\(\lim_ { x\rightarrow 0 }\frac { { e }^{ x }-1-x-{ x }^{ 2 } }{ { x }^{ 2 } } \)
\(\Rightarrow \lim_ { x\rightarrow 0 } \left[ \frac { \left( 1+x+\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 3 } }{ 3! } ... \right) -1-x }{ { x }^{ 2 } } -1 \right] \left[ \because { e }^{ x }=1+x+\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 3 } }{ 3! } ... \right] \)
\(\Rightarrow \lim_ { x\rightarrow 0 } \frac { \left( \frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 3 } }{ 3! } +.. \right) }{ { x }^{ 2 } } -1\)
\(\Rightarrow \lim_ { x\rightarrow 0 } \frac { { x }^{ 2 }\left( \frac { 1 }{ 2! } +\frac { { x } }{ 3! } +.... \right) }{ { x }^{ 2 } } -1\)
\(\Rightarrow \lim_{ x\rightarrow 0 } \frac { 1 }{ 2! } +\frac { { x } }{ 3! } -1\Rightarrow \frac { 1 }{ 2 } -1\Rightarrow -\frac { 1 }{ 2 } \)

3

Here \(f(x)=x^{5}(3-6x^{-9})\)
∴ \(f^{'}(x)=\frac{d}{dx}[x^{5}(3-6x^{-9})] \)
=\(x^{5}\frac{d}{dx}(3-6x^{-9})+(3-6x^{-9})\frac{d}{dx}(x^{5})\)
=\(x^{5}(54x^{-10})+(3-6x^{-9})\times 5x^{4}\)
=\(54x^{-5}+15x^{4}-30x^{-5}\)
=\(24x^{-5}+15x^{4}\)
=\(\frac{24}{x^{5}}+15x^{4}\).

Let ABC be an equilateral triangle inscribed in the circle
x²+y²+2gx+2fy+c=0.
Then center of circle isO(−g,−f).
\(\therefore OA=OB=OC=\sqrt { { { g }^{ 2 }{ +f }^{ 2 }-c } } \)
\(In\triangle OBD\)
\(\sin { 60^{ 0 } } =\frac { BD }{ OB } \Rightarrow \frac { \sqrt { 3 } }{ 2 } =\frac { BD }{ \sqrt { { g }^{ 2 }{ f }^{ 2 }-c } } \)
\(\therefore \quad BD=\frac { \sqrt { 3 } }{ 2 } \sqrt { { g }^{ 2 }{ f }^{ 2 }-c } \)

\(Also\quad BC=2BD\)
\( \Rightarrow BC=\sqrt { 3 } \sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
\( Area\quad \triangle ABC=\frac { \sqrt { 3 } }{ 4 } \times \left( BC \right) ^{ 2 }\)
\(=\frac { \sqrt { 3 } }{ 4 } \times 3({ g }^{ 2 }+{ f }^{ 2 }-c)\)
\(=\frac { 3\sqrt { 3 } }{ 4 } { (g }^{ 2 }{ +f }^{ 2 }-c)\) sq.units

(1, 1, -4)

\({\sqrt{3}-1\over 2\sqrt{2}}\)

0.97

Let x litre of 2% boric acid solution be added to 640litres of 8%boric acid solution. Then
Total quantity of mixture = (640 + x) litres
Total boric acid in (640 + x) litres of mixture
= \(\frac { 2x }{ 100 } +\frac { 8 }{ 100 } \times 640=\frac { x }{ 50 } +\frac { 256 }{ 5 } \)
It is given that the resulting mixture must be more than 4% but less than 6% boric acid
\(\therefore\) \(\frac { 4 }{ 100 } (640+x)<\frac { x }{ 50 } +\frac { 256 }{ 5 } <\frac { 6 }{ 100 } (640+x)\)
\(\Rightarrow\)  \(\frac { 640+x }{ 25 } <\frac { x+2560 }{ 50 } <\frac { 1920+3x }{ 50 } \)
\(\Rightarrow\) 1280 + 2x \(\Rightarrow\) 1280 + 2x \(\Rightarrow\) x < 1280 and -2x < -640
\(\Rightarrow\) x < 1280 and x > 320
\(\Rightarrow\) 320 Thus 2% boric acid solution must be more than 320 litres but less than 1280 litres.

