\((x+1)^{1\over 3} = (x-3)^{1\over 2}\)
L.C.M. of 2 and 3 is 6 & Raising to the power 6
\(\left\{ (x+1)^{ \frac { 1 }{ 3 } } \right\} ^{ 6 }=\left\{ (x-3)^{ \frac { 1 }{ 2 } } \right\} ^{ 6 }\)
(x+1)² = (x-3)³
x²+2x+1 =x³-9x²+27x-27
0 = x³-9x²+27x-27-x²-2x-1
x³-10x²+25x-28 =0
since constant term is - 28
we can have a factor as (x ± 2) or (x ± 4) or (x ± 7)
By trial and error method we find that (x - 7) is a factor
Using synthetic division

we get x³-10x²+25x-28=(x-7)(x²-3x+4)
solving x²-3x+4 =0
x=\(\frac { 3\pm \sqrt { 9-16 } }{ 2 } =\frac { 3\pm \sqrt { -7 } }{ 2 } \)
the roots are x =7 x=\(\frac { 3\pm \sqrt { -7 } }{ 2 } \)

Let, p(n) : = aⁿ - bⁿ, is divisible by a - b.
Substituting the value of n = 1, in the statement we get,
P(1) = a - b,
which is divisible by a-b. Hence, P(1) is true. Let us assume that the statement is true for n = k.
Then P(k) = a^k- b^k, is divisible by a - b. We can write
P(k) = a^k- b^k = \(\lambda\)(a - b), \(\lambda\)\(\in\) N.
We need to show that P(k + 1) = a^{k + 1}- b^k+1, is divisible by a - b.
P(k+1) = a^k+1-b^k+1
= a^{k+ 1}- ab^k + ab^k - b^k+1
a(a^k - b^k + b^k (a - b)
a(\(\lambda\)(a - b)) + b^k (a - b)
= (a - b)(a\(\lambda\) + b^k)
(a - b) \(\lambda_1,\) \(\lambda_1\) = a\(\lambda\) + b^k, \(\lambda_1\)\(\in\)N.
which is divisible by a-b. This implies that P(k + 1) is true. The validity of P(k + 1) follows from that of P(k).
Therefore by the principle of mathematical induction, aⁿ - bⁿ is divisible by a - b, where a > b, for all natural numbers n.

12x²-10xy+2y²+14x-5y+c=0
ax² +2hxy+ by² +2gx+2fy+c = 0
\(\left| \begin{matrix} 2h=-10 \\ h=-5 \end{matrix} \right| b=2\left| \begin{matrix} 2g=14 \\ g=7 \end{matrix} \right| \left| \begin{matrix} 2f=-5 \\ f=\frac { -5 }{ 2 } \end{matrix} \right| \)
af² +bg² +ch² -2fgh-abc=0 is the condition
12\(\left( \frac { 25 }{ 4 } \right) \)+2(7)²+c(-5)²-2\(\left( \frac { -5 }{ 2 } \right) \)(7)(-5)-12(2)(c)=0
75+98+25c-175-24c=0, c=2
The equation is 12x²-10xy+2y²+14x-5y+2=0
12x²-10xy+2y=(3x-y)(4x-2y)
Let 12x²-10xy+2y²+14x-5y+2=(3x-y+l)(4x-2y+m)
So that 4l+3m=14, -2l-m=-5
On solving we get t=\(\frac { 1 }{ 2 } \), m=4
∴ The separate equations are 3x-y+\(\frac { 1 }{ 2 } \)=0 ⇒ 6x-2y+1=0 and
4x-2y+4=0 ⇒ 2x-y+2=0
m₁=\(\frac { -6 }{ -2 } \)=3; m₂=\(\frac { -2 }{ -1 } \)=2
∴ tanθ=\(\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| =\left| \frac { 3-2 }{ 1+3.2 } \right| =\frac { 1 }{ 7 } \)
θ=tan^-1\(\left( \frac { 1 }{ 7 } \right) \).

