MCQ 1011 Mark
If the sum of the zeros of the polynomial $f(x)=2 x^3-3 k x^2+4 x-5$ is $6,$ then the value of $k$ is :
AnswerLet $\alpha,\beta$ be the zeros of the polynomial $f(x)=2 x^3-3 k x^2+4 x-5$ and we are given that
Then,
$\alpha+\beta+\gamma=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}$
It is given that
$\alpha+\beta+\gamma=6$
Substituting $\alpha+\beta+\gamma=\frac{3\text{k}}{2},$ we get
$\frac{+3\text{k}}{2}=6$
$+3\text{k}=6\times2$
$+3\text{k}=12$
$\text{k}=\frac{12}{+3}$
$\text{k}=+4$
The value of $k$ is $4$
Hence, the correct alternative is $(b)$
View full question & answer→MCQ 1021 Mark
The graph of a cubic polynomial $x^3-4 x$ meets the $x-$ axis at $(-2, 0), (0, 0) $ and $(2, 0),$ then the zeroes of the polynomial are :
- A
$2, 0$ and $0$
- B
$0, 0 $ and $2$
- ✓
$-2, 0$ and $2$
- D
AnswerCorrect option: C. $-2, 0$ and $2$
In each point $(-2, 0), (0, 0)$ and $(2, 0)$ all in the $x-$ axis and each point intersect the graph of cubic polynomials.
So, the zeroes are $-2, 0$ and $2\ ($which is coordinates of the $x-$ axis$)$
View full question & answer→MCQ 1031 Mark
The number of zeroes of a cubic polynomial is :
- ✓
At most $3$
- B
At least $3$
- C
$2$
- D
$3$
AnswerCorrect option: A. At most $3$
The number of zeroes of a cubic polynomial is at most $3$ because the highest power of the variable in cubic polynomial is $3,$
i.e. $a x^3+b x^2+c x+d$.
View full question & answer→MCQ 1041 Mark
The polynomial which when divided by $-x^2+ x - 1$ gives a quotient $x - 2$ and remainder $3,$ is :
- ✓
$ -x^3+3 x^2-3 x+5 $
- B
$ x^3-3 x^2-3 x+5 $
- C
$ x^3-3 x^2-3 x-5 $
- D
$ x^3-3 x^2+3 x-5 $
AnswerCorrect option: A. $ -x^3+3 x^2-3 x+5 $
We know that
$f(x) = g(x), q(x) + r(x)$
$ =\left(-x^2+x-1\right)(x-2)+3 $
$ =-x^3+x^2-x+2 x^2-2 x+2+3$
$ =-x^3+x^2+2 x^2-x-2 x+2+3 $
$= -x^3+3 x^2-3 x+5 $
$\therefore$ The polynomial which when divided by $-x^2+ x - 1$
gives a quotient $x - 2$ and remainder $3,$ is $ -x^3+3 x^2-3 x+5 .$
View full question & answer→MCQ 1051 Mark
Fig. 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 $c < 0$
AnswerCorrect option: B. $a < 0, b < 0$ and $c > 0$
Clearly, $f(x) = ax^2+ bx + c$ represent a parabola opening downwards.
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 $P$ are $(0, c).$ Clearly, $P$ lies on $OY$.
Therefore $c > 0$
The vertex $\Big(\frac{-\text{b}}{2\text{a}},\frac{-\text{D}}{4\text{a}}\Big)$ of the parabola is in the second quadrant.
Therefore $\frac{-\text{b}}{2\text{a}}<0,\text{b}<0$
Therefore $a < 0, b < 0,$ and $c > 0$
Hence, the correct choice is $(b)$ View full question & answer→MCQ 1061 Mark
Given that one of the zeroes of the cubic polynomial $ax^3+ bx^2+ cx + d$ is zero, the product of other two zeroes is :
- A
$\frac{-\text{c}}{\text{a}}$
- ✓
$\frac{\text{c}}{\text{a}}$
- C
$0$
- D
$\frac{-\text{b}}{\text{a}}$
AnswerCorrect option: B. $\frac{\text{c}}{\text{a}}$
Let $p(x)=a x^3+b x^2+c x+d$
Now $0$ is the zero of the polynomial.
