(4)

Ans.      neither one-one nor onto

Explanation:

We know

                            [2.2] = 2

                            [2.5] = 2

$\because$                       [2.2] = [2.5]

but                         2.2 ≠ 2.5

$\Rightarrow f$ is not one-one function.

Now, consider $0.6\in R$

It is known that $f\left(x\right)=\left[x\right]$ is always an integer. Thus, there does not exist any element $x\in R$ such that $f\left(x\right)=0.6$.

$\therefore$   $f$ is not onto.

Hence, greatest integer function is neither one-one nor onto.

(4)

Ans.         Neither one-one nor onto

Explanation:

Given,        $f\left(x\right)=\frac{x^2-8}{x^2+2}$

                  $f\left(3\right)=\frac{9-8}{9+2}=\frac{1}{11}$

              $f(-3)=\frac{9-8}{9+2}=\frac{1}{11}$

Here, $f\left(3\right)=f(-3)$ but $3\ne -3$.

$\Rightarrow f$ is not one-one.

Now,   $f\left(x\right)=\frac{x^2-8}{x^2+2}=1-\frac{10}{x^2+2}$

So, minimum value of $f\left(x\right)=1-5=-4$

$\therefore$  It is not onto.

Hence it is neither one-one nor onto.

(4)

Ans.     $\mathit{f}$ is neither one-one nor onto

Explanation:

$f:R\to R$ is defined as $f\left(x\right)=x^4$

Let $x,y\in R$ such that

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

$\Rightarrow$                          $x^4=y^4$

$\Rightarrow$                  $x^4-y^4=0$

$\Rightarrow$$(x^2+y^2)(x^2-y^2)=0$

$\Rightarrow$                           $x^2=y^2$

$\Rightarrow$                             $x=\pm y$

$\therefore$                        $f\left(x\right)=f\left(y\right)$ does not imply that $x=y$

For instance,

                              $f\left(1\right)=f(-1)=1$

Here $f\left(1\right)=f(-1)$ but $-1\ne 1$. Hence, $f$ is not one-one.

To check onto,

              $f\left(x\right)=x^4$

Let        $f\left(x\right)=y$ such that $y\in R$

$\Rightarrow$          $x^4=y$

$\Rightarrow$          $x=\pm y^{1/4}$

Note that $y$ is a real number, but it can also be negative.

For example, put     $y=-3$

$\Rightarrow$                          $x={(\pm 3)}^{1/4}$

$\Rightarrow$                          $x=\pm {(-\sqrt{3})}^{1/2}$

which is not possible, as root of a negative number is not a real i.e., $f\left(x\right)=x^4$ is not a onto function.

(4)

Ans.    ${\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{5}\mathit{x}\mathbf{+}\mathbf{6}$

Explanation:

                $f(x+1)=x^2-3x+2$

This function is valid for all real values of $x$

Hence, put $x-1$ in place of $x$,

        $f\left(x\right)={(x-1)}^2-3(x-1)+2$

               $=x^2+1-2x-3x+3+2$

               $=x^2-5x+6$

(1)

Ans.      one-one but not onto

Explanation:

Here,  $f:A\to B$  is defined as {(1, 4),(2, 5),(3, 6)}.

Since, the image of distinct elements of A under $f$ are distinct as:

                        $f\left(1\right)=4,$ $f\left(2\right)=5$ and $f\left(3\right)=6$

From above it is evident that

                        $x_1\ne x_2$ and $f\left(x_1\right)\ne f\left(x_2\right)$

$\therefore$  $f:A\to B\text{ is one-one.}$

Here,  Co-domain of $f\ne$Range of $f$

Hence, $f$ is not onto.

(3)

(3)

$\text{Let }y=f\left(x\right)={\log }_e\sqrt{4-x^2}$

$\Rightarrow \sqrt{4-x^2}=e^y\Rightarrow 4-x^2=e^{2y}\Rightarrow x^2=4-e^{2y}$

$\because$  $x^2\ge 0\Rightarrow 4-e^{2y}\ge 0$

$\Rightarrow e^{2y}\le 4\Rightarrow 2y\le {\log }_{e}4$

$\Rightarrow y\le {\log }_{e}2\Rightarrow -\infty <y\le {\log }_{e}2$

$\therefore$ $\text{Range of the function}=(-\infty ,{\log }_{e}2]$

(1)

$f\left(x\right)$$=\frac{1}{\sqrt{\left|x\right|-x}}$

$f\text{ is defined when }$$\left|x\right|$$-x>0\Rightarrow$$\left|x\right|$$>x$

$\text{Thus }x\text{ must be negative.}$

$\therefore$ $x\in (-\infty ,0)$

$\therefore$$\text{Both assertion and reason are true and reason is the correct explanation of assertion.}$

(3)

