\(\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{matrix} \right| =\left| \begin{matrix} 1 & 1 & 1 \\ 0 & x & 0 \\ 0 & 0 & y \end{matrix} \right| \begin{matrix} { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 } \\ { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 } \end{matrix}\)
= xy [∵ upper diagonal matrix]

4x - 3y + 1= 0; l ∈ R

(x- 2) (x + 3)² < 0 Critical numbers 2, - 3
We have three intervals (-\(\infty\), -3), (-3, 2), (2, \(\infty\))
 

The inequality is satisfied in the interval (∞, -3) and (-3, 2)
∴ Solution set is(∞, -3) ∪ (-3, 2)

Domain = R-{5}
Range = {-1}

Given f(0) =2
\(\lim _{ x\rightarrow 0 }{ f\left( x \right) } =\lim _{ x\rightarrow 0 }{ \left( \frac { \sin { x } }{ x } +\cos { x } \right) } =\lim _{ x\rightarrow 0 }{ \frac { \sin { x } }{ x } +\lim _{ x\rightarrow 0 }{ \cos { x } } } =1+1=2\)
\(\therefore \lim _{ x\rightarrow 0 }{ f\left( x \right) } =f\left( 0 \right) =2\)
\(\therefore f\left( x \right) is\quad continuous\quad at\quad x=2.\)

\(Given,\quad \lim _{ x\rightarrow 2 }{ \frac { { x }^{ n }-{ 2 }^{ n } }{ x-2 } } =80\)
\( \Rightarrow n.{ 2 }^{ n-1 }=80\)
By trial method, put n 5
\(\Rightarrow 5({ 2 }^{ 5-1 })=80\)
\(\Rightarrow 5({ 2 }^{ 4 })=80\)
\( \Rightarrow 5(16)=80\)
\( \Rightarrow 80=80\)
\(\therefore n=5\)

\(\lim _{ x\rightarrow 0 }{ \frac { { x }^{ \frac { 2 }{ 3 } }-9 }{ x-27 } } =\lim _{ x\rightarrow 0 }{ \frac { { x }^{ \frac { 2 }{ 3 } }-{ (27) }^{ \frac { 2 }{ 3 } } }{ x-27 } } =\frac { 2 }{ 3 } { (27) }^{ \frac { 2 }{ 3 } -1 }=\frac { 2 }{ 3 } { (27) }^{ \frac { -1 }{ 3 } }\)
\(=\frac { 2 }{ 3(27)^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ 3({ 3 }^{ 3 })^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ 3(3) } =\frac { 2 }{ 9 } \)

Let \(\triangle =\left| \begin{matrix} 1 & 1+p & 1+p+q \\ 2 & 3+2p & 4+4p+2q \\ 3 & 6+3p & 10+6p+3q \end{matrix} \right| \)
Applying C₂ ⟶ C₂ - pC₁ and C₃ ⟶ C₃ - qC₁, we get
\(\triangle =\left| \begin{matrix} 1 & 1 & 1+p \\ 2 & 3 & 4+3p \\ 3 & 6 & 10+6p \end{matrix} \right| \)
Applying C₃ ⟶ C₃ - pC₂, we get
= \(\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 3 & 6 & 10 \end{matrix} \right| \)
Applying C₂ ⟶ C₂ - Cland C₃ ⟶ C₃ - C, we get,
\(\triangle =\left| \begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 3 & 3 & 7 \end{matrix} \right| \)
Expanding along |Rl we get,
\(\triangle =1\left| \begin{matrix} 1 & 2 \\ 3 & 7 \end{matrix} \right| \)+ 0 + = 1(7-6) = 1
\(\therefore \triangle =1\)

