MCQ 511 Mark
If a polynomial $p(x)$ is given by $p(x)=x^2-5 x+6$, then the value of $p(1)+p(4)$ is
- A$0$
- ✓4
- C2
- D-4
Answer
View full question & answer→Correct option: B.
4
B
Questions · Page 2 of 3
M.C.Q (1 Marks)
MCQ 521 Mark
If $\alpha, \beta$ are the zeroes 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$\frac{3}{7}$
- ✓$-\frac{3}{7}$
Answer
View full question & answer→Correct option: D.
$-\frac{3}{7}$
D
MCQ 531 Mark
Which of the following is a quadratic polynomial having zeroes $-\frac{2}{3}$ and $\frac{2}{3}$ ?
- A$4 x^2-9$
- B$\frac{4}{9}\left(9 x^2+4\right)$
- C$x^2+\frac{9}{4}$
- ✓$5\left(9 x^2-4\right)$
Answer
View full question & answer→Correct option: D.
$5\left(9 x^2-4\right)$
D
MCQ 541 Mark
The graph of $y=p(x)$ is given, for a polynomial $p(x)$. The number of zeroes of $p(x)$ from the graph is
- A3
- B1
- ✓2
- D$0$
Answer
View full question & answer→Correct option: C.
2
C
MCQ 551 Mark
If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2-1$, then the value of $(\alpha+\beta)$ is
- A2
- B1
- C-1
- ✓$0$
Answer
View full question & answer→Correct option: D.
$0$
D
MCQ 561 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$a=2, b=-6$
- ✓$a=0, b=-6$
Answer
View full question & answer→Correct option: D.
$a=0, b=-6$
D
MCQ 571 Mark
The number of polynomials having zeros -3 and 5 is
- A1
- B2
- C3
- ✓more than 3
Answer
View full question & answer→Correct option: D.
more than 3
D
MCQ 581 Mark
If $\alpha, \beta$ are the zeros of the polynomial $f(x)=x^2+x-1$, then $\frac{1}{\alpha}+\frac{1}{\beta}=$
- ✓1
- B-1
- C$0$
- Dnone of these
Answer
View full question & answer→Correct option: A.
1
A
MCQ 591 Mark
The graph of a polynomial $f(x)$ is shown in Fig. The number of zeroes of $f(x)$ is
- ✓3
- B2
- C1
- D4
Answer
View full question & answer→Correct option: A.
3
(A) 3
We observe that the curve $y=f(x)$ crosses $x$-axis at three distinct points. So, the polynomial $f(x)$ has three zeros.
We observe that the curve $y=f(x)$ crosses $x$-axis at three distinct points. So, the polynomial $f(x)$ has three zeros.
MCQ 601 Mark
A quadratic polynomial, the sum of whose zeros 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
View full question & answer→Correct option: A.
$x^2+5 x+6$
A
MCQ 611 Mark
If one zero of the quadratic polynomial $x^2+3 x+k$ is 2 , then the value of $k$ is
- A10
- ✓-10
- C5
- D-5
Answer
View full question & answer→Correct option: B.
-10
B
MCQ 621 Mark
The zeros of the quadratic polynomial $f(x)=x^2+99 x+127$ are
- Aboth positive
- ✓both negative
- Cone positive and one negative
- Dboth equal
Answer
View full question & answer→Correct option: B.
both negative
(B) both negative
Let $\alpha, \beta$ be the zeros of the quadratic polynomial $f(x)=x^2+99 x+127$. Then, $\alpha+\beta=-99$ and $\alpha \beta=127$
Thus, the sum of the zeros is negative and product is positive. Therefore, $\alpha$ and $\beta$ both are negative.
Let $\alpha, \beta$ be the zeros of the quadratic polynomial $f(x)=x^2+99 x+127$. Then, $\alpha+\beta=-99$ and $\alpha \beta=127$
Thus, the sum of the zeros is negative and product is positive. Therefore, $\alpha$ and $\beta$ both are negative.
MCQ 631 Mark
If $\alpha, \beta, \gamma$ are the zeros of the polynomial $f(x)=a x^3-5 x+9$ such that $\alpha^3+\beta^3+\gamma^3=27$, then the value of $a$ is
- A-3
- B3
- C1
- ✓-1
Answer
View full question & answer→Correct option: D.