So $,\mathrm{p}(0)=0 $
$ \Rightarrow \mathrm{a}(0)^3+\mathrm{b}(0)^2+\mathrm{c}(0)+\mathrm{d}=0 $
$ \Rightarrow \mathrm{d}=0$
So,
$p(x)=a x^3+b x^2+c x=x\left(a x^2+b x+c\right)$
Putting $p(x)=0$, we get
$x=0 $ or $ a x^2+b x+c=0 ......(1)$
Let $\alpha,\beta$ be the other zeroes of $ax^2+ bx + c = 0$
So, $\alpha\beta=\frac{\text{c}}{\text{a}}$
Hence, the correct answer is option $(b)$
View full question & answer→MCQ 1071 Mark
If $x - 2$ is a factor of $x^3+ ax^2+ bx + 16$ and $a - b = 6$ then the values of $'a\ '$ and $'b\ '$ are :
- ✓
$a = -2$ and $b = -8$
- B
$a = 2$ and $b = 8$
- C
$a = 2$ and $b = -8$
- D
$a = -2$ and $b = 8$
AnswerCorrect option: A. $a = -2$ and $b = -8$
According to the question,
$ p(x)=x^3+a x^2+b x+16=0 \ p(x)$
$ \Rightarrow P(2)=(2)^3+a(2)^2+b \times 2+16=0 $
$\Rightarrow 8 + 4a + 2b + 16 = 0$
$\Rightarrow 4a + 2b = -24$
$\Rightarrow 2a + b = -12$
Also given $a - b = 6$
On solving $2a + b = -12$ and $a - b = 6,$
we have $a = -2$ and $b = -8$
View full question & answer→MCQ 1081 Mark
If a real number $\alpha , \alpha$ is a zero of a polynomial, then $........$ is a factor of $f(x)$ :
AnswerCorrect option: C. $\text{x} - \alpha$
If a real number $\alpha , \alpha$ is a zero of a polynomial, then $\text{x}-\alpha$ is a factor of that polynomial.
$($By Factor Theorem$)$
View full question & answer→MCQ 1091 Mark
If two of the zeros of the cubic polynomial $a x^3+b x^2+c x+d$ are each equal to zero, then the third zero is :
- A
$\frac{-\text{d}}{\text{a}}$
- B
$\frac{\text{c}}{\text{a}}$
- ✓
$\frac{-\text{b}}{\text{a}}$
- D
$\frac{\text{b}}{\text{a}}$
AnswerCorrect option: C. $\frac{-\text{b}}{\text{a}}$
Let $\alpha=0,\beta=0$ and $\gamma$ be the zeros of the polynomial
$f(x)=a x^3+b x^2+c x+d$
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 $(c)$
View full question & answer→MCQ 1101 Mark
If $\alpha,\ \beta,\ \gamma$ are the zeros of the polynomial $x^3-6 x^2-x+30$, then the value of $(\alpha\beta+\beta\gamma+\gamma\alpha)$ is :
AnswerHere, $p(x)=x^3-6 x^2-x+30$
Comparing the given polynomial with
$\text{x}^ 3-(\alpha+\beta+\gamma)\text{x}^2+(\alpha\beta+\beta\gamma+\gamma\alpha)\text{x}-\alpha\beta\gamma,$
we get : $(\alpha\beta+\beta\gamma+\gamma\alpha)=-1$
View full question & answer→MCQ 1111 Mark
A quadratic polynomial whose zeros are $5$ and $-3,$ is :
- A
$ x^2+2 x-15 $
- B
$ x^2-2 x+15 $
- ✓
$ x^2-2 x-15 $
- D
AnswerCorrect option: C. $ x^2-2 x-15 $
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
$\text{p}(\text{x})=\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$\Rightarrow\text{p}(\text{x})=\text{x}^2-(2)\text{x}+(-15)$
$=\text{x}^2-\text{2x}-15$
View full question & answer→MCQ 1121 Mark
If $f(x)=a x^2+b x+c$ has no real zeros and $a + b + c < 0,$ then :
- ✓
$C < 0$
- B
- C
$C = 0$
- D
$C > 0$
AnswerCorrect option: A. $C < 0$
We are given $a + b + c < 0$
$\Rightarrow \text{f}(1) < 0$
So, $f(x)$ must be negative for all $x$.
