Maths M.C.Q (1 Marks)

We know that, if $\text{x}=\alpha,$ is zero of a polynomial then $\text{x}-\alpha$ is a factor of $f(x)$
Since $15$ is zero of the polynomial $f(x) = x^2- 16x + 30,$ therefore $(x - 15)$ is a factor of $f(x)$
Now, we divide $f(x) = x^2- 16x + 30$ by $(x - 15)$ we get

Thus we should subtract the remainder $15$ from $x^2- 16x + 30,$
Hence, the correct choice is $(c)$.

Let the quadratic polynomial be $f(x) = x^2+ ax + b$
Now, the zeroes are $\alpha$ and $-\alpha$
So, the sum of the zeroes is zero.
$\therefore\ \alpha+(-\alpha)=\frac{-\text{a}}{1}=-\text{a}$
$\Rightarrow\text{a}=0$
So, the polynomial becomes $f(x) = x^2+ b,$ which is not linear
Also, the product of the zeros,
$\alpha\beta=\frac{\text{b}}{1}=\text{b}$
$\Rightarrow\alpha(-\alpha)=\text{b}$
$\Rightarrow\alpha^2=\text{b}$
Thus, the constant term is negative.
Hence, the correct answer is option $(a)$

We have to find the value of $\frac{1}{\alpha^2}+\frac{1}{\beta^2}$
Given $\alpha$ and $\beta$ be the zeros of the polynomial $f(x) = ax^3+ bx^2+ c$
$\alpha+\beta=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}$
$=\frac{-(\text{b})}{\text{a}}$
$\alpha\cdot\beta=\frac{\text{Constant term}}{\text{Coefficient of x}^2} $
$=\frac{\text{c}}{\text{a}}$
We have
$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\Big(\frac{1}{\alpha}+\frac{1}{\beta}\Big)^2-\frac{2}{\alpha\beta}$
$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\Big(\frac{\beta}{\alpha\beta}+\frac{\alpha}{\beta\alpha}\Big)^2-\frac{2}{\alpha\beta}$
$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\Big(\frac{\alpha+\beta}{\alpha\beta}\Big)^2-\frac{2}{\alpha\beta}$
$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\bigg(\frac{\frac{-\text{b}}{\text{a}}}{\frac{\text{c}}{\text{a}}}\bigg)^2-\frac{2}{\frac{\text{c}}{\text{a}}}$
$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\Big(\frac{-\text{b}}{\text{a}}\times\frac{\text{a}}{\text{c}}\Big)^2-\frac{2\text{a}}{\text{c}}$
$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\Big(\frac{-\text{b}}{\text{c}}\Big)^2-\frac{2\text{a}}{\text{c}}$
$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\Big(\frac{\text{b}^2}{\text{c}^2}\Big)-\frac{2\text{a}\times\text{c}}{\text{c}\times\text{c}}$
$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\Big(\frac{\text{b}^2}{\text{c}^2}\Big)-\frac{2\text{ac}}{\text{c}^2}$
$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\Big(\frac{\text{b}^2-2\text{ac}}{\text{c}^2}\Big)$
Hence, the correct choice is $(b)$

The given polynomial is $f(x) = (k - 1)x^2+ kx + 1$
Since -3 is one of the zeroes of the given polynomial, so $f(-3) = 0.$
$(k - 1)(-3)^2+ k(-3) + 1 = 0$
$\Rightarrow 9(k - 1) - 3k + 1 = 0$
$\Rightarrow 9k - 9 - 3k + 1 = 0$
$\Rightarrow 6k - 8 = 0$
$\Rightarrow\text{k}=\frac{4}{3}$
Hence, the correct answer is option $(a)$

Let the polynomial be $f(x) = ax^3+ bx^2+ cx + d$
Suppose the two zeroes of $f(x)$ are $\alpha=0$ and $\beta=0$
We know that,
Sum of the zeros,
$\alpha+\beta+\gamma=\frac{-\text{b}}{\text{a}}$
$\Rightarrow0+0+\gamma=\frac{-\text{b}}{\text{a}}$
$\Rightarrow\gamma=\frac{-\text{b}}{\text{a}}$
Hence, the correct answer is option (a)