$f\left(x\right)$ is to be defined when $x^2-1>0$

$\Rightarrow$$x^2>1\Rightarrow -1>x>1$ and $3+x>0$    $\therefore$  $x>-3$ and $x\ne -2$

$\therefore$$D_r=(-3,-2)\cup (-2,-1)\cup (1,\infty )$

(3)

$f\left(x\right)=\log \left(\frac{1+x}{1-x}\right)$

or  $f\left(y\right)=\log \left(\frac{1+y}{1-y}\right)$  (by changing $x$ to $y$)

Taking $f\left(x\right)+f\left(y\right)$,

$=\log \left(\frac{{\displaystyle 1+x}}{{\displaystyle 1-x}}\right)+\log \left(\frac{{\displaystyle 1+y}}{{\displaystyle 1-y}}\right)=\log \left(\right(\frac{{\displaystyle 1+x}}{{\displaystyle 1-x}})·(\frac{{\displaystyle 1+y}}{{\displaystyle 1-y}}\left)\right)$

$=\log \left(\frac{1+xy+x+y}{1-y-x+xy}\right)=\log \left(\frac{(1+xy)+(x+y)}{(1+xy)-(x+y)}\right)$

$=\log \left(\frac{1+{\displaystyle \frac{x+y}{1+xy}}}{1-{\displaystyle \frac{x+y}{1+xy}}}\right)=f\left(\frac{x+y}{1+xy}\right)$

(4)

(2)

$f\left(x\right)$$=\frac{x-1}{x+1}\Rightarrow x=\frac{f\left(x\right)+1}{1-f\left(x\right)}$

$\text{Now,}$$f\left(2x\right)$$=\frac{2x-1}{2x+1}=\frac{2\left[\frac{f\left(x\right)+1}{1-f\left(x\right)}\right]-1}{2\left[\frac{f\left(x\right)+1}{1-f\left(x\right)}\right]+1}=\frac{3f\left(x\right)+1}{f\left(x\right)+3}$

(2)

$f\left(x\right)=\frac{1}{\sqrt{9-x^2}}$

$\text{Clearly, }9-x^2>0\Rightarrow x^2-9<0$

$\Rightarrow$ $(x+3)(x-3)<0.$  Thus domain of $f\left(x\right)$ is $(-3,3)$.

(2)

(4)

      $x+3>0$$\Rightarrow x>-3\Rightarrow x\in (-3,\infty )$

      $x^2+3x+2\ne 0\Rightarrow (x+2)(x+1)\ne 0$

$\Rightarrow x\ne -1\text{ and }x\ne -2$

$\therefore$ $\text{Domain of }f\text{ is given by }(-3,\infty )-\{-1,-2\}$

(2)

        $x+2\ge 0\Rightarrow x\ge -2$

$\because$  ${\log }_{10}(1-x)\ne 0\Rightarrow 1-x\ne 1\Rightarrow x\ne 0$

Again,  $1-x>0\Rightarrow x<1$

All these can be combined as  $-2\le x<0\text{ or }0<x<1$

(4)

Given, $f\left(x\right)=2x^2$

$\Rightarrow$ $f(3.8)=2\times 3.8\times 3.8=28.88$

$\Rightarrow$ $f\left(4\right)=2\times 4\times 4=32$

$\therefore$$\frac{f(3.8)-f\left(4\right)}{3.8-4}=\frac{28.88-32}{3.8-4}=\frac{3.12}{0.2}=15.6$

(1)

(1)

Let $y=\sqrt{9-x^2}$$\Rightarrow y^2=9-x^2$

$\Rightarrow x^2=9-y^2\Rightarrow x=\sqrt{9-y^2}$

Clearly, $9-y^2\ge 0\Rightarrow y^2\le 9\Rightarrow -3\le y\le 3$

But $y\ge 0$. Hence, $0\le y\le 3$

(4)

Here, $f\left(x\right)={\log }_e\left(\frac{{\displaystyle 1-x}}{{\displaystyle 1+x}}\right)$

Now, $f\left(\frac{{\displaystyle 2x}}{{\displaystyle 1+x^2}}\right)={\log }_e\left(\frac{{\displaystyle 1-\frac{{\displaystyle 2x}}{{\displaystyle 1+x^2}}}}{{\displaystyle 1+\frac{{\displaystyle 2x}}{{\displaystyle 1+x^2}}}}\right)$

$={\log }_e\left(\frac{{\displaystyle 1+x^2-2x}}{{\displaystyle 1+x^2+2x}}\right)={\log }_e{\left(\frac{{\displaystyle 1-x}}{{\displaystyle 1+x}}\right)}^2$

$=2{\log }_e\left(\frac{{\displaystyle 1-x}}{{\displaystyle 1+x}}\right)=2f\left(x\right)$

