Question 12 Marks
The graph of the polynomial $f(x) = ax^2+ bx + c$ is as shown in Fig. Write the value of $b^2- 4ac$ and the number of real zeros of $f(x),$ write the sign of $c.$
AnswerThe parabola $y = ax^2+ bx + c$ cuts $y-$axis at point $P$ which lies on $OY.$
Putting $x = 0$ in $y = ax^2+ bx + c,$ We get $y = c.$
So, the coordinates of $P$ are $(0, c).$
Clearly, $P $ lies on $OY'$
Therefore $c < 0$ View full question & answer→Question 22 Marks
Write the standard form of a quadratic polynomial with real coefficients.
Answer$ax^2+ bx + c$ is a standard form of quadratic polynomial with real co-efficients and $a ≠ 0.$
View full question & answer→Question 32 Marks
The graph of the polynomial $f(x) = ax^2+ bx + c$ is as shown in Fig. Write the value of $b^2- 4ac$ and the number of real zeros of $f(x).$
AnswerThe graph of the polynomial $f(x) = ax^2+ bx + c$ or the curve touches $x-$axis at point $\Big(\frac{-\text{b}}{2\text{a}},0\Big).$
The $x-$coordinate of this point given two equal zeros of the polynomial and $b^2- 4ac = 0.$
Hence the number of real zeros of $f(x)$ is $2$ and $b^2- 4ac = 0$ View full question & answer→Question 42 Marks
If a quadratic polynomial $f(x)$ is factorizable into linear distinct factors, then what is the total number of real and distinct zeros of $f(x)?$
AnswerIn a quadratic polynomial $f(x)$ its degree is $2$ and it can be factorised in to two distinct linear factors.
$f(x)$ has two distinct zeros.
View full question & answer→Question 52 Marks
Write the family of quadratic polynomials having $-\frac{1}{4}$ and $1$ as its zeros.
AnswerWe know that, if $x = a$ is a zero of a polynomial then $x - 2$ is a factor of quadratic polynomials.
Since $\frac{-1}{4}$ and $1$ are zeros of polynomial.
Therefore $\big(\text{x}+\frac{1}{4}\big)(\text{x}-1)$
$=\text{x}^2+\frac{1}{4}\text{x}-\text{x}-\frac{1}{4}$
$=\text{x}^2+\frac{1}{4}\text{x}-\frac{1\times4}{1\times4}\text{x}-\frac{1}{4}$
$=\text{x}^2+\frac{1-4}{4}\text{x}-\frac{1}{4}$
$=\text{x}^2-\frac{3}{4}\text{x}-\frac{1}{4}$
Hence, the family of quadratic polynomials is $\text{f(x)}=\text{k}\Big(\text{x}^2-\frac{3}{4}\text{x}-\frac{1}{4}\Big),$ where $k$ is any non-zeros real number
View full question & answer→Question 62 Marks
Apply division algorithm to find the quotient $q(x)$ and remainder $r(x)$ in dividing $f(x)$ by $g(x)$ in the following:
$f(x)=10 x^4+17 x^3-62 x^2+30 x-3, g(x)=2 x^2+7 x+1$
Answer$f(x)=10 x^4+17 x^3-62 x^2+30 x-3, g(x)=2 x^2+7 x+1$
View full question & answer→Question 72 Marks
What must be subtracted from the polynomial $f(x)=x^4+2 x^3-13 x^2-12 x+21$ so that the resulting polynomial is exactly divisible by $x^2-4 x+3$ ?
Answer
We must subtract $[2x - 2] + 10m$ the given polynomial so as to get the resulting polynomial exactly divisible by $x^2- x + 3$ View full question & answer→Question 82 Marks
If $\alpha,\beta$ are the zeros of the polynomial $2y^2+ 7y + 5,$ write the value of $\alpha+\beta+\alpha\beta.$
Answer$\alpha$ and $\beta$ are the zeros of the polynomial $2y^2+ 7y + 5$
$\therefore\ \alpha+\beta=\frac{-\text{b}}{\text{a}}=\frac{-7}{2}$
$\alpha\beta=\frac{\text{c}}{\text{a}}=\frac{5}{2}$
$\therefore\ \alpha+\beta+\alpha\beta=\frac{-7}{2}+\frac{5}{2}=\frac{-2}{2}=-1$
View full question & answer→Question 92 Marks
State division algorithm for polynomials.
AnswerIf $f(x)$ is a polynomial and $g(x)$ is a non zero polynomial, there exist two polynomials $q(x)$ and $r(x)$ such that,
$f(x) = g(x) \times q(x) + r(x)$
where $r(x) = 0$ or degree $r(x) <$ degree $g(x)$
This is called division algorithm.