View full question & answer→MCQ 1131 Mark
If $\alpha, \beta$ and $' \gamma'$ are the zeroes of a cubic polynomial $a x^3+b x^2+c x+d,$ then $\alpha+\beta+\gamma$
- ✓
$\frac{-\text{b}}{\text{a}}$
- B
$\frac{\text{b}}{\text{a}}$
- C
$\frac{-\text{c}}{\text{a}}$
- D
$\frac{\text{c}}{\text{a}}$
AnswerCorrect option: A. $\frac{-\text{b}}{\text{a}}$
If $\alpha, \beta$ and $'\gamma '$ are the zeroes of a cubic $a x^3+b x^2+c x+d$
$\because$ Sum of the zeroes of a cubic polynomial $a x^3+b x^2+c x+d$
$=\frac{-\text{(Coefficient of x}^{2})}{\text{Coefficient of x}^{3}},$ Then $\alpha+\beta+\gamma = \frac{-\text{b}}{\text{a}}$
View full question & answer→MCQ 1141 Mark
The polynomial which when divided by $-x^2+ x - 1$ gives a quotient $x - 2$ and remainder $3,$ is :
- A
$ x^3-3 x^2+3 x-5 $
- B
$ -x^3-3 x^2-3 x-5 $
- ✓
$ -x^3+3 x^2-3 x+5 $
- D
$ x^3-3 x^2-3 x+5 $
AnswerCorrect option: C. $ -x^3+3 x^2-3 x+5 $
We know that
$f(x) = g(x) q(x) + r(x)$
$ =\left(-x^2+x-1\right)(x-2)+3 $
$ =-x^3+x^2-x+2 x^2-2 x+2+3 $
$ =-x^3+x^2+2 x^2-x-2 x+2+3 $
$ =-x^3+3 x^2-3 x+5 $
The polynomial which when divided by $-x^2+ x - 1$ gives a quotient $x - 2$ and remainder $3,$ is $-x^3+ 3x^2- 3x + 5$
Hence, the correct choice is $(c)$
View full question & answer→MCQ 1151 Mark
The zeroes of a polynomial $x^2+ 5x + 6$ are :
- A
- B
- ✓
- D
One positive and one negative
Answer$ x^2+5 x+6 $
$ =x^2+3 x+2 x+6 $
$= x(x + 3) + 2(x + 3)$
$= x(x + 3) + 2(x + 3)$
$= (x + 3) (x + 2)$
$= x + 3 = 0$ or $x + 2 = 0$
$\Rightarrow x = -3$ or $x = -2$
View full question & answer→MCQ 1161 Mark
If $\alpha$ and $\beta$ are the zeroes of the polynomial $2 x^2+5 x+1$, then the value of $\alpha+\beta+\alpha\beta$ is :
AnswerLet $\alpha,\beta$ are the zeroes of the given polynomial.
Since $\alpha+\beta+\alpha\beta$
$= \frac{\text{-b}}{\text{a}}+\frac{\text{c}}{\text{a}} = \frac{\text{-b+c}}{\text{a}}$
$\therefore\alpha+\beta+\alpha\beta= \frac{-5+1}{2}$
$= \frac{-4}{2}$
$=-{2}$
View full question & answer→MCQ 1171 Mark
The number of polynomials having zeroes $-2$ and $5$ is :
- A
$1$
- B
$2$
- C
$3$
- ✓
More than $3$
AnswerCorrect option: D. More than $3$
Polynomials having zeros $-2$ and $5$ will be of the form
$p(x)=a(x+2)^n(x-5)^m$
Here, $n$ and $m$ can take any value from $1, 2, 3, ...$
Thus, the number of polynomials will be more than 3.
Hence, the correct answer is option $(d)$
View full question & answer→MCQ 1181 Mark
Choose the correct answer from the given four options in the following questions : Which of the following is not the graph of a quadratic polynomial ?
AnswerFor any quadratic polynomial $a x^2+b x+c, a \neq 0$, the graph of the corresponding equation $y=a x^2+b x+c$ has one of the two shapes either open upwards like $u$ or open downwards like $\cap$ depandign on whether $a > 0 $ or $a a < 0. $
These curves are called parabolas.
So, aption $(d)$ cannot be possible. Also, the curve of a quadrativ pilynomial crosses the $x-$ axis on at most two points but in option $(d)$ the curve crosses the $x-$ axis on the three points, so it does not represent the quadratic polynomial.
View full question & answer→MCQ 1191 Mark
If $\alpha$ and $\beta$ are the zeroes of a quadratic polynomial $x^2+5 x-5$, then :
- A
$\alpha -\beta = \alpha\beta$
- ✓
$\alpha +\beta = \alpha\beta$
- C
$\alpha +\beta < \alpha\beta$
- D
$\alpha +\beta > \alpha\beta$
AnswerCorrect option: B. $\alpha +\beta = \alpha\beta$
$\alpha+\beta=\frac{\text{-b}}{\text{a}} = \frac{-5}{1}$
And $\alpha\beta = \frac{\text{c}}{\text{a}} = \frac{-5}{1}$
$\therefore \alpha+\beta = \alpha\beta$
View full question & answer→MCQ 1201 Mark
If two zeros of $x^3+x^2-5 x-5$ are $5-\sqrt{5}$ and $-5-\sqrt{5}$ then its third zero is :
AnswerLet $\alpha=\sqrt{5}$ and $\beta=-\sqrt{5}$ be the given zeroes and $y$ be third zero $x^3+x^2-5 x-5 = 0$ then by using
$\alpha+\beta+\text{y} = \frac{\text{Coefficient of }\text{x}^{2}}{\text{Coefficient of }\text{x}^{3}}$
$\alpha+\beta+\text{y}= \frac{(-1)}{1}$
$\alpha+\beta+\text{y} = -{1}$
By substituting $\alpha=\sqrt{5}$ and $\beta=-\sqrt{5}$ in $\alpha+\beta+\text{y} = -{1}$
$\sqrt{5}-\sqrt{5}+\text{y} = {-1} \text{ y}={-1}$
View full question & answer→MCQ 1211 Mark
The product of the zeros of $x^3+4 x^2+x-6$ is :
AnswerGiven $\alpha,\beta,\gamma$ be the zeros of the polynomial $f(x) =x^3+4 x^2+x-6$
Product of the zeros $=\frac{\text{Constant term}}{\text{Coefficient of x}^3}=\frac{-(-6)}{1}=6$
The value of Product of the zeros is $6$.
