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MATHS M.C.Q (1 Marks)

Polynomials · UP Board (UPMSP) STD 10 (UPMSP - English Medium). 203 questions.

Questions · Page 1 of 5

M.C.Q (1 Marks)

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MCQ 11 Mark
The degree of polynomial having zeroes $-3$ and $4$ only is :
  • ✓
    $2$
  • B
    $1$
  • C
    More than $3$
  • D
    $3$
Answer
Correct option: A.
$2$
We know that :
The polynomial is :
$\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$\therefore$ The polynomial having zeroes $-3$ and $4$.
$=x^2-(-3+4) x+(-3) 4$
$=x^2-(1) x-12$
$=x^2-x-12, $ the polynomial having degree $ 2$
Thus, the required "option $(a) \ 2$ " is correct.
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MCQ 21 Mark
On dividing a polynomial $p(x)$ by $x^2- 4,$ quotient and remainder are found to be $x$ and $3$ respectively. The polynomial $p(x)$ is :
  • A
    $ 3 x^2+x-12 $
  • ✓
    $ x^3-4 x+3 $
  • C
    $ x^2+3 x-4 $
  • D
    $ x^3-4 x-3 $
Answer
Correct option: B.
$ x^3-4 x+3 $
We have to find the polynomial $p(x),$ which is divided by $(x^2- 4)$.
Then, dividend $= p(x),$
divisor $= (x^2- 4)$,
quotient $= x$ and
remainder $= 3$
We know that,
dividend $=$ divisor $\times$ quotient $+$ remainder
i.e. $p(x) = (x^2- 4) \times (x + 3)$
$= (x^2\times x) - (4 \times x) + 3$
$= x^3-4 x+3 $
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MCQ 31 Mark
The number of zeroes for a polynomial $p(x)$ where graph of $y = p(x)$ is given in Figure, is :
  • ✓
    $3$
  • B
    $4$
  • C
    $0$
  • D
    $5$
Answer
Correct option: A.
$3$
$3$
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MCQ 41 Mark
In fig. the graph of the polynomial $p(x)$ is given. The number of zeroes of the polynomial is :
  • A
    $1$
  • ✓
    $2$
  • C
    $3$
  • D
    $0$
Answer
Correct option: B.
$2$
The number of zeroes is $2$ as the graph intersects the $x-$ axis at $2$ points.
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MCQ 51 Mark
The zeroes of the polynomial $x^2-3 x-m(m+3)$ are :
  • A
    $m, m + 3$
  • ✓
    $–m, m + 3$
  • C
    $m, – (m + 3)$
  • D
    $–m, – (m + 3)$
Answer
Correct option: B.
$–m, m + 3$
Given : equation $x^2-3 x-m(m+3)=0$,
where $m$ is a constant
To find the roots of the equation
The given equation is of form $a x^2+b x+c=0$
$\therefore a = 1, b = -3, c = -m(m + 3)$
We know the roots of the equation can be find out using the formula,
$\text{x} = \frac{-\text{b}\pm\sqrt{\text{b}^{2} - 4\text{ac}}}{2\text{a}}$
Substituting the values of $a, b, c,$ we get
$\text{x}=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-\text{m}(\text{m+3})}}{2}$
$\Rightarrow\text{x}=\frac{3\pm\sqrt{9 + 4\text{m}^2+12\text{m}}}{2}$
$\Rightarrow\text{x}=\frac{3\pm(2\text{m+3})}{2}$ or $\text{x}=\frac{3+(2\text{m+3})}{2},\text{x}=\frac{3-(2\text{m+3})}{2}$
$\Rightarrow x = m + 3, x = -m$ are the required roots of the equation.
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MCQ 61 Mark
The quadratic polynomial, the sum of whose zeroes is $-5$ and their product is $6,$ is :
  • ✓
    $ x^2+5 x+6 $
  • B
    $ x^2-5 x+6 $
  • C
    $ x^2-5 x-6 $
  • D
    $ -x^2+5 x+6 $
Answer
Correct option: A.
$ x^2+5 x+6 $
$= x^2- \ ($sum of zeroes$)\ x\  +$ product of zeroes
$= x^2- (-5)x + 6$
$= x^2+5 x+6 $
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MCQ 71 Mark
If one of the zeroes of the quadratic polynomial $x^2+3 x+k$ is $2,$ then the value of $k$ is :
  • A
    $10$
  • ✓
    $-10$
  • C
    $-7$
  • D
    $-2$
Answer
Correct option: B.
