Question

4. Which of the following statements are true and
which are false ? Justify your answers with a
short proof or counter-example. 5x2=10
(a) If A and B are two subsets of a universal set
U, then Ac \B = A\ Bc.
(b) The roots of a quadratic equation are always
real numbers.
(c) lx+yl = lx1 + ly1 for all x,yE R.
(d) The contrapositive of 'If two triangles have
the same area, then they are congruent' is 'If
two triangles are congruent, then they have
the same area'.
(e) If A is a square matrix with I A l = 0, then
two of its rows or two of its columns must be
the same.
5. (a) If x, y and z are positive real numbers,
show that
(x2y + y2z + z2x) (xy2+ yz2+ zx2) > 9x2y2z2.
(b) Use Cardano's method to obtain the roots of
x3— 3x + 2 = 0.

Answer 1

4.

a)  If A and B are two subsets of a universal set U, then $A^C\backslash B=A\backslash B^C$

$x\in A^C\backslash B: x\in U, x\notin A \ and\ x\notin B$

Hence $x\notin A\cup B$ and $x\in (A\cup B)^C$

$x\in A\backslash B^C: x\in A \ and\ x\in B$

Hence $x\in(A\cap B)$

In the left side $x\notin A.$ In the right side $x\in A.$ We have the contradiction/

The statement is False.

b) The roots of a quadratic equation are always real numbers.  

Discriminant: $D=b^2-4ac$.

Counter-example

If the discriminant $D<0,$ the quadratic equation has two complex roots.

The statement is False.

c) lx+yl = lx| + ly| for all x,yE R.  

Counter-example

Let $x=1, y=-1.$

Then $|x+y|=|1-1|=|0|, |x|=|1|=1, |y|=|-1|=1$

$|x|+|y|=1+1=2$

$|x+y|=0\not=2=|x|+|y|$

The statement is False.

(d) The contrapositive of "If two triangles have the same area, then they are congruent" is

"If two triangles are congruent, then they have the same area"  

Let " $p:$ two triangles have the same area" and "$q:$ they are congruent ". Then

"$\sim p:$ two triangles have not the same area"

" $\sim q:$ they are not congruent "

Therefore, the contrapositive of the given statement is

"If two triangles are not congruent, then they have not the same area. "

The statement is False.

e) If A is a square matrix with I A l = 0, then two of its rows or two of its columns must be the same.  

When the determinant of a matrix is zero, its rows are linearly dependent vectors, and its columns are linearly dependent vectors.

Counter-example

$$\begin{vmatrix} 1 & 1 & 2 \\ 2 & 2 & 4 \\ 1 & 2 & 3 \end{vmatrix}=1\begin{vmatrix} 2 & 4 \\ 2 & 3 \end{vmatrix}-1\begin{vmatrix} 2 & 4 \\ 1 & 3 \end{vmatrix}+2\begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix}=$$ $$=2(3)-4(2)-2(3)+4(1)+2(2(2)-2(1))=$$ $$=6-8-6+4+4=0$$

The statement is False.

5. (a) If x, y and z are positive real numbers, show that

$$(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)>9x^2y^2z^2$$

The inequality of arithmetic and geometric means, or more briefly the AM–GM inequality, states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list; and further, that the two means are equal if and only if every number in the list is the same.

$$(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)$$

$$=x^2y^2z^2({x \over z}+{y \over x}+{z \over y})xyz({y \over z}+{z \over x}+{x \over y})$$

Use $AM>GM$

$${{x \over z}+{y \over x}+{z \over y} \over 3}\geq\sqrt[3]{{x \over z}\cdot{y \over x}\cdot{z \over y}}$$

Hence

$${x \over z}+{y \over x}+{z \over y}\geq3$$

$${{y \over z}+{z \over x}+{x \over y} \over 3}\geq\sqrt[3]{{y \over z}\cdot{z \over x}\cdot{x \over y}}$$

Hence

$${y \over z}+{z \over x}+{x \over y}\geq3$$

$$\bigg({x \over z}+{y \over x}+{z \over y}\bigg)\bigg({y \over z}+{z\over x}+{x \over y}\bigg)\geq9$$

For $x>0, y>0, z>0$

$$x^2y^2z^2\bigg({x \over z}+{y \over x}+{z \over y}\bigg)\bigg({y \over z}+{z\over x}+{x \over y}\bigg)\geq9x^2y^2z^2$$

Therefore

$$(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\geq9x^2y^2z^2$$

(b) Use Cardano's method to obtain the roots of

$$x^3-3x+2=0$$

For the depressed cubic $x^3+px+q=0$ Cardano found the following formula for one solution:

$$x=\sqrt[3]{-{q \over 2}+\sqrt{{q^2\over 4}+{p^3 \over 27}}}+\sqrt[3]{-{q \over 2}-\sqrt{{q^2\over 4}+{p^3 \over 27}}}$$

$p=-3, q=2$

$$x=\sqrt[3]{-{2 \over 2}+\sqrt{{2^2\over 4}+{(-3)^3 \over 27}}}+\sqrt[3]{-{2 \over 2}-\sqrt{{2^2\over 4}+{(-3)^3 \over 27}}}$$

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

$$x^3-3x+2=x^3+8-3x-6=$$

$$=(x+2)(x^2-2x+4)-3(x+2)=$$

$$(x+2)(x^2-2x+4-3)=(x+2)(x^2-2x+1)=$$

$$=(x+2)(x-1)^2$$

$$x^3-3x+2=0=>(x+2)(x-1)^2=0$$

$$x_1=-2, x_2=x_3=1$$