightarrow x\ge a\) 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. \(2n\pi \pm \frac { \pi }{ 2 } \)

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

Step I Let P(n) be the given statement.
i.e P(n);\(2n+1 <{ 2 }^{ n }\)
Step II For n = 3,we have
(2 x 3 +1)<2³ \(\Rightarrow \) 7< 8, which is true.
Thus P(1) is true.
Step III Let us assume that P(k) is true.
i.e P(k):\(2k+1 <{ 2 }^{ k }\)
Step IV Now, we shall prove the statement for n=k+1.
for this, we have to show that \(2(k+1)+1<{ 2 }^{ k+1 }\)
from Eq.(i), \(2k+1 <{ 2 }^{ k }\)
So, (2k+1)+2 < 2^k +2 [adding 2 on both sides]
\(\Rightarrow 2 k+3<2^{k} \cdot 2 \quad {\left[\because 2^{k}+2<2^{k} \cdot 2\right]} \)
\(\Rightarrow 2 k+3<2^{k+1} \Rightarrow 2(k+1)+1<2^{k+1}\)
thus, P(k+1) is true, whenever P(k) is true.
hence, by principle of mathematical induction P(n) is true for all natural numbers, \(n\ge 3\).

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

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.

\(LHS=sin2x+2sin4x+sin6x\)
\(=\quad sin2x+sin6x+2sin4x\)
\(=\quad 2sin\left( \frac { 2x+6x }{ 2 } \right) cos\left( \frac { 2x-6x }{ 2 } \right) +2sin4x\)
\(\left[ \because sinA+sinB=2sin\left( \frac { A+B }{ 2 } \right) .cos\left( \frac { A-B }{ 2 } \right) \right] \)
\(=\quad 2sin4x.cos(-2x)+2sin4x\)
\(=\quad 2sin4x.cos2x+2sin4x\quad \quad \left[ \because cos(-\theta )=cos\theta \right] \)
\(=\quad 2sin4x(cos2x+1)\)
\(=\quad 2sin4x(2cos^{ 2 }x)=4sin4x.cos^{ 2 }x\quad \left[ \because cos2x=2cos^{ 2 }x-1 \right] \)
Hence proved.
\(\because \) LHS = RHS

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 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]
\(=(A'\cup B)\cup (B\cup C)\)   \([\quad \because (B')'=B]\)
\(=(A'\cup B)\cup B)\cap (A'\cup B)\cup C\)
\(=(A'\cup (B\cup B))\cap (A'\cup (B\cup C))\)
\(=(A'\cup B)\cap (A'\cup B\cup C)\)
\(=(A'\cup B)\quad =RHS\)