Maths M.C.Q (1 Marks)

$5x^3$ is a monomial as it contains only one term.

Here, $p(x) = x^2 - 2x + 3k$
Comparing the given polynomial with $ax^2+ bx + c$, we get:
$a = 1, b = -2$ and $c = 3k$
It is given that $\alpha$ and $\beta$ are the roots of the polynomial.
$\therefore\alpha+\beta=-\frac{\text{b}}{\text{a}}$
$\Rightarrow\alpha+\beta=-\Big(\frac{-2}{1}\Big)$
$\Rightarrow\alpha+\beta=2\dots(\text{i})$
Also,
$\alpha\beta=\frac{\text{c}}{\text{a}}$
$\Rightarrow\alpha\beta=\frac{\text{5k}}1{}$
$\Rightarrow\alpha\beta=\text{3k}\dots(\text{ii})$
Now, $\alpha+\beta=\alpha\beta$
$\Rightarrow2=\text{3k} $[Using $(i)$ and $(ii)]$
$\Rightarrow\text{k}=\frac{2}{3}$

$f(x)=x^2-2 x-3$
$=x^2-3 x+x-3$
$=x(x-3)+1(x-3)$
$=(x-3)(x+1)$
$\therefore f(x)=0$
$(x-3)(x+1)=0$
$x-3=0 \text { or } x+1=0$
$x=3 \text { or } x=-1$
So, the zeros of given polynomial are $3$ and $-1$

Let $\alpha$ and $\beta$ be the zeros of the $x^2 + 5x + 8$
Then, we have
$\alpha+\beta=-\frac{\text{b}}{\text{a}}$
$=-\frac{5}{1}=-5$

Here, $p(x) = x^2 - 2x - 3$
Let $x^2- 2x - 3 = 0$
$\Rightarrow x^2- (3 - 1)x - 3 = 0$
$\Rightarrow x^2- 3x + x - 3 = 0$
$\Rightarrow x(x - 3) + 1(x - 3) = 0$
$\Rightarrow (x - 3)(x + 1) = 0$
$\Rightarrow x = 3, -1$

Let $0,\ \beta,\ \gamma$ be the zeros of the cubic polynomial $ax^3 + bx^2 + cx + d.$
Now, sum of the products of zeros, taken two at a time is given by
$\Rightarrow0\times\beta+\beta\gamma+\gamma\times0=\frac{\text{c}}{\text{a}}$
$\Rightarrow\beta\gamma=\frac{\text{c}}{\text{a}}$

Since $\alpha,\ \beta,\ \gamma$ are the zeros of $2x^3 + x^2 - 13x + 6$, we have
$\alpha\beta\gamma=-\frac{\text{d}}{\text{a}}$
$=-\frac{6}{2}=-3$

Let the zeroes of the polynomial be $\alpha$ and $\alpha+4.$
Here, $p(x) = 4x^2- 8kx + 9$
Comparing the given polynomial with $ax^2+ bx + c$, we get:
$a = 4, b = -8k$ and $c = 9$
Now, sum of the roots $=-\frac{\text{b}}{\text{a}}$
$\Rightarrow\alpha+\alpha+4=\frac{-(-8\text{k})}{4}$
$\Rightarrow2\alpha=4=\text{2k}$
$\Rightarrow\alpha+2=\text{k}$
$\Rightarrow\alpha=(\text{k}-2)\dots(\text{i})$
Also, product of the roots, $\alpha\beta=\frac{\text{c}}{\text{a}}$
$\Rightarrow\alpha(\alpha+4)=\frac{9}{4}$
$\Rightarrow(\text{k}-2)(\text{k}-2+4)=\frac{9}{4}$
$\Rightarrow(\text{k}-2)(\text{k}+2)=\frac{9}{2}$
$\Rightarrow\text{k}^2-4=\frac{9}{4}$
$\Rightarrow\text{4k}^2-16=9$
$\Rightarrow\text{4k}^2=25$
$\Rightarrow\text{k}^2=\frac{25}{4}$
$\Rightarrow\text{k}=\frac{5}{2}$
$(\because\text{k} > 0)$

