Maths

$f:R\to R$ is given by,

$\begin{array}{l}f\left(x\right)=4x+3\\ One-one:\\ Let,f\left(x\right)=f\left(y\right).\\ \Rightarrow 4x+3=4y+3\\ \Rightarrow 4x=4y\\ \Rightarrow x=y\end{array}$

$\therefore$ f is a one-one function.

Onto:

$\begin{array}{l}For,y\in R,let,y=4x+3.\\ \Rightarrow x=\frac{y-3}{4}\in R\end{array}$

Therefore, for any $y\in R$ , there exists $x=\frac{y-3}{4}\in R$ such that

$f\left(x\right)=f\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)+3=y$

$\therefore$ f is onto.

Thus, f is one-one and onto and therefore, $f^{-1}$ exists.

Let us define $g:R\to R$ by $g\left(x\right)=\frac{y-3}{4}$

$\begin{array}{l}Now,\left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)=g\left(4x+3\right)=\frac{\left(4x+3\right)-3}{4}=x\\ \left(fog\right)\left(y\right)=f\left(g\left(y\right)\right)=f\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)+3=y-3+3=y\\ \therefore gof=fog=I_R\end{array}$

Hence, f is invertible and the inverse of f is given by

$f^{-1}=g\left(y\right)=\frac{y-3}{4}$

$f:\left[-1,1\right]\to R$ is given as $f\left(x\right)=\frac{x}{x+2}$

$\begin{array}{l}Let,f\left(x\right)=f\left(y\right).\\ \Rightarrow \frac{x}{x+2}=\frac{y}{y+2}\\ \Rightarrow xy+2x=xy+2y\\ \Rightarrow 2x=2y\\ \Rightarrow x=y\end{array}$

$\therefore$ f is a one-one function.

It is clear that $f:\left[-1,1\right]\to$ Range f is onto.

$\therefore$ $f:\left[-1,1\right]\to$ Range f is one-one onto and therefore, the inverse of the function:

$f:\left[-1,1\right]\to$ Range f exists.

Let g: Range $f\to \left[-1,1\right]$ be the inverse of f.

Let y be an arbitrary element of range f.

Since $f:\left[-1,1\right]\to$ Range f is onto, we have:

$\begin{array}{l}\Rightarrow y=\frac{x}{x+2}\\ \Rightarrow xy+2y=x\\ \Rightarrow x\left(1-y\right)=2y\\ \Rightarrow x=\frac{2y}{1-y},y\ne 1\\ g\left(y\right)=\frac{2y}{1-y},y\ne 1\\ Now,\left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)=g\left(\frac{x}{x+2}\right)=\frac{2\left(\frac{x}{x+2}\right)}{1-\frac{x}{x+2}}=\frac{2x}{x+2-x}=2\frac{x}{2}=x\\ \left(fog\right)\left(y\right)=f\left(g\left(y\right)\right)=f\left(\frac{2y}{1-y}\right)=\frac{\frac{2y}{\left(1-y\right)}}{\left(\frac{2y}{1-y}\right)+2}=\frac{2y}{2y+2-2y}=\frac{2y}{2}=y\\ \therefore gof=I_{-1,1},and,fog-I_{Range,f}\\ \therefore f^{-1}=g\\ \therefore f^{-1}\left(y\right)=\frac{2y}{1-y},y\ne 1\end{array}$

It is given that $f\left(x\right)=\frac{4x+3}{6x-4},x\ne \frac{2}{3}$

$\begin{array}{l}\left(fof\right)\left(x\right)=f\left(f\left(x\right)\right)=f\left(\frac{4x+3}{6x-4}\right)\\ =\frac{4\left(\frac{4x+3}{6x-4}\right)+3}{6\left(\frac{4x+3}{6x-4}\right)-4}=\frac{16x+12+18x-12}{24x+18-24x+16}=\frac{34x}{34}=x\end{array}$

Therefore $fof\left(x\right)=x$ for all $x\ne \frac{2}{3}$

$\Rightarrow fof=1$

Hence, the given function f is invertible and the inverse of f is itself.

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.