Maths

Define $f:N\to Z$ as $f\left(x\right)$ and $g:Z\to Z$ as $g\left(x\right)=\left|x\right|$

We first show that g is not injective.

It can be observed that:

$\begin{array}{l}g\left(-1\right)=\left|-1\right|=1\\ g\left(1\right)=\left|1\right|=1\\ \therefore g\left(-1\right)=g\left(1\right),but,-1\ne 1\end{array}$

$\therefore g$ is not injective.

Now, $gof:N\to Z$ is defined as

$gof\left(x\right)=y\left(f\left(x\right)\right)=y\left(x\right)=\left|x\right|$

Let $x,y\in N$ such that $gof\left(x\right)-gof\left(y\right)$

$\Rightarrow \left|x\right|=\left|y\right|$

Since $x,y\in N$ , both are positive.

$\therefore \left|x\right|=\left|y\right|\Rightarrow x=y$

Hence, gof is injective

61. The given Eqⁿ of the line is $\frac{x}{4}+\frac{y}{6}$ = 1 ______ (1)

so, Slope of line = -$\frac{6}{4}=-\frac{3}{2}.$

The line ⊥ to line (1) say l₂ has

Slope of l₂ = $\frac{-1}{(-3/2)}=\frac{2}{3}.$

Let P (0, y) be the point of on y-axis where it is cut by the line (1)

Then,  $\frac{0}{4}+\frac{y}{6}=1$

y = 6

i.e, the point P has co-ordinate (0, 6)

Eqⁿ of line ⊥ to $\frac{x}{4}+\frac{y}{6}=1$ and cuts y-axis at P (0,6) is

y – 6 = $\frac{2}{3}$ (x – 0)

3y – 18 = 2x

2x – 3y + 18 = 0

f: R → R is given as f (x) = x³.

Suppose f (x) = f (y), where x, y ∈ R.

⇒ x³ = y³  … (1)

Now, we need to show that x = y.

Suppose x ≠ y, their cubes will also not be equal.

⇒ x³ ≠ y³

However, this will be a contradiction to (1).

∴ x = y

Hence, f is injective.

26. Given, f(x)=x².

$\frac{f(1.1)-f(1)}{1.1-1}=\frac{{(1.1)}^2-1^2}{1.1-1}=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=2.1$

60. The given eqn of lines are

x - 7y + 5 = 0 ______ (1) ⇒ x = 7y - 5

and 3x + y = 0 _________ (2)

Solution (1) and (2) we get,

3 [7y – 5] + y = 0 .

⇒ 21y - 15 + y = 0

⇒ 22y = 15

It is given that $f:R\to R$ is defined as $f\left(x\right)=x^2-3x+2.$

$\begin{array}{l}f\left(f\left(x\right)\right)=f\left(x^2-3x+2\right)\\ ={\left(x^2+3x+2\right)}^2-3\left(x^2-3x+2\right)+2\\ =x^4+9x^2+4-6x^2-12x+4x^2-3x^2+9x-6+2\\ =x^4-6x^2+10x^2-3x\end{array}$

25. Given, f(x)= $\{\begin{array}{ll}{x}^2,\hfill & 0\le x\le 3\hfill \\ 3x,\hfill & 3\le x\le 10\hfill \end{array}$

f(x)={(0,0),(1,1),(2,4),(3,9),(4,12),(5,15),(6,18),(7,21),(8,24),(9,27),(10,30)}

So, the elements in domain of f  has one and only one image.

 ? f(x) is a function.

Given, g(x)= $\{\begin{array}{ll}{x}^2,\hfill & 0\le x\le 2\hfill \\ 3x,\hfill & 2\le x\le 10\hfill \end{array}$ .

g(x)={(0,0),(1,1),(2,4),(2,6),(3,9),(4,12),(5,15),(6,18),(7,21),(8,24),(9,27),(10,30)}

So, the element 2 of the domain has more than one image i.e., 4 and 6.  

? g(x) is not a function.

It is given that $f:R\to R$ is defined as $f\left(x\right)=10x+7.$

One-one:

$\begin{array}{l}Let,f\left(x\right)=f\left(y\right),where,x,y\in R\\ \Rightarrow 10x+7=10y+7\\ \Rightarrow x=y\end{array}$

$\therefore f$ is a one-one function.

Onto:

$\begin{array}{l}For,y\in R,let,y=10x+7\\ \Rightarrow x=\frac{y-7}{10}\in R\end{array}$

Therefore, for any $y\in R$ ,there exists $x=\frac{y-7}{10}\in R$

Such that

$f\left(x\right)=f\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)+7=y-7+7=y$

$\therefore f$ is onto.

Therefore, f is one-one and onto.

Thus, f is an invertible function.

Let us define $g:R\to R$ as $g\left(y\right)=\frac{y-7}{10}$

Now, we have

$\begin{array}{l}gof\left(x\right)=g\left(f\left(x\right)\right)=g\left(10x+7\right)=\frac{\left(10x+7\right)-7}{10}=10\frac{x}{10}=10\\ And\\ fog\left(y\right)=f\left(g\left(y\right)\right)=f\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)+7=y-7+7=y\end{array}$

Hence, the required function $g:R\to R$ is defined as $g\left(y\right)=\frac{y-7}{10}$