-1
(D) -1
It is given that $\alpha, \beta, \gamma$ are zeros of $f(x)=a x^3-5 x+9$. Therefore,
$
\alpha+\beta+\gamma=-\frac{\text { Coeff. } x^2}{\text { Coeff. } x^3}=0 \text { and, } \alpha \beta \gamma=-\frac{9}{a}
$
Now, $\quad \alpha+\beta+\gamma=0 \Rightarrow \alpha^3+\beta^3+\gamma^3=3 \alpha \beta \gamma \Rightarrow 27=3 \times-\frac{9}{a} \Rightarrow a=-1$
It is given that $\alpha, \beta, \gamma$ are zeros of $f(x)=a x^3-5 x+9$. Therefore,
$
\alpha+\beta+\gamma=-\frac{\text { Coeff. } x^2}{\text { Coeff. } x^3}=0 \text { and, } \alpha \beta \gamma=-\frac{9}{a}
$
Now, $\quad \alpha+\beta+\gamma=0 \Rightarrow \alpha^3+\beta^3+\gamma^3=3 \alpha \beta \gamma \Rightarrow 27=3 \times-\frac{9}{a} \Rightarrow a=-1$
MCQ 641 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 $c=$
- ✓1
- B$0$
- C-1
- D2
Answer
View full question & answer→Correct option: A.
1
(A) 1
It is given that $\alpha, \beta$ are the zeroes of $f(x)=x^2-p x-(p+c)$.
$\therefore \quad \alpha+\beta=p$ and $\alpha \beta=-(p+c)$
Now, $\quad(\alpha+1)(\beta+1)=0 \Rightarrow \alpha \beta+(\alpha+\beta)+1=0 \Rightarrow-p-c+p+1=0 \Rightarrow c=1$
It is given that $\alpha, \beta$ are the zeroes of $f(x)=x^2-p x-(p+c)$.
$\therefore \quad \alpha+\beta=p$ and $\alpha \beta=-(p+c)$
Now, $\quad(\alpha+1)(\beta+1)=0 \Rightarrow \alpha \beta+(\alpha+\beta)+1=0 \Rightarrow-p-c+p+1=0 \Rightarrow c=1$
MCQ 651 Mark
If the polynomial $f(x)=a x^3+b x-c$ is divisible by the polynomial $g(x)=x^2+b x+c$ and $c \neq 0$, then $a b=$
- A1
- B$\frac{1}{c}$
- C-1
- D$-\frac{1}{c}$
Answer
View full question & answer→A highway underpass is parabolic in shape as shown in the following picture.
(i) If the highway overpass is represented by $x^2-2 x-8$. Then its zeros are
(a) $2,-4$ $\qquad$ (b) $4,-2$ $\qquad$ (c) $-2,-2$ $\qquad$ (d) $-4,-4$
(ii) Number of zeros of the polynomial representing highway overpass is equal to number of points where the graph of polynomial
(a) intersects $x$-axis
(b) intersects $y$-axis
(c) intersects $y$-axis or $x$-axis
(d) none of these
(iii) Graph of quadratic polynomial is a
(a) straight line $\qquad$ (b) circle $\qquad$ (c) parabola $\qquad$ (d) ellipse
(iv) The representation of Highway Underpass whose one zero is 6 and sum of the zeros is 0 , is
(a) $x^2-6 x+2$ $\qquad$ (b) $x^2-36$ $\qquad$ (c) $x^2-6$ $\qquad$ (d) $x^2-3$
(i) If the highway overpass is represented by $x^2-2 x-8$. Then its zeros are
(a) $2,-4$ $\qquad$ (b) $4,-2$ $\qquad$ (c) $-2,-2$ $\qquad$ (d) $-4,-4$
(ii) Number of zeros of the polynomial representing highway overpass is equal to number of points where the graph of polynomial
(a) intersects $x$-axis
(b) intersects $y$-axis
(c) intersects $y$-axis or $x$-axis
(d) none of these
(iii) Graph of quadratic polynomial is a
(a) straight line $\qquad$ (b) circle $\qquad$ (c) parabola $\qquad$ (d) ellipse
(iv) The representation of Highway Underpass whose one zero is 6 and sum of the zeros is 0 , is
(a) $x^2-6 x+2$ $\qquad$ (b) $x^2-36$ $\qquad$ (c) $x^2-6$ $\qquad$ (d) $x^2-3$
MCQ 661 Mark
Given that one of the zeroes of the cubic polynomial $a x^3+b x^2+c x+d$ is zero, the product of other two zeroes is
- A$-\frac{c}{a}$
- ✓$\frac{c}{a}$
- C$0$
- D$-\frac{b}{a}$
Answer
View full question & answer→Correct option: B.