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

{2,4,6,8,10}

Exclusive 'OR' because students cannot take both Hindi and Sanskrit as their language

The equation of the lines x - y + 3 = 0 can be rewritten as y = x + 3.
Here m=tan\(\theta \)=1  \(\theta \)=45⁰.

We know that AM between a and b is \( \frac{a+b}{2}\)
 \(\therefore\)  \(\frac{a^n+b^n}{a^{n-1+b^{n-1}}}\) =  \(\frac{a+b}{2}\)
\(\Rightarrow\) 2a\(^n\) +2b\(^n\) = a\(^n\) +ab\(^{n-1}\)+ba\(^{n-1}\)+b\(^n\)
\(\Rightarrow\) 2a\(^n\) -a\(^n\) +2b\(^n\) -b\(^n\) = ab\(^{n-1}\) +ba\(^{n-1}\)
\(\Rightarrow\) a\(^n\)+b\(^n\)= ab\(^{n-1}\)+ba\(^{n-1}\)
\(\Rightarrow\) a\(^n\) - ba\(^{n-1}\) = ab\(^{n-1}\) - bn
\(\Rightarrow\) a\(^{n-1}\)[a-b] = b\(^{n-1}\) [a-b]
\(\Rightarrow\) a\(^{n-1}\) =  b\(^{n-1}\)   [ \(\because\) a \(\neq\) b]
\(\Rightarrow\) \((\frac{a}{n})^{n-1}\) = 1 = \( (\frac{a}{n})^0\) [ \(\because\)x\(^0\) = 1 ]
On comparing the exponential powersd, we get 
n - 1 = 0
\(\Rightarrow\) n = 1

\({ T }_{ 7 }=^{ n }{ { C }_{ 6 } }{ \left( \sqrt [ 3 ]{ 2 } \right) }^{ n-6 }{ \left( \frac { 1 }{ \sqrt [ 3 ]{ 3 } } \right) }^{ 6 }=^{ n }{ { C }_{ 6 } }{ 2 }^{ \frac { n-6 }{ 3 } }{ \left( \frac { 1 }{ 3 } \right) }^{ 2 }\) 
and 7th term from the end = (n + 1) - (7 - 1)th term from beginning =(n - 5)th term from beginning.
\(=^{ n }{ { C }_{ n-6 } }{ \left( \sqrt [ 3 ]{ 2 } \right) }^{ n-n+6 }{ \left( \frac { 1 }{ \sqrt [ 3 ]{ 3 } } \right) }=^{ n }{ { C }_{ n-6 } }{ 2 }^{ 2 }{ \left( \frac { 1 }{ 3 } \right) }^{ \frac { n-6 }{ 3 } }\) 
Now, \(\frac { ^{ n }{ { C }_{ 6 } }{ 2 }^{ \frac { n-6 }{ 3 } }{ \left( \frac { 1 }{ 3 } \right) }^{ 2 } }{ ^{ n }{ { C }_{ n-6 } }{ 2 }^{ 2 }{ \left( \frac { 1 }{ 3 } \right) } } =\frac { 1 }{ 6 } \Rightarrow \frac { N-12 }{ 3 } =-1\Rightarrow N=9\)

Here, we need to perform two operations:
(i)  Selecting a door to enter.
(ii)  Selecting a door to come out.
Clearly, the man can enter the room through anyone of the seven doors. So, there are seven ways of entering into the room. Note that the man can come out through anyone of the remaining six doors. So, he can come out through a different door in 6 ways. Hence, by fundamental principle of counting, required number of ways = 7 \(\times\) 6 = 42  

\(\underset { x\longrightarrow 0 }{ lim } \frac { { tan\quad x }^{ 0 } }{ { x }^{ 0 } } =\underset { x\longrightarrow 0 }{ lim } \cfrac { tan\frac { \pi x }{ 180 } }{ \frac { \pi x }{ 180 } } =1\)