View full question & answer→Question 102 Marks
If the sum of the zeros of the quadratic polynomial $f(x) = kx^2- 3x + 5$ is $1,$ write the value of $k.$
Answer$f(x) = kx^2- 3x + 5$
Here $a = k, b = -3, c = 5$
$\therefore$ Sum of zeros $=\frac{-\text{b}}{\text{a}}=-\Big(\frac{-3}{\text{k}}\Big)=\frac{3}{\text{k}}$
$\therefore \frac{3}{\text{k}}=1 \Rightarrow\text{k}=3$
View full question & answer→Question 112 Marks
If the product of zeros of the quadratic polynomial $f(x) = x^2- 4x + k$ is $3,$ find the value of $k.$
AnswerProduct of zeros $=\frac{\text{c}}{\text{a}}$
$\alpha\cdot\beta=\frac{\text{K}}{1}$
$3=\frac{\text{K}}{1}$
$3=\text{K}$
$\text{K}=3$
View full question & answer→Question 122 Marks
Define a polynomial with real coefficients.
AnswerIn the polynomial $\text{f(x)}=\text{a }_\text{n}\text{x}^\text{n}+\text{a}_{\text{n}-1}\text{x}^{\text{n}-1}+...+\text{a}_1\text{x}+\text{a}_0,\text{a}_\text{n}\text{x}^\text{n},\text{a}_{\text{n}-1}\text{x}^{\text{n}-1},...,\text{a}_1\text{x},$ and $a_0$ are known as the terms of the polynomial and $\text{a}_\text{n},\text{a}_{\text{n}-1},...,\text{a}_1$ $a_0$ are their real coefficients.
For example, $p(x) = 3x - 2$ is a polynomial and $3$ is a real coefficient
View full question & answer→Question 132 Marks
The Sum and product of the zeros of a quadratic polynomial are $-\frac{1}{2}$ and -3 respectively. What is the quadratic polynomial?
AnswerSum of zeros $(\alpha+\beta)=\frac{-1}{2}$
and product of zeros $(\alpha\cdot\beta)=-3$
$\therefore$ Quadratic polynomial with be
$k(x^2-$ {sum of zeros$) $ x + product of zeros}
$\Rightarrow\text{k}\Big[\text{x}^2-\Big(-\frac{1}{2}\Big)\text{x}+(-3)\Big]$
$\Rightarrow \text{k}\Big[\text{x}^2+\frac{1}{2}\text{x}-3\Big]$
View full question & answer→Question 142 Marks
The graph of a polynomial $y = f(x),$ shown in Fig. Find the number of real zeros of $f(x).$
AnswerThe curve touches $x-$axis at one point and also intersects at one point So number of zeros will be $3,$ two equal and one distinct.
View full question & answer→Question 152 Marks
If the graph of quadratic polynomial $ax^2+ bx + c$ cuts negative direction of $y-$axis, then what is the sign of $c?$
AnswerSince graph of quadratic polynomial $f(x) = ax^2+ bx + c$ cuts negative direction of $y-$axis
So, put $x = 0$ to find the intersection point on $y-$axis $y = 0 + 0 + c = c$
So, the point is $(0, c)$
Now it is given that the quadratic polynomial cuts negative direction of $y$
So, $c < 0$
View full question & answer→Question 162 Marks
Write a quadratic polynomial, sum of whose zeros is $2\sqrt{3}$ and their product is $2.$
AnswerSum of zeros $=2\sqrt{3}$
and product of zeros $= 2$
Quadratic polynomial will be $f(x) = x^2- ($sum of zeros$) x +$ product of zeros $ = \text{x}^2 – 2\sqrt{3}\text{x} + 2$
View full question & answer→Question 172 Marks
Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:
$g(x)=2 x^2-x+3, f(x)=6 x^5-x^4+4 x^3-5 x^2-x-15$
Answer$g(x)=2 x^2-x+3, f(x)=6 x^5-x^4+4 x^3-5 x^2-x-15$
View full question & answer→Question 182 Marks
Write the standard form of a cubic polynomial with real coefficients.
AnswerThe most general form of a cubic polynomial with coefficients as real numbers is of the from $f(x)=a x^3+b x^2+c x+d$ where $a, b, c, d$ are real number and $a ≠ 0.$
View full question & answer→Question 192 Marks
Write the standard form of a linear polynomial with real coefficients.