The given series is 1 + 5(-1)+9(1)+13(-1)+....
= 1 + 5(-1)+9(-1)²+13(-1)³+....
This is an arithmetic - geometric series with corresponding A.P. 1, 5, 9, 13.... and G.P. 1, -1, (-1)², (-1)³, ....
\(\therefore\) Tn of A.G. series = (Tn of A.P) (Tn of G.P)
= [ 1 + ( n - 1)4 ]  [ 1 (-1)^n-1]  [ \(\because\) T_n in A.P. is a + (n - 1)d, T_n in G.P. is a r^n-1 ]
= (4n-3)(-1)^n-1
Let Sn be the Sum of the first n terms of AG series
S_n = T₁ + T₂ + T₃ + ... T_n-1 + Tn
S_n = 1 + 5(-1)+9(-1)² + .. + (4n-7)(-1)^n-2 + (4n-3) (-1)^n-1  ...(2)
Multiplying by (-1) we get,
-1 S_n = 1 + 5(-1)+9(-3)³ + .. + (4n-7)(-1)^n-1 + 4n-3(-1)ⁿ
(2)-(3) we get,
S_n + S_n = 1 + [4(-1) + 4(-1)²]+...+4(-1)^n-1]-(4n-3)(-1)ⁿ
\(\Rightarrow\)   \(2S_n=1+{4(-1)[1-(-1)^{n-1}]\over1-(-1)}-(4n-3)(-1)^n\)    \(\begin{bmatrix} \because For\ GP, S_N={a(1-r^n)\over 1-r} \\a=1,r=-1 \end{bmatrix}\)
\(\Rightarrow\)    \(2S_n=1-2()1-(-1)^{n-1}-(4n-3){(-1)}^{n}\)
\(\Rightarrow\)    \(S_n={1\over 2}-1+{(-1)}^{n-1}-{4n-3\over 2}{(-1)}^{n-1}(-1)\)
\(\Rightarrow\)   \(S_n={-{1\over 2}}+{2+4n-3\over 2}{(-1)}^{n-1}\)
\(\Rightarrow\)   \(S_n={1\over 2}[-1+(4n-14){(--1)}^{n-1}].\)

Let p(n) be the statement
x²ⁿ - y²ⁿ is divisible
Step 1: Putting n = 1, we get
x² - y² = (x +y) (x - y) which is divisible by (x +y).
∴ p(1) is true
Step 2: Let us assume that p(k) is true
∴. x^2k - y^2k is divisible by (x +y)
⇒ x^2k = m(x +y) + y^2k for some scalars
Step 3: To prove thatp(k+ 1) is true.
i.e to P. T. x^2(k+1) - y^2(k+1) is divisible by (x + y).
Consider x^2(k+l) - y^2(k+l)
X^2(k+1) - y^2(k+2) = x². x^2k - y^2k+2
x² [m (x +y) +y^2k) - y^2k+2
m (x+ y) (x²) + x² y^2k - y^2k.y²
m (x+ y) x² + y^2k (x² - y²)
m (x+ y) x² + y^2k (x + y) (x - y)
(x + y) [mx² + (x - y) y2k] which is divisible by (x + y).
∴ x^2(k+1) - y^2(k+1) is divisible by (x +y).
∴ p(k+ 1) is true.
Hence by the principle of mathematical induction,p(n) is true for all values of n.

Given R = {(a, b) : |a - b| is a multiple of 4}
Reflexivity: Where a, b ∈ A = {0, 1, 2, ... 12}. For any a ∈ A, we have |a - a| = 0 which is a multiple of 4.
⇒ (a, a) ∈ A for all a ∈ A
∴ R is reflexive
Symmetry: Let (a, b) ∈ R. Then
(a, b) ∈ R
⇒ |a - b| is a multiple of 4.
⇒ |a - b| = 4⋋- for some ⋋∈N.
⇒ Ib - al = 4⋋- for some ⋋∈N.
⇒ (b, a) ∈N
∴ R is symmetricTransitivity: Let (a, b) ∈ Rand (b, c) ∈ R
Then (a, b) ∈ R and (b, c) ∈ R
⇒ |a - b| is a multiple of 4 and Ib - c| is a multiple of 4.
⇒ |a - b| = 4⋋ and |b - c| = 4μ for some ⋋ μ∈ N
⇒ a-b = ±4 -and b-c = ±4μ for some ⋋, μ ∈ N
⇒ a-c = ±4⋋ ±4μ for some ⋋, μ ∈ N
⇒ |a - c| is a multiple of 4.
⇒ (a-c)∈R
∴ R is transitive.
Hence, R is an equivalence relation.

