21 Real Numbers
If $a$ and $b$ are any two positive integers, then:
Concept:
1
Fundamental Property of Integers: For any two positive integers, the product of their Highest Common Factor (HCF) and their Least Common Multiple (LCM) is strictly equal to the product of the numbers themselves.
Hint:
1
Look for the option that shows multiplication ($\times$) between the HCF and LCM, and multiplication ($\times$) between the two integers $a$ and $b$.
Mistakes to Avoid:
!
Three Number Trap: Remember that this exact rule $\text{HCF} \times \text{LCM} = \text{Product of Numbers}$ ONLY applies to exactly two numbers. It does not work for three numbers ($a, b, c$).
👉 Scroll horizontally to read the full handwritten solution.
A fundamental property of numbers states that for any two
positive integers $a$ and $b$, the product of their HCF and LCM
is strictly equal to the product of the two numbers themselves.
Mathematically, this relationship is expressed as:
$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$ [1 M]
Looking at the options, Option (C) perfectly matches this theorem.
Correct Option: (C)
22 Real Numbers
A rational number between $\sqrt{2}$ and $\sqrt{3}$ is:
Concept:
1
Rationalizing Decimal Approximations: A rational number has a terminating or repeating decimal expansion. We know $\sqrt{2} \approx 1.414$ and $\sqrt{3} \approx 1.732$. We need a terminating decimal that fits strictly between these two values.
Hint:
1
Look at the decimal options. Is $1.3$ between $1.414$ and $1.732$? Is $1.8$ between them? (Note: The official paper had a discrepancy here. See the solution for details).
Mistakes to Avoid:
!
Averaging Irrationals: Option A calculates the exact midpoint between the two numbers. However, the sum of two irrational numbers is (usually) irrational, and dividing by 2 keeps it irrational! It does not satisfy the "rational" requirement.
👉 Scroll horizontally to read the full handwritten solution.
Let us analyze the given numbers and the options.
We know the approximate decimal values for the square roots:
$\sqrt{2} \approx 1.414$
$\sqrt{3} \approx 1.732$
We need to find a rational number that falls strictly between $1.414$ and $1.732$.
Evaluating the options:
- (A) $\frac{\sqrt{2} + \sqrt{3}}{2}$: This is the average, but it is an IRRATIONAL number.
- (B) $\frac{\sqrt{2} - \sqrt{3}}{2}$: This is irrational and negative.
- (C) $1.3$: This is rational, but $1.3 < 1.414$ (not between).
- (D) $1.8$: This is rational, but $1.8 > 1.732$ (not between).
Note: Since none of the provided options correctly represent a rational number between $\sqrt{2}$ and $\sqrt{3}$, the question contains a discrepancy. Therefore, full marks are awarded to all candidates regardless of their choice.
All Options Awarded Marks [1 M]
23 Real Numbers
If two positive integers $a$ and $b$ are written as $a = x^3 y^2$ and $b = x y^3$, where $x$ and $y$ are prime numbers, then LCM (a, b) is:
Concept:
1
LCM of Variables: To find the Least Common Multiple (LCM) of numbers expressed in prime factored form (like variables $x$ and $y$), you must identify every prime base present and take the highest power of that base found in either expression.
Hint:
1
Look at the base $x$. The powers are $x^3$ and $x^1$. The highest is $x^3$. Now do the same for the base $y$. Combine them!
Mistakes to Avoid:
!
Mixing up HCF and LCM: If you took the lowest powers ($x^1$ and $y^2$), you would get $xy^2$. This is the Highest Common Factor (HCF), not the LCM. Always remember: LCM = Largest powers!
👉 Scroll horizontally to read the full handwritten solution.
We are given the prime factorizations of two integers:
$a = x^3 y^2$
$b = x^1 y^3$
To find the Least Common Multiple (LCM), we must take the
highest power of each prime factor that appears.
1. The prime factor $x$ has powers $3$ (in $a$) and $1$ (in $b$).
The highest power is $x^3$.
