L.H.S The given expression can be writen as \(\frac { 1 }{ { 9 }^{ 3 } } +\frac { 1 }{ 9 } +\frac { 1 }{ 27 } +..\infty \)
Let us find the sum \(\frac { 1 }{ 3 } +\frac { 1 }{ 9 } +\frac { 1 }{ 27 } +.....\infty \)
Here \(a=\frac { 1 }{ 3 } \), \(r=\frac { 1 }{ 3 } \)
\(\therefore\)  \({ s }_{ \infty }=\frac { a }{ 1-r } =\frac { \frac { 1 }{ 3 } }{ 1-\frac { 1 }{ 3 } } \Rightarrow \frac { 1 }{ 2 } \)
\(\therefore\)  \(\frac { 1 }{ { 9 }^{ 3 } } +\frac { 1 }{ 9 } +\frac { 1 }{ 27 } +..\infty \) = \(\frac { 1 }{ { 9 }^{ 2 } } \) = 3
Hence Proved

41

(-15, -10, 16)

15, 5, - 5 or 3, 5, 7

\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 34 } \)=1

2x - 3y - 6 = 0.

(1-i)⁴ =[(1-i)²]² = (1+i2-2i)²
= (1-1-2i)² = (-2i)²
= 4i²=4\(\times\)-1 =-4

Here f(x)=\(\frac{2x+3}{3x+2}\)
Then \(f(x+h)=\frac{2(x+h)+3}{3(x+h)+2}=\frac{2x+2h+3}{3x+3h+2}\)
We know that \(f^{'}(x)=\lim_{h\rightarrow0} \frac{f(x+h)-(x)}{h}\)
\(f^{'}(x)=\lim_{h\rightarrow0} \frac{\frac{2x+2h+3}{3x+3h+2}-\frac{2x+3}{3x+2}}{h}\)
=\(\lim_{h\rightarrow0} \frac{(2x+2h+3)(3x+2)-(2x+3)(3x+3h+2)}{h(3x+3h+2)(3x+2)}\)
=\(\lim_{h\rightarrow 0} \frac{(2x+3)(3x+2)+2h(3x+2)-(2x+3)(3x+2)-3h(2x+3)}{h(3x+3h+2)(3x+2)}\)
=\(\lim_{h\rightarrow0}\frac{h[6x+4-6x-9]}{h(3x+3h+2)(3x+2)}=\frac{-5}{(3x+2)^{2}}\)

10

The given inequality is 2x - 3y > 6.
Draw the graph of line 2x - 3y = 6.

table of values satisfying the equation 2x - 3y = 6

Putting (0,0)in the given inequation, we have
2 x 0 - 3 x 0 > 6 \(\Rightarrow\) 0 > 6, which is false
\(\therefore\) Half plane of 2x - 3y > 6 is away from origin

Here  3(1-x)< 2 (x+4)
\(\Rightarrow\) 3 - 3x < 2x + 8 \(\Rightarrow\) -3x - 2x < 8 - 3
\(\Rightarrow\) -5x < 5
Dividing both sides by - 5, we have x > -1
the solution set is (-1,\(\infty\))
The representation of the solution set on the number line is

{1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}

Let S be the sample space. Then, n(s) = 52
Let A, B and C be the event of getting a king, a heart and a red card, respectively.
Then,
\(n(A)=4,n(B)=13,n(C)=26\)
\(\therefore \ P(A)=\frac { n(A) }{ n(S) } =\frac { 4 }{ 52 } =\frac { 1 }{ 13 } \)
\(P(B)=\frac { n(B) }{ n(S) } =\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \)
\(P(C)=\frac { n(C) }{ n(S) } =\frac { 26 }{ 52 } =\frac { 1 }{ 2 } \)
Clearly, \((A\cap B)\) is the event of getting a king among hearts.
\((B\cap C)\) is the event of getting a heart among red cards.
\((A\cap C)\) is the event of getting the king among red cards.
\((A\cap B\cap C)\) is the event of getting a king among heart and the red cards.
\(\Rightarrow n(A\cap B)=1,n(B\cap C)=13,n(A\cap C)=2\ and\ n(A\cap B\cap C)=1\)
\( \therefore \ P(A\cap B)=\frac { n(A\cap B) }{ n(S) } =\frac { 1 }{ 52 } ,\)
\(P(B\cap C)=\frac { n(B\cap C) }{ n(S) } =\frac { 13 }{ 52 } =\frac { 1 }{ 4 }\)
\(P(A\cap C)=\frac { n(A\cap C) }{ n(S) } =\frac { 2 }{ 52 } =\frac { 1 }{ 26 } \)
\(and\ P(A\cap B\cap C)=\frac { n(A\cap B\cap C) }{ n(S) } =\frac { 1 }{ 52 } \)
\(Now,\ P(getting\ a\ king\ or\ heart\ or\ red\ card)\)
\( =P(AorBorC)=P(A\cup B\cup C)\)}
\(=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)\)
\(=\frac { 1 }{ 13 } +\frac { 1 }{ 4 } +\frac { 1 }{ 2 } -\frac { 1 }{ 52 } -\frac { 1 }{ 26 } -\frac { 1 }{ 4 } +\frac { 1 }{ 52 } =\frac { 28 }{ 52 } =\frac { 7 }{ 13 } \)