LHS = \(\sin ^{ 2 }{ \frac { \pi }{ 18 } } +\sin ^{ 2 }{ \frac { \pi }{ 9 } } +\sin ^{ 2 }{ \frac { 7\pi }{ 18 } } +\sin ^{ 2 }{ \frac { 4\pi }{ 9 } } \)
\(=​​​​\sin ^{ 2 }{ \left( \frac { \pi }{ 18 } \times \frac { 180 }{ \pi } \right) } +\sin ^{ 2 }{ \left( \frac { \pi }{ 9 } \times \frac { 180 }{ \pi } \right) } +\sin ^{ 2 }{ \left( \frac { 7\pi }{ 18 } \times \frac { 180 }{ \pi } \right) } +\sin ^{ 2 }{ \left( \frac { 4\pi }{ 9 } \times \frac { 180 }{ \pi } \right) } \)
= sin² 10^o + sin² 20^o + sin² 70^o + sin² 80^o
= [sin (90 - 80^o)]² + [sin (90 - 70)]² + sin² 70^o + sin² 80^o
= cos² 80^o + cos² 70^o + sin² 70^o +sin² 80^o
= (cos² 80^o + sin² 80^o) + (cos² 70 + sin² 70)
= 1 + 1 = 2 = RHS
Hence proved.

\(\tan \theta=\frac{2 \sqrt{b^{2}-a b}}{a+b}=\frac{2 \sqrt{\left(\frac{-1}{2}\right)^{2}+6}}{-5}\)
\(=\left|\frac{2 \sqrt{\frac{1}{4}+6}}{-5}\right|\)
\(=\frac{2\left(\frac{5}{2}\right)}{5}=1 \Rightarrow \theta=45^{\circ}\)
\(\frac{2 \cos \theta+3 \sin \theta}{4 \sin \theta+5 \cos \theta}=\frac{\frac{2}{\sqrt{2}}+\frac{3}{\sqrt{2}}}{\frac{4}{\sqrt{2}}+\frac{5}{\sqrt{2}}}=\frac{5}{9}\)

\(\mathrm{a} =1, \quad \mathrm{~d}=6, \quad \mathrm{r}=\frac{1}{2} \)
\(\mathrm{~S}_{\infty} =\frac{a}{1-\mathrm{r}}+\frac{\mathrm{dr}}{(1+\mathrm{r})^{2}} \)
\(=\frac{1}{1-\frac{1}{2}}+\frac{6 \times \frac{1}{2}}{\left(\frac{1}{2}\right)^{2}} \)
\(=2+(3 \times 4)=14 \)

\(x^{2}+a x+c=0 \)
\(x^{2}+d x+b=0 \)
\(8 \& 2 \text { are the roots }\)\(\text { 3. } 3 \text { are the roots }\)
\(\therefore a=-10 ; c=16 \quad d=-6, \quad b=9\)
\(x^{2}+a x+b =0 \)
\(x^{2}-10 x+9 =0 \)
\(\Rightarrow(x-1)(x-9) =0 \)
\(\therefore x =1 \text { (or) } 9 \)

We have f(x) = \(\frac { { x }^{ 2 }-9 }{ x-3 } \)
Domain of f : Clearly f(x) is not defined for x - 3 = 0 i.e. x = 3. Therefore, Domain (f) = R- {3}
Range of f: Let f(x) = y. Then,
f(x) = y ⇒ \(\frac { { x }^{ 2 }-9 }{ x-3 } \)=y ⇒ x+3 = y
it follows from the above relation that y takes all real values except 6 when x takes values in the ser R - {3}. Therefore, Range (f) = R {6}.