Let $f(x) = x^2+ 99x + 127$
Product of the zeroes of $f(x) = 127 × 1 = 127 [$Product of zeroes = $\frac{c}{a}$ when $f(x) = ax^2+ bx + c]$
Since the product of zeroes is positive, we can say that it is only possible when both zeroes are positive or both zeroes are negative.
Also, sum of the zeroes $= -99$ $\Big[\text{Sum of zeroes}=-\frac{\text{b}}{\text{a}}\text{ when f(x) = ax}^2+\text{bx + c}\Big]$
The sum being negative implies that both zeroes are positive is not correct.
So, we conclude that both zeroes are negative.
Hence, the correct answer is option $(b)$.

Clearly, $f(x) = ax^2+ bx + c$ represent a parabola opening upwards.

Therefore, $a > 0 y = ax^2+ bx + c$ cuts $Y$ axis at $P$ which lies on $OY$.
 Putting $x = 0 $in $y = ax^2+ bx + c,$ we get $y = c$. So the coordinates of $P$ is $(0, c)$.
Clearly, $P$ lies on $OY$. Therefore $c > 0$
Hence, the correct choice is $(a)$

1. Graph cuts $x$-axis at two distinct points (two zeroes)
2. Graph cuts the $x$-axis at exactly one point (one zero)
3. The graph is either completely above the $x$-axis or completely below the $x$-axis (no zeroes)
4. The graph is cutting the $x$-axis at three distinct points and it is not a parabola opening either upwards or downwards.

If one zero of the polynomial $f(x) = 5x^2+ 13x + k$ is reciprocal of the other. So $\beta=\frac{1}{\alpha}\Rightarrow\alpha\beta=1$
Now we have
$\alpha\times\beta=\frac{\text{Constant term}}{\text{Coefficient of x}^2}$
$=\frac{\text{k}}{5}$
Since $\alpha\beta=1$
Therefore we have
$\alpha\beta=\frac{\text{k}}{5}$
$1=\frac{\text{k}}{5}$
$\Rightarrow\text{k}=5$
Hence, the correct choice is $(b)$

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x) = x^2+ x + 1$
$\alpha+\beta=-\frac{\text{coefficient of x}}{\text{coefficient of x}^2}$
$=\frac{-1}{1}=-1$
$\alpha\times\beta=\frac{\text{constant term}}{\text{coefficient of x}^2}$
$=\frac{1}{1}=1$
We have
$=\frac{1}{\alpha}+\frac{1}{\beta}$
$=\frac{\beta+\alpha}{\alpha\beta}$
$=\frac{-1}{1}$
$=-1$
The value of $\frac{1}{\alpha}+\frac{1}{\beta}$ is $-1$
Hence, the correct choice is $(b)$.

Since $\alpha$ and $\beta$ are the zeros of quadratic polynomial $f(x) = ax^2- 6x^2+ 11x - 6$
$\alpha\beta=\frac{-\text{Constant term}}{\text{Coefficient of x}^2}$
So we have
$4=-\Big(\frac{-6}{\text{a}}\Big) $
$4=\frac{6}{\text{a}}$
$4\text{a}=6$
$ \text{a}=\frac{6}{4}$
$ \text{a}=\frac{3\times2}{2\times2}$
$\text{a}=\frac{3}{2}$
The value of $\alpha$ is $\frac{3}{2}$
Hence, the correct alternative is $(a)$