On rolling a die, we get outcomes as odd number.So,sample space,  S = (1,3,5} \(\Rightarrow \) n(S) = 3
Let E be the event of getting a prime number then,
E={3,5} \(\Rightarrow \) n(E) = 2
Required probability = \(\frac { n(E) }{ n(S) } =\frac { 2 }{ 3 } \)

Here, we observe that average monthly wages in both the plants is same i.e 3000. Therefore, the plant with greater variance will have more variability. Hence, the plant B has greater variability in individual wages.

The component statements are
p :Number 3 is prime
q :Number 3 is odd
Both p and q are true. So \(p\vee q\) is true.

(i) The negation of the given statements is It is false that sum of 2 and 3 is 4
or
The sum of 2 and 3 is not equal to 4
This statements is true because 2+3= 5 \(\neq \) 4
(ii) The negation of the given statements is
It is false that \(\sqrt { 3 } \) is rational
or
\(\sqrt { 3 } \) is not rational
This statements is true because \(\sqrt { 3 } \) is rational
(iii) The negation of the given statements is It is false thet New Delhi is a city
or
New Delhi is not a city
This statements is false.
(iv) The negation of the given staement is 
It is false that every natural number is greater than 0.
This is a false statement.

\(LHL=\lim _{ X\rightarrow { -1 }^{ - } }{ (x+2)= } \lim _{ h\rightarrow 0 }{ (-1-h+2)=1 } \)
\(RHL=\lim _{ x\rightarrow { 1 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { -1 }^{ + } }{ { cx }^{ 2 } } =\lim _{ h\rightarrow 0 }{ c{ (-1+h })^{ 2 } } \)
Ans. c = 1

Yes, ;et A(0, 7, 10), B(-1, 6, 6) and C(-4, 9, 6) be the vertices of a triangle. Then,
Side \(AB=\sqrt { { \left( 0+1 \right) }^{ 2 }+{ \left( 7-6 \right) }^{ 2 }+{ \left( 10-6 \right) }^{ 2 } }\)
\( [\therefore \quad distance\quad =\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-{ y }_{ 2 } \right) }^{ 2 }+{ \left( { z }_{ 1 }-{ z }_{ 2 } \right) }^{ 2 } } ]\)
\(\Rightarrow AB=\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 4 }^{ 2 } } =\sqrt { 1+1+16 } =\sqrt { 18 } =\sqrt [ 3 ]{ 2 } units\)
\(Side\quad BC=\sqrt { { \left( -1+4 \right) }^{ 2 }+{ \left( 6-9 \right) }^{ 2 }+{ \left( 6-6 \right) }^{ 2 } } =\sqrt { { 3 }^{ 2 }+{ 3 }^{ 2 }+0 } \)
\(\Rightarrow BC=\sqrt { 9+9+0 } =\sqrt { 18 } =\sqrt [ 3 ]{ 2 } units\)
\(and\quad side\quad CA=\sqrt { { \left( -4-0 \right) }^{ 2 }+{ \left( 9-7 \right) }^{ 2 }+{ \left( 6-10 \right) }^{ 2 } } \)
\(=\sqrt { { 4 }^{ 2 }+{ 2 }^{ 2 }+{ 4 }^{ 2 } } \)
\( \Rightarrow CA=\sqrt { 16+4+16 } =\sqrt { 36 } =6\quad units\)
\(Now, { AB }^{ 2 }+{ BC }^{ 2 }={ \left( \sqrt [ 3 ]{ 2 } \right) }^{ 2 }+{ \left( \sqrt [ 3 ]{ 2 } \right) }^{ 2 }{ =6 }^{ 2 }={ CA }^{ 2 }\)
Hence, \(\Delta ABC\) is right angled triangle at B.
Yes, this question gives us message to protect our environment and help us to follow these in our daily lives to make ourselvegs healthy.