Given \(\left[ \begin{matrix} { x }^{ 2 } \\ { y }^{ 2 } \end{matrix} \right] -3\left[ \begin{matrix} x \\ 2y \end{matrix} \right] =\left[ \begin{matrix} -2 \\ 9 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} { x }^{ 2 } & 3x \\ { y }^{ 2 } & -6y \end{matrix} \right] =\left[ \begin{matrix} -2 \\ 9 \end{matrix} \right] \)
\(\Rightarrow\)x²-3x =-2 and y²-6y=9
\(\Rightarrow\)x²-3x+2 =0
\(\Rightarrow\)(x-1)(x-2) = 0
\(\Rightarrow\)x = 1,2
Also, y²-6y-9 = 0
y =\(\frac { 6\pm \sqrt { 36+36 } }{ 2 } =\frac { 6+6\sqrt { 2 } }{ 2 } =\frac { 3(3\pm 3\sqrt { 2 } ) }{ 2 } \)
y=3 \(\pm \) 3\(\sqrt { 2 } \)
x =1, 2 and y= 3\(\pm \)3\(\sqrt { 2 } \)

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 that \(f(x)=\frac { x-1 }{ x+1 } \)
\(f\left( \frac { 1 }{ x } \right) =\frac { \frac { 1 }{ x } -1 }{ \frac { 1 }{ x } +1 } =\frac { 1-x }{ 1+x } =\frac { -(x+1) }{ x+1 } =-f(x)\)
Hence, \(f\left( \frac { 1 }{ x } \right) =-f(x)\)

This relation is reflexive, symmetric and transitive. Thus it is an equivalence relation.

Then clearly this relation is reflexive as every number in S divides itself.
(i.e.,) 1R1 as 1 divides 1; 2R2 as 2 divides 2 and so on.

9

x>\(\frac{5}{6}\)

x = 1 or \(\frac{3}{2}\)

Given T_r = \(\frac{1}{3}r+\frac{1}{6}\)
T₁ = \(\frac{1}{3}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\)
T₂ = \(\frac{1}{3}(2)+\frac{1}{6}=\frac{5}{6}\)
T₃ = \(\frac{1}{3}(3)+\frac{1}{6}=\frac{7}{6}\)
The series is \(\frac{1}{2}+\frac{5}{6}+\frac{7}{6}+..\)
Here T₂ - T₁ = \(\frac{5}{6}-\frac{1}{2}=\frac{5-3}{6}=\frac{2}{6}=\frac{1}{3}\)
and T₃ - T₂ = \(\frac{7}{6}-\frac{5}{6}=\frac{2}{6}=\frac{1}{3}\)
The series is an arithmetic series with common difference \(\frac{1}{3}\).

In (1+2x)⁸ , we have n = 8, x = 1, a = 2x.
T_r+1 = nC_r x^n-r a^r
⇒ T_r+1 = 8Cr(1)^8-r.(2x)^r = 8Cr 2^r.x^r  --- (1)
and T_r-1 = 8C_r-1.(2x)^r-1 = 8C_r-1.2^r-1x^r-1 ----- (2)
Dividing (2) ÷ (1) we get,
\(\frac{T_{r+1}}{T_r}=\frac{8C_r.2^{r}.x^{r}}{8C_{r-1}.2^{r-1}.x_{r-1}}=\frac{8!}{r!(8-r)!}.\frac{(r-1)!(8-r+1)}{8!}2x\)
= \(\frac{8-r+1}{r}.2r=\frac{9-r}{r}.2(2)\)   [since x=2]
= \(\frac{36-4r}{r}\)
Now T_r+1 ≥T _r if \(\frac{T_{r+1}}{T_{r}}\ge1\)
⇒ \(\frac{36-4r}{r}\ge1\)
⇒ 36 - 4r≥1
⇒ 5r ≤ 36
⇒ r ≤ \(\frac{36}{5}\)
⇒ r ≤ 7.2
∴ the greatest possible value of r is 7.
The greatest possible value of r is 7.