Let $\alpha,\beta,\gamma$ be the zeros of polynomial $f(x)=2 x^3+6 x^2-4 x+9$ such that $\alpha\beta=3$
We have,
$\alpha\beta\gamma=\frac{\text{Constant term}}{\text{Coefficient of x}^2}$
$=\frac{-9}{2}$
Putting $\alpha\beta=3\text{ in }\alpha\beta\gamma=\frac{-9}{2},$ we get
$\alpha\beta\gamma=\frac{-9}{2}$
$3\gamma=\frac{-9}{2}$
$\gamma=\frac{-9}{2}\times\frac{1}{3}$
$\gamma=\frac{-3}{2}$
Therefore, the value of third zero is $\frac{-3}{2}$
Hence, the correct alternative is $(b)$

Since $\alpha$ and $\beta$ are the zeros of quadratic polynomial
$\text{f(x)}=\text{x}^2-\text{p(x}+1)-\text{c}$
$=\text{x}^2-\text{px}-\text{p}-\text{c}$
$\alpha+\beta=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}$
$=-\Big(\frac{-\text{p}}{1}\Big)$
$=\text{p}$
$\alpha\times\beta=\frac{\text{Constant term}}{\text{Coefficient of x}^2}$
$=\frac{-\text{p}-\text{c}}{1}$
$=-\text{p}-\text{c}$
We have
$(\alpha+1)(\beta+1)$
$=\alpha\beta+\beta+\alpha+1$
$=\alpha\beta+(\alpha+\beta)+1$
$=-\text{p}-\text{c}+(\text{p})+1$
$=-\text{p}-\text{c}+\text{p}+1$
$=-\text{c}+1$
$=1-\text{c}$
The value of $(\alpha+1)(\beta+1)$is $1 - c$
Hence, the correct choice is $(b)$

Let $f(x) = x^2+ ax + a$
Product of the zeroes of $f(x) = a$ $\Big[\text{Product of zeroes}=\frac{\text{c}}{\text{a}}\text{ when f(x) = ax}^2+\text{bx + c}\Big]$
Since the product of zeroes is positive, so the zeroes must be either both positive or both negative.
Also, sum of the zeroes $= -a$ $\Big[\text{Sum of zeroes}=-\frac{\text{b}}{\text{a}}\text{when f(x) = ax}^2+\text{bx + c}\Big]$
So, the sum of the zeroes is negative, so the zeroes cannot be both positive.
Hence, the correct answer is option $(b)$

Let $\alpha,\beta$ be the zeros of the polynomial $f(x) = x^2+ px + q.$
Then,
$\alpha+\beta=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}$
$=-\frac{\text{p}}{1}$
$=-\text{p}$
And
$\alpha\beta=\frac{\text{Constant term}}{\text{Coefficient of x}^2}$
$=\frac{\text{q}}{1}$
$=\text{q}$
Let $S$ and $R$ denote respectively the sum and product of the zeros of a polynomial
Whose zeros are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ then
$\text{S}=\frac{1}{\alpha}+\frac{1}{\beta}$
$=\frac{\alpha+\beta}{\alpha\beta}$
$=\frac{-\text{p}}{\text{q}}$
$\text{R}=\frac{1}{\alpha}\times\frac{1}{\beta}$
$=\frac{1}{\alpha\beta}$
$=\frac{1}{\text{q}}$
Hence, the required polynomial g(x) whose sum and product of zeros are S and R is given by
$\text{x}^2-\text{Sx}+\text{R}=0$
$\text{x}^2+\frac{\text{p}}{\text{q}}\text{x}+\frac{1}{\text{q}}=0$
$\frac{\text{qx}^2+\text{qx}+1}{\text{q}}=0$
$\Rightarrow\ \text{qx}^2+\text{qx}+1$
So, $g(x) = qx^2+ Px + 1$
Hence, the correct choice is $(c)$

Let the given polynomial be $f(x) = x^2+ 3x + k$
Since $2$ is one of the zero of the given plynomial, so $(x - 2)$ will be a factor of the given polynomial.
Now, $f(2) = 0$
$\Rightarrow 2^2+ 3 \times 2 + k = 0$
$\Rightarrow 4 + 6 + k = 0$
$\Rightarrow k = -10$
Hence, the correct answer is option (b)