On comparing the given equation with (x - h)² + (y - k)² = r² , we get
\(h=\frac { 1 }{ 2 } ,k=-\frac { 1 }{ 3 } \)
\(and\quad r=\frac { 1 }{ 2 } \)
\((\frac { 1 }{ 2 } ,-\frac { 1 }{ 3 } )\quad and\quad \frac { 1 }{ 2 } \)

Let the given points are \(A({ x }_{ 1 },{ y }_{ 1 })\equiv A(-1,1)\) and,\(B({ x }_{ 2 },{ y }_{ 2 })\equiv B(2,-4)\) then equation of line AB is
\(y-1=\frac { -4-1 }{ 2+1 } (x+1)\)
\( \left[ \because y-{ y }_{ 1 }=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } (x-{ x }_{ 1 }) \right]\)
\( \Rightarrow y-1=\frac { -5 }{ 3 } (x+1) \Rightarrow 3y-3=-5x-5\)
\(\Rightarrow 5x+3y+2=0\)

Here, a = 2x, b = 3y and n = 5
Given, \(\left( 2x-3y \right) ^{ 4 }\)
\(=^{ 4 }{ C }_{ o }\left( 2x \right) ^{ 4 }-^{ 4 }{ C }_{ 1 }\left( 2x \right) ^{ 3 }\times \left( 3y \right) ^{ 1 }+^{ 4 }{ C }_{ 2 }\left( 2x \right) ^{ 2 }\times \left( 3y \right) ^{ 2 }-^{ 4 }{ C }_{ 3 }\left( 2x \right) ^{ 1 }\times \left( 3y \right) ^{ 3 }+^{ 4 }{ C }_{ 4 }\left( 2x \right) ^{ o }\left( 3y \right) ^{ 4 }\)
Ans. \(16{ x }^{ 4 }-96{ x }^{ 3 }y+216{ x }^{ 2 }{ y }^{ 2 }-216{ xy }^{ 3 }+81{ y }^{ 4 }\)

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

We have,
\(^{13}{C}{_{6}}\)+ \(^{13}{C}{_{5}}\)  = \(^{14}{C}{_{6}} \)   \([ \because ^{n}{C}{_{r}} + ^{n}{C}{_{r-1}} = ^{n+1}{C}{_{r}}]\)
=\(\frac {14!}{6!(14-6)} \)  \(\left[ \because \ \ ^{ n }{ C }{ _{ r } }= \frac { n! }{ r!(n-r)! } \right] \)
=\(\frac{14!}{6 \times 8}\) = \(\frac {14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8!}{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 8!} \) 
=7\(\times\)13\(\times\)11\(\times\)3 = 3003

\({ i }^{ 29 }+\frac { 1 }{ { i }^{ 29 } } =\frac { { i }^{ 29 }.{ i }^{ 29 }+1 }{ { i }^{ 29 } } =\frac { { (i }^{ 2 })^{ 29 }+1 }{ { i }^{ 29 } } \)
\(=\frac { { (-1) }^{ 29 }+1 }{ { i }^{ 29 } } =\frac { -1+1 }{ { i }^{ 29 } } =0=0+0i\)

Consider P(k):2+4+6+8+.....+2k = k(k+1)
Now, P(k+1):2+4+6+.....2(k)+2(k+1)
=k(k+1)+2(k+1)=k² + 3k+2
=(k+1)(k+1)

\(We\quad known\quad that,\quad tan(\alpha +\beta )=\frac { tan\alpha +tan\beta }{ 1-tan\alpha \quad tan\beta }\)
\( \Rightarrow tan(\alpha +\beta )=\frac { \frac { 1 }{ \sqrt { x\left( { x }^{ 2 }+x+1 \right) } } +\frac { \sqrt { x } }{ \sqrt { { x }^{ 2 }+x+1 } } }{ 1-\frac { \sqrt { x } }{ \sqrt { x\left( { x }^{ 2 }+x+1 \right) } \sqrt { { x }^{ 2 }+x+1 } } } \)
\(\Rightarrow tan(\alpha +\beta )=\frac { \frac { 1+x }{ \sqrt { x\left( { x }^{ 2 }+x+1 \right) } } }{ 1-\frac { 1 }{ { x }^{ 2 }+x+1 } }\)
\(\Rightarrow tan(\alpha +\beta )=\frac { 1+x }{ \sqrt { x\left( { x }^{ 2 }+x+1 \right) } } \times \frac { { x }^{ 2 }+x+1 }{ { x }^{ 2 }+x+1-1 } \)
\(\Rightarrow tan(\alpha +\beta )=\frac { \left( 1+x \right) \left( { x }^{ 2 }+x+1 \right) }{ x\left( x+1 \right) \sqrt { x\left( { x }^{ 2 }+x+1 \right) } } \)
\(\Rightarrow tan(\alpha +\beta )=\frac { \sqrt { x\left( { x }^{ 2 }+x+1 \right) } }{ { x }^{ 2 } }\)
\(=\frac { \sqrt { \left( { x }^{ 3 }+{ x }^{ 2 }+x \right) } }{ { x }^{ 4 } } =\sqrt { { x }^{ -1 }+{ x }^{ -2 }+{ x }^{ -3 } }\)
\(\Rightarrow tan(\alpha +\beta )=\quad tan\gamma \therefore \quad \alpha +\beta =\gamma \)

