JEE MAIN MATHS – LECTURE 01: Identities & Number System
Topic 1: Number System & Prime Numbers
- Number System: It comprises various types of numbers including natural numbers ($\mathbb{N}$), integers ($\mathbb{Z}$), and rational numbers ($\mathbb{Q}$).
- Rational Numbers ($\mathbb{Q}$): Any number that can be expressed as the fraction $p/q$, where $p$ and $q$ are integers and $q \neq 0$.
- Prime Numbers: A natural number greater than 1 that has exactly two distinct positive divisors: 1 and the number itself. (e.g., 2, 3, 5, 7…). Note: 2 is the only even prime number.
- સંખ્યા પદ્ધતિ (Number System): તેમાં પ્રાકૃતિક સંખ્યાઓ ($\mathbb{N}$), પૂર્ણાંકો ($\mathbb{Z}$) અને સંમેય સંખ્યાઓ ($\mathbb{Q}$) નો સમાવેશ થાય છે.
- સંમેય સંખ્યાઓ (Rational Numbers): એવી સંખ્યા જેને $p/q$ સ્વરૂપે દર્શાવી શકાય, જ્યાં $p$ અને $q$ પૂર્ણાંકો છે અને $q \neq 0$.
- અવિભાજ્ય સંખ્યાઓ (Prime Numbers): 1 કરતાં મોટી એવી પ્રાકૃતિક સંખ્યા જેના માત્ર બે જ ધન અવયવો હોય: 1 અને સંખ્યા પોતે. (દા.ત., 2, 3, 5, 7…). નોંધ: 2 એ એકમાત્ર બેકી (યુગ્મ) અવિભાજ્ય સંખ્યા છે.
Extra Illustration
Identify the sum of the first 3 prime numbers.
પ્રથમ 3 અવિભાજ્ય સંખ્યાઓનો સરવાળો શોધો.
Solution:
The first 3 prime numbers in the natural number system are 2, 3, and 5.
Therefore, the sum is: $2 + 3 + 5 = 10$.
The first 3 prime numbers in the natural number system are 2, 3, and 5.
Therefore, the sum is: $2 + 3 + 5 = 10$.
Linked Homework Questions:
HW Q1, HW Q2, HW Q9
HW Q1, HW Q2, HW Q9
Topic 2: Algebraic Identities & Manipulations
- Algebraic Identities: Standard equations that remain true for all values of their variables.
- Key JEE Expansions:
$\bullet$ $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$
$\bullet$ $a^3+b^3 = (a+b)(a^2-ab+b^2)$ OR $(a+b)^3 – 3ab(a+b)$
$\bullet$ $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ - Sum of Squares Property: If $x^2 + y^2 + z^2 = 0$ for real numbers $x, y, z$, then strictly $x=0$, $y=0$, and $z=0$.
- બૈજિક નિત્યસમો: પ્રમાણિત સમીકરણો જે તેમના ચલોની તમામ કિંમતો માટે સમાન અને સાચા રહે છે.
- અગત્યના વિસ્તરણો:
$\bullet$ $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$
$\bullet$ $a^3+b^3 = (a+b)(a^2-ab+b^2)$ અથવા $(a+b)^3 – 3ab(a+b)$
$\bullet$ $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ - વર્ગોના સરવાળાનો ગુણધર્મ: વાસ્તવિક સંખ્યાઓ માટે, જો વર્ગોનો સરવાળો શૂન્ય થાય ($x^2 + y^2 + z^2 = 0$), તો દરેક પદ ફરજિયાતપણે શૂન્ય જ હોય ($x=0$, $y=0$, $z=0$).
Extra Illustration
If $x – \frac{1}{x} = 4$, find the value of $x^2 + \frac{1}{x^2}$.
જો $x – \frac{1}{x} = 4$ હોય, તો $x^2 + \frac{1}{x^2}$ ની કિંમત શોધો.
Solution:
Given: $x – \frac{1}{x} = 4$
Squaring both sides:
$\left(x – \frac{1}{x}\right)^2 = 4^2$
$x^2 + \frac{1}{x^2} – 2(x)\left(\frac{1}{x}\right) = 16$
$x^2 + \frac{1}{x^2} – 2 = 16$
$x^2 + \frac{1}{x^2} = 18$
Given: $x – \frac{1}{x} = 4$
Squaring both sides:
$\left(x – \frac{1}{x}\right)^2 = 4^2$
$x^2 + \frac{1}{x^2} – 2(x)\left(\frac{1}{x}\right) = 16$
$x^2 + \frac{1}{x^2} – 2 = 16$
$x^2 + \frac{1}{x^2} = 18$
Classwork Q1
If $x+\frac{1}{x}=2$, then $x^{2}+\frac{1}{x^{2}}$ is equal to
જો $x+\frac{1}{x}=2$, તો $x^{2}+\frac{1}{x^{2}}=\_$
- (A) 0
- (B) 1
- (C) 2
- (D) 3
Solution:
Given $x + \frac{1}{x} = 2$.
Squaring both sides:
$\left(x + \frac{1}{x}\right)^2 = (2)^2$
$x^2 + \frac{1}{x^2} + 2(x)\left(\frac{1}{x}\right) = 4$
$x^2 + \frac{1}{x^2} + 2 = 4$
$x^2 + \frac{1}{x^2} = 2$
Alternatively, observe that $x=1$ perfectly satisfies the initial equation. Substituting $x=1$ yields $1^2 + 1/1^2 = 2$.