Given \({ \left( x-\frac { 1 }{ 2y } \right) }^{ 10 }\)
Here n = 10, x = x and \(a=\left( \frac { -1 }{ 2y } \right) \)
Middle term = \({ T }_{ \frac { 10+2 }{ 2 } }={ T }_{ 6 }\)
General term is \({ T }_{ r+1 }=nCr{ x }^{ n-r }{ a }^{ r }\)
Putting r = 5 we get,
\({ T }_{ 6 }=10{ C }_{ 5 }{ x }^{ 10-5 }{ \left[ -\frac { 1 }{ 2y } \right] }^{ 5 }=\frac { 10\times 9\times 8\times 7\times 6 }{ 5\times 4\times 3\times 2\times 1 } .{ x }^{ 5 }\left( \frac { -1 }{ 32.{ y }^{ 5 } } \right) \)
\(=-225.{ x }^{ 5 }.\frac { 1 }{ 32{ y }^{ 5 } } { T }_{ 6 }=\frac { -63{ x }^{ 5 } }{ 8{ y }^{ 5 } } \)

Since the given line, is perpendicular to y-axis, it will be parallel to y-axis.
\(\therefore\) Equation of the line is x = -2

sin\(\left( \frac { -11\pi }{ 3 } \right) =-sin\frac { 11\pi }{ 3 } \)
= \(-sin\left( \frac { 11\times 180 }{ 3 } \right) \) = -sin(660⁰)
= -sin(2 \(\times\) 360⁰ - 60⁰)
= -(-sin(60⁰))   [Angle is in the IV quadrant and sine is negative
= sin 60⁰ = \(\frac { \sqrt { 3 } }{ 2 } \).

Let p = {-1, 1}
\(\Rightarrow\) P = {x \(\in \) R : x is a root of the equation x²-1 = 0}

= cos 20° cos 40° \((\frac{1}{2})\)cos 80°
\(=\frac{1}{4}\)(cos 20° cos 40° cos 80°)
\(=\frac{1}2(\frac{1}{8})\)
\(=\frac{1}{16}\)

Let the roots be α and α²
sum of the roots α + α² = \(\frac { -k }{ 3 } \)   ...(1)
product of the roots α(α²) = (α³) = \(\frac { -81 }{ 3 } \) = -27
α³ = -27=(-3)³ ⇒ α = -3
substituting α value in (1) we get
-3 + (-3)² = \(\frac { -k }{ 3 } \)
(i.e) -3 + 9 = \(\frac { -k }{ 3 } \)
(i.e) \(\frac { -k }{ 3 } \) = 6 ⇒ k = -18

Total number of points = 18
Out of 18 numbers, 5 are collinear and we get a straight line by joining any two points.
\(\therefore\) Total number of straight line formed by joining 2 points out of 18 points = ¹⁸C₂
Number of straight lines formed by joining 2 points out of 5 points = ⁵C₂
But 5 points are collinear and we get only one line when they are joined pairwise.
So, the required number of straight lines are
=¹⁸C₂ -⁵C₂ +1 = \({18 ·17\over2.1}-{5·4\over2.1}+1= 153 -10 + 1-144\)
Hence, the total number of straight lines = 144

Here a₁ = 4, and a_n+1 = 2na_n.
Putting n = 1, a₂ = 2 \(\times\) 1\(\times\) a₁ = 2 \(\times\) 1\(\times\) 4 = 8
Putting n = 2, a₃ =2\(\times\)2\(\times\)a₂ = 4 \(\times\) 8 = 32
Putting n = 3, a₄ = 2\(\times\)3\(\times\)a₃ = 6 \(\times\) 32 = 192
Putting n = 4, a₅ = 2\(\times\)4\(\times\)a₄ = 8 \(\times\) 192 = 1536
Putting n = 5, a₆ = 2\(\times\) 5\(\times\) a₅ = 10\(\times\)1536 = 15360