\((30)''=\left( \frac { 1 }{ 2 } \right) ^{ ' }\Rightarrow 37'30''=\left( 37\frac { 1 }{ 2 } \right) ^{ ' }=\left( \frac { 75 }{ 2 } \times \frac { 1 }{ 60 } \right) ^{ o }=\left( \frac { 5 }{ 8 } \right) ^{ o }\)
\({ 5 }^{ 0 }37'30''={ 5 }^{ 0 }+\frac { { 5 }^{ o } }{ 8 } =\frac { 45 }{ 8 } \times \frac { \pi }{ 180 } \)
Ans. \(\frac { \pi }{ 32 } rad\)

We have, \(y=f(x)=\frac { 1-x }{ 1+x } \)
Then, \(f(y)=\frac { 1-y }{ 1+y } =\frac { 1-\left( \frac { 1-x }{ 1+x } \right) }{ 1+\left( \frac { 1-x }{ 1+x } \right) } \)  [from Eq. (i)]
\(\frac { \frac { 1+x-(1-x) }{ 1+x } }{ \frac { 1+x+1-x }{ 1+x } } =\frac { 2x }{ 2 } \)
\(\Rightarrow \) f(y) = x
\(\therefore \)  x = f(y), when \(y=f(x)=\frac { 1-x }{ 1+x } \)

f₁ is a function.

Given sets are A = {1,3,5,6,7} and B={3,7,8,9}.
Now, A-B={1,3,5,6,7} - {3,7,8,9} ={1,5,6}
[set of those elements of A, which are not present in B]
and, B-A={3,7,8,9} - {1,3,5,6,7} ={8,9}
[set of those elements of B, which are not present in A]
\(\because \) Required symmetric difference,
\(A\triangle B=(A-B)\cup (B-A)\quad =\{ 1,5,6\} U\{ 8,9\} =\{ 1,5,6,8,9\} \)

52

Let us make the following table from the given data.

\(\therefore \ (\bar { x } )=\frac { \sum { { f }_{ i }{ x }_{ i } } }{ N } =\frac { 760 }{ 40 } =19\)
\(and\ standard\ deviation=\sqrt { \frac { 1 }{ N } [\sum { { f }_{ i }({ x }_{ i }-\bar { x } )^{ 2 }] } } \)
\(=\sqrt { \frac { 1736 }{ 40 } } =\sqrt { 434 } =6.59\)
Ans.19,6.59

 \(f\prime (x)=\lim_ { h\rightarrow 0 }{ lim } \frac { log(x+k+h)-(logx+k) }{ h } \)
\(=\lim_ { h\longrightarrow 0 }{ lim } \frac { log\left[ 1+\frac { h }{ x+k } \right] }{ \frac { h }{ (x+k) } (x+k) } \)
\(=\frac { 1 }{ x+k }\)

\(\left( \frac { -3 }{ 10 } ,\frac { 5 }{ 2 } ,\frac { -1 }{ 5 } \right) \)

Given equation of ellipse is \(​​\frac { { x }^{ 2 } }{ 49 } +\frac { { y }^{ 2 } }{ 16 } =1.\)
\(\text{On comparing with} \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\text{,we get }a=7,b=4\)

Here, a>b, so major axis is along X-axis.
\(Eccentricity,e=\frac { c }{ a } =\frac { \sqrt { 33 } }{ 7 } \)

\(\frac { x }{ 4 } +\frac { y }{ 4 } =1,\quad a=4\quad and\quad b=\frac { 4 }{ 3 } \)

Let the GP be A,AR,AR²,AR³....
Given , first term , 
A = a ....(i)
and nth term , AR^n-1 = b ....(ii)
Now , p = Product of n terms
p = A\(\times\)AR¹\(\times\)AR²\(\times\)AR³\(\times\).....\(\times\)n terms
p = A^{1+1+1+1+.....+n} terms R^{1+2+3+....(n-1)}
p = AⁿR \(\frac{n(n-1)}{2}\)
p2 = Aⁿ Aⁿ R^n(n-1) = Aⁿ (AR^n-1)n
p2 = aⁿbⁿ  [using Eqs.(i)  and (ii)]

11ⁿ - 10ⁿ = (1+10)ⁿ - 10n
= 1 + 10n + ⁿC₂(10)² + ⁿC₃(10)³ + -10n
= 1 + 100 {ⁿC₂ + ⁿC₃ 10+...+ ⁿC_n10^n-2}
= 100 x an integer + 1
\(\Rightarrow \) 11ⁿ - 10n leaves remainder 1 when divided by 100.