AnswerAny linear polynomial in variable $x$ with real coefficients is of the form $f(x) = ax + b$ where $a, b$ are real numbers and $a ≠ 0.$
View full question & answer→Question 202 Marks
If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = ax^2+ bx + c,$ then evaluate:
$\frac{1}{\alpha}+\frac{1}{\beta}-2\alpha\beta$
Answer$\frac{1}{\alpha}+\frac{1}{\beta}-2\alpha\beta$
$\Rightarrow\ \Big[\frac{\alpha+\beta}{\alpha\beta}\Big]-2\alpha\beta$
$ \Rightarrow\ \frac{-\text{b}}{\text{a}}\times\frac{\text{a}}{\text{c}}-2\frac{\text{c}}{\text{a}}=-2\frac{\text{c}}{\text{a}}-\frac{\text{b}}{\text{c}}=\frac{-\text{ab}-2\text{c}^2}{\text{ac}}-\Big[\frac{\text{b}}{\text{c}}+\frac{2\text{c}}{\text{a}}\Big]$
View full question & answer→Question 212 Marks
If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = ax^2+ bx + c,$ then evaluate:
$\alpha-\beta$
AnswerThe two zerose of the polynomials are,
$\frac{-\text{b}+\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}-\bigg(\frac{-\text{b}-\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}\bigg)$
$=\frac{-\text{b}+\sqrt{\text{b}^2-4\text{ac}}+\text{b}+\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
$=\frac{2\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}=\frac{\sqrt{\text{b}^2-4\text{ac}}}{\text{a}}$
View full question & answer→Question 222 Marks
Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:
$g(x)=x^3-3 x+1, f(x)=x^5-4 x^3+x^2+3 x+1$
Answer$g(x)=x^3-3 x+1, f(x)=x^5-4 x^3+x^2+3 x+1$
View full question & answer→Question 232 Marks
If $f(x) = x^3+ x^2- ax + b$ is divisible by $x^2- x$ write the value of a and b.
Answer$ f(x)=x^3+x^2-a x+b$
$ g(x)=x^2-x=x(x-1)$
Let $x(x-1)=0$, there $x=0$ or $x-1=0$
$\Rightarrow x=1$
$\therefore 0,1$ are the zeros of $f(x)$
Now, $f(0)=0+0-0+b=b$
$ \therefore b=0 $
$ \text { and } f(1)=(1)^3+(1)^2-a(1)+b$
$ =1+1-a+b $
$ =2-a+b$
$ \therefore 2-a+b=0 \Rightarrow 2-a+0=0$
$⇒ 2 - a = 0 ⇒ a = 2$
Hence $a = 2, b = 0$
View full question & answer→Question 242 Marks
Define value of polynomial at a point.
AnswerIf $f(x)$ is a polynomial and a is any real number then the real number obtained by replacing $x$ by α in $f(x)$ is called the value of $f(x)$ at $\text{x}=\alpha$ and is denoted by $\text{f}(\alpha).$
View full question & answer→Question 252 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
$6x^2- 3 - 7x$
Answer$6x^2- 3 - 7x = 6x^2- 7x - 3 = (3x + 1) (2x - 3)$
Zeros of polynomials are $\frac{+3}{2}$ and $ \frac{-1}{3}$
Sum of zeros $= \frac{-1}{3} + \frac{3}{2} = \frac{7}{6} = \frac{-7}{6} = \frac{-(\text{coefficient of x)} }{\text{coefficient of x}^{2}}$
Productor zeroes $= \frac{-1}{3} \times \frac{3}{2} = \frac{-1}{2} = \frac{-3}{6} = \frac{+\text{Constant term}}{\text{Coefficient of x}^{2}}$
$\therefore $ Hence, relationship varified.
View full question & answer→Question 262 Marks
The graph of the polynomial $f(x) = ax^2+ bx + c$ is as shown below Fig. Write the signs of $'a'$ and $b^2- 4ac,$ write the sign of $c.$
AnswerThe parabola $y = ax^2+ bx + c$ cuts y-axis at point P which lies on $y-$axis Putting $x = 0$ in $y = ax^2+ bx + c,$ we get $y = c,$
So the coordinates of $P$ are $(0, c).$
Clearly, $P$ line on $OY. $ Therefore $c > 0$
Hence, the sign of $c$ is $c > 0$ View full question & answer→Question 272 Marks
If graph of quadratic polynomial $ax^2+ bx + c$ cuts positive direction of $y-$axis, then what is the sign of $c?$
AnswerThe graph of quadratic polynomial $ax^2+ bx + c$ cuts positive direction of $y-$axis Then sign of constant term $c$ will be also positive.