Given, R = \(\left\{ \left( x,y \right) :x\in A,y\in A\quad and\quad x+2y=10 \right\} \)
Here, x+2y=10
\(\Rightarrow \) \(y=\frac { 10-x }{ 2 } \forall x,y\in A\) .....(i)
and A={1,2,3,4,5,6,7,8,9,10}
From Eq. (i),
When x = 2,y = 4;   When x = 4,y = 3
When x = 6,y = 2;   When x = 8,y = 1
\(\therefore \) R = {(2,4),(4,3),(6,2),(8,1)}
\(\Rightarrow \) \({ R }^{ -1 }\)= {(4,2),(3,4),(2,6),(1,8)}
Clearly, domain (R) = {2,4,6,8} = Range(\({ R }^{ -1 }\)) and range (R) = {4,3,2,1} = Domain(\({ R }^{ -1 }\))

Given, A= {x : x is a letter of the word "LOYAL"} = {L, O, Y, A}
and B= {x : x is a letter of the word "ALLOY"} = {A, L, O, Y}
Here, we see that both sets have exactly the same elements.
\(\therefore \)  A = B

sin x + sin 3x + sin 5x = 0
\(\Rightarrow\)(sin 5x + sin x) + sin 3x = 0
\(\Rightarrow\)2sin \(({5x+x\over 2})cos ({5x+x\over 2})+sin 3x=0\)
\(\Rightarrow\)2 sin 3x cos 2x + sin 3x = 0
\(\Rightarrow\)sin 3x (2 cos 2x + 1) = 0
\(\Rightarrow\)Either sin 3x = 0 or 2 cos 2x + 1 = 0
\(\Rightarrow\)3x = n\(\pi\) or cos 2x=\(-{1\over2}=cos {2\pi\over3},n\in Z\)
\(\Rightarrow\) \(x={n\pi\over3} or 2x=2n\pi\pm{2\pi\over3},n\in z.\)
\(\Rightarrow\) \(x={n\pi\over3} or x=n\pi\pm{\pi\over3},n\in z.\)

Number of favourable outcomes = \(^{ 2 }{ C }_{ 1 }\)x \(^{ 2 }{ C }_{ 1 }\)
= \(\frac{2}{3}\)

cos x + sin x.

\(PQ=QR\quad and\quad { PQ }^{ 2 }+{ QR }^{ 2 }={ PR }^{ 2 }\)

Let p(x,y) be any point on the parabola.
=Distance of p(x,y) from the directrix x-2y-6=0  [by definition of parabola]
PF=PM
PF²=PM²

\(\Rightarrow (x-2)^{ 2 }+(y-3)^{ 2 }=\left| \frac { x-2y-6 }{ \sqrt { 1^{ 2 }+2^{ 2 } } } \right| ^{ 2 }\)
\(\Rightarrow (x-2)^{ 2 }+(y-3)^{ 2 }=\frac { (x-2y-6)^{ 2 } }{ 5 } \)
\(\Rightarrow 5[(x-2)^{ 2 }+(y-3)^{ 2 }]=(x-2y-6)^{ 2 }\)
\(\Rightarrow 5(x^{ 2 }-4x+4+y^{ 2 }-6y+9)\)
\(=x^{ 2 }+4y^{ 2 }+36-12x-4xy+24y\)
\(\Rightarrow { 4x }^{ 2 }+{ y }^{ 2 }+4xy-8x-54y+29=0\)
\(which\quad is\quad the\quad required\quad equation\quad of\quad parabola.\)

Here, r>1, n>2
\(\therefore \)  T_3r = ²ⁿC_3r-1 x^3r-1; T_{r+2 =} ²ⁿC_r+1 x^r+1
Then, ²ⁿC_3r-1= ²ⁿC_r+1 
\(\Rightarrow \) 3r - 1 + r + 1 = 2n \(\Rightarrow \) n = 2r

Every number between 100 and 1000 is a 3-digit number. We, first, have to count the permutations of 6 digits taken 3 at a time. This number would be ⁶P₃. But, these permutations will include those also where 0 is at the 100’s place. For example, 092, 042, . . ., etc are such numbers which are actually 2-digit numbers and hence the number of such numbers has to be subtracted from ⁶P₃ to get the required number. To get the number of such numbers, we fix 0 at the 100’s place and rearrange the remaining 5 digits taking 2 at a time. This number is ⁵P₂. So,The required number \(={ }^{6} \mathrm{P}_{3}-{ }^{5} \mathrm{P}_{2}=\frac{6 !}{3 !}-\frac{5 !}{3 !} \)
\(=4 \times 5 \times 6-4 \times 5=100\)