$-10$
Let the given quadratic polynomial be $p(x)=x^2+3 x+k$
Given one of the zero of the quadratic polynomial is $2$ .
Hence $p(2)=0$
Put $x=2$ in $p(x)$, we get
$p(2)=2^2+3(2)+k$
$\Rightarrow 0=4+6+k$
$\Rightarrow 0=10+k$
$\Rightarrow k=-10$
 
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MCQ 81 Mark
If $\alpha, \beta$ and $\gamma$ are the zeroes of a cubic polynomial $a x^3+b x^2+c x+d,$ then $\alpha\beta+\beta\text{y}+\text{y}\alpha$ is :
  • A
    $\frac{-\text{b}}{\text{a}}$
  • B
    $\frac{-\text{c}}{\text{a}}$
  • ✓
    $\frac{\text{c}}{\text{a}}$
  • D
    $\frac{\text{b}}{\text{a}}$
Answer
Correct option: C.
$\frac{\text{c}}{\text{a}}$
$\alpha,\beta$ and $\gamma$ are the zeroes of a cubic polynomial $a x^3+b x^3+c x+d$
$\therefore$ Sum of the product of zeroes of a cubic $a x^3+b x^2+c x+d$
$=\frac{ {(\text{Coefficient of x)}}}{\text{Coefficient of x}^{3 }}$ then $\alpha\beta+\beta\text{y}+\text{y}\alpha=\frac{\text{c}}{\text{a}}$
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MCQ 91 Mark
Which of the following is a true statement ?
  • A
    $x^2+ 5x - 3$ is a linear polynomial.
  • B
    $x^2+ 4x - 1$ is a binomial.
  • C
    $x + 1$ is a monomial.
  • ✓
    $5x^3$ is a monomial.
Answer
Correct option: D.
$5x^3$ is a monomial.
$5x^3$ is a monomial as it contains only one term.
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MCQ 101 Mark
If $\alpha$ and $\beta$ are the zero of $2 x^2+5 x-8$, then the value of $(\alpha\beta)$ is :
  • A
    $\frac{-5}{2}$
  • B
    $\frac{5}{2}$
  • ✓
    $\frac{-9}{2}$
  • D
    $\frac{9}{2}$
Answer
Correct option: C.
$\frac{-9}{2}$
Let $\alpha$ and $\beta$ be the zeros of the $2 x^2+5 x-9$
Then, we have $\alpha\beta=\frac{\text{c}}{\text{a}}=\frac{-9}{2}$
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MCQ 111 Mark
What should be subtracted to the polynomial $x^2-16 x+30,$ so that $15$ is the zero of the resulting polynomial ?
  • A
    $30$
  • B
    $14$
  • ✓
    $15$
  • D
    $16$
Answer
Correct option: C.
$15$
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 $f(x) = x^2-16 x+30,$
therefore $(x - 15)$ is a factor of $f(x)$
Now, we divide $f(x) = x^2-16 x+30$ by $(x - 15)$ we get

Thus we should subtract the remainder $15$ from $x^2-16 x+30$,
Hence, the correct choice is $(c)$
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MCQ 121 Mark
Choose the correct answer from the given four options in the following questions : The number of polynomials having zeroes as $-2$ and $5$ is :
  • A
    $1.$
  • B
    $2.$
  • C
    $3.$
  • ✓
    More than $3.$
Answer
Correct option: D.
More than $3.$
Let $p(x) = ax^2+ bx + c$ be the required polynimial whose zeroes are $-2$ and $5$.
$\therefore$ Sum of zeroes $=\frac{-\text{b}}{\text{a}}$
$\Rightarrow\ \frac{-\text{b}}{\text{c}}=-2+5=\frac{3}{1}=\frac{-(-3)}{1}\ .....(\text{i})$
and Procudt of zeroes $=\frac{\text{c}}{\text{a}}$
$\Rightarrow\ \frac{\text{c}}{\text{a}}=-2\times5=\frac{-10}{1}\ .....(\text{ii})$
From Eqs. $(i)$ and $(ii)$
$a = 1, b = -3$ and $c = -10$
$\therefore p(x) = ax^2+ bx + c $
$= 1.x^2- 3x - 10$
$= x^2- 3x - 10$
But we know that. if we multiply/divide any polynomial by any arbitraru constant.
Then, the zeroes of polynomial never change.