Hence, the correct choice is $(c)$
View full question & answer→MCQ 1221 Mark
If the zeroes of the quadratic polynomial $x^2+(a+1) x+b$ are $2$ and $-3,$ then
- A
$a = -7,b = -1$
- B
$a = 5, b = -1$
- C
$a = 2, b = -6$
- ✓
$a = 0, b = -6$
AnswerCorrect option: D. $a = 0, b = -6$
The given quadratic equation is $x^2+(a+1) x+b=0$
Since the zeroes of the given equation are $2$ and $-3$.
So,
$\alpha=2$ and $\beta=-3$
Now,
Sum of zeroes $=-\frac{\text{Coefficient of x}}{\text{Coefficient of x}^2}$
$\Rightarrow2+(-3)=-\frac{(\text{a}+1)}{1}$
$\Rightarrow-1=-\text{a}-1$
$\Rightarrow\text{a}=0$
Product of zeroes $=\frac{\text{Constant of x}}{\text{Coefficient of x}^2}$
$\Rightarrow 2\times(-3)=\frac{\text{b}}{1}$
$\Rightarrow\text{b}=-6$
So, $a = 0$ and $b = -6$
Hence, the correct answer is option $(d)$
View full question & answer→MCQ 1231 Mark
If the polynomial $3x^3- 4x^2- 17x - k$ is exactly divisible by $x - 3,$ then the value of $k$ is :
AnswerIf the polynomial $3x^3- 4x^2- 17x - k$ is exactly divisible by $x - 3,$ then
$p(3) = 0 \ ($By factor theorem$)$
$\Rightarrow 3(3)^3- 4(3)^2- 17 \times 3 - k = 0$
$\Rightarrow 81 - 36 - 51 - k = 0$
$\Rightarrow -6 - k = 0$
$\Rightarrow k = -6$
View full question & answer→MCQ 1241 Mark
If one zero of the quadratic polynomial $kx^2+ 3x + k$ is $2,$ then the value of $k$ is :
- A
$\frac{5}{6}$
- B
$\frac{-5}{6}$
- C
$\frac{6}{5}$
- ✓
$\frac{-6}{5}$
AnswerCorrect option: D. $\frac{-6}{5}$
Since $2$ is a zero of $f(x) = kx^2+ 3x + k,$ we have
$f(2) = 0$
$\Rightarrow k(2)^2+ 3(2) + k = 0$
$\Rightarrow 4k + 6 + k = 0$
$\Rightarrow 5k + 6 = 0$
$\Rightarrow 5k = -6$
$\Rightarrow\text{k}=-\frac{6}{5}$
View full question & answer→MCQ 1251 Mark
If $2$ is the zero of both the polynomials $3 x^2+m x-14$ and $2 x^3+n x^2+x-2$, then the value of $m-2 n$ is :
AnswerAccording to the question, $p(2)=3 x^2+m x-14=0$
$\Rightarrow 3(2)^2+ {m} \times 2-14=0$
$\Rightarrow 12 + 2m - 14 = 0$
$\Rightarrow m = 1$
Also $p(2)=2 x^3+n x^2+x-2=0$
$\Rightarrow 2 \times(2)^3+ {n} \times(2)^2+2-2=0$
$\Rightarrow 16 + 4n = 0$
$\Rightarrow n = -4$
$\therefore m - 2n = 1 - 2 \times (-4) $
$= 1 + 8 = 9$
View full question & answer→MCQ 1261 Mark
A quadratic polynomial whose zeros are $\frac{3}{5}$ and $\frac{-1}{2}$ is :
- A
$ 10 x^2+x+3 $
- B
$ 10 x^2+x-3 $
- C
$ 10 x^2-x+3 $
- ✓
$ 10 x^2-x-3 $
AnswerCorrect option: D. $ 10 x^2-x-3 $
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 $10\text{x}^2-\text{x}-3$
View full question & answer→MCQ 1271 Mark
If $\alpha,\beta$ are the zeros of the polynomial $f(x)=x^2-p(x+1)-c$ such that $(\alpha+1)(\beta+1)=0,$ then $c =$
AnswerSince $\alpha$ and $\beta$ are the zeros of quadratic polynomial
$\text{f(x)}=\text{x}^2-\text{p(x}+1)-\text{c}$
$\text{f(x)}=\text{x}^2-\text{px}-\text{p}-\text{c}$
$\alpha+\beta=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}$
$=-\Big(\frac{-\text{p}}{1}\Big)$
$=\text{p}$
$\alpha\beta=\frac{-\text{Constant term}}{\text{Coefficient of x}^2}$
$=\frac{-\text{p}-\text{c}}{1}$
$=-\text{p}-\text{c}$
We have
$0=(\alpha+1)(\beta+1)$
$0=\alpha\beta+(\alpha+\beta)+1$
$0=-\text{p}-\text{c}+\text{p}+1$
$0=-\text{c}+1$
$\text{c}=1$
Hence, the correct alternative is $(a)$
View full question & answer→MCQ 1281 Mark
What should be added to the polynomial $x^2-5 x+4$, so that $3$ is the zero of the resulting polynomial ?