$\frac{c}{a}$
(B) $\frac{c}{a}$
Let $\alpha, \beta, 0$ be three zeroes of $f(x)=a x^3+b x^2+c x+d$. Then,
$
\alpha \beta+\beta \times 0+0 \times \alpha=\frac{c}{a} \Rightarrow \alpha \beta=\frac{c}{a} \quad\left[\because \alpha \beta+\beta \gamma+\gamma \alpha=(-1)^2 \frac{\text { Coeff.of } x}{\text { Coeff. of } x^3}\right]
$
Let $\alpha, \beta, 0$ be three zeroes of $f(x)=a x^3+b x^2+c x+d$. Then,
$
\alpha \beta+\beta \times 0+0 \times \alpha=\frac{c}{a} \Rightarrow \alpha \beta=\frac{c}{a} \quad\left[\because \alpha \beta+\beta \gamma+\gamma \alpha=(-1)^2 \frac{\text { Coeff.of } x}{\text { Coeff. of } x^3}\right]
$
MCQ 671 Mark
If one of the zeroes of the cubic polynomial $x^3+a x^2+b x+c$ is -1 , then the product of other two zeroes is
- ✓$b-a+1$
- B$b-a-1$
- C$a-b+1$
- D$a-b-1$
Answer
View full question & answer→Correct option: A.
$b-a+1$
(A) $b-a+1$
Let $\alpha, \beta,-1$ be three zeroes of $f(x)=x^3+a x^2+b x+c$. Then,
$
\text { Product of zeroes }=-\frac{c}{1} \Rightarrow \alpha \beta \times-1=-c \Rightarrow \alpha \beta=c\qquad\ldots(i)
$
It is given that -1 is a zero of $f(x)$. Therefore,
$
f(-1)=0 \Rightarrow-1+a-b+c=0 \Rightarrow c=b-a+1 \Rightarrow \alpha \beta=b-a+1\qquad[Using (i)]
$
Let $\alpha, \beta,-1$ be three zeroes of $f(x)=x^3+a x^2+b x+c$. Then,
$
\text { Product of zeroes }=-\frac{c}{1} \Rightarrow \alpha \beta \times-1=-c \Rightarrow \alpha \beta=c\qquad\ldots(i)
$
It is given that -1 is a zero of $f(x)$. Therefore,
$
f(-1)=0 \Rightarrow-1+a-b+c=0 \Rightarrow c=b-a+1 \Rightarrow \alpha \beta=b-a+1\qquad[Using (i)]
$
MCQ 681 Mark
If two zeroes of the cubic polynomial $f(x)=a x^3+b x^2+c x+d$ are each equal to zero, then third zero is
- A$-\frac{d}{a}$
- B$\frac{c}{a}$
- ✓$-\frac{b}{a}$
- D$\frac{b}{a}$
Answer
View full question & answer→Correct option: C.
$-\frac{b}{a}$
(C) $-\frac{b}{a}$
Let $\alpha, \beta=0, \gamma=0$ be three zeroes of $f(x)$. Then,
Sum of the zeroes $=-\frac{b}{a} \Rightarrow \alpha+0+0=-\frac{b}{a} \Rightarrow \alpha=-\frac{b}{a}$
Hence, third zero is $-\frac{b}{a}$.
Let $\alpha, \beta=0, \gamma=0$ be three zeroes of $f(x)$. Then,
Sum of the zeroes $=-\frac{b}{a} \Rightarrow \alpha+0+0=-\frac{b}{a} \Rightarrow \alpha=-\frac{b}{a}$
Hence, third zero is $-\frac{b}{a}$.
MCQ 691 Mark
If $\alpha, \beta, \gamma$ are the zeroes of the polynomial $f(x)=x^3-p x^2+q x-r$, then $\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}$ is equal to
- A$\frac{r}{p}$
- ✓$\frac{p}{r}$
- C$-\frac{p}{r}$
- D$-\frac{r}{p}$
Answer
View full question & answer→Correct option: B.
$\frac{p}{r}$
(B) $\frac{p}{r}$
It is given that $\alpha, \beta, \gamma$ are zeroes of $f(x)=x^3-p x^2+q x-r$.
$
\therefore \quad \alpha+\beta+\gamma=p, \alpha \beta+\beta \gamma+\gamma \alpha=q \text { and } \alpha \beta \gamma=r
$
Hence, $\quad \frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}=\frac{\alpha+\beta+\gamma}{\alpha \beta \gamma}=\frac{p}{r}$
It is given that $\alpha, \beta, \gamma$ are zeroes of $f(x)=x^3-p x^2+q x-r$.