$\therefore p(x) = kx^2- 3kx - 10k \ [$where, $k$ is a real number$]$
$\therefore\ \text{p}(\text{x})=\frac{\text{x}^2}{\text{k}}-\frac{3}{\text{k}}\text{x}-\frac{10}{\text{k}}, [$where $k$ is a nonzero real number$].$
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MCQ 131 Mark
If the graph of a polynomial intersects the $x-$ axis at three points, then the number of zeroes $=$
  • A
    At most three
  • B
    At least three
  • ✓
    $3$
  • D
    $0$
Answer
Correct option: C.
$3$
If the graph of a polynomial intersects the $x-$ axis at three points, then the number of zeroes are $3$ because the $x-$ axis is intersected three times by the three coordinates.
So, number of zeroes of the polynomial $=$ number of the coordinates of the points $($where its graph intersects the $x-$ axis.$)$
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MCQ 141 Mark
If $\alpha$ and $\beta$ are the zeroes of a quadratic polynomial $a x^2+b x+c$, then $\alpha +\beta = $
  • A
    $\frac{\text{c}}{\text{a}}$
  • B
    $\frac{\text{b}}{\text{a}}$
  • ✓
    $\frac{-\text{b}}{\text{a}}$
  • D
    $\frac{-\text{c}}{\text{a}}$
Answer
Correct option: C.
$\frac{-\text{b}}{\text{a}}$
If $\alpha$ and $\beta$ are the zeroes of a quadratic polynomial $a x^2+b x+c$,
$\because$ Sum of the zeroes of a quadratic polynomial $a x^2+b x+c$
$=\frac{\text{-(Coefficient of x)}}{\text{Coefficient of x}^{2}} $ then $ \alpha+\beta = \frac{\text{-b}}{\text{a}}$
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MCQ 151 Mark
If one of the zeroes of a quadratic polynomial of the form $x^2+a x+b$ is the negative of the other, then it :
  • ✓
    Has no linear term and constant term is negative.
  • B
    Has no linear term and the constant term is position.
  • C
    Can have a linear term but the constant term is negative.
  • D
    Can have a linear term but the constant term is positive.
Answer
Correct option: A.
Has no linear term and constant term is negative.
Let the quadratic polynomial be $f(x)=x^2+a x+b$
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 $(a)$
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MCQ 161 Mark
The zeros of the polynomial $\text{x}^2+\frac{1}{6}\text{x}-2$ are :
  • A
    $-3,\ 4$
  • ✓
    $\frac{-3}{2},\ \frac{4}{3}$
  • C
    $\frac{-4}{3},\ \frac{3}{2}$
  • D
    None of these.
Answer
Correct option: B.
$\frac{-3}{2},\ \frac{4}{3}$
$\text{f}(\text{x})=\text{x}^2+\frac{1}{6}\text{x}-2$
Now $, f(x) = 0$
$\Rightarrow\text{x}^2+\frac{1}{6}\text{x}-2=0$
$\Rightarrow 6 x^2+x-12=0$
$\Rightarrow 6 x^2+9 x-8 x-12=0$
$\Rightarrow 3 x(2 x+3)-4(2 x+3)=0$
$\Rightarrow(2 x+3)(3 x-4)=0$
$\Rightarrow 2 x+3=0 $ or $ 3 x-4=0$
$\Rightarrow\text{x}=-\frac{3}{2}$ or $\text{x}=\frac{4}{3}$
So, the zeros of given polynomial are $-\frac{3}{2}$ and $-\frac{4}{3}$
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MCQ 171 Mark
if the sum of the zeroes of the cubic polynomial $4 x^3-k x^2-8 x-12$ is $\frac{-3}{4}$ then the value of $'k\ ' $is :
  • A
    $3$
  • B
    $\frac{-1}{3}$
  • C
    $\frac{1}{3}$
  • ✓
    $-3$
Answer
Correct option: D.
$-3$
Let $\alpha, \beta, \gamma$ are the zeroes of the given polynomial.
we know that, $\alpha +\beta +\gamma = \frac{-3}{4}$
$\Rightarrow \frac{-\text{b}}{\text{a}} = \frac{-3}{4}$
$\Rightarrow\frac{\text{k}}{4} = \frac{3}{-4}$
$\Rightarrow\text{k} = {-3}$
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MCQ 181 Mark
Choose the correct answer from the given four options in the following questions : If the zeroes of the quadratic polynomial $ax^2+ bx + c, c \neq 0$ are equal, then :
  • ✓
    $c$ and $a$ have opposite signs.