|x| <4=-4
Ans.] -4,4 [

Let z = x + iy, then
\(\frac { z-1 }{ z+1 } =\frac { ({ x }^{ 2 }-1)+{ y }^{ 2 }+i[y(x+1)-y(x-1)] }{ ({ x }^{ 2 }+1)^{ 2 }+y^{ 2 } } \\ \)
\(\because \frac { z-1 }{ z+1 } \)  is purely imaginary.
\(\therefore Re(\frac { z-1 }{ z+1 } ) =0\ i.e.\ \frac { ({ x }^{ 2 }-1)+{ y }^{ 2 } }{ ({ x }^{ 2 }+1)^{ 2 }+y^{ 2 } } =0\)
\({ x }^{ 2 }-1-{ y }^{ 2 }=0\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }=1= |z|=1\)

From arrow diagram,we have
\(P=\{ 9,4,25\} \) and \(Q=\{ 5,4,3,2,1,-2,-3,-5\} \)
Here, the relation R is 'x' is the square of 'y' where \(x\in P\)  and \(y\in Q\)
In roster form, it can be written as
\(R={ (9,3)(9,-3),(4,2),(4,-2)(25,5)(25,-5)} \)
In set-builder from R cam be written as
\(R=\{ x,y):x\in P,y\in Q\) and x is the square of y}

(i) No month has 35 days. Thus the sentence is false declarative sentence. Hence it is a statement. (ii)Here the correctness of the sentence depends upon the observer. It may be easy for someone and may be difficult for other one. Hence it is not a statement.
(iii)The sentence is true. Hence it is a statement.
(iv) Here the correctness of the sentence depends upon the number that is squared. Hence it is not a statement.
(v) This sentence is sometimes true and sometimes false. For example sides in squares and rhombuses have equal length whereas, in a rectangle and trapezium, they have unequal length. Hence it is not a statement.
(vi) This sentence is an order. Hence it is not a statement.
(vii)The sentence is true. Hence it is a statement.
(viii) The sentence is false. Hence it is a statement.
(ix) It is not clear from the context which day is referred. Hence it is not a statement.
(x) The sentence is true because all real numbers can be written in the form a + i x O.Hence it is a statement.

Step I Let P(n) be the given statement.
i.e.P(n):\({ 3 }^{ n }>{ 2 }^{ n }\)
Step II For =1,we have \({ 3 }^{ 1 }>{ 2 }^{ 1 }\)
\(\Rightarrow \) 3 > 2, Which is true.
Thus P(1) is true.
Step III Let us assume that P(k) is true.
i.e  P(k):\({ 3 }^{ k }>{ 2 }^{ k }\)
Step IV Now, we shall prove the statement for n=k+1.For this,we have to show \({ 3 }^{ k+1 }>{ 2 }^{ k+1 }\)
from Eq.(i) we have \({ 3 }^{ k }>{ 2 }^{ k }\)
\({ 3 }^{ k }.3>{ 2 }^{ k }.3\) [multiplying both sides by 3]
\(\Rightarrow \)  \({ 3 }^{ k+1 }>{ 2 }^{ k }.3\)
\({ 2 }^{ k }.3>{ 2 }^{ k }.3\) \(\Rightarrow \)  \({ 3 }^{ k }.3>{ 2 }^{ k }.2\) = \({ 2 }^{ k+1 }\)
Thus ,P(K+1) is true whenever P(k) is true. Hence, by principle of mathematical induction,P(n) is true for all \(n\in N\)

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 .

\({ 1 }^{ 2 }{ +2 }^{ 2 }+....+{ k }^{ 2 }>\frac { { k }^{ 3 } }{ 3 } \)
\(\Rightarrow \) \( ​​​​{ 1 }^{ 2 }{ +2 }^{ 2 }+....+{ k }^{ 2 }+{ (k+1) }^{ 2 }>\frac { { k }^{ 3 } }{ 3 } +{ (k+1) }^{ 2 }\)
\(=\frac { 1 }{ 3 } [{ k }^{ 3 }+{ S(k+1) }^{ 2 }]=\frac { 1 }{ 3 } [{ k }^{ 3 }{ +3k }^{ 2 }+6k+3]\)
\(=\frac { 1 }{ 3 } [({ k }+1)^{ 3 }+3k+2]\)
\({ 1 }^{ 2 }{ +2 }^{ 2 }+....+{ k }^{ 2 }>\frac { 1 }{ 3 } (k+1){k}^{3}\)