$
\therefore \quad \alpha+\beta+\gamma=p, \alpha \beta+\beta \gamma+\gamma \alpha=q \text { and } \alpha \beta \gamma=r
$
Hence, $\quad \frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}=\frac{\alpha+\beta+\gamma}{\alpha \beta \gamma}=\frac{p}{r}$
MCQ 701 Mark
If $\alpha, \beta, \gamma$ are the zeroes of the polynomial $f(x)=a x^3+b x^2+c x+d$, then $\alpha^2+\beta^2+\gamma^2=$
- A$\frac{b^2-a c}{a^2}$
- B$\frac{b^2-2 a c}{a}$
- C$\frac{b^2+2 a c}{b^2}$
- ✓$\frac{b^2-2 a c}{a^2}$
Answer
View full question & answer→Correct option: D.
$\frac{b^2-2 a c}{a^2}$
(D) $\frac{b^2-2 a c}{a^2}$
It is given that $\alpha, \beta, \gamma$ are the zeroes of $f(x)=a x^3+b x^2+c x+d$. Therefore,
$
\begin{array}{ll}
& \alpha+\beta+\gamma=-\frac{b}{a} \text { and } \alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a} \\
\therefore \quad & \alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha) \Rightarrow \alpha^2+\beta^2+\gamma^2=\left(-\frac{b}{a}\right)^2-\frac{2 c}{a}=\frac{b^2-2 a c}{a^2}
\end{array}
$
It is given that $\alpha, \beta, \gamma$ are the zeroes of $f(x)=a x^3+b x^2+c x+d$. Therefore,
$
\begin{array}{ll}
& \alpha+\beta+\gamma=-\frac{b}{a} \text { and } \alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a} \\
\therefore \quad & \alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha) \Rightarrow \alpha^2+\beta^2+\gamma^2=\left(-\frac{b}{a}\right)^2-\frac{2 c}{a}=\frac{b^2-2 a c}{a^2}
\end{array}
$
MCQ 711 Mark
If two zeros of the polynomial $f(x)=x^3+x^2-5 x-5$ are $\sqrt{5}$ and $-\sqrt{5}$, then its third zero is
- A1
- ✓-1
- C2
- D-2
Answer
View full question & answer→Correct option: B.
-1
(B) -1
Let the third zero of $f(x)$ be $\alpha$. Then,
Sun of the zeroes $=-\frac{\text { Coeff.of } x^2}{\text { Coeff.of } x^3} \Rightarrow \sqrt{5}+(-\sqrt{5})+\alpha=-\frac{1}{1} \Rightarrow \alpha=-1$
Let the third zero of $f(x)$ be $\alpha$. Then,
Sun of the zeroes $=-\frac{\text { Coeff.of } x^2}{\text { Coeff.of } x^3} \Rightarrow \sqrt{5}+(-\sqrt{5})+\alpha=-\frac{1}{1} \Rightarrow \alpha=-1$
MCQ 721 Mark
If the product of two zeros of the polynomial $f(x)=2 x^3+6 x^2-4 x+9$ is 3 , then its third zero is
- A$\frac{3}{2}$
- ✓$-\frac{3}{2}$
- C$\frac{9}{2}$
- D$-\frac{9}{2}$
Answer
View full question & answer→Correct option: B.
$-\frac{3}{2}$
(B) $-\frac{3}{2}$
Let $\alpha, \beta, \gamma$ be three zeroes of $f(x)=2 x^3+6 x^2-4 x+9$ such that $\alpha \beta=3$. But it is given that
Product of zeroes $=-\frac{9}{2} \Rightarrow \alpha \beta \gamma=-\frac{9}{2} \Rightarrow 3 \gamma=-\frac{9}{2} \Rightarrow \gamma=-\frac{3}{2}$
Hence, the third zero is $-\frac{3}{2}$.
Let $\alpha, \beta, \gamma$ be three zeroes of $f(x)=2 x^3+6 x^2-4 x+9$ such that $\alpha \beta=3$. But it is given that
Product of zeroes $=-\frac{9}{2} \Rightarrow \alpha \beta \gamma=-\frac{9}{2} \Rightarrow 3 \gamma=-\frac{9}{2} \Rightarrow \gamma=-\frac{3}{2}$
Hence, the third zero is $-\frac{3}{2}$.
MCQ 731 Mark
If zeros of the polynomial $f(x)=x^3-3 p x^2+q x-r$ are in A.P., then
- ✓$2 p^3=p q-r$
- B$2 p^3=p q+r$
- C$p^3=p q-r$
- Dnone of these
Answer
View full question & answer→Correct option: A.