Let $\alpha$ and $\beta$ be the zeros of the required quadratic polynomial.
Then, we have
$\alpha+\beta=5+(-3)=2$ and $\alpha\beta=5\times(-3)=-15$
Now, a quadratic polynomial whose zeros are $\alpha$ and $\beta$ is given by
$\text{p}(\text{x})=\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$\Rightarrow\text{p}(\text{x})=\text{x}^2-(2)\text{x}+(-15)$
$=\text{x}^2-\text{2x}-15$

Since $2$ is a zero of $f(x) = kx^2 + 3x + k,$ we have
$f(2) = 0$
$\Rightarrow k(2)^2 + 3(2) + k = 0$
$\Rightarrow 4k + 6 + k = 0$
$\Rightarrow 5k + 6 = 0$
$\Rightarrow 5k = -6$
$\Rightarrow\text{k}=-\frac{6}{5}$

Let $\alpha$ and $\beta$ be the zeros of the required quadratic polynomial.
Then, we have
$\alpha+\beta=\frac{3}{5}+\Big(-\frac{1}{2}\Big)$
$=\frac{6-5}{10}=\frac{1}{10}$
$\alpha\beta=\frac{3}{5}\times\Big(-\frac{1}{2}\Big)=-\frac{3}{10}$
Now, a quadratic polynomial whose zeros are $\alpha$ and $\beta$ is given by
$\text{p}(\text{x})=\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$\Rightarrow\text{p}(\text{x})=\text{x}^2-\Big(\frac{1}{10}\Big)\text{x}+\Big(-\frac{3}{10}\Big)$
$=\text{x}^2-\frac{1}{10}\text{x}-\frac{3}{10}$ or $10\text{x}^2-\text{x}-3$

Since $\alpha$ and $\beta$ are the zeros of $2x^2 + 5x + k$, we have
$\alpha+\beta=-\frac{5}{2}$ and $\alpha\beta=\frac{\text{k}}{2}$
Now, $\alpha^2+\beta^2+\alpha\beta=\frac{21}{4}$
$\Rightarrow(\alpha+\beta)^2-\alpha\beta=\frac{21}{4}$
$\Rightarrow\Big(\frac{-5}{2}\Big)^2-\frac{\text{k}}{2}=\frac{21}{4}$
$\Rightarrow\frac{25}{4}-\frac{\text{k}}2{}=\frac{21}{4}$
$\Rightarrow\frac{\text{k}}2{}=\frac{25}{4}-\frac{21}{4}=1$
$\Rightarrow\text{k}=2$

Since $-1$ is a zero of the cubic polynomial $x^3 + ax^2 + bx + c,$
$(-1)^3 + a(-1)^2 + b(-1) + c = 0$
$\Rightarrow -1 + a - b + c = 0$
$\Rightarrow c = 1 - a + b$
Now, product of all zeros is given by:
$-1\times\beta\times\gamma=-\text{c}$
$\Rightarrow-\beta\gamma=-(1-\text{a}+\text{b})$
$\Rightarrow\beta\gamma=1-\text{a}+\text{b}$

Let $\alpha$ and $\frac{1}{\alpha}$ be the roots of $3x^2 + 8x + k.$
Then, we have
$\alpha\times\frac{1}{\alpha}=\frac{\text{k}}{3}$
$\Rightarrow1=\frac{\text{k}}{3}$
$\Rightarrow\text{k}=3$

Since $\alpha$ and $\beta$ are the zeros of $x^2 + 6x + 2$, we have
$\alpha+\beta=-\frac{6}{1}=-6$
$\alpha\beta=\frac{2}{1}=2$
$\therefore\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha\beta}$
$=\frac{-6}{2}=-3$