View full question & answer→Question 282 Marks
If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = ax^2+ bx + c,$ then evaluate:
$\alpha^2\beta+\alpha\beta^2$
Answer$\alpha^2\beta+\alpha\beta^2$
$\alpha\beta(\alpha+\beta)$
$ =\frac{\text{c}}{\text{a}}\Big(\frac{-\text{b}}{\text{a}}\Big)$
$=\frac{-\text{bc}}{\text{a}^2}$
View full question & answer→Question 292 Marks
The graph of the polynomial $f(x) = ax^2+ bx + c$ is as shown below Fig. Write the signs of 'a' and $b^2- 4ac.$
AnswerClearly, $f(x) = ax2 + bx + c$ represent a parabola opening upwards. Therefore, $a > 0$
Since the parabola cuts x-axis at two points, this means that the polynomial will have two real solutions
Hence $b2 - 4ac > 0$
Hence $a > 0$ and $b2 - 4ac > 0$ View full question & answer→Question 302 Marks
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
$h(t) = t^2- 15$
Answer$\text{h(t)}=\text{t}^2-15=(\text{t}^2)-\big(\sqrt{15}\big)^2=\big(\text{t}+\sqrt{15}\big)\big(\text{t}-\sqrt{15}\big)$
Zeroes of the polynomials are $-\sqrt{15}$ and $\sqrt{15}$
Sum of zeroes $=0$
$-\sqrt{15}+\sqrt{15}=0$
$0=0$
Now,
Product of zeroes $=\frac{-\text{Constant term}}{\text{Coefficient of x}^2}$
We have to products of polynomials are $-\sqrt{15}$ and $\sqrt{15}$
So, Product of zeroes $=\frac{-\sqrt{15}\times\sqrt{15}}{1}$
$= -15$
View full question & answer→Question 312 Marks
Apply division algorithm to find the quotient $q(x)$ and remainder $r(x)$ in dividing $f(x)$ by $g(x)$ in the following:
$f(x)=x^3-6 x^2+11 x-6, g(x)=x^2+x+1$
Answer$f(x)=x^3-6 x^2+11 x-6,$
$ g(x)=x^2+x+1$
View full question & answer→Question 322 Marks
Define degree of a polynomial.
AnswerThe exponent of the highest degree term in a polynomial is known as its degree. A polynomial of degree O is called a constant polynomial.
View full question & answer→Question 332 Marks
In Fig. the graph of a polynomial $p(x)$ is given. Find the zeros of the polynomial.
AnswerThe graph of the given polynomial meets the $x-$axis at $-1$ and $-3$
Zero will be $-1$ and $-3$
Zero of polynomial is $3$
View full question & answer→Question 342 Marks
If $f(x)$ is a polynomial such that $f(a) f(b) < 0,$ then what is the number of zeros lying between $a$ and $b?$
AnswerIf $f(x)$ is a polynomial such that $f(a) f(b) < 0$ then this means the value of the polynomial are of different sign for a to $b$.
Hence, at least one zero will be lying between $a$ and $b$
View full question & answer→Question 352 Marks
If $x = 1$ is a zero of the polynomial $ f(x)=x^3-2 x^2-4 x+k $, write the value of k.
Answer$ f(x)=x^3-2 x^2-4 x+k $
$ \because x=1 \text { is the zero of } f(x) $
$ \therefore f(x)=0 $
$ \therefore f(1)=(1)^3-2(1)^2+4 \times 1+k $
$ =1-2+4+k=3+k $
$ \therefore 3+k=0 \Rightarrow k=-3$
View full question & answer→Question 362 Marks
The graph of a polynomial $f(x)$ is as shown in Fig. Write the number of real zeros of $f(x).$
AnswerThe graph of a polynomial $f(x)$ touches $x-$axis at two points
We know that if a curve touches the $x-$axis at two points then it has two common zeros of $f(x).$
Hence the number of zeros of $f(x),$ in this case is $2.$ View full question & answer→Question 372 Marks
Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:
$g(t)=t^2-3, f(t)=2 t^4+3 t^3-2 t^2-9 t-12$
Answer$g(t)=t^2-3, f(t)=2 t^4+3 t^3-2 t^2-9 t-12$
View full question & answer→Question 382 Marks
Define the zero of a polynomial.
AnswerThe zero of a polynomial $f(x)$ is defined as any real number $\alpha$ such that $\text{f}(\alpha)=0.$
View full question & answer→Question 392 Marks
If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = ax^2+ bx + c,$ then evaluate:
$\frac{1}{\alpha}-\frac{1}{\beta}$
Answer$\Rightarrow\ \frac{1}{\alpha}-\frac{1}{\beta}=\frac{\beta-\alpha}{\alpha\beta}$
$\Rightarrow\ \frac{1}{\alpha}-\frac{1}{\beta}=\frac{-(\alpha-\beta)}{\alpha\beta}$
$\Rightarrow\ \frac{1}{\alpha}-\frac{1}{\beta}=\frac{-\frac{1}{\text{a}}\sqrt{\text{b}^2-4\text{ac}}}{\frac{\text{c}}{\text{a}}}$
$\Rightarrow\ \frac{1}{\alpha}-\frac{1}{\beta}=\frac{-\sqrt{\text{b}^2-4\text{ac}}}{\text{c}}$
Thus, $\frac{1}{\alpha}-\frac{1}{\beta}=\frac{-\sqrt{\text{b}^2-4\text{ac}}}{\text{c}}$
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