$2 p^3=p q-r$
(A) $2 p^3=p q-r$
Let $a-d, a, a+d$ be the zeroes of $f(x)=x^3-3 p x^2+q x-r$. Then,
$
(a-d)+a+(a+d)=3 p \Rightarrow 3 a=3 p \Rightarrow a=p
$
But, $a$ is a zero of $f(x)$. Therefore,
$
f(a)=0 \Rightarrow a^3-3 p a^2+q a-r=0 \Rightarrow p^3-3 p^3+p q-r=0 \Rightarrow 2 p^3=p q-r
$
Let $a-d, a, a+d$ be the zeroes of $f(x)=x^3-3 p x^2+q x-r$. Then,
$
(a-d)+a+(a+d)=3 p \Rightarrow 3 a=3 p \Rightarrow a=p
$
But, $a$ is a zero of $f(x)$. Therefore,
$
f(a)=0 \Rightarrow a^3-3 p a^2+q a-r=0 \Rightarrow p^3-3 p^3+p q-r=0 \Rightarrow 2 p^3=p q-r
$
MCQ 741 Mark
If $f(x)=a x^2+b x+c$ has no real zeroes and $a+b+c<0$, then
- A$c=0$
- B$c>0$
- ✓$c<0$
- Dc can be positive or negative
Answer
View full question & answer→Correct option: C.
$c<0$
(C) $c<0$
If $f(x)=a x^2+b x+c$ has no ral zero, then its graph does not cross $x$-axis. Therefore, either the complete curve is above $x$-axis or it lies below $x$-axis. This means that either $f(x)>0$ or $f(x)<0$ for all values of $x$. It is given that $f(1)=a+b+c<0$. Therefore,
$
f(x)<0 \text { for all values of } x \Rightarrow f(0)<0 \Rightarrow c<0
$
If $f(x)=a x^2+b x+c$ has no ral zero, then its graph does not cross $x$-axis. Therefore, either the complete curve is above $x$-axis or it lies below $x$-axis. This means that either $f(x)>0$ or $f(x)<0$ for all values of $x$. It is given that $f(1)=a+b+c<0$. Therefore,
$
f(x)<0 \text { for all values of } x \Rightarrow f(0)<0 \Rightarrow c<0
$
MCQ 751 Mark
The zeroes of the polynomial $x^2+\frac{1}{6} x-2$ are
- A$-3,4$
- ✓$-\frac{3}{2}, \frac{4}{3}$
- C$-\frac{4}{3}, \frac{3}{2}$
- D$-\frac{4}{3},-\frac{3}{2}$
Answer
View full question & answer→Correct option: B.
$-\frac{3}{2}, \frac{4}{3}$
(B) $-\frac{3}{2}, \frac{4}{3}$
Let $f(x)=x^2+\frac{1}{6} x-2$. Then,
$
f(x)=\frac{1}{6}\left(6 x^2+x-12\right)=\frac{1}{6}\left(6 x^2+9 x-8 x-12\right)=\frac{1}{6}\{3 x(2 x+3)-4(2 x+3)\}=\frac{1}{6}(2 x+3)(3 x-4)
$
Zeroes of $f(x)$ are given by
$
f(x)=0 \Rightarrow \frac{1}{6}(2 x+3)(3 x-4)=0 \Rightarrow 2 x+3=0 \text { or } 3 x-4=0 \Rightarrow x=-\frac{3}{2}, \frac{4}{3}
$
Let $f(x)=x^2+\frac{1}{6} x-2$. Then,
$
f(x)=\frac{1}{6}\left(6 x^2+x-12\right)=\frac{1}{6}\left(6 x^2+9 x-8 x-12\right)=\frac{1}{6}\{3 x(2 x+3)-4(2 x+3)\}=\frac{1}{6}(2 x+3)(3 x-4)
$
Zeroes of $f(x)$ are given by
$
f(x)=0 \Rightarrow \frac{1}{6}(2 x+3)(3 x-4)=0 \Rightarrow 2 x+3=0 \text { or } 3 x-4=0 \Rightarrow x=-\frac{3}{2}, \frac{4}{3}
$
MCQ 761 Mark
If zeroes of the quadratic polynomial $f(x)=\left(k^2+4\right) x^2+7 x+4 k$ are reciprocal of each other, then the value (s) of $k$ is (are)
- A1
- B-1
- ✓2
- D-2
Answer
View full question & answer→Correct option: C.