AnswerIf $\text{x}=\alpha$ is a zero of a polynomial then $\text{x}-\alpha$ is a factor of $f(x)$
Since $3$ is the zero of the polynomial $f(x)=x^2-5 x+4$,
Therefore $x - 3$ is a factor of $f(x)$
Now, we divide $f(x)=x^2-5 x+4$ by $(x - 3)$ we get
Therefore we should add $2$ to the given polynomial
Hence, the correct choice is $(b)$ View full question & answer→MCQ 1291 Mark
A quadratic polynomial, the sum of whose zeroes is $0$ and one zero is $3,$ is :
- ✓
$ x^2-9 $
- B
$ x^2+9 $
- C
$ x^2+3 $
- D
$ x^2-3 $
AnswerCorrect option: A. $ x^2-9 $
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 $(a)$
View full question & answer→MCQ 1301 Mark
If the product of zeros of the polynomial $f(x)=a x^3-6 x^2+11 x-6$ is $4,$ then $a =$?
- A
$-\frac{3}{2}$
- B
$\frac{2}{3}$
- C
$-\frac{2}{3}$
- ✓
$\frac{3}{2}$
AnswerCorrect option: D. $\frac{3}{2}$
Since $\alpha$ and $\beta$ are the zeroes of quadratic polynomial $f(x)=a x^3-6 x^2+11 x-6$
$\alpha\beta =\frac{ \text{Constant term}}{\text{Coefficient of }{\text{x}^{2}}}$
So we have
${4} = \big(\frac{-6}{\text{a}}\big)$
${4}=\frac{6}{\text{a}}$
$\text{a} = \frac{3}{2}$
$\therefore$ value of $a$ is $\frac{3}{2}$
View full question & answer→MCQ 1311 Mark
Given that one of the zeroes of the cubic polynomial $a x^3+b x^2+c x+d$ is zero then the product of the other two zeroes is :
- A
$\frac{-\text{b}}{\text{a}}$
- B
$\frac{-\text{c}}{\text{a}}$
- ✓
$\frac{\text{c}}{\text{a}}$
- D
$\frac{\text{b}}{\text{a}}$
AnswerCorrect option: C. $\frac{\text{c}}{\text{a}}$
Let $\alpha,\beta,\gamma$ are the zeroes of the given polynomial.
given : $\alpha={0}$
To find $\beta\gamma$
Since $,\alpha\beta+\beta\gamma+\gamma\alpha=\frac{\text{c}}{\text{a}}$
$\therefore 0\times\beta+\beta\gamma+\gamma\times0 = \frac{\text{c}}{\text{a}}$
$\Rightarrow\beta \gamma = \frac{\text{c}}{\text{a}}$
View full question & answer→MCQ 1321 Mark
If $\sqrt{5}$ and $-\sqrt{5}$ are two zeroes of the polynomial $x^3+3 x^2-5 x-15$, then its third zero is :
AnswerLet $\alpha=\sqrt{5}$ and $\beta=-\sqrt{5}$ be the given zeros and $\gamma$ be the third zero of the polynomial $x^3+3 x^2-5 x-15$ 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 $(b)$
View full question & answer→MCQ 1331 Mark
The maximum number of zeroes that a polynomial of degree $3$ can have is :
AnswerThe maximum number of zeroes that a polynomial of degree $3$ can have is three because the number of zeroes of a polynomial is equals to the degree of that polynomial.
View full question & answer→MCQ 1341 Mark
The zeroes of the quadratic polynomial $x^2+ 9x + 20$ are :
- ✓
$-4$ and $-5$
- B
$4$ and $+5$
- C
$-4$ and $5$
- D
$4$ and $5$
AnswerCorrect option: A. $-4$ and $-5$
$(x^2+ 9x + 20) = 0$ Splitting the middle term, we get
$x^2+ 5x + 4x + 20 = 0$
$= (x + 5) + 4 (x + 5) = 0$
$= (x + 5) (x + 4) = 0$
$\because x + 5 = 0$ and $x + 4 = 0$
$\Rightarrow x = -5$ and $x = -4$
View full question & answer→MCQ 1351 Mark
The zeroes of a polynomial $x^2+ 5x - 24$ are :
- A
- ✓
One positive and one negative.
- C
- D
AnswerCorrect option: B. One positive and one negative.