0 do not comes in hundred's place, so only 4 numbers are posiible in hundred's place. Since, the digits are repeated, so remaining two digits, 5 numbers are possible in each place.
\(\therefore \) Total number of ways = 4 x 5 x 5 = 100

Consider the equation of the line x+2y=8. We opbserve that the shaded region and the origin lie on the same side of the line.So, the corresponding inequality is \(x+2y<8.\)
Now, consider the equation of the line x-y=1.
We observe that the shaded region and orgin both lie on the same side. So, corresponding inequality is \(x-y<1.\)
Consider the equation of the line 2x+y=2.We observe that the shaded region and origin lie on opposite side of and is on the right side of Y-axis.So, \(x\ge 0\ and\ y\ge 0.\)
Thus, the linear inequalities corresponding to the given solution ser are
\(x+2y<8,2x+y>2\)
\(x-y>1\)
\(x\ge 0,y\ge 0\)

Since, -3 + ix²y and x² + y + 4i are conjugate of each other \(-3+i{ x }^{ 2 }y=\overline { { x }^{ 2 }+y+4i } x = {-b \pm \sqrt{b^2-4ac} \over 2a}\).
After this, equate real and imaginary parts, to get the values of x and y.  (x = 1, y = -4) or (x = -1, y = -4)

\(RHS=4sinAcosA(cos^{ 2 }A-sin^{ 2 }A)\)
\(= 2sin2A(cos2A)=sin4A\)

Given,  f(x)=x²
\(\therefore \frac{\mathrm{f}(1.1)-\mathrm{f}(1)}{(1.1-1)}=\frac{(1.1)^2-(1)^2}{(1.1-1)}=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=2.1\)

{0,1,2}

Direct Method For any real number x,y, it is given
x = y \(\Rightarrow \) 2x = 2y
\(\Rightarrow \) 2x + a = 2y + a for some a \(\in\) Z.
Contrapositive Method The contrapositive statement of 'p' is 'For any real numbers x,y, if 2x+a\(\neq \)2y+a, where a\(\epsilon \)Z.then x\(\neq \)yⁿ .

2k<(k+2)!\(\Rightarrow \)2k+2(k+2)!+2]
\(\Rightarrow \) (k+1)² < 2^k + 2^k    [(2k+1)<2^k for k\(\le \)3]
\(\Rightarrow \) (k+1)² < 2^k+1

Step I: Let P(n) be the given statement
i.e. P(n): \(1\times 1!+2\times 2!+3\times 3!+....+n\times n!=(n+1)!-1\)
Step II : For n=1, we have
LHS=\(1\times 1!\)=1
and RHS=(1+1)!-1=2!-1=2-1=LHS
\(\because \)  LHS=RHS
\(\therefore \)  P(1) is true
Step III  Let us assume that P(n) is true for n=k
Then, we have
P(k): \(1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)!-1\quad \quad ....(i)\)
Step IV  Now, we shall prove the statement for n=k+1. For this we have to show that
\(1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)\times (k+1)!=(k+1+1)!-1\)
Then, LHS \(=1\times 1!+2\times 2!+3\times 3!+....+k\times k!+(k+1)\times (k+1)!\)
=(k+1)!-1+(k+1)!\(\times \)(k+1)      [from Eq.(1)]
=(k+1+1)(k+1)!-1=(k+2)(k+1)!-1
=(k+2)!-1   [\(\because \)n(n-1)!=n]
Thus, P(k+1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all natural numbers n.

Let P(n) = (2n + 7) < (n + 3)2
For n =1
P(1) =  (2\(\times\)1+7)<(1+3)²
⇒ 9 < 16
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = (2k+7)<(k+3)²    ...(i)
For n = k+1
P(k+1) =2(k+1)+7<(k+1+3)²
⇒ 2(k+1)+7<(k+4)²​        From (i)​
2k+7<(k+3)²​​​​​
Adding 2 on both sides
2k+7+2<(k+3)²+2
⇒  2(k+1)+72+9+6k+2
⇒ 2(k+1)+72+6k+112+8k+16
⇒ 2(k+1)+7<(k+4)²
∴ P(k +1) is true
Thus P(k) is true
⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n∊N.