2
(C) 2
Let $\alpha$ and $\frac{1}{\alpha}$ be zeroes of $f(x)$. Then,
$
\alpha \times \frac{1}{\alpha}=\frac{4 k}{k^2+4} \Rightarrow k^2+4=4 k \Rightarrow(k-2)^2=0 \Rightarrow k=2
$
Let $\alpha$ and $\frac{1}{\alpha}$ be zeroes of $f(x)$. Then,
$
\alpha \times \frac{1}{\alpha}=\frac{4 k}{k^2+4} \Rightarrow k^2+4=4 k \Rightarrow(k-2)^2=0 \Rightarrow k=2
$
MCQ 771 Mark
The quadratic polynomial whose zeroes are reciprocal of the zeroes of quadratic polynomial $a x^2+b x+c, a \neq 0, c \neq 0$, are given by
where $k$ is a non-zero real number.
where $k$ is a non-zero real number.
- A$k\left(c x^2+a x+b\right)$
- ✓$k\left(c x^2+b x+a\right)$
- C$k\left(c x^2-b x+a\right)$
- D$k\left(c x^2+b x-a\right)$
Answer
View full question & answer→Correct option: B.
$k\left(c x^2+b x+a\right)$
(B) $k\left(c x^2+b x+a\right)$
Let $\alpha, \beta$ be the zeroes of quadratic polynomial $a x^2+b x+c$. Then,
$
\alpha+\beta=-\frac{b}{a} \text { and } \alpha \beta=\frac{c}{a}
$
Quadratic polynomials having $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ as zeroes are given by
$
\begin{array}{l}
\quad f(x)=\lambda\left\{x^2-x\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+\frac{1}{\alpha \beta}\right\}, \text { where } \lambda \neq 0 \\
\Rightarrow \quad f(x)=\lambda\left\{x^2-x\left(\frac{\alpha+\beta}{\alpha \beta}\right)+\frac{1}{\alpha \beta}\right\}=\lambda\left\{x^2+\frac{b}{c} x+\frac{a}{c}\right\}=\frac{\lambda}{c}\left(c x^2+b x+a\right)=k\left(c x^2+b x+a\right), \\
\\
\text { where } k=\lambda / c \neq 0
\end{array}
$
REMARK To find a quadratic polynomial having zeroes as the reciprocal of zeros of a given quadratic simply interchange the coefficient of $x^2$ and constant term.
Let $\alpha, \beta$ be the zeroes of quadratic polynomial $a x^2+b x+c$. Then,
$
\alpha+\beta=-\frac{b}{a} \text { and } \alpha \beta=\frac{c}{a}
$
Quadratic polynomials having $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ as zeroes are given by
$
\begin{array}{l}
\quad f(x)=\lambda\left\{x^2-x\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+\frac{1}{\alpha \beta}\right\}, \text { where } \lambda \neq 0 \\
\Rightarrow \quad f(x)=\lambda\left\{x^2-x\left(\frac{\alpha+\beta}{\alpha \beta}\right)+\frac{1}{\alpha \beta}\right\}=\lambda\left\{x^2+\frac{b}{c} x+\frac{a}{c}\right\}=\frac{\lambda}{c}\left(c x^2+b x+a\right)=k\left(c x^2+b x+a\right), \\
\\
\text { where } k=\lambda / c \neq 0
\end{array}
$
REMARK To find a quadratic polynomial having zeroes as the reciprocal of zeros of a given quadratic simply interchange the coefficient of $x^2$ and constant term.
MCQ 781 Mark
If $(a-2) x^2+3 x-5$ is a quadratic polynomial, then
- Aa can take any real value
- Ba can take any non-zero value
- ✓$a \neq 2$
- D$a=2$
Answer
View full question & answer→Correct option: C.
$a \neq 2$
(C) $a \neq 2$
If $(a-2) x^2+3 x-5$ is a quadratic polynomial, then $a-2 \neq 0$ i.e. $a \neq 2$.
If $(a-2) x^2+3 x-5$ is a quadratic polynomial, then $a-2 \neq 0$ i.e. $a \neq 2$.
MCQ 791 Mark
If $\alpha$ and $\beta$ are the zeros of the polynomial $f(x)=p x^2-2 x+3 p$ and $\alpha+\beta=\alpha \beta$, then the value of $p$ is
- A$-\frac{2}{3}$
- ✓$a^2+2 b$
- C$\frac{1}{3}$
- D$-\frac{1}{3}$
Answer
View full question & answer→Correct option: B.
$a^2+2 b$
(B) $a^2+2 b$
Clearly, $\alpha+\beta=\frac{2}{p}$ and $\alpha \beta=3$. It is given that
$\alpha+\beta=\alpha \beta \Rightarrow \frac{2}{p}=3 \Rightarrow p=\frac{2}{3}$
Clearly, $\alpha+\beta=\frac{2}{p}$ and $\alpha \beta=3$. It is given that
$\alpha+\beta=\alpha \beta \Rightarrow \frac{2}{p}=3 \Rightarrow p=\frac{2}{3}$
MCQ 801 Mark
If the product of the zeroes of the quadratic polynomial $3 x^2+5 x+k$ is $-\frac{2}{3}$, then the value of $k$ is
- A-3
- ✓-2
- C2
- D3
Answer
View full question & answer→Correct option: B.