Let p be the length of the perpendicular drawn from the origin to the required line.
The perpendicular makes 45° with the x-axis.
The equation of the required line is of the form,
x cos \(\alpha\) + y sin \(\alpha\) = p
⇒ x cos 45° +y sin 45° = p
x + y = √2p
This equation cuts the coordinate axes at A(√2p, 0) and B(0, √2p)
Area of the ΔOAB is \(\frac{1}{2}\times\sqrt{2}p\times\sqrt{2}p=36\)
p = 6
Therefore the equation of the required line is x + y = 6\(\sqrt{2}\)

Let f(x) = \(\frac { 1 }{ 1-2sinx } \)
When the denominator is 0,
1-2 sin x = 0
\(\Rightarrow\) 1 = 2 sin x
\(\Rightarrow sin\quad x=\frac { 1 }{ 2 } \)
\(\Rightarrow sin\quad x=sin\frac { \pi }{ 6 } \)
\(\Rightarrow x=n\pi +{ (-1) }^{ n }\frac { \pi }{ 6 } n\in Z\)   \(\left[ \because sin\quad x=sin\alpha \Rightarrow x=n\pi +{ (-1) }^{ n }\alpha \quad n\in Z \right] \)
Domain of f(x) is R - \(\left( n\pi +{ (-1) }^{ n }\frac { \pi }{ 6 } \right) ,n\in Z\)

Let f = {(1, b) (2, b) (3, c) (4, e)}
Different elements in A does not have different images in B
∴ f is not one- one
Now, Co-domain = {a, b, e, d}, Range = {b, e}
Co-domain ≠ range
∴ f is not onto. Hence f is neither one - one and nor onto.

Area of ΔABC = \(\frac{1}{2}\)ab sin C

= \(\frac{1}{2}\) \(\times\) 12 \(\times\) 8 sin 30°
Δ = 6 \(\times\) 8 \(\times\) \(\frac{1}{2}\)= 24 sq.cm

Given \(sin\theta =\frac { 8 }{ 9 } \)
\(cos\theta =\sqrt { 1-{ sin }^{ 2 }\theta } =\sqrt { 1-\frac { 64 }{ 81 } } =\sqrt { \frac { 81-64 }{ 81 } } =\frac { \sqrt { 17 } }{ 9 } \)
\(sin\frac { \theta }{ 2 } =\sqrt { \frac { 1-cos\theta }{ 2 } } =\sqrt { \frac { 1-\frac { \sqrt { 17 } }{ 9 } }{ 2 } } =\sqrt { \frac { 9-\sqrt { 17 } }{ 18 } } \)
\(cos\frac { \theta }{ 2 } =\sqrt { \frac { 1+cos\theta }{ 2 } } =\sqrt { \frac { 1+\frac { \sqrt { 17 } }{ 9 } }{ 2 } } =\sqrt { \frac { 9+\sqrt { 17 } }{ 18 } } \)
consider \(cos\left( \frac { \pi }{ 4 } -\frac { \theta }{ 2 } \right) =cos\frac { \pi }{ 4 } cos\frac { \theta }{ 2 } -sin\frac { \pi }{ 4 } sin\frac { \theta }{ 2 } \)
\(\frac { 1 }{ \sqrt { 2 } } cos\frac { \theta }{ 2 } -\frac { 1 }{ \sqrt { 2 } } sin\frac { \theta }{ 2 } =\frac { 1 }{ \sqrt { 2 } } \left( cos\frac { \theta }{ 2 } -sin\frac { \theta }{ 2 } \right) \)
= \(\frac { 1 }{ \sqrt { 2 } } \left( \sqrt { \frac { 9+\sqrt { 17 } }{ 18 } } -\sqrt { \frac { 9-\sqrt { 17 } }{ 18 } } \right) =\frac { 1 }{ 3\sqrt { 2 } } \left( \sqrt { \frac { 9+\sqrt { 17 } }{ 18 } } -\sqrt { \frac { 9-\sqrt { 17 } }{ 18 } } \right) \)