  • B
    $c$ and $b$ have opposite signs.
  • C
    $c$ and $a$ have the same sign.
  • D
    $c$ and $b$ have the same sign.
Answer
Correct option: A.
$c$ and $a$ have opposite signs.
The zeroes of the given quadratic polynomial $ax^2+ bx + c, c \neq 0$ are equal, if coefficient of $x^2$ and constant term have the same sign
i.e., $c$ and $a$ have the same sign.
While $b$ i.e., coefficient of $x$ can be positive/negative but not zero.
e.g., $ (i) x^2+ 4x + 4 = 0 $
$(ii) x^2- 4x + 4 = 0$
$\Rightarrow (x + 2)^2= 0 $
$\Rightarrow (x - 2)^2= 0$
$\Rightarrow x = -2, -2 $
$\Rightarrow x = 2, 2$
Alternate Answer
Given that, the zeroes of the quadratic polynomial $ax^2+ bx + c$ where $c \neq 0,$ are equal
i.e., discriminant $(D) = 0$
$\Rightarrow b^2- 4ac = 0$
$\Rightarrow b^2+ 4ac$
$\Rightarrow\ \text{ac}=\frac{\text{b}^2}{4}$
$\Rightarrow\ \text{ac}>0$
Which is olny possible when $a$ and $c$ have the same signs.
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MCQ 191 Mark
If a polynomial of degree five is divided by a quadratic polynomial, then the degree of the quotient polynomial is :
  • A
    $4$
  • B
    $2$
  • ✓
    $3$
  • D
    $1$
Answer
Correct option: C.
$3$
If a polynomial of degree five is divided by a quadratic polynomial, then the degree of the quotient polynomial is three.
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MCQ 201 Mark
A polynomial of degree $..........$ is called a quadratic polynomial :
  • A
    $1$
  • B
    $3$
  • C
    $0$
  • ✓
    $2$
Answer
Correct option: D.
$2$
A polynomial of degree two is called a quadratic polynomial.
An equation involving a quadratic polynomial is called a quadratic equation.
A quadratic equation is an equation that can be written in the form $a x^2+b x+c=0$, where $\text{a}\neq{0}.$
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MCQ 211 Mark
The product of the zeroes of $x^3+4 x^2+x-6$ is :
  • A
    $-6$
  • B
    $4$
  • C
    $-4$
  • ✓
    $6$
Answer
Correct option: D.
$6$
Given $\alpha,\beta, \gamma$ be the zeroes of the polynomial $ f(x) = x^3+4 x^2+x-6$
Product of the zeroes $= \frac{\text{Constant term}}{\text{Coefficient of }{\text{x}^{3}}} = \frac{-(6)}{1} = {6}$
The value of product of the zeroes is $6$.
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MCQ 221 Mark
The zero of the polynomial $p(x) = ax + b$ is :
  • A
    $\frac{\text{b}}{\text{a}}$
  • ✓
    $\frac{-\text{b}}{\text{a}}$
  • C
    $\frac{\text{a}}{\text{b}}$
  • D
    $\frac{-\text{a}}{\text{b}}$
Answer
Correct option: B.
$\frac{-\text{b}}{\text{a}}$
Given polynomial $p(x) = ax + b$
put $p(x) = 0$
$ax + b = 0$
$ax = -b$
$\text{x} = \frac{-\text{b}}{\text{a}}$
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MCQ 231 Mark
If one zero of the polynomial $p(x)=(k+4) x^2+13 x+3 k$ is reciprocal of the other, then the value of $k$ is :
  • A
    $5$
  • ✓
    $2$
  • C
    $3$
  • D
    $4$
Answer
Correct option: B.
$2$
$2$
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MCQ 241 Mark
If $\alpha,\ \beta$ are the zeros of $kx^2- 2x + 3k$ such that $\alpha+\beta=\alpha\beta,$ then $ k =$ ?
  • A
    $\frac{1}{3}$
  • B
    $\frac{-1}{3}$
  • ✓
    $\frac{2}{3}$
  • D
    $\frac{-2}{3}$
Answer
Correct option: C.
$\frac{2}{3}$
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 $(ii)]$
$\Rightarrow\text{k}=\frac{2}{3}$
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MCQ 251 Mark
A polynomial whose sum and product of zeroes are $-4$ and $3$ is :
  • A
    $x^2- 4x + 3$
  • B
    $x^2- 4x - 3$
  • C
    None of these
  • ✓
    $x^2+ 4x + 3$
Answer
Correct option: D.
