Maths

To prove:

$\begin{array}{l}\left(f+g\right)oh=foh+goh\\ consider:\\ \left(\left(f+g\right)oh\right)\left(x\right)\\ =\left(f+g\right)\left(h\left(x\right)\right)\\ =f\left(h\left(x\right)\right)+g\left(h\left(x\right)\right)\\ =\left(foh\right)\left(x\right)+\left(goh\right)\left(x\right)\\ =\left\{\left(foh\right)+\left(goh\right)\right\}\left(x\right)\\ \therefore \left(\left(f+g\right)oh\right)\left(x\right)=\left\{\left(foh\right)+\left(goh\right)\right\}\left(x\right),\forall x\in R\\ Hence,\left(f+g\right)oh=foh+goh\end{array}$

To prove

$\begin{array}{l}\left(f.g\right)oh=\left(foh\right).\left(goh\right)\\ Consider\\ \left(\left(f.g\right)oh\right)\left(x\right)\\ =\left(f.g\right)\left(h\left(x\right)\right)\\ =f\left(h\left(x\right)\right).g\left(h\left(x\right)\right)\\ =\left(foh\right)\left(x\right).\left(goh\right)\left(x\right)\\ =\left\{\left(foh\right).\left(goh\right)\right\}\left(x\right)\\ \therefore \left(\left(f.g\right)oh\right)\left(x\right)=\left\{\left(foh\right).\left(goh\right)\right\}\left(x\right),\forall x\in R\\ Hence,\left(f.g\right)oh=\left(foh\right).\left(goh\right)\end{array}$

52. The given equation lines are.

line 1: xcosθ-y sin θcos 2θ

⇒ xcosθ-y sin θ - kcos 2θ = 0

The perpendicular distance from origin (0,0) to line 1 is

The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as

f = { (1, 2), (3, 5), (4, 1)} and g = { (1, 3), (2, 3), (5, 1)}.

gof (1) = g (f (1) = g (2) = 3 [f (1) = 2 and g (2) = 3]

gof (3) = g (f (3) = g (5) = 1 [f (3) = 5 and g (5) = 1]

gof (4) = g (f (4) = g (1) = 3 [f (4) = 1 and g (1) = 3]

$\therefore$ gof = { (1, 3), (3, 1), (4, 3)}

51. 

Let 0 (o, o) be the origin and P (-1, 2) be the given point on the line y = mx + c.

Then, slope of OP, = $=\frac{2-0}{-1-0}$

Slope of OP = -2

As the line y = mx + c is ⊥ to OP we can write

Given,  $f:R\to R$ defined as $f(x)=3x$

For $x_1,x_2\in R$ such that $f(x_1)=f(x_2)$

$\Rightarrow 3x_1^{}=3x_2^{}$

$\Rightarrow x_1^{}=x_2^{}$

So,  $f$ is one-one

And for $y\in R$ , there exist $\frac{y}{3}\in R$ such that

$f\left(\frac{y}{3}\right)=\frac{3*y}{3}=y$

$\therefore f$ is onto

Hence, option (A) is correct.

Given,  $f:R\to R$ defined by $f(x)=x_{}^4$

For $x_1,x_2\in R$ such that $f(x_1)=f(x_2)$

$\Rightarrow x_1^4=x_2^4$

$\Rightarrow x_1^{}=\pm x_2^{}$

$\Rightarrow x_1^{}=x_2^{}$ or $x_1^{}=-x_2^{}$

So,  $f$ is not one-one

The range of $f(x)$ is a set of all positive real numbers which is not equal to co-domain $R$

So,  $f$ in not onto

$\therefore$ Option (D) is correct

Given, $f:A\to B$ defined by $f(x)=\left(\frac{x-2}{x-3}\right)$

Let $x_1,x_2\in A=R-\left\{3\right\}$ such that

$f(x_1)=f(x_2)$

$\Rightarrow \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}$

$\Rightarrow (x_1-2)(x_2-3)=(x_2-2)(x_1-3)$

$\Rightarrow x_1{x}_2-3x_1-2x_2+6=x_2{x}_1-3x_2-2x_1+6$

$\Rightarrow 2x_1-3x_1=2x_2-3x_2$

$\Rightarrow -x_1=-x_2$

$\Rightarrow x_1=x_{}$

So, $f$ is one-one

For $y\in B=R-\left\{1\right\}$ there exist $f(x)=y$ such that

$\frac{x_{}-2}{x_{}-3}=y$

$\Rightarrow x-2=xy-3y$

$\Rightarrow x-xy=2-3y$

$\Rightarrow x(1-y)=2-3y$

$\Rightarrow x=\frac{2-3y}{1-y}$ where $y\ne 1.\in A$

Thus, $f\left(\frac{2-3y}{1-y}\right)=\frac{\left(\frac{2-3y}{1-y}\right)-2}{\left(\frac{2-3y}{1-y}\right)-3}=\frac{2-3y-2+2y}{2-3y-3+3y}$

$\Rightarrow \frac{-y}{-1}=y$

$\therefore f$ is onto

Given, $f:N\to N$ defined $f(x)=\left(\begin{array}{cc}\hfill \frac{x+1}{2},if\hfill & \hfill xisodd\hfill \\ \hfill \frac{x}{2},if\hfill & \hfill xiseven\hfill \end{array}\right)$ $\cup x\in N$

Let $x_1=1$ and $x_2=2\in N,$

$f(x_1)=f(x_2)\Rightarrow f(1)=f(2)$

$\Rightarrow \frac{1+1}{2}=\frac{2}{2}$

$\Rightarrow 1=1$ but $1\ne 2$

So, $f$ is not one-one

For $x=$ odd and $x\in N$ , say $x=2C+1$ where $C\in N$

There exist $(4C+1)$ $\in N$ such that

$(4C+1)=\frac{4C+1+1}{2}=2C+1.\in N$

And for $x=$ even $\in N$ , say $x=2C$ where $C\in N$

There exist $(4C)\in N$ such that

$f(4C)=\frac{4C}{2}=2C.\in N$

So, $f$ is onto

But, $f$ is not bijective

Given, $f:A*B\to B*A$ defined as $f(a,b)=(b,a)$

Let $(a_1,b_1),(a_2,b_2)\in A*B$ such that

$f(a_1,b_1)=f(a_2,b_2)$

$\Rightarrow (b_1,a_1)=(b_2,a_2)$

So, $b_1=b_2$ and $a_1=a_2$

$\Rightarrow (a_1,b_1)=(a_2,b_2)$

$\therefore f$ is one-one

For $(a,b)\in B*A$

There exist $(a,b)\in A*B$ such that $f(a,b)=(b,a)$

$\therefore f$ is onto

Hence, $f$ is bijective