Let $\alpha=0,\beta=0$ and $\gamma$ be the zeros of the polynomial
$f(x) = ax^3+ bx^2+ cx + d$
Therefore
$\alpha+\beta+\gamma=\frac{-\text{Coefficient of x}^2}{\text{Coefficient of x}^3}$
$=-\Big(\frac{\text{b}}{\text{a}}\Big)$
$\alpha+\beta+\gamma=-\frac{\text{b}}{\text{a}}$
$0+0+\gamma=-\frac{\text{b}}{\text{a}}$
$\gamma=-\frac{\text{b}}{\text{a}}$
The value of $\gamma=-\frac{\text{b}}{\text{a}}$
Hence, the correct choice is $(c)$

Let $p(x) = x^3+ ax^2+ bx + c$
Now, -1 is a zero of the polynomial
So, $p(0) = 0$
$\Rightarrow (-1)^3+ a(-1)^2+ b(-1) + c = 0$
$\Rightarrow -1 + a - b + c = 0$
$\Rightarrow a - b + c = 1$
$\Rightarrow c = 1 - a + b$
Now, if $\alpha,\beta,\gamma$ are the zeroes of the cubic polynomial $ax^3+ bx^2+ cx + d$, then product of zeroes is given by
$\alpha\beta\gamma=-\frac{\text{d}}{\text{a}}$
So, for the given polynomial, $p(x) = x^3+ ax^2+ bx + c$
$\alpha\beta(-1)=\frac{-\text{c}}{1}=\frac{-(1-\text{a}+\text{b})}{1}$
$\Rightarrow\alpha\beta=1-\text{a + b}$
Hence, the correct answer is option $(a)$

If $\text{x}=\alpha$ is a zero of a polynomial then $\text{x}-\alpha$ is a factor of $f(x)$
Since $3$ is the zero of the polynomial $f(x) = x^2- 5x + 4,$
Therefore $x - 3$ is a factor of f(x)
Now, we divide $f(x) = x^2- 5x + 4$ by $(x - 3)$ we get

Therefore we should add $2$ to the given polynomial
Hence, the correct choice is $(b)$

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomials such that
$0=\alpha+\beta$
If one of zero is 3 then
$\alpha+\beta=0$
$3+\beta=0$
$\beta=0-3$
$\beta=-3$
Substituting $\beta=-3$ in $\alpha+\beta=0$ we get
$\alpha-3=0$
$\alpha=3$
Let S and P denote the sum and product of the zeros of the polynomial respectively then
$\text{S}=\alpha+\beta$
$\text{S}=0$
$\text{p}=\alpha\beta$
$\text{p}=3\times-3$
$\text{p}=-9$
Hence, the required polynomials is
$=(\text{x}^2-\text{Sx}+\text{p})$
$=(\text{x}^2-0\text{x}-9)$
$=\text{x}^2-9$
Hence, the correct choice is $(a)$

Let $\alpha=\sqrt{5}$ and $\beta=-\sqrt{5}$ be the given zeros and $\gamma$ be the third zero of the polynomial $x^3+ 3x^2- 5x - 15$ Then,
By using $\alpha+\beta+\gamma=\frac{-\text{Coefficient of x}^2}{\text{Coefficient of x}^3}$
$\alpha+\beta+\gamma=\frac{-3}{1}$
$\alpha+\beta+\gamma=-3$
Substituting $\alpha=\sqrt{5}$ and $\beta=-\sqrt{5}$ in $\alpha+\beta+\gamma=-3$
We get
$\sqrt{5}-\sqrt{5}+\gamma=-3$
$\gamma=-3$
Hence, the correct choice is $(b)$

We have to find the value of $ab$
Given $f(x) = ax^3+ bx - c$ is divisible by the polynomial $g(x) = x^2+ bx + c$