$x^2+ 4x + 3$
$x^2- \ ($Sum the Zeroes $) \ x \ +\ ($Product of Zeroes$)$
$x^2- (-4) x + 3$
$= x^2+ 4x + 3$
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MCQ 261 Mark
If $\alpha,\beta$ are the zeros of the polynomial $f(x)=a x^2+b x+c$, then $\frac{1}{\text{a}^2}+\frac{1}{\beta^2}=$
  • A
    $\frac{\text{b}^2-2\text{ac}}{\text{a}^2}$
  • ✓
    $\frac{\text{b}^2-2\text{ac}}{\text{c}^2}$
  • C
    $\frac{\text{b}^2+2\text{ac}}{\text{a}^2}$
  • D
    $\frac{\text{b}^2+2\text{ac}}{\text{c}^2}$
Answer
Correct option: B.
$\frac{\text{b}^2-2\text{ac}}{\text{c}^2}$
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 $f(x)=a x^3+b x^2+c$
$\alpha+\beta=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2} =\frac{-(\text{b})}{\text{a}}$
$\alpha\cdot\beta=\frac{\text{Constant term}}{\text{Coefficient of x}^2} =\frac{\text{c}}{\text{a}}$
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 $(b)$
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MCQ 271 Mark
If $2, -7$ and $-14$ are the sum, sum of the product of its zeroes taken two at a time and the product of its zeroes of a cubic polynomial, then the cubic polynomial is :
  • A
    $ x^3+2 x^2+7 x+14 $
  • ✓
    $ x^3-2 x^2-7 x+14 $
  • C
    $ x^3-2 x^2+7 x+14 $
  • D
    $ x^3-2 x^2-7 x-14 $
Answer
Correct option: B.
$ x^3-2 x^2-7 x+14 $
Let $\alpha,\beta,\gamma$ are the zeroes of the given polynomial.
$\alpha+\beta+\text{y} = {2}$ and $\alpha\beta+\beta\gamma+\text{y}\alpha = -7$ and $\alpha \beta\gamma = -{14}$
required polynomials $= [\text{x}^{3}-(\alpha+\beta+\gamma)\text{ x}^{2}+(\alpha\beta+\beta\gamma+\gamma\alpha)\text{ x}-\alpha\beta\gamma]$
$= [x^3- 2x^2+ (-7)x - (-14)]$
$= [x^3- 2x^2- 7x + 14]$
Required polynomial is $ x^3-2 x^2-7 x+14 $
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MCQ 281 Mark
If $x - 2$ is a factor of the polynomial $3 x^3-7 x^2+k x-16,$ then the value of $k$ is :
  • ✓
    $10$
  • B
    $-2$
  • C
    $2$
  • D
    $-10$
Answer
Correct option: A.
$10$
If the polynomial $3 x^3-7 x^2+k x-16$ is exactly divisible by $x-2$, then $p(2)=0$
$\Rightarrow 3(2)^3-7(2)^2+\mathrm{k} \times 2-16$
$\Rightarrow 24-28+2 \mathrm{k}-16=0$
$\Rightarrow-20+2 \mathrm{k}=0$
$\Rightarrow \mathrm{k}=10$
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MCQ 291 Mark
If one of the zeroes of the cubic polynomial $x^3-7 x+6$ is $2,$ then the product of the other two zeroes is :
  • A
    $3$
  • B
    $2$
  • C
    $-2$
  • ✓
    $-3$
Answer
Correct option: D.
$-3$
Let $\alpha, \beta,\gamma$ are the zeroes of the given polynomial.
Given: $\alpha={2}$
Since $\alpha\beta\gamma = \frac{\text{Constant term C}}{\text{Coefficient of x}^{3}} = \frac{\text{- c}}{\text{a}}$
$\Rightarrow 2\times\beta\gamma = \frac{-6}{1}$
$\Rightarrow\beta\gamma = \frac{-6}{2} = {-3}$
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MCQ 301 Mark
The zeroes of a polynomial $x^2+ 4x + 4$ are :
  • A
    Both positive
  • ✓
    Both equal
  • C
    One positive and one negative
  • D
    Both unequal
Answer
Correct option: B.
