Section A (1 Mark each)
Q1. (d) $x^3 - 4x^2 - x + 1 = (x-2)^3$
LHS is cubic. RHS $(x-2)^3 = x^3 - 6x^2 + 12x - 8$. The $x^3$ terms cancel out, leaving a quadratic equation $-4x^2... = -6x^2... \Rightarrow 2x^2...$.
Q2. (a) -8
$D = b^2 - 4ac = (-4)^2 - 4(2)(3) = 16 - 24 = -8$.
Q3. (b) $\pm 0.2$
$x^2 = 0.04 \Rightarrow x = \sqrt{0.04} = \pm 0.2$.
Q4. (a) Both A and R are true and R is the correct explanation of A.
$D = 4^2 - 4(1)(5) = 16 - 20 = -4 < 0$. Since $D < 0$, no real roots.
Section B (2 Marks each)
Q5. $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$.
Split middle term: $2x + 5x$. Product $2 \times 5 = 10 = \sqrt{2} \times 5\sqrt{2}$.
$\sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0 \Rightarrow \sqrt{2}x(x+\sqrt{2}) + 5(x+\sqrt{2}) =
0$.
$(x+\sqrt{2})(\sqrt{2}x+5) = 0$.
Roots: $-\sqrt{2}, -\frac{5}{\sqrt{2}}$ [2]
Q6. For equal roots, $D = 0$.
$k^2 - 4(2)(3) = 0 \Rightarrow k^2 - 24 = 0 \Rightarrow k^2 = 24$.
$k = \pm \sqrt{24} = \pm 2\sqrt{6}$. [2]
Q7. $(x-2)(x+1) = x^2 - x - 2$.
$(x-1)(x+3) = x^2 + 2x - 3$.
$x^2 - x - 2 = x^2 + 2x - 3 \Rightarrow -3x + 1 = 0$.
It is a linear equation, not quadratic. No [2]
Section C (3 Marks each)
Q8. Let integers be $x, x+1$.
$x^2 + (x+1)^2 = 365 \Rightarrow 2x^2 + 2x + 1 = 365 \Rightarrow 2x^2 + 2x - 364 = 0$.
$x^2 + x - 182 = 0$. Factors of 182: 13, 14.
$(x+14)(x-13) = 0 \Rightarrow x = 13$ (since positive).
Integers are 13 and 14. [3]
Q9. $\frac{(x-7) - (x+4)}{(x+4)(x-7)} = \frac{11}{30} \Rightarrow
\frac{-11}{x^2-3x-28} = \frac{11}{30}$.
$-1 = \frac{1}{30}(x^2-3x-28) \Rightarrow -30 = x^2 - 3x - 28 \Rightarrow x^2 - 3x + 2 =
0$.
$(x-1)(x-2) = 0 \Rightarrow x = 1, 2$. [3]
Section D (5 Marks)
Q10. Let base = $x$ cm. Altitude = $x-7$ cm.
$x^2 + (x-7)^2 = 13^2 \Rightarrow x^2 + x^2 - 14x + 49 = 169$.
$2x^2 - 14x - 120 = 0 \Rightarrow x^2 - 7x - 60 = 0$. [2]
$(x-12)(x+5) = 0 \Rightarrow x = 12$ (side can't be negative).
Base = 12 cm, Altitude = $12-7 = 5$ cm. [3]
Section E (Case Study - 4 Marks)
(i) Let breadth $x$. Length $2x+1$. Area $x(2x+1) = 300 \Rightarrow 2x^2 + x - 300 = 0$. [1]
(ii) $D = 1^2 - 4(2)(-300) = 1 + 2400 = 2401$. [1]
(iii) $x = \frac{-1 \pm \sqrt{2401}}{4} = \frac{-1 \pm 49}{4}$.
$x = 48/4 = 12$ (ignore negative).
Breadth = 12 m, Length = $2(12)+1 = 25$ m. [2]