\(Step\quad I\quad Let\quad P(n)\quad be\quad the\quad given\quad statement. i.e.P(n):{ 7 }^{ n }-{ 2 }^{ n }is\quad divisible\quad by\quad 5.\)
\(Step\quad II\quad For\quad n=1,\quad we\quad have\quad P(1):{ 7 }^{ 1 }-{ 2 }^{ 1 }=5\)
Which is divisible by 5.
Thus,P(1) is true.
\(Step\quad III\quad Let\quad us\quad assume\quad that\quad P(n)\quad is\quad true\quad for\quad n=k. i.e.P(k):{ 7 }^{ k }-{ 2 }^{ k }is\quad divisible\quad by\quad 5.\)
\(Then\quad { 7 }^{ k }-{ 2 }^{ k }\quad =5d\quad for\quad some\quad d\in N.\)
\(\Rightarrow { 7 }^{ k }=5d+{ 2 }^{ k },for\quad some\quad d\in N.\quad ...(i)\)
Step IV Now, we shall prove the statement for n=k+1.
For this, we have to show that N.
\({ 7 }^{ k+1 }-{ 2 }^{ k+1 }is\quad divisible\quad by\quad 5.\)
\(Then,\quad { 7 }^{ k+1 }-{ 2 }^{ k+1 }={ 7 }^{ k }.7-{ 2 }^{ k+1 }\)
\(=7(5d+{ 2 }^{ k })-{ 2.2 }^{ k }\quad \quad [from\quad Eq.(i)]\)
\(=35d+7.2k-{ 2.2 }^{ k }=35d+{ 5.2 }^{ k }\)
\(=5(7d+{ 2 }^{ k }),which\quad s\quad divisible\quad by\quad 5.\)

Thus,P(k+1) is true, when ever P(k) is true.Hence,by principle of mathematical induction,P(n) is true for all n∈N.

Given,    A = {e, f, g} and B = \(\phi\)  
\(\Rightarrow A \cap B = \phi\)  [Since, there is no common element]

Let F,E and S denote the student studying French, English and Sanskrit, respectively.
Here, \(n(F)=17,n(E)=13,n(5),n(F\cap E)=9\)
\(n(E\cap S)=4,n(F\cap S)=5,n(F\cap E\cap S)=3\)
The number of students studying French only,
=\(n(F)-n(F\cap E)-n(F\cap S)+n(F\cap E\cap S)\)
=17-95+3=6

In this case, two classes meet at different hours and 30 students are enrolled in both the courses.
\(n(M\cap P)\) = 30
 Now, \(n(M\cup P)=n(M)+n(P)-n(M\cap P)\)
=60+9030=120

Let M be the set of students in Mathematics class and P be the set of students in Physics class.
Given that, n(M)=60 and n(P)=90
Two classes meet at same hour
\(M\cap P=\phi \Rightarrow n(M\cap P)=0\)
Now, \(n(M\cup P)=n(M)+n(P)-n(M\cap P)\)
=60+90-=150

{x : x \(\in\) R, 0 \(\le\) x < 7} is the set that contain 0 but not 7. So, it can be represented as an interval whose first end is closed and the other end is open.
So, the interval is [0,7).

{x : x \(\in\) R, -12< x < -10} is the set that neither contains -12 nor -10. So, it can be represented as an open interval.
So, the interval is (-12, -10).

Given, A = {1,3,5}   n(A) = 3
Number of elements in P(A) = 2³ = 8
\(\therefore\)   Number of elements in P(P(A)) = 2⁸ = 256

Let A = {a}, B = {{a},b}, C = {{a},b,c}
Clearly, A\(\in\) B and B\(\subset\)  C. But A \(\nsubseteq\) C as a \(\in\) A but a \(\notin\)  C
Thus, the given statement is not true.

n(MUP) = 20, n(M) = 12, \(n(P\cap M)=4\)
\(\because n(M\cup p=n(M)+n(P)n(M\cap p)\)
Ans.60

U

{d,e,f}

All rectangles, Rhombus and square are parallelograms because its opposite sides are equal and parallel.
Therefore \({ F }_{ 2 }\subset { F }_{ 1 },{ F }_{ 3 }\subset { F }_{ 1 }\) and \({ F }_{ 4 }\subset { F }_{ 1 }\)
\({ F }_{ 1 }={ F }_{ 2 }\cup { F }_{ 3 }\subset { F }_{ 4 }\)

\(n({ A }^{ ' }\cap { B }^{ ' })=n(U)-n(A\cup B)\)
Ans.300

Let H be the set of people speaking Hindi and E be the set of people speaking English.
\(\therefore\) n(H) = 250, n(E) = 200 and n(H \(\cup\) E) = 400
We have to find n(H \(\cap\) E).
We know that
n(H \(\cup\) E) = n(H) + n(E) - n(H \(\cap\) E)
\(\therefore\) 400 = 250 + 200 - n(H \(\cap\) E)
\(\therefore\) n(H \(\cap\) E) = 450 - 400 = 50.

x:x∈R,-5 is the set that does not contain -5 but contains 6. So, it can be written as a semi-closed interval whose first end point is open and last end point is close i.e(-5,6)
On the real line, (-5,6) can be graphed aa shown in figure given below

The dark portion on the number line represent (-5,6).
 