Since the numbers between 100 and 1000 are all three digit numbers. So these three digit numbers can be formed by using the digits 0, 1, 2, 3,..., 8, 9.
Here the unit place and tens place can be filled in 10 ways each and hundreds place can be filled in 9 ways.
So total number of three digit numbers between 100 and 1000 = 9 x 10 x 10 = 900
We have to find out those numbers between 100 and 1000 which do not have 7 as a digit.
Here the unit place and tens place can be filled in 9 ways each and hundreds place can be filled in 8 ways.
So total number of three digit numbers between 100 and 1000 which do not have 7 as a digit = 8 x 9 x 9 = 648
Thus total number of three digit numbers between 100 and 1000 which having at least one of their digits as 7 = 900 - 648 = 252.

Let Q divides PR in the ratio k : l. Thus co-ordinates of Q are
\(\left[ \frac { 11k+5 }{ k+1 } ,\frac { 10k+4 }{ k+1 } ,\frac { -8k-2 }{ k+1 } \right] \)
It is given that coordinates of Q are (7, 6, - 4).
\(\therefore \frac { 11k+5 }{ k+1 } =7,\frac { 10k+4 }{ k+1 } =6,\frac { -8k-2 }{ k+1 } =-4\)
Now solving these, we get k=\(\frac { 1 }{ 2 } \) .
Thus Q divides PR in the ratio \(\frac { 1 }{ 2 } \) : 1 or 1 : 2.

Let  \(x+yi=\sqrt {3-4\sqrt 7}i\)
Squaring both sides, we get
x²- y² + 2xyi = \(3-4\sqrt 7i\)
Equating the real and imaginary parts
x² - y² = 3.......(i)
\(2xy=-4\sqrt 7\Rightarrow\ xy=-2\sqrt 7\)
Now from the identity, we know
(x²+y²)²=(x²-y²)²+4x²y²
\(=(3)^2+4(-2\sqrt 7)^2\)
=9+112=121
\(\therefore\) x² + y² = 11........(ii)   [Neglecting (-) sign as x²+y²>0]
Solving (i) and (ii) we get
x²=7 and y²=4
\(\therefore\) \(x=\pm\sqrt 7,y=\pm 2\)
Since the sign of xy is (-)
\(\therefore\) if \(x=\sqrt 7 y=-2\)
and if \(x=-\sqrt 7\ y=2\)
\(\therefore \sqrt {3-4\sqrt7}i=\pm(\sqrt 7-2i).\)

Let (x', y') be the new coordinates of the point (x, y) in new system.
Origin is shifted to a new point (-3, 2) by a translation.
\(\therefore\) h = -3, k = 2
x = x' + h = x' - 3
and y = y' + k = y' + 2
Substituting these vales of x and y in the given equation  \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) , we get
\(\frac { ({ x }^{ ' }-3)^{ 2 } }{ { a }^{ 2 } } +\frac { { (y'+2) }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\Rightarrow \frac { { x' }^{ 2 }+9-6x' }{ { a }^{ 2 } } +\frac { { y' }^{ 2 }+4+{ 4y }^{ ' } }{ { b }^{ 2 } } =1\)
\(\Rightarrow\) b²x'²+9b²-6b²x'+a²y'2+4a²+4a²y'=a²b²
\(\Rightarrow\) b²x'²+a²y'²-6b²x'+4a²y'+4a²+9b²-a²b²=0
Hence the new equation of the ellipse is b²x²+a²y²-6b²x+4a²y+4a²+9b²-a²b² = 0.