2. The prime factor $y$ has powers $2$ (in $a$) and $3$ (in $b$).
The highest power is $y^3$.
Multiplying these highest powers together gives the LCM:
$\text{LCM}(a, b) = x^3 y^3$ [1 M]
Correct Option: (C)
24 Polynomials
The zeroes of the polynomial $p(x) = x^2 - x - 6$ are:
Concept:
1
Splitting the Middle Term: To find the zeroes of a quadratic $ax^2 + bx + c$, we find two numbers that multiply to $a \times c$ and add up to $b$. Then we factor the polynomial and set each bracket to 0.
Hint:
1
You need two numbers that multiply to $-6$ and add up to $-1$. Those numbers are $-3$ and $+2$. Write $(x - 3)(x + 2) = 0$ and solve for $x$.
Mistakes to Avoid:
!
Sign Flipping Error: Students correctly find the factors are $-3$ and $2$ but forget to flip the signs when pulling them out of the brackets $(x - 3) = 0$. The roots have the opposite sign of the factors!
👉 Scroll horizontally to read the full handwritten solution.
To find the zeroes of $p(x) = x^2 - x - 6$, set it to zero:
$x^2 - x - 6 = 0$
We need two numbers that multiply to $-6$ and add to $-1$.
These two numbers are $-3$ and $2$ (since $-3 \times 2 = -6$ and $-3 + 2 = -1$).
Rewrite the middle term:
$x^2 - 3x + 2x - 6 = 0$
Factor by grouping:
$x(x - 3) + 2(x - 3) = 0$
$(x - 3)(x + 2) = 0$
Setting each factor to zero gives the zeroes:
$x - 3 = 0 \implies x = 3$
$x + 2 = 0 \implies x = -2$ [1 M]
Correct Option: (C) $3, -2$
25 Polynomials
If $\alpha$ and $\beta$ are the zeroes of a polynomial $f(x) = px^2 - 2x + 3p$ and $\alpha + \beta - \alpha\beta = 0$, then the value of $p$ is:
Concept:
1
Relationship between Zeroes and Coefficients: For a quadratic polynomial $ax^2 + bx + c$, the sum of the zeroes ($\alpha + \beta$) is equal to $\frac{-b}{a}$ and the product of the zeroes ($\alpha\beta$) is equal to $\frac{c}{a}$.
Hint:
1
Find the values: $\alpha + \beta = \frac{-(-2)}{p}$ and $\alpha\beta = \frac{3p}{p} = 3$. Plug these two results into the given equation $\alpha + \beta - \alpha\beta = 0$ and solve for $p$.
Mistakes to Avoid:
!
Double Negative Errors: When calculating the sum $\frac{-b}{a}$, students often write $\frac{-2}{p}$ instead of $\frac{-(-2)}{p} = \frac{2}{p}$. Watch your signs closely!
👉 Scroll horizontally to read the full handwritten solution.
Given polynomial: $f(x) = px^2 - 2x + 3p$
Comparing with standard form $ax^2 + bx + c$:
$a = p$, $b = -2$, and $c = 3p$.
Sum of zeroes ($\alpha + \beta$):
$\alpha + \beta = \frac{-b}{a} = \frac{-(-2)}{p} = \frac{2}{p}$
Product of zeroes ($\alpha\beta$):
$\alpha\beta = \frac{c}{a} = \frac{3p}{p} = 3$
We are given the condition: $\alpha + \beta - \alpha\beta = 0$
Substitute the expressions we found:
$\frac{2}{p} - 3 = 0$
$\frac{2}{p} = 3 \implies 3p = 2 \implies p = \frac{2}{3}$ [1 M]
Correct Option: (B)
Disclaimer: This content is provided solely for educational purposes to help students with their CBSE Board Exam preparations. While we strive for maximum accuracy, this guide does not claim absolute authenticity and has no official affiliation with the CBSE board setting authorities.
