Maths M.C.Q

2. $-\frac{5}{2}$

Solution:

Finding a zero of p(x) is the same as solving an equation p(x) = 0.

Now, p(x) = 0 ⇒ 2x + 5 = 0,

2x = -5

Which give us $\text{x}=-\frac{5}{2}.$

Therefore, $-\frac{5}{2}$ is the zero of the polynomial.

2. $1$

Solution:

We have,

$\text{p}\text{(x)}=\text{x}^2-2\sqrt{2}\text{x}+1$

$\text{p}(2\sqrt{2})=(2\sqrt{2})^2-2\sqrt{2}(2\sqrt{2})+1$

$= 8 - 8 + 1$

$= 1$

3. 0

Solution:

Given, $\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{x}}=-1$

$\Rightarrow\frac{\text{x}^2+\text{y}^2}{\text{xy}}=-1$

$\Rightarrow\text{x}^2+\text{y}^2=-\text{xy}$

$\Rightarrow\text{x}^2+\text{y}^2+\text{xy}=0$

Now, $\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{x}^2+\text{xy}+\text{y}^2) \ ...(\text{i})$

$[\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)]$

$=(\text{x}-\text{y})\times0=0$ [From Eq. (i)]

3. $\frac{1}{4}$

Solution:

$49\text{x}^2 -\text{b}=\Big(7\text{x}+\frac{1}{2}\Big)\Big(7\text{x}-\frac{1}{2}\Big)$

$\Rightarrow49\text{x}^2 -\text{b}=\Big(7\text{x}\Big)^2-\Big(\frac{1}{2}\Big)^2$

$49^2-\frac{1}{4} [\therefore(\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2]$

So, we get $\text{b}=\frac{1}{4}.$

3. ​​​​​​$\text{x}^2+\frac{3\text{x}^{\frac{3}{2}}}{\sqrt{\text{x}}}$

Solution:

1. Now, $\frac{\text{x}^2}{2}-\frac{2}{\text{x}^2}=\frac{\text{x}^2}{2}-2\text{x}^{-2},$ it is not a polynomial, because exponent of x is -2 which is not a whole number.

2. Now, $\sqrt{2\text{x}-1}=\sqrt{\text{2}\text{x}}^{\frac{1}{2}}-1, $ it is not a polynomial, because exponent of x is $-\frac{1}{2}$ which is not a whole number.

3. Now, $\text{x}^2+\frac{3\text{x}^{\frac{3}{2}}}{\sqrt{\text{x}}}=\text{x}^2+3\text{x}^{\frac{3}{2}-\frac{1}{2}}=\text{x}^2+3\text{x}^{\frac{2}{2}}=\text{x}^2+3\text{x},$ it is not a polynomial, because exponent of x is which is a whole number.

4. $\frac{\text{x}-1}{\text{x}+1},$ it is not a polynomial because it is a rational function.

2. ​​​​​​0

Solution:

$\sqrt{2}$ is a constant polynomial. The only term here is $\sqrt{2}$ which can be written as $\sqrt{2}\text{x}^\circ.$ So, the exponent of x is zero. Therefore, the degree of the polynomial is 0.

2. $\frac{1}{2}$

Solution:

Let p(x) = 2x² + 7x - 4

= 2x² + 8x - x- 4 [by splitting middle term]

= 2x(x + 4) -1(x + 4)

=(2x - 1)(x + 4)

For zeroes of p(x), put p(x) = 0 ⇒ (2x - 1)(x + 4) = 0

⇒ 2x - 1 = 0 and x + 4 = 0

$\Rightarrow\text{x}=\frac{1}{2}$ and $\text{x}=-4$

Hence, one of the zeroes of the polynomial p(x) is $\frac{1}{2}.$

4. 2abc

Solution:

Now, a³ + b³ + c³ = (a + b + c)(a² + b² + c² - ab - be - ca) + 3abc

[Using identity, a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² - ab - be - ca)] = 0 + 3abc

$\therefore$ a + b + c = 0, given

a³ + b³ + c³ = 3abc