The three examples of sentences, which are not statements, are as follows.
(i) He is a doctor.
It is not evident from the sentence as to whom ‘he’ is referred to. Therefore, it is not a statement.
(ii) Geometry is difficult.
This is not a statement because for some people, geometry can be easy and for some others, it can be difficult.
(iii) Where is she going?
This is a question, which also contains ‘she’, and it is not evident as to who ‘she’ is. Hence, it is not a statement.

Here n = 7, which is odd.
So the middle terms are \(\left( \frac { 7+1 }{ 2 } \right) th,\left( \frac { 7+1 }{ 2 } +1 \right) th\) are 4^th and 5^th terms.
The general term in the expansion of\(\left( 3-\frac { { x }^{ 3 } }{ 6 } \right) ^{ 7 }\) is
\({ T }_{ r+1 }=^{ 7 }{ C }_{ r }{ (3) }^{ 7-r }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ r }....(i)\)
Putting r = 3 and 4 in (i)
\(\therefore \quad { T }_{ 4 }=^{ 7 }C_{ 3 }{ (3) }^{ 7-3 }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ 3 }\)
\(=^{ 7 }C_{ 3 }{ (3) }^{ 4 }.{ (-1) }^{ 3 }.\frac { { x }^{ 9 } }{ { (6) }^{ 3 } } \)
\(=35\times 81\times -\frac { { x }^{ 9 } }{ 216 } =-\frac { 105 }{ 8 } { x }^{ 9 }\)
Now \({ T }_{ 5 }=^{ 7 }C_{ 4 }{ (3) }^{ 7-4 }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ 3 }\)
\(=^{ 7 }C_{ 4 }{ (3) }^{ 3 }.(-1)^{ 4 }\frac { { x }^{ 12 } }{ { (6) }^{ 4 } } \)
\(=35\times 27\times \frac { { x }^{ 12 } }{ 1296 } =\frac { 35 }{ 48 } { x }^{ 12 }\) 

(2n + 1)\({\pi\over 4},2m\pi;m,n \in Z\)

Let A(2, 0, 3), B(O, 3, 2) and C(O, 0,1) be given points.
Let P(x, y, 0) be any point in XY plane such that PA = PB = PC.
Now PA = PB => PA2 = PB2
\(\therefore \) (x - 2)² + (y - 0)² + (- 3)²
= (x - 0)² + (y - 3)² + (- 2)²
=> x²+4-4x+y²+9
=x²+y²+9-6y+4
=> 4x - 6y = 0 => 2x - 3y = 0 ....(i)
Also PB = PC => PB² = PC²
(x - 0)² + (y - 3)² + (0 - 2)² = (x - 0)² + (y - 3)² + (0 - 2)²
\(\therefore \)(x - 0)² + (y - 3)² + (0 - 2)² = (x - 0)² + (y - 3)² + (0 - 1)²
=> x²+ y² + 9 - 6y + 4
=x²+y²+1
=> 6y = 12 => y = 2
Putting value of y in (i), we have
2x-3x2=0 => x=3
Thus co-ordinates ofrequired point are (3, 2, 0).

Here foci are (0, ±\(\sqrt10\)) which lie on y-axis.
So the equation of hyperbola in standard form is \(\frac { { y }^{ 2 } }{ { a }^{ 2 } } -\frac { { x }^{ 2 } }{ { b }^{ 2 } } \)=1
∴ foci (0, ± c) is (0, ± \(\sqrt10\))  a = \(\sqrt10\)
We know that c² = a² + b²
∴ (\(\sqrt10\))² = a² + b² ⇒ b² = 10 - a²
Since the hyperbola passes through (2, 3)
∴ \(\frac { 9 }{ a^{ 2 } } -\frac { 4 }{ { b }^{ 2 } } \)=1
⇒ \(\frac { 9 }{ a^{ 2 } } -\frac { 4 }{ 10-{ a }^{ 2 } } \)=1
⇒ \(\frac { 9(10-{ a }^{ 2 })-4{ a }^{ 2 }={ a }^{ 2 }(10-{ a }^{ 2 }) }{ { a }^{ 2 }(10-{ a }^{ 2 }) } \)
⇒ a⁴ - 23a² + 90 = 0
⇒ a⁴ - 18a² - 5a² + 90 = 0
⇒ (a² - 18) (a² - 5) = 0
⇒ a² = 18 or a² = 5
When a²= 18 then b²= 10-18=-8 (which is not possible)
When a² = 5 then b² = 10 - 5 = 5
Thus required equation of hyperbola is
\(\frac { { y }^{ 2 } }{ 5 } -\frac { { x }^{ 2 } }{ 5 } \)=1