(4)

Clearly, $f\left(x\right)$ is defined for all $x\ge 0$ and $g\left(x\right)$ is defined for all $x\le 1$

In the common domain of $f+g,f-g,\frac{f}{g},\frac{g}{f},g-f,$ $f\left(x\right)\ne 0$ and $g\left(x\right)\ne 0$$\Rightarrow x\ne 0,1$

Hence, $x\in (0,1)$

(4)

For $f\left(x\right)$ to be defined, $1-$$\left|x\right|$$\ne 0$

$\Rightarrow$$\left|x\right|$$\ne 1\Rightarrow Domain\left(f\right)=R-\{-1,1\}$

(2)

     ${\left[x\right]}^2-3\left[x\right]-10>0$

$\Rightarrow {\left[x\right]}^2-5\left[x\right]+2\left[x\right]-10>0$$\Rightarrow \left(\right[x]-5)\left(\right[x]+2)>0$

$\Rightarrow \left[x\right]<-2\text{ or }\left[x\right]>5$$\Rightarrow x<-2\text{ or }x\ge 6$

$\Rightarrow x\in (-\infty ,-2)\cup [6,\infty )$

(3)

$f\left(x\right)+f\left(\frac{{\displaystyle 1}}{{\displaystyle 1-x}}\right)=1+x$

$x=2\Rightarrow f\left(2\right)+f(-1)=3$              $\cdots \left(i\right)$

$x=-1\Rightarrow f(-1)+f\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)=0$       $\cdots \left(ii\right)$

$x=\frac{1}{2}\Rightarrow f\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)+f\left(2\right)=\frac{3}{2}$          $\cdots \left(iii\right)$

$\left(i\right)+\left(iii\right)-\left(ii\right)\Rightarrow 2f\left(2\right)=\frac{9}{2}$

$\therefore$$f\left(2\right)=\frac{9}{4}$

(3)

We have, $f\left(x\right)=x^2+9$

The graph of $f\left(x\right)$ is the same as the graph of $x^2$ shifted up by 9 units.

We know that the range of $g\left(x\right)=x^2$ is $[0,\infty )$.

$\therefore$  Range of $f\left(x\right)=x^2+9$ is $[9,\infty )$.

(2, 3, 4)

$domf=\mathrm{ℝ}\text{ and }domg=\{x:x>1\}.$ 

Since $x^2\ge 0$ for any $x$, so $rangef=[4,\infty )\text{ and }rangeg=(0,\infty ).$

Hence, $Dom(f+g)=\mathrm{ℝ}\cap (1,\infty )=(1,\infty ),$ $Rangef\cap Rangeg=[4,\infty )$ and $Rangef\cup Rangeg=(0,\infty ).$

(2)

$y=f\left(e^x\right)+f\left(\ln \right|x\left|\right)$

$\text{domain }f\left(x\right)=(0,1)\Rightarrow 0<e^x<1\Rightarrow x<0$$\dots \dots \left(1\right)$

$\text{and }0<\ln$$\left|x\right|$$<1\Rightarrow 1<$$\left|x\right|$$<e\Rightarrow x\in (-e,-1)\cup (1,e)$$\dots \dots \left(2\right)$

$\text{Taking intersection }x\in (-e,-1)$

(2)

$x^2-16\ge 0,x\ne 2,-2,(x+5)(x-2)\ge 0$

$x\in (-\infty ,-5)\cup (2,\infty )$

(2)

We have,

Figure

Total functions = $3^4$ = 81

Number of onto functions = $\frac{4!}{2!2!}\times 3!$ = 36

$\therefore$  Number of not onto functions = 81 – 36 = 45

(1)

$f\left(x\right)=f^{–1}\left(x\right)$

$\Rightarrow f\left(x\right)=x \Rightarrow {(x–1)}^4+1=x$

$\Rightarrow {(x–1)}^4–(x–1)=0$

Figure

$\Rightarrow {(x–1)\left[\right(x–1)}^3–1]=0$

$\Rightarrow x=1, x=2$

So, set S has exactly 2 elements

$\therefore$  (I) is true.

Now, $f\left(x\right)={(x–1)}^4+1$

$\Rightarrow y–1={(x–1)}^4 \Rightarrow x={(y–1)}^{1/4}+1$

$\Rightarrow f^{–1}\left(y\right)={(y–1)}^{1/4}+1$

i.e., $f^{–1}\left(x\right)={(x–1)}^{1/4}+1$

$\therefore f^{–1}(x+1)=x^{1/4}+1$

Now, $f\left(x\right)=f^{–1}(x+1)$

$\Rightarrow {(x–1)}^4+1=x^{1/4}+1$

$\Rightarrow {(x–1)}^4=x^{1/4}$

$\therefore$  (II) is not true.