We must have
$bx - acx + ab^2x + abc - c = 0,$ for all x
$x(b - ac + ab^2) + c(ab - 1) = 0 .....(1)$
$c(ab - 1) = 0$
Since $c ≠ 0,$ so
$ab - 1 = 0$
$\Rightarrow ab = 1$
Now in the equation $(1)$ the condition is true for all $x$. So put $x = 1$ and also we have $ab = 1$
Therefore we have
$ b-a c+a b^2=0 $
$ b+a b^2-a c=0$
$ b(1+a b)-a c=0 $
Substituting $\text{a}=\frac{1}{\text{b}}$ and $ab = 1$ we get,
$\text{b}(1+1)-\frac{1}{\text{b}}\times\text{c}=0$
$2\text{b}-\frac{1}{\text{b}}\times\text{c}=0$
$-\frac{1}{\text{b}}\times\text{c}=-2\text{b}$
$\text{c}=2\text{b}\times\frac{\text{b}}{1}$
Hence, the correct alternative is $(c)$

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $p(x)=4 x^2+3 x+7$
$\alpha+\beta=-\frac{\text{coefficient of x}}{\text{coefficient of x}^2}$
$=\frac{-3}{4}$
$\alpha\beta=\frac{\text{constant term}}{\text{coefficient of x}^2}$
$=\frac{7}{4}$
We have
$=\frac{1}{\alpha}+\frac{1}{\beta}$
$=\frac{\beta+\alpha}{\alpha\beta}$
$=\frac{\frac{-3}{4}}{\frac{7}{4}}$
$=\frac{-3}{4}\times\frac{4}{7}$
$=\frac{-3}{7}$
The value of $\frac{1}{\alpha}+\frac{1}{\beta}$ is $\frac{-3}{7}$
Hence, the correct choice is $(d)$.

We have to find the value of $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$
Given $\alpha,\beta,\gamma$ be the zeros of the polynomial $f(x) = ax^3+ bx^2+ cx + d$
We know that
$\alpha\beta+\beta\gamma+\gamma\alpha=\frac{\text{Coefficient of x}}{\text{Coefficient of x}^3}$
$=\frac{\text{c}}{\text{a}}$
$\alpha\beta\gamma=\frac{-\text{Constant term}}{\text{Coefficient of x}^3} $
$=\frac{-\text{d}}{\text{a}}$
So
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma} $
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\frac{\text{c}}{\text{a}}}{-\frac{\text{d}}{\text{a}}}$
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\text{c}}{\text{a}}\times\Big(-\frac{\text{a}}{\text{d}} \Big)$
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=-\frac{\text{c}}{\text{d}}$
Hence, the correct choice is $(c)$

We have to find the value of $\frac{1}{\alpha\beta}+\frac{1}{\beta\gamma}+\frac{1}{\gamma\alpha}$
Given $\alpha,\beta,\gamma$ be the zeros of the polynomial $f(x) = x^3- px^2+ qx - r,$
$\alpha+\beta+\gamma=\frac{-\text{Coefficient of x}^2}{\text{Coefficient of x}^3}$
$=\frac{-(-\text{p})}{1}$
$=\text{p}$
$\alpha\cdot\beta\cdot\gamma=\frac{\text{Constant term}}{\text{Coefficient of x}^3} $
$=\frac{-\text{r}}{1}$
$=-\text{r}$
Now we calculate the expression
$\frac{1}{\alpha\beta}+\frac{1}{\beta\gamma}+\frac{1}{\gamma\alpha}=\frac{\gamma}{\alpha\beta\gamma}+\frac{\alpha}{\alpha\beta\gamma}+\frac{\beta}{\alpha\beta\gamma}$
$\frac{1}{\alpha\beta}+\frac{1}{\beta\gamma}+\frac{1}{\gamma\alpha}=\frac{\alpha+\gamma+\beta}{\alpha\beta\gamma}$
$\frac{1}{\alpha\beta}+\frac{1}{\beta\gamma}+\frac{1}{\gamma\alpha} = \frac{\text{p}}{-\text{r}}$
Hence, the correct choice is $(c)$