Let two remaining observations be x and y. Then
\(\frac { 6+7+10+12+12+13+x+y }{ 8 } =9\)
\(\therefore\) 60 + x + y = 72  \(\Rightarrow\) x + y = 12  ......(i)
Also \(\frac { 1 }{ 8 } ({ 6 }^{ 2 }+{ 7 }^{ 2 }+{ 10 }^{ 2 }+{ 12 }^{ 2 }+{ 12 }^{ 2 }+{ 13 }^{ 2 }+{ x }^{ 2 }+{ y }^{ 2 })-{ (9) }^{ 2 }=9.25\)
\(\Rightarrow \frac { 1 }{ 8 } (36+49+100+144+144+169+{ x }^{ 2 }+{ y }^{ 2 })-81=9.25\)
\(\Rightarrow\)  642 + x² + y² = 722
\(\Rightarrow\) x² + y² = 80 ......(ii)
Now (x+y)² + (x-y)² = 2(x²+y²)
\(\Rightarrow\) (12)² + (x - y)² = 2 x 80
\(\Rightarrow\) (x - y)² = 160 - 144
\(\Rightarrow\) (x - y)² = 16 \(\Rightarrow\) x - y = \(\pm \) 4
When x - y = 4
Solving x + y = 12 and x - y = 4 we get x = 8 and y = 4
When x - y = -4
Solving x + y = 12 and x - y = -4 we get x = 4 and y = 8.

Let the number of sides of polygon be n. The interior angles of the polygon form an A.P.
Here, a = 120° and d=5°
We know that sum of interior angles of a polygon with n sides is (n-2) \(\times\)180°
S_n = (n-2) \(\times\) 180°
\(\Rightarrow \frac{n}{2}[2 a+(n-1) d]=180^{\circ}(n-2) \)
\(\Rightarrow \frac{n}{2}\left[240^{\circ}+(n-1) 5^{\circ}\right]=180(n-2) \)
\(\Rightarrow n[240+(n-1) 5]=360(n-2) \)
\(\Rightarrow 240 n+5 n^{2}-5 n=360 n-720 \)
\(\Rightarrow 5 n^{2}+235 n-360 n+720=0 \)
\(\Rightarrow 5 n^{2}-125 n+720=0 \)
\(\Rightarrow n^{2}-25 n+144=0 \)
\(\Rightarrow n^{2}-16 n-9 n+144=0 \)
\(\Rightarrow n(n-16)-9(n-16)=0 \)
\(\Rightarrow(n-9)(n-16)=0 \)
\(\Rightarrow n=9 \text { or } 16\)

Here A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}
(i) A \(\cap\) B = {3, 5, 7,9, 11} \(\cap\) {7, 9, 11, 13}
= {7, 9, 11}
(ii) B \(\cap\) C = {7, 9, 11, 13} \(\cap\) {11, 13, 15}
= {11, 13}
(iii) A\(\cap\)C\(\cap\)D = {3, 5, 7, 9, 11} \(\cap\) {11, 13, 15} \(\cap\){15,17}=ф
(iv) A\(\cap\) C = {3, 5, 7, 9, 11} \(\cap\) {11, 13, 15}
= {11}
(v) B \(\cap\) D = {7, 9, 11, 13} \(\cap\) {15, 17} = ф
(vi) A\(\cap\)(B \(\cup\)C) ={3, 5, 7, 9, 11} \(\cap\) ({7, 9,11, 13} \(\cup\) {11, 13, 15})
= {3, 5, 7, 9, 11} \(\cap\) {7, 9, 11, 13, 15}
= {7, 9, 11}
(vii) A \(\cap\) D
= {3, 5, 7, 9, 11}\(\cap\) {15, 17} = ф
(viii) A \(\cap\) (B \(\cup\)D)
= {3, 5, 7, 9, 11}\(\cap\) ({7,9,11,13}\(\cup\){15,17}
= {3, 5, 7, 9, 11} \(\cap\) {7, 9,11,13,15,17} = {7, 9, 11}
(ix) (A \(\cap\) B)\(\cap\) (B \(\cup\)C)
= ({3, 5, 7, 9, 11} \(\cap\) {7, 9,11,13}) \(\cap\) ({7, 9, 11, 13} \(\cup\) {11, 13, 15})
= {7, 9, 11} \(\cap\) {7, 9, 11, 13, 15}
= {7, 9, 11}
(x) (A \(\cup\)D) \(\cap\) (B \(\cup\)C)
= ({3, 5, 7, 9, 11}\(\cup\) {15, 17}) \(\cap\) ({7, 9, 11, 13}\(\cup\){11, 13, 15})
= {3, 5, 7, 9, 11, 15, 17}\(\cap\) {7, 9, 11, 13, 15}
= {7, 9, 11, 15}