SECTION - C (3 Marks Each)
Q.26
Real Numbers
[3 Marks]
Given that $\sqrt{3}$ is an irrational number, prove that $(2 + 3\sqrt{3})$ is an irrational number.
💡 Hint: Use the method of contradiction. Assume the number is rational (equal to $\frac{p}{q}$). Do basic algebra to get $\sqrt{3}$ completely alone on one side. Then, write a strong concluding statement! (Ref: www.maheshkumarsir.in)
⚠️ Common Mistake: Just writing "Rational + Irrational = Irrational, hence proved." For a 3-mark question, you must write the full $\frac{p}{q}$ proof steps.
www.maheshkumarsir.in
Let us assume, to the contrary, that $2 + 3\sqrt{3}$ is a rational number.
Then, we can find coprime integers $p$ and $q$ ($q \neq 0$) such that:
$2 + 3\sqrt{3} = \frac{p}{q}$ [1 Mark]
Rearranging the equation to isolate $\sqrt{3}$:
$3\sqrt{3} = \frac{p}{q} - 2$
$3\sqrt{3} = \frac{p - 2q}{q}$
$\sqrt{3} = \frac{p - 2q}{3q}$ [1 Mark]
Since $p$ and $q$ are integers, $\frac{p - 2q}{3q}$ is a rational number.
This implies that $\sqrt{3}$ is a rational number.
However, this contradicts the given fact that $\sqrt{3}$ is an irrational number.
This contradiction has arisen because of our incorrect assumption.
$\therefore 2 + 3\sqrt{3}$ is an irrational number. [1 Mark]
Q.27
Polynomials
[3 Marks]
If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $2x^2 - 8x + 5$, then find the value of $\left(\alpha + \frac{1}{\beta}\right) \times \left(\beta + \frac{1}{\alpha}\right)$.
💡 Hint: Do not find the actual roots using the quadratic formula! First, simplify the expression by taking the LCM inside the brackets. Then multiply them together. You will get an expression that only needs $\alpha+\beta$ and $\alpha\beta$.
Solution 27
www.maheshkumarsir.in
Given polynomial: $P(x) = 2x^2 - 8x + 5$
Comparing with $ax^2 + bx + c$, we have $a=2, b=-8, c=5$.
Sum of zeroes $(\alpha + \beta) = -\frac{b}{a} = -\frac{-8}{2} = 4$
Product of zeroes $(\alpha\beta) = \frac{c}{a} = \frac{5}{2}$ [1 Mark]
We need to find the value of: $\left(\alpha + \frac{1}{\beta}\right) \left(\beta + \frac{1}{\alpha}\right)$
Taking LCM inside each bracket:
$= \left(\frac{\alpha\beta + 1}{\beta}\right) \left(\frac{\alpha\beta + 1}{\alpha}\right)$
$= \frac{(\alpha\beta + 1)^2}{\alpha\beta}$ [1 Mark]
Now, substituting $\alpha\beta = \frac{5}{2}$:
$= \frac{\left(\frac{5}{2} + 1\right)^2}{\frac{5}{2}}$
$= \frac{\left(\frac{7}{2}\right)^2}{\frac{5}{2}}$
$= \frac{\frac{49}{4}}{\frac{5}{2}} = \frac{49}{4} \times \frac{2}{5} = \frac{49}{10} = 4.9$
$\therefore \text{The required value is } 4.9 \text{ (or } \frac{49}{10})$ [1 Mark]
Q.28
Arithmetic Progressions
[3 Marks]
(A) Ridhi drew a polygon with $n$ sides. The smallest exterior angle is $8^\circ$ and each subsequent exterior angle is $4^\circ$ more than the previous exterior angle. Find the number of sides of the polygon that Ridhi had drawn.
OR
(B) Find the sum of integers between $1$ and $400$ that are multiples of $4$ as well as of $5$.
💡 Hint (A): The sum of the exterior angles of ANY polygon is always exactly $360^\circ$. Form an A.P. where $a=8$ and $d=4$, and set $S_n = 360$. Solve the quadratic equation for $n$.
⚠️ Common Mistake: Confusing exterior angles with interior angles! If it were interior angles, the sum would be $(n-2) \times 180^\circ$. But for exterior angles, the sum is a fixed $360^\circ$.
Solution 28(A)
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The exterior angles form an A.P.
First term ($a$) = $8^\circ$
Common difference ($d$) = $4^\circ$
We know the sum of all exterior angles of any polygon is $360^\circ$.
$\therefore S_n = 360$ [1 Mark]
Using the sum formula: $S_n = \frac{n}{2} [2a + (n-1)d]$
$360 = \frac{n}{2} [2(8) + (n-1)4]$
$720 = n [16 + 4n - 4]$
$720 = n [4n + 12]$
$720 = 4n^2 + 12n$
Dividing by $4$:
$n^2 + 3n - 180 = 0$ [1 Mark]
Factorizing the quadratic equation:
$n^2 + 15n - 12n - 180 = 0$
$n(n + 15) - 12(n + 15) = 0$
$(n - 12)(n + 15) = 0$
$n = 12$ or $n = -15$.
Since the number of sides cannot be negative, $n = -15$ is rejected.
$\therefore \text{The polygon has } 12 \text{ sides.}$ [1 Mark]
💡 Hint (B): "Multiples of 4 as well as of 5" means you need the Lowest Common Multiple (LCM) of 4 and 5, which is 20. Also, "between" 1 and 400 means you must strictly exclude the endpoints 1 and 400!
⚠️ Common Mistake: Including $400$ in your A.P. series! The word between $A$ and $B$ mathematically means $A < x < B$. So the last term is $380$, not $400$.
Solution 28(B)
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Numbers that are multiples of $4$ as well as $5$ are multiples of their LCM.
$\text{LCM}(4, 5) = 20$.
The integers between $1$ and $400$ that are multiples of $20$ form an A.P.:
$20, 40, 60, \dots, 380$ [1 Mark]
(Note: 400 is excluded because the question says "between").
Here, $a = 20$, $d = 20$, and last term $a_n = 380$.
We know, $a_n = a + (n-1)d$
$380 = 20 + (n-1)20$
$360 = 20(n-1)$
$n - 1 = 18 \Rightarrow n = 19$. [1 Mark]
Now, find the sum using $S_n = \frac{n}{2} (a + l)$
$S_{19} = \frac{19}{2} (20 + 380)$
$S_{19} = \frac{19}{2} (400)$
$S_{19} = 19 \times 200 = 3800$
$\therefore \text{The required sum is } 3800$ [1 Mark]