$ x^2+5 x-24 $
$ =x^2+8 x-3 x-24 $
$= x (x + 8) - 3 (x + 8) = 0$
$= (x + 8) (x - 3) = 0$
$= x + 8 = 0$ or $x - 3 = 0$
$\Rightarrow x = -8$ or $x = 3$
View full question & answer→MCQ 1361 Mark
A quadratic polynomial whose product and sum of zeroes are $\frac{1}{3}$ and $\sqrt{2}$ respectively is :
- A
$\text{3x}^{2}+\text{x }-3\sqrt{2}\text{x}$
- B
$\text{3x}^{2}-\text{x+3 }\sqrt{2}\text{x}$
- C
$\text{3x}^{2}+3 \sqrt{ 2}\text{x}+{1}$
- ✓
$\text{3x}^{2}-3 \sqrt{ 2}\text{x}+{1}$
AnswerCorrect option: D. $\text{3x}^{2}-3 \sqrt{ 2}\text{x}+{1}$
Given : $\alpha+\beta=\frac{\sqrt{2}}{1} $
$=\frac{ (-\sqrt{2})}{1} = \frac{(-3\sqrt{2})}{3}$
And $ \alpha\beta = \frac{\text{c}}{\text{a}} = \frac{1}{3}$ on comparing,
we get $\text{a = 3, b = }-3\sqrt{2}, \text{c} = {1}$
Putting these values in the general form of a quadratic polynomial $a x^2+b x+c,$
we have ${3}\text{x}^{2} - {3}\sqrt{2}+{1}$
View full question & answer→MCQ 1371 Mark
A polynomial of degree $..........$ is called a cubic polynomial :
AnswerA polynomial of degree $3$ is called a cubic polynomial.
A univariate cubic polynomial has the form $f(x)=a^3 x^3+a^2 x^2+a_1 x+a 0$.
An equation involving a cubic polynomial is called a cubic equation.
View full question & answer→MCQ 1381 Mark
If $\alpha,\ \beta$ be the zero of the polynomial $2 x^2+5 x+k$ such that $\alpha^2+\beta^2+\gamma^2=\frac{21}{4}$ then $k =$ ?
AnswerSince $\alpha$ and $\beta$ are the zeros of $2 x^2+5 x+k,$ we have
$\alpha+\beta=-\frac{5}{2}$ and $\alpha\beta=\frac{\text{k}}{2}$
Now, $\alpha^2+\beta^2+\alpha\beta=\frac{21}{4}$
$\Rightarrow(\alpha+\beta)^2-\alpha\beta=\frac{21}{4}$
$\Rightarrow\Big(\frac{-5}{2}\Big)^2-\frac{\text{k}}{2}=\frac{21}{4}$
$\Rightarrow\frac{25}{4}-\frac{\text{k}}2{}=\frac{21}{4}$
$\Rightarrow\frac{\text{k}}2{}=\frac{25}{4}-\frac{21}{4}=1$
$\Rightarrow\text{k}=2$
View full question & answer→MCQ 1391 Mark
Choose the correct answer from the given four options in the following questions : If one of the zeroes of the cubic polynomial $x^3+ ax^2+ bx + c is –1$, then the product of the other two zeroes is :
- ✓
$b – a + 1.$
- B
$b – a – 1.$
- C
$a – b + 1.$
- D
$a – b – 1.$
AnswerCorrect option: A. $b – a + 1.$
Let $p(x) = x^3+ ax^2+ bx + c$
Let $a, p $ and $y$ be the zeroes of the given cubic polynomial $p(x)$.
$\therefore\ \alpha= -1\ \ [$given$]$
and $p(-1) = 0$
$\Rightarrow (-1)^3+ a(-1)^2+ b(-1) + c = 0$
$\Rightarrow -1 + a - b + c = 0$
$\Rightarrow c = 1 - a + b .....(i)$
We know that,
Product of all zeroes $=(-1)^3\frac{\text{Constant term}}{\text{Coefficient of x}^3}=-\frac{\text{c}}{1}$
$\alpha\beta\gamma = -\text{c}$
$\Rightarrow\ (-1)\beta\gamma= -c\ \ \big[\therefore\ \alpha=-1\big]$
$\Rightarrow\ \beta\gamma=\text{c}$
$\Rightarrow\ \beta\gamma= 1\text{ a}+\text{b}\ \ \big[$ From Eq. $(i)\big]$
Hecne, product of the other two roots is $1 - a + b$.
