True False[1 Marks ]

Question 11 Mark

Are the following statements 'True' or 'False'? Justify your answers.
If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.

Answer

True:
Let $\beta = 0, \gamma = 0$
$\text{f(x)}=(\text{x}-\alpha)(\text{x}-\beta)(\text{x}-\gamma)$
$=(\text{x}-\alpha)\text{x . x}$
$\Rightarrow\ \text{f(x)}=\text{x}^3-\alpha\text{x}^2$
which has no linear (coefficient of x) and constant terms.

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Question 21 Mark

Are the following statements 'True' or 'False'? Justify your answers.
If the graph of a polynomial intersects the x-axis at only one point, it cannot be a quadratic polynomial.

Answer

False:
The given statement is false, because when two zeroes of a quadratic polynomial are equal, then two intersecting ponits coincide to become one point.

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Question 31 Mark

Are the following statements 'True' or 'False'? Justify your answers.
If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign.

Answer

True:
$\alpha,$ $\beta,$ and $\gamma$ are all (-)ive for cubic polynomial $ax^3 + bx^2 + cx + d.$
$\alpha +\beta +\gamma=\frac{-\text{b}}{\text{c}}\ .....\text{(i)}$
$\alpha\beta+\beta\alpha+\gamma\alpha=\frac{\text{c}}{\text{a}}\ .....\text{(ii)}$
$\alpha\beta\gamma=\frac{-\text{d}}{\text{a}}\ .....(\text{iii})$
$\because\ \alpha,\beta,\gamma,$ are all negative so,
$\alpha+\beta+\gamma=-\text{x}$ (any negative number)
$\Rightarrow\ \frac{-\text{b}}{\text{c}}=-\text{x}$ [From (i)]
$\Rightarrow\ \frac{\text{b}}{\text{a}}=\text{x}$
So, a, b, have same sign and product of any two zeroes will be positive.
So, $\alpha\beta+\beta\gamma+\gamma\alpha=+\text{y}$ (Any positive number)
$\Rightarrow\ \frac{\text{d}}{\text{a}}=\text{z}$
So, d and a will have same sign.
Hence, signs of b, c, d will be same as of a.
So, signs of a, b, c, d will be same either positive of negative.

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Question 41 Mark

Are the following statements 'True' or 'False'? Justify your answers.
If the zeroes of a quadratic polynomial $ax^2 + bx + c$ are both positive, then a, b and c all have the same sign.

Answer

False:
Let $\alpha$ and $\beta$ be the roots of the quadratic polynomial. if a and $\beta$ are positive then $\alpha + \beta=\frac{-\text{b}}{\text{a}}$ it shows that $\frac{-\text{b}}{\text{a}}$ is negative nut sum of two positive numbers $(\alpha,\beta)$ must be +ive i.e. either b or a must be negative. So a, b and c will have different signs.

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Question 51 Mark

Are the following statements 'True' or 'False'? Justify your answers.
If all three zeroes of a cubic polynomial $x^3+a x^2-b x+c$ are positive, then at least one of $a, b$ and $c$ is non-negative.

Answer

True:As all zeroes of cubic polynomial are positive
Let $f(x) = x^3 + ax^2 - bx + c$
$\therefore\ \alpha+\beta+\gamma=+\text{ ive say}+\text{x}$
$\Rightarrow\ \frac{-\text{b}}{\text{a}}=\text{x}$
⇒ a and b has opposite signs .....(i)
$\alpha\beta+\beta\gamma+\gamma\alpha=+\text{y}$
$\Rightarrow\ \frac{\text{c}}{\text{a}}=\text{y}$
So, Signs of a and c are same. .....(ii)
Now, $\alpha\beta\gamma=+\text{ive}=+\text{z}$
$\Rightarrow\ \frac{-\text{d}}{\text{a}}=\text{z}$
⇒ a and d have opposite signs. [From (i)]
From (i), if a is positive, then b is negative.
From (ii) if a is positive, then c is also positive.
From (iii) if a is positive, then d is negative.
Hence, if zeroes $\alpha,$ $\beta,$ $\gamma$ of cubic polynomial are positive then out of a, b, c at least one is positive.

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Question 61 Mark

Are the following statements 'True' or 'False'? Justify your answers.
If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.

Answer

True:
If a polynomial of degree more than two has real zeroes and other zeroes are not real or are imaginary, then graph of the polynomial will intersect at two points on x-axis.

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Question 71 Mark

Are the following statements 'True' or 'False'? Justify your answers.
The only value of k for which the quadratic polynomial $kx^2 + x + k$ has equal zeros is $\frac{1}{2}$.

Answer

False:$f(x) = kx^2 + x + k (a = k, b = 1, c = k)$
For equal roots
$b^2 - 4ac = 0$
$\Rightarrow (1)^2 - 4(k)(k) = 0$
$\Rightarrow 4k^2 = 1$
$\Rightarrow\ \text{k}^2=\frac{1}{4}$
$\Rightarrow\ \text{k}=\pm\frac{1}{2}$
So, there are $\frac{1}{2}$ and $\frac{-1}{2}$ values of k so that the given equation has equal rooks.

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Maths True False[1 Marks ]

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