We have,
A\(\cap \) =(A\(\cap \)U) = A\(\cap \)(B\(\cup \)B')
=(A\(\cap \)B)\(\cup \)(A\(\cap \)B')\(\cup \)]
=(A\(\cap \)B)\(\cup \)\(\phi \)
\(\Rightarrow \) A =A\(\cap \)B\(\Rightarrow \)A\(\subseteq \)B

To prove, that \(A-(B\cap C)=(A-B)\cup (A-C).\)
Let  \(x\in A-(B\cap C)\)
\(\Rightarrow\quad x\in A\quad and\quad x\notin (B\cap C)\)
\(\Rightarrow \quad x\in A\quad and\quad (x\notin B\quad or\quad x \notin C)\)
\(\Rightarrow \quad (x\in A\quad and\quad x\notin B)\quad or\quad (x\in A\quad and\quad (x\notin C)\)
\(\Rightarrow \quad x\in (A-B)\quad or\quad x\in (A-c)\)
\(\Rightarrow \quad x\in (A-B)\cup (A-C)\)
\(\because \quad A-(B\cap C)\subseteq (A-B)\cup (A-C)\quad \quad \quad .....(i)\)
Again, let y \(\in\) \((A-B)\cup (A-C)\)
\(\Rightarrow \quad y\in (A-B)\quad or\quad y\in (A-c)\)
\(\Rightarrow \quad (y\in A\quad and\quad y\notin B)\quad or(y\in A\quad and\quad y\notin C)\)
\(\Rightarrow \quad y\in A\quad and\quad (y\notin B\quad or\quad y\notin C)\)
\(\Rightarrow\quad y\in A\quad and\quad y\notin (B\cap C)\Rightarrow y\epsilon A-(B\cap C)\Rightarrow y\epsilon A-(B\cap C)\)
\(\Rightarrow \quad (A-B)\cup (A-C)\subseteq A-(B\cap C) \quad \quad ........(ii)\)
From Eqs. (i) and (ii), we get
\(A-(B\cap C)=(A-B)\cup (A-C).\)

\(f(f(x))=f\left( \frac { 1 }{ 2x+1 } \right) =\frac { 1 }{ 2\left( \frac { 1 }{ 2x+1 } \right) +1 } =\frac { 2x+1 }{ 2x+3 } \)
\(\text{Clearly, it is not defined at} x\neq -\frac { 3 }{ 2 } .\)

\(\left\{ x : x \in N\quad and\quad x\le \frac { 9 }{ 2 } \right\} \)

A\(\cup\)B = {1, 2, 3, 4, 5, ........16, 17, 18}
A' = {2, 4, 6, 8, 10,..........18, 19, 20, 21, 22, ,......}
B' = {1, 3, 5, 7, 9, .......17, 19, 20, 21, 22, ,.......}
\(\therefore\) (A\(\cup\)B)\(\cap\)B'= {1, 3, 5, 7, 9, ......17}

Required number of individuals=\(n({ C }_{ 1 }^{ ' }\cap { C }_{ 2 })\)
=\(n({ C }_{ 1 }) -n({ C }_{ 1 }\cap { C }_{ 2 })\) =50-30 =20

Step I Let P(n) be the given statement
i.e.P(n) : n(n + 1) (n + 5) is a multiple of 3.
Step II Forn = 1,we have P(1) : 1(1 + 1)(1 + 5) = 1 × 2 × 6 =12=3×4
Which is a multiple of 3.
So, P(1) is true.
Step III Let P(k) be true form = k.
i.e.P(k) : k(k + 1)(k + 5) is a multiple of 3.
Then,k(k + 1)(k + 5) = 3λ
⇒k(k² + 5k + k5) = 3λ
⇒k³ + 6k² + 5k = 3λ...(i)
Step IV Now,we shall prove the statement form = k + 1.
For this,we have to show that (k + 1)(k + 1 + 1)(k + 1+5)is a multiple of 3.
Then,(k+1)(k+1+1)(k+1+5)=(k+1)(k+2)(k+6)
=(k²+2k+k+2)(k+6)=(k²+3k+2)(k+6)
=k³+6k²+3k²+18k+2k+12
=k³+9k²+20k+12
=(3λ−6k²−5k)+9k²+20k+12 [fromEq.(i),k³=3λ−6k²−5k]
=3λ+3k²+15k+12=3(λ+k²+5k+4)
Which is a multiple of 3 .
Thus,P(k+1) is true, whenever P(k) is true.
Hence,by principle of mathematical induction,the statement is true for all natural numbers n.