Given p(n) :"7 divides 2³ⁿ- 1"
\(\Rightarrow\) 2³ⁿ- 1 = 7k    [where k is a constant]
\(\Rightarrow\) 2³ⁿ = 7k+1 ...(1)
Now, p(n+1) is 7 divides 2³⁽ⁿ⁺¹⁾-1
2³⁽ⁿ⁺¹⁾-1 = 7k₁
2³ⁿ.2³-1 = 7k₁
8(2³ⁿ) = 7k₁+1 which is the required statement.

LHS = \(\frac { 1+sinx-cosx }{ 1+sinx+cosx } =\frac { (1-cosx)+sinx }{ (1+cosx)+sinx } \)
= \(\frac { 2sin^{ 2 }\left( \frac { x }{ 2 } \right) +2sin\frac { x }{ 2 } .cos\frac { x }{ 2 } }{ 2cos^{ 2 }\left( \frac { x }{ 2 } \right) +2sin\frac { x }{ 2 } .cos\frac { x }{ 2 } } =\frac { 2sin\frac { x }{ 2 } \left( sin\frac { x }{ 2 } +cos\frac { x }{ 2 } \right) }{ 2cos\frac { x }{ 2 } \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) } \)    
\(\left[ \because 1-cosx=2sin^{ 2 }\frac { x }{ 2 } \\ 1+cosx=2cos^{ 2 }\frac { x }{ 2 } \\ sinx=2sin\frac { x }{ 2 } cos\frac { x }{ 2 } \right] \)
= \(\frac { sin\frac { x }{ 2 } }{ cos\frac { x }{ 2 } } =tan\frac { x }{ 2 } \) = RHS

LHS = \(cos\left( \frac { \pi }{ 13 } \right) cos\left( \frac { 9\pi }{ 13 } \right) +cos\frac { 3\pi }{ 13 } +cos\frac { 5\pi }{ 13 } \)
\(cos\left( \frac { 9\pi }{ 13 } +\frac { \pi }{ 13 } \right) +cos\left( \frac { 9\pi }{ 13 } -\frac { \pi }{ 13 } \right) +cos\left( \frac { 3\pi }{ 13 } \right) +cos\left( \frac { 5\pi }{ 13 } \right) \)    [2cos A cos B = cos(A + B) + cos(A - B)
= \(cos\left( \frac { 10\pi }{ 13 } \right) +cos\left( \frac { 8\pi }{ 13 } \right) +cos\left( \frac { 3\pi }{ 13 } \right) +cos\left( \frac { 5\pi }{ 13 } \right) \)
= \(cos\left( \pi -\frac { 3\pi }{ 13 } \right) +cos\left( \pi -\frac { 5\pi }{ 13 } \right) +cos\left( \frac { 3\pi }{ 13 } \right) +cos\left( \frac { 3\pi }{ 13 } \right) =-cos\left( \frac { 3\pi }{ 13 } \right) -cos\left( \frac { 5\pi }{ 13 } \right) +cos\left( \frac { 3\pi }{ 13 } \right) +cos\left( \frac { 5\pi }{ 13 } \right) \)
= 0 = RHS 
\(\left[ \because cos(\pi -A)=-cosA \right] \)

f(x₁) = f(x₂)
⇒ 2x₁= 2x₂
⇒ x₁ = x₂
ஃ f is one-one.
For 1∈N, there does not exist any x in N such that f(x) = 2x = 1
ஃ f is not onto.

Since A = {1, 2, 3}
(1,1) (2, 2) (3, 3) ∈ R ⇒ is reflexive.
Also (1, 2) ∈ R but (2, 1) ∉ R ⇒ R is not symmetric.
And (1, 2) ∈ R, (2, 3) ∈ R but (1, 3) ∉ R ⇒ R is not transitive.
R is reflexive but neither symmetric nor transitive.

Given A⊂B

(i) From the diagram A ∩ B =A
(ii) A\B = Ø

Given P = {1, 2, 3, 5, 7, 11, 13}
and Q = {2, 4, 6, 8, 10, 12, 14}
Now P⋂Q = {2}
Since P⋂Q ≠ Ø, the given sets are not disjoint.