-2
(B) -2
Let $\alpha, \beta$ be zeroes of $3 x^2+5 x+k$. Then, product of its zeroes is $\frac{k}{3}$.
But, $\quad$ Product of zeroes $=-\frac{2}{3} \Rightarrow \frac{k}{3}=-\frac{2}{3} \Rightarrow k=-2$
Let $\alpha, \beta$ be zeroes of $3 x^2+5 x+k$. Then, product of its zeroes is $\frac{k}{3}$.
But, $\quad$ Product of zeroes $=-\frac{2}{3} \Rightarrow \frac{k}{3}=-\frac{2}{3} \Rightarrow k=-2$
MCQ 811 Mark
The maximum number of zeroes a cubic polynomial can have, is
- A1
- B4
- C2
- ✓3
Answer
View full question & answer→Correct option: D.
3
(D) 3
The graph of a cubical polynomial crosses/cuts $x$-axis at most at three points. So, cubic polynomial can have at most three zeroes.
The graph of a cubical polynomial crosses/cuts $x$-axis at most at three points. So, cubic polynomial can have at most three zeroes.
MCQ 821 Mark
If one zero of the quadratic polynomial $k x^2+3 x+k$ is 2 , then the value of $k$ is
- A$\frac{5}{6}$
- B$-\frac{5}{6}$
- C$\frac{6}{5}$
- ✓$-\frac{6}{5}$
Answer
View full question & answer→Correct option: D.
$-\frac{6}{5}$
(D) $-\frac{6}{5}$
It is given that 2 is a zero of $f(x)=k x^2+3 x+k$. Therefore,
$
f(2)=0 \Rightarrow k \times 2^2+3 \times 2+k=0 \Rightarrow 5 k+6=0 \Rightarrow k=-\frac{6}{5}
$
It is given that 2 is a zero of $f(x)=k x^2+3 x+k$. Therefore,
$
f(2)=0 \Rightarrow k \times 2^2+3 \times 2+k=0 \Rightarrow 5 k+6=0 \Rightarrow k=-\frac{6}{5}
$
MCQ 831 Mark
Which of the following is a polynomial?
- A$2 x^2+\frac{3}{x}-5$
- B$-3 x^2+\sqrt{2 x}+4$
- ✓$\sqrt{2} x^3+\sqrt{3} x^2+\sqrt{5} x-3$
- D$\frac{5}{x^3}+2 x^2-3 x+\frac{1}{7}$
Answer
View full question & answer→Correct option: C.
$\sqrt{2} x^3+\sqrt{3} x^2+\sqrt{5} x-3$
(C) $\sqrt{2} x^3+\sqrt{3} x^2+\sqrt{5} x-3$
An expression of the form $a_0 x^n+a_1 x^{n-1}+a_2 x^{n-2}+\ldots+a_{n-1} x+a_n, a_0 \neq 0$ is a polynomial of degree $n$ (a natural number).
Clearly, in expression given in option (a) has a term containing $x^{-1}$. So, it is not a polynomial. Expression in option (b) has a term containing $x^{1 / 2}$, so it is not a polynomial. The expression in option (d) has a term containing $x^{-3}$, so it is not a polynomial. Expression in option (c) satisfies the definition of a polynomial.
An expression of the form $a_0 x^n+a_1 x^{n-1}+a_2 x^{n-2}+\ldots+a_{n-1} x+a_n, a_0 \neq 0$ is a polynomial of degree $n$ (a natural number).
Clearly, in expression given in option (a) has a term containing $x^{-1}$. So, it is not a polynomial. Expression in option (b) has a term containing $x^{1 / 2}$, so it is not a polynomial. The expression in option (d) has a term containing $x^{-3}$, so it is not a polynomial. Expression in option (c) satisfies the definition of a polynomial.
MCQ 841 Mark
If 2 and $\frac{1}{2}$ are zeros of $p x^2+5 x+r$, then
- A$p=r=2$
- ✓$p=r=-2$
- C$p=2, r=-2$
- D$p=-2, r=2$
Answer
View full question & answer→Correct option: B.
$p=r=-2$
B
MCQ 851 Mark
If the zeros of the quadratic polynomial $a x^2+b x+c, c \neq 0$ 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
Answer
View full question & answer→Correct option: C.