Here a+ib =\(\frac { ({ x }^{ 2 }+i)^{ 2 } }{ { 2x }^{ 2 }+1 } =\frac { { x }^{ 2 }+i+2ix }{ { 2x }^{ 2 }+1 } \)
=  \(\frac { { x }^{ 2 }-1 }{ 2{ x }^{ 2 }+1 } +i\frac { 2x }{ { 2x }^{ 2 }+1 } \)
Comparing both sides, we have
a= \(\frac { { x }^{ 2 }-1 }{ 2{ x }^{ 2 }+1 } \)and b= \(\frac { 2x }{ { 2x }^{ 2 }+1 } \)
∴ a²+b²= \(\left( \frac { { x }^{ 2 }-1 }{ 2{ x }^{ 2 }+1 } \right) ^{ 2 }+\left( \frac { 2x }{ { 2x }^{ 2 }+1 } \right) ^{ 2 }\)
= \(\frac { { (x }^{ 2 }-1)^{ 2 } }{ (2{ x }^{ 2 }+1)^{ 2 } } +\frac { (2x)^{ 2 } }{ ({ 2x }^{ 2 }+1)^{ 2 } } \)
= \(\frac { { (x }^{ 2 }-1)^{ 2 }+(2x)^{ 2 } }{ (2{ x }^{ 2 }+1)^{ 2 } } \)
= \(\frac { { x }^{ 4 }+1-{ 2x }^{ 2 }+{ 4x }^{ 2 } }{ ({ 2x }^{ 2 }+1)^{ 2 } } \)
= \(\frac { { x }^{ 4 }+1+{ 2x }^{ 2 } }{ ({ 2x }^{ 2 }+1)^{ 2 } } =\frac { ({ x }^{ 2 }+1)^{ 2 } }{ ({ 2x }^{ 2 }+1)^{ 2 } } \)

\(\therefore \frac { 5! }{ (5-r)! } =\frac { 6! }{ (7-r)! }\)
\( \Rightarrow \frac { 5! }{ (5-r)! } =\frac { 6\times 5! }{ (7-r)(6-r)(5-r)! }\)
\( \Rightarrow 1=\frac { 6 }{ (7-r)(6-r) } \)
\(\Rightarrow { r }^{ 2 }-13r+42=6\)
\(\Rightarrow { r }^{ 2 }-13r+36=0\)
\( \Rightarrow { r }^{ 2 }-{ 9r }-4r+36=0\)
\(\Rightarrow r(r-9)-4(r-9)=0\)
\(\Rightarrow (r-9)(r-4)=0\)
\( \Rightarrow r=9\quad or\quad r=4\)
\(Now\quad r=9\quad is\quad not\quad possible\quad because\quad r>n\)
\(Thus\quad r=4\)

\(\text { | et } A B C \text { be the riaht analed trianale where } \angle \mathrm{C}=90^{0}\)
\(\text { There are infinitely many such lines. }\)
\(\text { Let } m \text { be the slope of } A C \text { . }\)
\(\text { Slope of } B C=-\frac{1}{m}\)
\(\text { Equation of } A C: \quad y-3=m(x-1)\)
\(\Rightarrow x-1=\frac{1}{m}(y-3)\)
\(\text { Equation of BC: } \quad y-1=-\frac{1}{m}(x+4)\)
\(\Rightarrow x+4=-m(y-1)\)
\(\text { For a given value of } m \text { , we can get theseequations }\)
\(\text { For } m=0, \quad y-3=0 ; x+4=0\)
\(\text { For } m \rightarrow \infty, \quad x-1=0 : y-1=0\)

The given inequality is 4x +3y\(\le\)60
Draw the graph of the line 4x +3y = 60

table of values satisfying the equation 4x +3y=60

 Putting (0, 0) in the given inequation, we have 4 x 0 + 3 x 0 \(\le\)60 \(\Rightarrow\)0 \(\le\)60, which is true.
. . Half plane of 4x +3y \(\le\) 60 is towards origin.
Also the given inequality is 2x-y\(\le\)0
Draw the graph of the line 2x - y = O.
Table of values satisfying the equation 2x-y=0

Putting (10,0) in the given inequation, we have 2 x 10-0\(\le\) 0 \(\Rightarrow\) 20\(\le\)0, which is false.
\(\therefore\)  Half plane of 2x -y \(\le\)0 does not contain (10,0).
The given inequality is x \(\ge\)3.
Draw the graph of the line x = 3.
Putting (0,0) in the given inequation, we have o \(\ge\)3, which is false.
\(\therefore\) Half plane of x \(\ge\)3 is away from origin

