Solutions: Polynomials - Test 1

Q1. (b) -10

Put $x=2$ in $x^2 + 3x + k = 0$.
$(2)^2 + 3(2) + k = 0 \Rightarrow 4 + 6 + k = 0 \Rightarrow k = -10$.

Q2. (c) $x^2 - x - 12$

Sum = $-3 + 4 = 1$. Product = $-3 \times 4 = -12$.
Polynomial = $k[x^2 - (\text{Sum})x + \text{Product}] = k(x^2 - x - 12)$.

Q3. (d) more than 3

There are infinite polynomials with given zeroes, as we can multiply by any non-zero constant $k$.

Q4. (d) A is false but R is true.

Discriminant $D = b^2 - 4ac = 16 - 20 = -4 < 0$. No real zeroes. So A is false.
Degree is 2, so at most 2 zeroes. R is true.

Q5. $6x^2 - 7x - 3 = 6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3)$.
Zeroes: $x = 3/2, -1/3$. [1]
Sum = $3/2 - 1/3 = 7/6 = -(-7)/6$. Product = $(3/2)(-1/3) = -1/2 = -3/6$. Verified. [1]

Q6. $\alpha + \beta = 1$, $\alpha\beta = -4$.
Expression = $\frac{\alpha + \beta}{\alpha\beta} - \alpha\beta = \frac{1}{-4} - (-4) = -\frac{1}{4} + 4 = \frac{15}{4}$. [2]

Q7. Sum $S = \sqrt{2}$, Product $P = -3/2$.
$p(x) = k(x^2 - Sx + P) = k(x^2 - \sqrt{2}x - \frac{3}{2})$.
For $k=2$, $2x^2 - 2\sqrt{2}x - 3$. [2]

Q8. $\alpha + \beta = -5/5 = -1$, $\alpha\beta = 1/5$. [1]
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.
$= (-1)^2 - 2(1/5) = 1 - 2/5 = 3/5$. [2]

Q9. Sum of zeroes = $-\frac{b}{a} = -\frac{-2}{k^2 - 14} = \frac{2}{k^2 - 14}$. [1]
Given Sum = 1. So, $\frac{2}{k^2 - 14} = 1 \Rightarrow k^2 - 14 = 2 \Rightarrow k^2 = 16$.
$k = \pm 4$. [2]

Q10. $\alpha + \beta = p$, $\alpha\beta = q$. [1]
LHS = $\frac{\alpha^4 + \beta^4}{\alpha^2\beta^2}$.
$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = p^2 - 2q$. [1]
$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2 = (p^2 - 2q)^2 - 2q^2$.
$= p^4 + 4q^2 - 4p^2q - 2q^2 = p^4 - 4p^2q + 2q^2$. [2]
Divide by $\alpha^2\beta^2 = q^2$:
$\frac{p^4}{q^2} - \frac{4p^2q}{q^2} + \frac{2q^2}{q^2} = \frac{p^4}{q^2} - \frac{4p^2}{q} + 2$. [1]

(i) Since coefficient of $x^2$ is $1 > 0$, the parabola opens upwards. [1]

(ii) $x^2 - 8x + 12 = 0 \Rightarrow (x-6)(x-2) = 0$.
Zeroes are 2 and 6. [2]

(iii) At $x=0$, $p(0) = 0 - 0 + 12 = 12$. [1]