Alternate Answer
Since $, -1$ is one of the zeroes of the cubic polynomial $f(x) = x^2+ ax^2+ bx + c$
i.e. $, (x + 1)$ is a factor of $f(x),$
Now, using division algorithm,
$ \Rightarrow x^2+a x^2+b x+c$
$=(x+1) x\left\{x^2+(a-1) x+(a-a+1) > +(c-b+a-1)\right\}. $
$ \Rightarrow x^2+a x^2+b x+(b-a+1)$
$=(x+1)\left\{x^2+(a-1) x+(b-a+1)\right.\} $
Let $a$ and $p$ be the other two zeroes of the given polynomial, then
Product of zeroes $=(-1)\alpha.\beta=\frac{\text{-Constant term}}{\text{Coefficient of x}^3}$
$\Rightarrow\ -\alpha.\beta=\frac{-(\text{b}-\text{a}+1)}{1}$
$\Rightarrow\ \alpha\beta=-\text{a}+\text{b}+1$
Hence, the required product of other two roots is $(-a + b + 1).$ View full question & answer→MCQ 1401 Mark
If zeros of the polynomial $f(x) = x^3- 3px^2+ qx - r$ are in $A.P.,$ then :
- ✓
$ 2 p^3=p q-r $
- B
$ 2 p^3=p q+r $
- C
$ p^3=p q-r $
- D
AnswerCorrect option: A. $ 2 p^3=p q-r $
Let $a - d, a, a + d$ be the zeros of the polynomial $f(x) = x^3- 3px^2+ qx - r$ then
Sum of zeros $=\frac{-\text{Coefficient of x}^2}{\text{Coefficient of x}^3}$
$(\text{a}-\text{d})+\text{a}+(\text{a}+\text{d})=\frac{-(-3\text{p})}{1}$
$\text{a}-\text{d}+\text{a}+\text{a}+\text{d}=3\text{p}$
$3\text{a}=3\text{p}$
$\text{a}=\frac{3}{3}\text{p}$
$\text{a}=\text{p}$
Since a is a zero of the polynomial $f(x)$
Therefore,
$f(a) = 0$
$a^3-3 p a^2+q a-r=0$
Substituting a = p we get
$p^3-3 p(p)^2+q \times p-r=0 $
$ p^3-3 p^3+q p-r=0 $
$-2 p^3+q p-r=0 $
$ q p-r=2 p^3$
Hence, the correct choice is $(a)$
View full question & answer→MCQ 1411 Mark
If the polynomial $f(x)=a x^3+b x-c$ is divisible by the polynomial $g(x)=x^2+b x+c$, then $a b=$
AnswerCorrect option: C. $2b^2$
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 $c \neq 0,$ 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
$b - ac + ab^2= 0$
$b + ab^2- ac = 0$
$b(1 + ab) - ac = 0$
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 $(c)$ View full question & answer→MCQ 1421 Mark
If $f(x)=a x^2+b x+c$ has no real zeros and $a + b + c = 0,$ then :
- A
$c = 0$
- B
$c > 0$
- ✓
$c < 0$
- D
AnswerCorrect option: C. $c < 0$
If $f(x)=a x^2+b x+c$ has no real zeros and $a + b + c < 0$
then $c < 0$
Hence, the correct choice is $(c)$
View full question & answer→MCQ 1431 Mark
The sum and product of the zeroes of the polynomial $x^2-6 x+8$ are respectively :
- A
$\frac{-3}{2}$ and $-1$
- ✓
$6 $ and ${8}$
- C
$\frac{3}{2 }$ and ${1}$
- D
$\frac{+3}{2} $ and $-1$
AnswerCorrect option: B. $6 $ and ${8}$
Sum of the zeroes of the polynomial $=\frac{-\text{b}}{\text{a}} = \frac{6}{1} = {6}$
And product of the zeroes of the polynomial $= \frac{\text{c}}{\text{a}} = \frac{8}{1} = {8}$
View full question & answer→MCQ 1441 Mark
If $\alpha,\beta$ are the zeros of the polynomial $p(x)=4 x^2+3 x+7$, then $\frac{1}{\alpha}+\frac{1}{\beta}$ is equal to :
- A
$\frac{7}{3}$
- B
$\frac{-7}{3}$
- C
$\frac{3}{7}$
- ✓
$\frac{-3}{7}$
AnswerCorrect option: D. $\frac{-3}{7}$
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 $(d).$
View full question & answer→MCQ 1451 Mark
If the zeroes of a quadratic polynomial $a x^2+b c+c, c \neq 0$ are equal, then:
- A
$C$ and $a$ have opposite signs.
- B
$C$ and $b$ have opposite signs.
- ✓
$C$ and $a$ have the same sign.
- D
$C$ and $b$ have the same sign.
AnswerCorrect option: C. $C$ and $a$ have the same sign.
Let the given quadratic polynomial be $f(x)=a x^2+b x+c$
Suppose $\alpha$ and $\beta$ be the zeroes of the given polynomial.
Since $\alpha$ and $\beta$ are equal so they will have the same sign
i.e., either both are positive or both are negative.
