2 Marks Questions

Question 12 Marks

Very-Short-Answer Question:
If $3$ is a zero of the polynomial $2 x^2+x+k$, find the value of $k$.

Answer

Since $3$ is a zero of $f(x)=2 x^2+x+k$, we have
$ f(3)=0 $
$\Rightarrow 2(3)^2+3+k=0 $
$ \Rightarrow 18+3+k=0$
$ \Rightarrow 21+k=0$
$ \Rightarrow k=-21$

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Question 22 Marks

Very-Short-Answer Question:
State division algorithm for polynomials.

Answer

If $f(x)$ and $g(x)$ are any two polynomials with $g(x) ≠ 0,$ then we can always find polynomials $q(x)$ and $r(x)$ such that $f(x) = q(x)g(x) + r(x)$,
where $r(x) = 0$ or degree $r(x) <$ degree $g(x)$.

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Question 32 Marks

Very-Short-Answer Question:
If $\alpha$ and $\beta$ are the zeroes of a polynomial $2x^2+ 7x + 5,$ write the value of $\alpha+\beta+\alpha\beta.$

Answer

Let $\alpha,\ \beta$ are the zeros of $2x^2+ 7x + 5$
Then we have
$\alpha+\beta=-\frac{7}2{}$
$\alpha\beta=\frac{5}{2}$
Hence,
$\alpha+\beta+\alpha\beta=(\alpha+\beta)+\alpha\beta$
$=-\frac{7}{2}+\frac{5}2{}$
$=\frac{-7+5}{2}$
$=\frac{-2}{2}$
$=-1$

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Question 42 Marks

Very-Short-Answer Question:
If the sum of the zeros of the quadratic polynomial $kx^2- 3x + 5$ is $1$, write the value of $k$.

Answer

Let $f(x) = kx^2- 3x + 5$
Sum of zeros $=\frac{-(\text{coefficient of }\text{x}^2)}{\text{coefficient of x}}$
$=\frac{-(-3)}{\text{k}}=\frac{3}{\text{k}}$
Given, sum of zeros $= 1$
$\therefore\frac{3}{\text{k}}=1$
$\Rightarrow\text{k}=3$

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Question 52 Marks

Very-Short-Answer Question:
Find the zeros of the polynomial $x^2+x-p(p+1) $.

Answer

We have,
$ f(x)=x^2+x-p(p+1) $
$ =x^2+(p+1) x-p x-p(p+1) $
$ =x[x+(p+1)]-p[x+(p+1)] $
$ =(x-p)[x+(p+1)] $
$ \therefore f(x)=0 $
$ \Rightarrow(x-p)[x+(p+1)]=0 $
$ \Rightarrow x-p=0 \text { or } x+(p+1)=0 $
$ \Rightarrow x=p \text { or } x=-(p+1)$
so, the zeros of $f(x)$ are p and $-(p + 1).$

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Question 62 Marks

Very-Short-Answer Question:
If $-2$ is a zero of the polynomial $3x^2+ 4x + 2k$ then find the value of $k$.

Answer

Since $-2$ is a zero of $f(x) = 3x^2+ 4x + 2K,$ we have
$f(2) = 0$
$\Rightarrow 3(-2)^2- 4(-2) + 2k = 0$
$\Rightarrow 12 - 8 + 2k = 0$
$\Rightarrow 4 + 2k = 0$
$\Rightarrow 2k = -4$
$\Rightarrow k = -2$

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Question 72 Marks

If one zero of the polynomial $(a^2+ 9)x^2+ 13x + 6a$ is the reciprocal of the other, find the value of $a$.

Answer

$(a^2+ 9)x^2+ 13x + 6a = 0$
Here, $A = (a^2+ 9), B = 13$ and $C = 6a$
Let $\alpha$ and $\frac{1}{\alpha}$ be the two zeros.
Then, product of the zeros $=\frac{\text{C}}{\text{A}}$
$\Rightarrow\alpha,\ \frac{1}{\alpha}=\frac{\text{6a}}{\text{a}^2+9}$
$\Rightarrow1=\frac{\text{6a}}{\text{a}^2+9}$
$ \Rightarrow a^2+9=6 a $
$ \Rightarrow a^2-6 a+9=0 $
$ \Rightarrow a^2-2 \times a \times 3+3^2=0 $
$ \Rightarrow(a-3)^2=0 $
$ \Rightarrow a-3=0 $
$ \Rightarrow a=3$

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Question 82 Marks

Very-Short-Answer Question:
If one zero of the quadratic polynomial $k x^2+3 x+k$ is $2$ then find the value of $k$.

Answer

Since $2$ is a zero of $f(x)=k x^2+3 x+k$, we have
$ f(2)=0 $
$ \Rightarrow k(2)^2+3(2)+k=0$
$\Rightarrow 4k + 6 + k = 0$
$\Rightarrow 5k + 6 = 0$
$\Rightarrow 5k = -6$
$\Rightarrow\text{k}=-\frac{6}{5}$

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Question 92 Marks

Very-Short-Answer Question:
If $1$ is a zero of the polynomial $ax^2- 3(a - 1)x - 1,$ then find the value of $a$.

Answer

Since $1$ is a zero of $f(x) = ax^2- 3(a - 1)x - 1,$ we have
$f(1) = 0$
$\Rightarrow a(1)^2- 3(a - 1)1 - 1 = 0$
$\Rightarrow a - 3a + 3 - 1 = 0$
$\Rightarrow -2a + 2 = 0$
$\Rightarrow 2a = 2$
$\Rightarrow a = 1$

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Question 102 Marks

Very-Short-Answer Question:
Write the zeros of the polynomial $x^2- x - 6$

Answer

We have,
$ f(x)=x^2-x-6 $
$ =x^2-3 x+2 x-6 $
$= x(x - 3) + 2(x - 3)$
$= (x + 2)(x - 3)$
$\therefore f(x) = 0$
$\Rightarrow (x + 2)(x - 3) = 0$
$\Rightarrow x + 2 = 0 or x - 3 = 0$
$\Rightarrow x = -2$ or $x = 3$
So, the zeros of $f(x)$ are $-2$ and $3$.

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Question 112 Marks

Very-Short-Answer Question:
If one zero of the polynomial $x^2- 4x + 1$ is $2+\sqrt3.$ Write the other zero.

Answer

Let the other zero of the polynomial $(x^2- 4x + 1)$ be $\alpha.$
Then, sum of roots $=\frac{-(-4)}{1}$
$\therefore2+\sqrt3+\alpha=4$
$\Rightarrow\alpha=2-\sqrt3$
Thus, the other zero is $\big(2-\sqrt3\big).$

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Maths 2 Marks Questions

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