Both equal
$x^2+ 4x + 4$
$= x^2+ 2x + 2x + 4 = 0$
$= x(x + 2) + 2(x + 2) = 0$
$= (x + 2) (x + 2) = 0$
$\therefore x + 2 = 0$ or $x + 2 = 0$
$\Rightarrow x = -2$ or $x = -2$
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MCQ 311 Mark
If one of the zeroes of the quadratic polynomial $(k - 1)x^2+ kx + 1 is -3$, then the value of $k$ is :
  • ✓
    $\frac{4}{3}$
  • B
    $\frac{-4}{3}$
  • C
    $\frac{2}{3}$
  • D
    $\frac{-2}{3}$
Answer
Correct option: A.
$\frac{4}{3}$
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 $(a)$
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MCQ 321 Mark
The zeros of the polynomial $x^2- 2x - 3$ are :
  • A
    $-3, 1$
  • B
    $-3, -1$
  • ✓
    $3, -1$
  • D
    $3, 1$
Answer
Correct option: C.
$3, -1$
$f(x) = x^2- 2x - 3$
$= x^2- 3x + x - 3$
$= x(x - 3) + 1(x - 3)$
$= (x - 3)(x + 1)$
$\therefore f(x) = 0$
$(x - 3)(x + 1) = 0$
$x - 3 = 0$ or $x + 1 = 0$
$x = 3$ or $x = - 1$
So, the zeros of given polynomial are $3$ and $-1$
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MCQ 331 Mark
Given that of the zeroes of the cubic polynomial $a x^3+b x^2+c x+d$ are $0,$ the third zero is :
  • ✓
    $\frac{-\text{b}}{\text{a}}$
  • B
    $\frac{\text{b}}{\text{a}}$
  • C
    $\frac{\text{c}}{\text{a}}$
  • D
    $\frac{-\text{d}}{\text{a}}$
Answer
Correct option: A.
$\frac{-\text{b}}{\text{a}}$
Let the polynomial be $f(x) = a x^3+b x^2+c x+d$
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 $(a)$
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MCQ 341 Mark
The zeroes of the quadratic polynomial $x^2+99 x+127$ are :
  • A
    Both positive.
  • ✓
    Both negative.
  • C
    Both equal.
  • D
    One positive and one negative.
Answer
Correct option: B.
Both negative.
Let $ f(x) = x^2+99 x+127$
Product of the zeroes of $f(x) = 127 \times 1 = 127 \ [$Product of zeroes $= \frac{c}{a}$ when $f(x)=a x^2+b x+c]$
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 $(b).$
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MCQ 351 Mark
If the diagram in Fig. shows the graph of the polynomial $f(x) = ax^2+ bx + c,$ then :
  • ✓
    $a > 0, b < 0$ and $c > 0$
  • B
    $a < 0, b < 0$ and $c < 0$
  • C
    $a < 0, b > 0$ and $c > 0$
  • D
    $a < 0, b > 0$ and $c < 0$
Answer
Correct option: A.
$a > 0, b < 0$ and $c > 0$
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 $(a)$
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MCQ 361 Mark
The zeros of the quadratic polynomial $x^2+88 x+125$ are :
  • A
    Both positive.
  • ✓
    Both negative.
  • C
    One positive and one negative.
  • D
    Both equal.
Answer
Correct option: B.
Both negative.
Let $\alpha$ and $\beta$ be the zeros of the $x^2+88 x+125$
Then, we have
$\alpha+\beta=-\frac{\text{b}}{\text{a}}=-88$ and $\alpha\beta=\frac{\text{c}}{\text{a}}=125$
Now, this is applicable only if $\alpha$ and $\beta$ are both negative.
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MCQ 371 Mark
If the zeroes of the quadratic polynomial $x^2+ (a + 1) x + b$ are $2$ and $-3,$ then :
  • A
    $a = 2, b = –6$
  • B
    $a = -7, b = -1$
  • ✓
    $a = 0, b = -6$
  • D
    $a = 5, b = -1$
Answer
Correct option: C.
$a = 0, b = -6$
Zeroes of a polynomial are the values of $x$ at which the polynomial is equal to zero.
$2$ and $-3$ are the zeroes of the polynomial $p(x) = x^2+ (a + 1) x + b$
i.e. $p(2) = 0$ and $p(-3) = 0$
$p(2) = (2)^2+ (a + 1) (2) + b = 0$
$\Rightarrow 4 + 2a + 2 + b = 0$
$\Rightarrow 6 + 2a + b = 0 ...(1)$
$P(-3) = (-3)^2+ (a + 1) (-3) + b = 0$
$\Rightarrow 9 - 3a - 3 + b = 0$
$\Rightarrow 6 - 3a + b = 0 ...(2)$
Equating $(1) $ and $(2),$ as both the equations are equal to zero.