So, $\alpha\beta>0$
But $\alpha\beta=\frac{\text{c}}{\text{a}}$
$\therefore\ \frac{\text{c}}{\text{a}}>0,$ which is possible only when both have same sign
Hence, the correct answer is option $(c)$
View full question & answer→MCQ 1461 Mark
A quadratic polynomial whose zeroes are $-3$ and $6, $ is :
- A
$\text{x}^{2}-{3}\text{x}+{18}$
- B
$\text{x}^{2}+{3}\text{x}+{18}$
- ✓
$\frac{\text{x}^{2}}{6}-\frac{\text{x}}{2}-{3}$
- D
$\text{x}^{2}+{3}\text{x}-{18}$
AnswerCorrect option: C. $\frac{\text{x}^{2}}{6}-\frac{\text{x}}{2}-{3}$
Here $\alpha+\beta = -{3}+{6} = \frac{3}{1}=\frac{-(-3)}{1} = \frac{\text{b}}{\text{a}}$
And $\alpha\beta = (-3)\times{6} = \frac{-18}{1}=\frac{\text{c}}{\text{a}}$
on comparing we get $a = 1, b = -3, c = -18$
putting these values in the general from of quadratic polynomial
$\text{ax}^{2}+\text{bx}+\text{c} = \frac{\text{x}^{2}-{3}\text{x}-{18}}{6} $
$= \frac{\text{x}^{2}}{6} -\frac{\text{x}}{2}-{3}\ [$Dividing all terms by $6]$
View full question & answer→MCQ 1471 Mark
If one of the zeros of the cubic polynomial $x^3+ ax^2+ bx + c is -1,$ then the product of the other two zeros is :
- A
$a - b - 1$
- B
$b - a - 1$
- ✓
$1 - a + b$
- D
$1 + a - b$
AnswerCorrect option: C. $1 - a + b$
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 :
$-1\times\beta\times\gamma=-\text{c}$
$\Rightarrow-\beta\gamma=-(1-\text{a}+\text{b})$
$\Rightarrow\beta\gamma=1-\text{a}+\text{b}$
View full question & answer→MCQ 1481 Mark
If $\alpha.\beta,\gamma$ are the zeros of the polynomial $f(x)=a x^3+b x^2+c x+d$, then $\alpha^2+\beta^2+\gamma^2=$
- A
$\frac{\text{b}^2-\text{ac}}{\text{a}^2}$
- B
$\frac{\text{b}^2-2\text{ac}}{\text{a}}$
- C
$\frac{\text{b}^2+2\text{ac}}{\text{b}^2}$
- ✓
$\frac{\text{b}^2-2\text{ac}}{\text{a}^2}$
AnswerCorrect option: D. $\frac{\text{b}^2-2\text{ac}}{\text{a}^2}$
We have to find the value of $\alpha^2+\beta^2+\gamma^2$
Given $\alpha,\beta,\gamma$ be the zeros of the polynomial $f(x)=a x^3+b x^2+c x+d$
We know that
$\alpha+\beta+\gamma=\frac{\text{Coefficient of x}}{\text{Coefficient of x}^3}$
$=\frac{-\text{b}}{\text{a}}$
$\alpha\beta+\beta\gamma+\gamma\alpha=\frac{-\text{Constant term}}{\text{Coefficient of x}^3} $
$=\frac{\text{c}}{\text{a}}$
Now
$\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)$
$\alpha^2+\beta^2+\gamma^2=\Big(\frac{-\text{b}}{\text{a}}\Big)^2-2\Big(\frac{\text{c}}{\text{a}}\Big)$
$\alpha^2+\beta^2+\gamma^2=\frac{\text{b}^2}{\text{a}^2}-\frac{2\text{c}\times\text{a}}{\text{a}\times\text{a}}$
$\alpha^2+\beta^2+\gamma^2=\frac{\text{b}^2}{\text{a}^2}-\frac{2\text{ca}}{\text{a}^2}$
$\alpha^2+\beta^2+\gamma^2=\frac{\text{b}^2-2\text{ac}}{\text{a}^2}$
The value of $\alpha^2+\beta^2+\gamma^2=\frac{\text{b}^2-2\text{ac}}{\text{a}^2}$
Hence, the correct choice is $(d)$
View full question & answer→MCQ 1491 Mark
If $\alpha,\beta,\gamma$ are the zeros of the polynomial $f(x)=a x^3+b x^2+c x+d$, then $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=$
- A
$-\frac{\text{b}}{\text{d}}$
- B
$\frac{\text{c}}{\text{d}}$
- ✓
$-\frac{\text{c}}{\text{d}}$
- D
$-\frac{\text{c}}{\text{a}}$
AnswerCorrect option: C. $-\frac{\text{c}}{\text{d}}$
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 $f(x)=a x^3+b x^2+c x+d$
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 $(c)$
View full question & answer→MCQ 1501 Mark
A polynomial of degree $n$ has :
- A
- B
At least $n$ zeroes
- ✓
At most $n$ zeroes
- D
$N$ zeroes
AnswerCorrect option: C. At most $n$ zeroes
A polynomial of degree $n$ has at most $n$ zeroes because the degree of a polynomial is equal to the zeroes of that polynomial only.
View full question & answer→