\(\text { The given function is }\)
\(f(x)=\left\{\begin{array}{ll} a+b x, & x<1 \\ 4, & x=1 \\ b-a x & x>1 \end{array}\right.\)
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}(a+b x)=a+b \)
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(b-a x)=b-a \)
\(f(1)=4\)
\(\text { It is given that } \lim _{x \rightarrow 1} f(x)=f(1) \text { . }\)
\(\therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x)=f(1) \)
\(\Rightarrow a+b=4 \text { and } b-a=4\)
\(\text { On solving these two equations, we obtain } a=0 \text { and } b=4 \text { . }\)
\(\text { Thus, the respective possible values of } a \text { and } b \text { are } 0 \text { and } 4\)

Here A = {1, 2, 3, 4} and B = {5, 6, 7}, a \(\in\) A, b \(\in\) B.
\(\therefore\) a - b = 1 - 5, 1 - 6, 1 - 7, 2 - 5, 2 - 6, 2 - 7, 3 - 5, 3 - 6, 3 - 7, 4 - 5, 4-6,4-7
= -4, -5, -6, -3, -4, -5, -2, -3, -4, -1, -2,-3
R = {(1, 5), (1, 7), (2, 6), (3,5), (3, 7), (4,6)}.

Taking A, B, C events from question 6 above we have
\(i)\ A\cap B=\phi \)
Thus A and B are mutually exclusive and exhaustive events.
∴ True.
\( ii)\ A\cap B\ =\phi \ and\ A\cap B=S\)
Thus A and B are mutually exclusive and exhaustive events.
∴ True.
(iii) B′={(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}=A.
∴ True.
(iv) A∩C = {(2,1)(2,2)(2,3)(4,1)}=ϕ
Thus A and C are not mutually exclusive events.
∴ False
(v) A∩B′ = A ≠ ϕ
Thus A and B are not mutually exclusive events.
(vi) Since A′ = B and B′ = A, A∩B = ϕ
B∩C = {(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)} ≠ ϕ
A∩C = {(2,1)(2,2)(2,3)(4,1)} = ϕ
Thus A′, B′ and C are not mutually exclusive.
∴ False.

Let 'a' be the first term and 'd' be the common difference of given A.P.
\(\because \quad { S }_{ m }=\frac { m }{ 2 } [2a+(m-1)d]\)
and \({ S }_{ n }=\frac { n }{ 2 } [2a+(n-1)d]\)
\(\therefore \quad \frac { { S }_{ m } }{ { S }_{ n } } =\frac { \frac { m }{ 2 } [2a+(m-1)d] }{ \frac { n }{ 2 } [2a+(n-1)d] } \)
But \(\frac { { S }_{ m } }{ { S }_{ n } } =\frac { { m }_{ 2 } }{ { n }_{ 2 } } \)   [Given]
\(\Rightarrow \frac { \frac { m }{ 2 } [2a+(m-1)d] }{ \frac { n }{ 2 } [2a+(n-1)d] } =\frac { { m }^{ 2 } }{ { n }^{ 2 } } \)
\(\Rightarrow \frac { 2a+(n-1)d }{ 2a+(n-1)d] } =\frac { { m }^{ 2 } }{ { n }^{ 2 } } \times \frac { n }{ 2 } \times \frac { 2 }{ m } =\frac { m }{ n } \)
\(\Rightarrow\) 2an + n(m-1)d = 2am + m(n-1)d
\(\Rightarrow\) 2an-2am= (mn-m)d-(mn-n)d
\(\Rightarrow\) 2a[n-m]=[mn-m-mn+n]d
\(\Rightarrow\) 2a[n-m]=[n-m]d
\(\Rightarrow d=\frac { 2a[n-m] }{ [n-m] } =2a\)
Now, \(\frac { { a }_{ m } }{ { a }_{ n } } =\frac { a+(m-1)d }{ a+(n-1)d } =\frac { a+(m-1)\times 2a }{ a+(n-1)\times 2a } \)
\(=\frac { a[1+2m-2] }{ a[1+2n-2] } =\frac { 2m-1 }{ 2n-1 } \)
Thus, the ratio of 'm^th' and 'n^th' term is (2m+1): (2m-1).

Arrange the data in ascending order, we have
10,11,12,13,13,14,16,16,17,17,18
Here n = 12 (which is even)
So median is average of 6th and 7th observations
∴ Median =\(\frac { 13+14 }{ 2 } =\frac { 27 }{ 2 } =13.5\)

M.D. about median =\(\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ \left| { x }_{ i }-M \right| } \)
\(=\frac{1}{12}\times28=2.33\)