$\therefore 6 + 2a + b = 6 - 3a + b$
$\Rightarrow 5a = 0$
$\Rightarrow a = 0$
Putting the value of $'a\ ' $ in $(1)$
$6 + 2 (0) + b = 0$
$\Rightarrow b = -6$
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MCQ 381 Mark
Which of the following is not the graph of a quadratic polynomial?
  • A
  • ✓
  • C
  • D
Answer
Correct option: B.
For a quadratic polynomial, $a x^2+b x+c$, the zeros are precisely the $x-$coordinates of the points where the graph representing $y=a x^2+b x+c$ intersects the $x$-axis.
The graph has one of the two shapes either open upwards like $u ($parabolic shape$)$ or open downwards like $\cap ($parabolic shape$)$ depending on whether $a > 0$ or $a < 0$.
Three cases are thus possible:
Graph cuts $x-$axis at two distinct points $($two zeroes$)$
Graph cuts the $x-$axis at exactly one point $($one zero$)$
. The graph is either completely above the $x -$axis or completely below the $x -$axis $($no zeroes$)$
. The graph is cutting the $x -$axis at three distinct points and it is not a parabola opening either upwards or downwards.
So, option $(d)$ does not represent the graph of a quadratic polynomial.
Hence, the correct answer is option $(b)$
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MCQ 391 Mark
If two of the zeroes of a cubic polynomial $a x^3+b x^2+c x+d$ are zero, then the third zero is :
  • A
    $\frac{\text{d}}{\text{a}}$
  • B
    $\frac{\text{c}}{\text{a}}$
  • ✓
    $\frac{\text{-b}}{\text{a}}$
  • D
    $\frac{\text{b}}{\text{a}}$
Answer
Correct option: C.
$\frac{\text{-b}}{\text{a}}$
Given : $\alpha=0, \beta=0 $ and $\text{ y}=?$
Since $ \alpha+\beta+\text{y} =\frac{\text{-b}}{\text{a}}$
$\therefore 0+0+\text{y} = \frac{\text{-b}}{\text{a}}$
$\Rightarrow\text{y} = \frac{\text{-b}}{\text{a}}$
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MCQ 401 Mark
The number polynomials having zeroes as $-2$ and $5$ is :
  • A
    $2$
  • ✓
    More than $3$
  • C
    $3$
  • D
    $1$
Answer
Correct option: B.
More than $3$
The number polynomials having zeroes as $-2$ and $5$ is more than $3$.
If $'S\ '$ is the sum and $'P\ ' $ is the product of the zeroes then the corresponding family of quadratic polynomial is given by $p(x)=k\left(x^2-S x+P\right)$ where $k$ is any real number.
$\therefore$ putting different values of $k,$ we can make more than $3$ numbers of polynomials.
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MCQ 411 Mark
The degree of a biquadratic polynomial is :
  • A
    $2$
  • ✓
    $4$
  • C
    $3$
  • D
    $1$
Answer
Correct option: B.
$4$
The biquadratic polynomial is a polynomial of the fourth degree.
Biquadratic polynomial $= a\left(x^2\right)^2+b(x)^2+c$
$=a x^4+b x^2+c$.
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MCQ 421 Mark
Degree of the polynomial $2 x^4+3 x^3-5 x^2+9 x+1$ is :
  • A
    $3$
  • B
    $1$
  • C
    $2$
  • ✓
    $4$
Answer
Correct option: D.
$4$
The highest power of the variable is $4$.
So, the degree of the polynomial is $4.$
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MCQ 431 Mark
The zeros of the polynomial $x^2-2 x-3$ are :
  • ✓
    $3, -1$
  • B
    $3, 1$
  • C
    $-3, 1$
  • D
    $-3, -1$
Answer
Correct option: A.
$3, -1$
$x^2-2 x-3$
$= x(x - 3) + (x - 3)$
$\therefore (x - 3) (x + 1) = 0$
$\Rightarrow x = 3$ or $x = -1$
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MCQ 441 Mark
If one root of the polynomial $f(x)=5 x^2+13 x+k$ is reciprocal of the other, then the value of $k$ is :
  • A
    $0$
  • ✓
    $5$
  • C
    $\frac{1}{5}$
  • D
    $6$
Answer
Correct option: B.
