Mathematics Test - 15

X = Set of all citizens of India

R = {(x, y) : x, y ∈ X, lx-yl =5}

$\left|x-x\right|=0\ne 5$ (R is not reflexive)

$xRy\Rightarrow \left|x-y\right|=5$

$\Rightarrow \left|y-x\right|=5$ (R is symmetric)

$xRy\Rightarrow \left|x-y\right|=5$

$yRz\Rightarrow \left|y-z\right|=5$

But $\left|x-z\right|\ne 5$ (R is not transitive)

Given that, A = {u, v, x, y, z}; B = {p, q, r, s}

As we know, a mapping f : x → y is said to be a function, if each element in the set x has its image in set y. It is also possible that these are few elements in set y which are not the image of any element in set x. Every element in set x should have one and only one image.

Given equation ax² + bx + c = 0 (where a ≠ 0) α and β are roots of given equation.

$(a\alpha +b)(a\beta +b)$

$=a^2\alpha \beta +ab(\alpha +\beta )+b^2$

From the given quadratic equation

$\alpha +\beta =\frac{-b}{a},\alpha \beta =\frac{c}{a}$

$a^2\times \frac{c}{a}+ab\times -\frac{b}{a}+b^2=ac$

S = Set of all integers and

R = {(a, b), a ,b∈S and ab≥0}

For reflexive : aRa ⇒ a.a = a²≥0

For all integers a. a ≥ 0

For symmetric : aRb ⇒ ab ≥0 V a, b∈S

If a b ≥ 0, then ba≥ 0 ⇒ bRa

For transitive

If ab ≥ 0, bc ≥ 0, then also ac ≥ 0

Relation R is reflexive, symmetric and transitive.

Therefore relation is equivalence.

We have, 2a²x² — 2abx + b² = 0

Discriminant, D = (-2ab)² - 4(2a²)(b²)

= 4a²b² — 8a²b² = —4a²b² < 0

Roots are always complex.

$f$ : Z N and $f$ (x) = $\left|x\right|$

When we draw a parallel line to x-axis.

It cuts the curve into more than one point.

Therefore, $f$(x) =$\left|x\right|$ is not one-one.

The two complex number are

P = x + iy and Qα + iβ

Quotient =$=\frac{P}{Q}=\frac{x+iy}{\alpha +i\beta },\left|\frac{{\displaystyle P}}{{\displaystyle Q}}\right|=\left|\frac{{\displaystyle x+iy}}{{\displaystyle \alpha +i\beta }}\right|$

=$=\frac{\sqrt{x^2+y^2}}{\sqrt{{{\displaystyle \alpha }}^2+{\beta }^2}}=\sqrt{\frac{x^2+y^2}{{{\displaystyle \alpha }}^2+{\beta }^2}}=\left|\frac{{\displaystyle z_1}}{{\displaystyle z_2}}\right|=\left|\frac{{\displaystyle z_1}}{{\displaystyle z_2}}\right|$

Hence, the quotient of their modulus is equal to the quotient of their moduli.

|2z-1|=|z-2|

|2(x+iy)-1|=|x+iy-2|

|(2x-1)+2yi|= |(x-2)+iy|

$\sqrt{{(2x-1)}^2+4y^2}=\sqrt{{(x-2)}^2+y^2}$

on squaring both sides,

4x²+1-4x+4y²=x²+4-4x+y²

3x²+3y²=3

x²+y²=1

It is the equation of circle.

So, the point z describes a circle.

GP = x

$\frac{a}{1-r}$= x(where, a = Ist term and r = common ratio)

$\Rightarrow \frac{2}{1-r}$= x ... (i) (∵ Given a = 2 and IrI < 1)

⇒$-1<r<1\Rightarrow 1>-r>-1$

⇒$1+1>1-r>1-1$

⇒$0<1-r<2$

$\Rightarrow \frac{1}{1-r}>\frac{1}{2},\frac{2}{1-r}>1$

From equation (i) x > 1

Hence, 1 < x < ∞.