$c$ and $a$ have the same sign
C
MCQ 861 Mark
The zeros of the quadratic polynomial $x^2+a x+a, a \neq 0$,
- ✓cannot both be positive
- Bcannot both be negative
- Care always unequal
- Dare always equal
Answer
View full question & answer→Correct option: A.
cannot both be positive
A
MCQ 871 Mark
If $a-b, a$ and $a+b$ are zeroes of the polynomial $x^3-3 x^2+x+1$, then the value of $a+b$ is
- A$\sqrt{2}-1$
- B$\sqrt{2}$
- C$-\sqrt{2}-1$
- ✓$1 \pm \sqrt{2}$
Answer
View full question & answer→Correct option: D.
$1 \pm \sqrt{2}$
D
MCQ 881 Mark
If the polynomial $f(x)=2 x^3-k x^2+5 x+9$ is exactly divisible by $x+2$, then $k=$
- A$\frac{17}{4}$
- ✓$-\frac{17}{4}$
- C$-\frac{15}{4}$
- D$\frac{15}{4}$
Answer
View full question & answer→Correct option: B.
$-\frac{17}{4}$
B
MCQ 891 Mark
If the sum of two zeroes of the polynomial $x^3-2 x^2+q x-r$ is zero, then
- A$q=2 r$
- ✓$r=2 q$
- C$q=r$
- D$r=4 q$
Answer
View full question & answer→Correct option: B.
$r=2 q$
B
MCQ 901 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$-\frac{b}{d}$
- B$\frac{c}{d}$
- ✓$-\frac{c}{d}$
- D$-\frac{c}{a}$
Answer
View full question & answer→Correct option: C.
$-\frac{c}{d}$
C
MCQ 911 Mark
If the zeroes of the polynomial $x^3-12 x^2+44 x+c$ are in A.P., then the value of $c$ is
- A44
- B48
- C-44
- ✓-48
Answer
View full question & answer→Correct option: D.
-48
D
MCQ 921 Mark
If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2-(k+6) x+2(2 k-1)$ such that $\alpha+\beta=\frac{\alpha \beta}{2}$, then the value of $k$ is
- A6
- B2
- C14
- ✓7
Answer
View full question & answer→Correct option: D.
7
D
MCQ 931 Mark
If zeros of the polynomial $f(x)=x^3-3 p x^2+q x-r$ are in A.P., then
- ✓$2 p^3=p q-r$
- B$2 p^3=p q+r$
- C$p^3=p q-r$
- Dnone of these
Answer
View full question & answer→Correct option: A.
$2 p^3=p q-r$
A
MCQ 941 Mark
If two zeroes of the polynomial $x^3+7 x^2-2 x-14$ are $\sqrt{2}$ and $-\sqrt{2}$, then the third zero is
- A7
- ✓-7
- C-14
- D14
Answer
View full question & answer→Correct option: B.
-7
B
MCQ 951 Mark
If $\alpha, \beta$ and $\gamma$ are the zeroes of the polynomial $x^3-x^2-10 x-8$, then $\alpha \beta+\beta \gamma+\gamma \alpha+\alpha \beta \gamma$ is equal to
- ✓-2
- B2
- C18
- D-18
Answer
View full question & answer→Correct option: A.
-2
A
MCQ 961 Mark
If $\alpha$ and $\beta$ are the zeroes of the polynomial $a x^2-5 x+c$ and $\alpha+\beta=\alpha \beta=10$, then
- A$a=5, c=\frac{1}{2}$
- B$a=1, c=\frac{5}{2}$
- C$a=\frac{5}{2}, c=1$
- ✓$a=\frac{1}{2}, c=5$
Answer
View full question & answer→Correct option: D.
$a=\frac{1}{2}, c=5$
D
MCQ 971 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$\frac{-d}{a}$
- B$\frac{c}{a}$
- ✓$\frac{-b}{a}$
- D$\frac{b}{a}$
Answer
View full question & answer→Correct option: C.
$\frac{-b}{a}$
C
MCQ 981 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$x^3-3 x^2-3 x+5$
Answer
View full question & answer→Correct option: C.
$-x^3+3 x^2-3 x+5$
C
MCQ 991 Mark
If $x+2$ is a factor of $x^2+a x+2 b$ and $a+b=4$, then
- A$a=1, b=3$
- ✓$a=3, b=1$
- C$a=-1, b=5$
- D$a=5, b=-1$
Answer
View full question & answer→Correct option: B.
$a=3, b=1$
B
MCQ 1001 Mark
What should be subtracted to the polynomial $x^2-16 x+30$, so that 15 is the zero of the resulting polynomial?
- A30
- B14
- ✓15
- D16
Answer
View full question & answer→Correct option: C.
15
C