Answer: (C)
Given $x + \frac{1}{x} = 2$.
Squaring both sides:
$\left(x + \frac{1}{x}\right)^2 = (2)^2$
$x^2 + \frac{1}{x^2} + 2(x)\left(\frac{1}{x}\right) = 4$
$x^2 + \frac{1}{x^2} + 2 = 4$
$x^2 + \frac{1}{x^2} = 2$
Alternatively, observe that $x=1$ perfectly satisfies the initial equation. Substituting $x=1$ yields $1^2 + 1/1^2 = 2$.
Answer: (C)
Classwork Q2
If $\frac{(2+1)(2^{2}+1)(2^{4}+1)(2^{8}+1)}{(2^{8}-1)}=4^{n}+1$, then $n$ is
જો $\frac{(2+1)(2^{2}+1)(2^{4}+1)(2^{8}+1)}{(2^{8}-1)}=4^{n}+1$, તો $n$ મેળવો.
- (A) 4
- (B) 3
- (C) 2
- (D) 1
Solution:
Let the numerator be $N = (2+1)(2^2+1)(2^4+1)(2^8+1)$.
Multiply and divide the expression by $(2-1)$ which is just 1:
$N = \frac{(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)}{1}$
Using the identity $(a-b)(a+b) = a^2-b^2$ repeatedly:
Step 1: $(2-1)(2+1) = (2^2-1)$
Step 2: $(2^2-1)(2^2+1) = (2^4-1)$
Step 3: $(2^4-1)(2^4+1) = (2^8-1)$
Step 4: $(2^8-1)(2^8+1) = (2^{16}-1)$
So, the entire fraction becomes:
$\frac{2^{16}-1}{2^8-1} = \frac{(2^8-1)(2^8+1)}{2^8-1} = 2^8+1$
We are given that $2^8 + 1 = 4^n + 1$.
$2^8 = 4^n \implies (2^2)^4 = 4^n \implies 4^4 = 4^n \implies n = 4$.
Answer: (A)
Let the numerator be $N = (2+1)(2^2+1)(2^4+1)(2^8+1)$.
Multiply and divide the expression by $(2-1)$ which is just 1:
$N = \frac{(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)}{1}$
Using the identity $(a-b)(a+b) = a^2-b^2$ repeatedly:
Step 1: $(2-1)(2+1) = (2^2-1)$
Step 2: $(2^2-1)(2^2+1) = (2^4-1)$
Step 3: $(2^4-1)(2^4+1) = (2^8-1)$
Step 4: $(2^8-1)(2^8+1) = (2^{16}-1)$
So, the entire fraction becomes:
$\frac{2^{16}-1}{2^8-1} = \frac{(2^8-1)(2^8+1)}{2^8-1} = 2^8+1$
We are given that $2^8 + 1 = 4^n + 1$.
$2^8 = 4^n \implies (2^2)^4 = 4^n \implies 4^4 = 4^n \implies n = 4$.
Answer: (A)
Classwork Q3
If $\left(a+\frac{1}{a}\right)^{2}=3$, $a^{3}+\frac{1}{a^{3}}=\_.$
જો $\left(a+\frac{1}{a}\right)^{2}=3$, તો $a^{3}+\frac{1}{a^{3}}=\_.$
- (A) 4
- (B) 3
- (C) 2
- (D) 1
Solution:
Given $\left(a+\frac{1}{a}\right)^2 = 3$.
Taking the square root on both sides: $a+\frac{1}{a} = \pm\sqrt{3}$.
We need the value of $a^3 + \frac{1}{a^3}$.
Using the algebraic identity: $a^3 + b^3 = (a+b)^3 – 3ab(a+b)$.
$a^3 + \frac{1}{a^3} = \left(a+\frac{1}{a}\right)^3 – 3(a)\left(\frac{1}{a}\right)\left(a+\frac{1}{a}\right)$
$a^3 + \frac{1}{a^3} = (\pm\sqrt{3})^3 – 3(\pm\sqrt{3})$
$a^3 + \frac{1}{a^3} = \pm3\sqrt{3} \mp 3\sqrt{3} = 0$.
Given $\left(a+\frac{1}{a}\right)^2 = 3$.
Taking the square root on both sides: $a+\frac{1}{a} = \pm\sqrt{3}$.
We need the value of $a^3 + \frac{1}{a^3}$.
Using the algebraic identity: $a^3 + b^3 = (a+b)^3 – 3ab(a+b)$.
$a^3 + \frac{1}{a^3} = \left(a+\frac{1}{a}\right)^3 – 3(a)\left(\frac{1}{a}\right)\left(a+\frac{1}{a}\right)$
$a^3 + \frac{1}{a^3} = (\pm\sqrt{3})^3 – 3(\pm\sqrt{3})$
$a^3 + \frac{1}{a^3} = \pm3\sqrt{3} \mp 3\sqrt{3} = 0$.
Linked Homework Questions:
HW Q3, HW Q4, HW Q5, HW Q6, HW Q7, HW Q8
HW Q3, HW Q4, HW Q5, HW Q6, HW Q7, HW Q8