**Question:**

Which of the following statements are true and which are false? In each case give a valid reason for your answer.

(i) p: $\sqrt{11}$ is an irrational number

(ii) q: Circle is a particular case of an ellipse.

(iii) r: Each radius of a circle is a chord of the circle

(iv) S: The center of a circle bisects each chord of the circle

(v) $\mathrm{t}$ : If $\mathrm{a}$ and $\mathrm{b}$ are integers such that $\mathrm{a}<\mathrm{b}$, then $-\mathrm{a}>-\mathrm{b}$.

(vi) $y$ : The quadratic equation $x^{2}+x+1=0$ has no real roots

**Solution:**

(i) $\mathbf{p}: \sqrt{1} 1$ is an irrational number is a TRUE statement.

An irrational number is any number which cannot be expressed as a fraction of two integers. $\sqrt{1} 1$ cannot be expressed as a fraction of two integers, so $\sqrt{11}$ is an irrational number.

(ii) $\mathrm{q}$ : Circle is a particular case of an ellipse is a TRUE statement.

A circle is a particular case of an ellipse with the same radius in all points.

The equation of an ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

When $a=b$, we will get the equation of the circle, $x^{2}+y^{2}=1$

(iii) $\mathrm{r}$ : Each radius of a circle is a chord of the circle is a FALSE statement.

A chord intersects the circle at two points, but radius intersects the circle only at one point. So the radius is not a chord of the circle.

(iv) S: The center of a circle bisects each chord of the circle is a FALSE statement.

The only diameter of a circle is bisected by the center of the circle. Except for diameter, no other chords are bisected the center of the circle. The only center lies on the diameter of the circle.

(v) $\mathrm{t}$ : If $\mathrm{a}$ and $\mathrm{b}$ are integers such that $\mathrm{a}<\mathrm{b}$, then $-\mathrm{a}>-\mathrm{b}$ is a TRUE statement. $a**-b$, is TRUE by the rule of inequality.**

(vi) $y$ : The quadratic equation $x^{2}+x+1=0$ have no real roots is a TRUE statement.

General form of a quadratic equation is $a x^{2}+b x+c=0$.

If $b^{2}-4 a c<0$, there is no real solution.

In the given equation; $x^{2}+x+1=0$

$a=1 ; b=1 ; c=1$

$b^{2}-4 a c=1-4 \times 1 \times 1=-3<0$

So, there is no real root.

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