$5$
If one zero of the polynomial $f(x)=5 x^2+13 x+k$ 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 $(b)$
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MCQ 451 Mark
Figure show the graph of the polynomial $f(x) = ax^2+ bx + c$ for which :
  • A
    $a < 0, b < 0$ and $c < 0$
  • ✓
    $a > 0, b < 0$ and $c > 0$
  • C
    $a < 0, b > 0$ and $c > 0$
  • D
    $a > 0, b > 0$ and $c < 0$
Answer
Correct option: B.
$a > 0, b < 0$ and $c > 0$
Clearly, $f(x) = ax^2+ bx + c$ represent a parabola opening upwards.
$\therefore a > 0 $ The vertex of the parabola is in the fourth quadrant,
$\therefore b < 0 y = ax^2+ bx + c$ 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 $OY$.
$\Rightarrow c > 0$
$a > 0, b < 0 $ and $c > 0$
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MCQ 461 Mark
Which of the following is a polynomial ?
  • A
    $\text{x}\frac{3}{2}- \text{x}+\text{x}\frac{1}{2}+{1}$
  • B
    $\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$
  • ✓
    $\sqrt{2\text{x}}^{2}-{3}\sqrt{3}\text{x}+\sqrt{6}$
  • D
    $\text{x}^{2}-{5}\text{x}+{4}\sqrt{\text{x}}+{3}$
Answer
Correct option: C.
$\sqrt{2\text{x}}^{2}-{3}\sqrt{3}\text{x}+\sqrt{6}$
Clearly, $\sqrt{2\text{x}}^{2}-{3}\sqrt{3}\text{x}+\sqrt{6}$ is a polynomial.
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MCQ 471 Mark
The degree of a constant polynomial is :
  • A
    $2$
  • B
    $1$
  • C
    $3$
  • ✓
    $0$
Answer
Correct option: D.
$0$
The degree of a polynomial is the highest degree of its individual terms with non zero coefficients.
Since in a constant polynomial, the degree of its monomials is zero.
The degree of polynomial is zero.
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MCQ 481 Mark
If $\alpha$ and $\beta$ are zeros of $x^2+5 x+8,$ then the value of $(\alpha+\beta)$ is :
  • A
    $8$
  • B
    $5$
  • ✓
    $-5$
  • D
    $-8$
Answer
Correct option: C.
$-5$
$\text{x}^{2}+{5}\text{x}+{8}$
$\alpha+\beta=\frac{\text{Coefficient of x}}{\text{Coefficient of }{\text{x}}^{2}}$
$=\frac{-5}{1} = {-5}$
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MCQ 491 Mark
If‘ $\alpha$ and $\beta$ are the zeroes of the polynomial $3 x^2+11 x-4$, then the value of $\alpha^{2}+\beta^{2}$ is :
  • ✓
    $\frac{145}{9}$
  • B
    $\frac{150}{9}$
  • C
    $\frac{152}{9}$
  • D
    $\frac{144}{9}$
Answer
Correct option: A.
$\frac{145}{9}$
Here $a = 3, b = 11, c = -4$
Since $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2} - {2}\alpha\beta$
$=(\frac{\text{b}}{\text{a}})^{2} - {2}\times\frac{\text{c}}{\text{a}} = \frac{\text{b}^{2}}{\text{a}^{2}} - \frac{\text{2c}}{\text{a}} = \frac{\text{b}^{2}-\text{2ac}}{\text{a}^{2}}$
Putting the values of $a, b$ and $c,$ we get
$=\frac{(11)^{2}-2\times3\times(-4)}{(3)^{2}} $
$=\frac{121+24}{9}$
$=\frac{145}{9}$
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MCQ 501 Mark
The sum of two zeroes of the polynomial $f(x)=2 x^2+(p+3) x+5$ is zero, then the value of $p$ is :
  • A
    $4$
  • B
    $4$
  • C
    $3$
  • ✓
    $-3$
Answer
Correct option: D.
$-3$
Let zeroes of the given polynomial be $\alpha$ and $\beta$.
According to the question, $\alpha +\beta ={0}$
$\frac{-\text{b}}{\text{a}} = {0}$
$\frac{-(\text{p}+{3})}{2} = {0}$
$\Rightarrow -(p + 3) ($by cross multiplication$)$
$\Rightarrow P = -3$
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M.C.Q (1 Marks) - MATHS STD 10 Questions - Vidyadip