Maths M.C.Q (1 Marks)

$a=b$

$R_1$ is a reflexive symmetric and transitive. ( as we know $a=b$ )

$\therefore(a, a)$ is present

$\therefore(a, b)(b, a)$ will also be present.

So it is an equivalence relation .

Therefore, total number of reflexive relation set with $4$ elements $ = {2^4}$.

Since $2 | 6$ but $6 |2$, therefore $R$ is not symmetric.

Let $n\ R\ m$ and $m\ R\ p$ ==> $n|m$ and $m|p ==> n|p ==> nRp$.

So, $R$ is transitive.

${R^{ - 1}} = \{ (5,\,4);\,(4,\,1);\,(6,\,4);\,(6,\,7);\,(7,\,3)\} $.

We now obtain the elements of ${R^{ - 1}}oR$ we first pick the element of

$R$ and then of ${R^{ - 1}}$. Since $(4,\,5) \in R$ and $(5,\,4) \in {R^{ - 1}}$, we have $(4,\,4) \in {R^{ - 1}}oR$

Similarly, $(1,\,4) \in R,\,(4,\,1) \in {R^{ - 1}} \Rightarrow \,(1,\,1) \in {R^{ - 1}}oR$

$(4,\,6) \in R,\,(6,\,4) \in {R^{ - 1}} \Rightarrow \,(4,\,4) \in {R^{ - 1}}oR,$

$(4,\,6) \in R,\,(6,\,7) \in {R^{ - 1}} \Rightarrow \,(4,\,7) \in {R^{ - 1}}oR$

$(7,\,6) \in R,\,(6,\,4) \in {R^{ - 1}} \Rightarrow \,(7,\,4) \in {R^{ - 1}}oR,$

$(7,\,6) \in R,\,(6,\,7) \in {R^{ - 1}} \Rightarrow \,(7,\,7) \in {R^{ - 1}}oR$

$(3,\,7) \in R,\,(7,\,3) \in {R^{ - 1}} \Rightarrow \,(3,\,3) \in {R^{ - 1}}oR,$

Hence,${R^{ - 1}}oR = \{(1, 1); (4, 4); (4, 7); (7, 4), (7, 7); (3, 3)\}.$

$(1,\,2)\,\, \in R,\,(2,\,3) \in R$

$\therefore $ $R$ is symmetric if $(2, 1), (3, 2) \in R.$

Now, $R = \{ (1,\,1),\,(2,\,2),\,(3,\,3),\,(2,\,1),\,(3,\,2),\,(2,\,3),\,(1,\,2)\} $

$R$ will be transitive if $(3, 1); (1, 3)$ $\therefore$  $R$.

Thus, $R$ becomes an equivalence relation by adding $(1, 1) \,(2, 2)\, (3, 3)\, (2, 1)\, (3,2)\, (1, 3)\, (3, 1)$.

Hence, the total number of ordered pairs is $7$.

$\therefore$  $R = \{ (2,\,1);(3,\,1)\} $

Hence, range of $R = \{ 1\} $.

${R_1}$ is not symmetric because $(p,\,q) \in {R_1}$ but $(q,\,p) \notin {R_1}$

${R_2}$ is not symmetric because $(r,\,q)\, \in {R_2}$ but $(q,\,r)\, \notin {R_2}$

${R_3}$is not symmetric because $(p,\,q) \in {R_3}$ but $(q,\,p) \notin {R_3}$.

Hence, ${R_1},\,{R_2},\,{R_3}$ are not equivalence relation.

${l_1}$ is parallel ${l_2}$ and also ${l_2}$ is parallel to ${l_1}$, so it is symmetric.

Clearly, it is also reflexive and transitive.

Hence it is equivalence relation.

$f(x/y) = \sin \log (x/y) = \sin \,(\log x - \log y)$.....$(ii)$

$\therefore \,\,f(xy) + f(x/y) = 2\,\sin \log x\cos \,\,\log y$

Hence required value of the function is

$2\,\sin \,\,\log x\,\cos \,\log y - 2\,\sin \,\log x\,\cos \,\log y = 0$.

$ \Rightarrow \,\,[b\,{(x + 1)^2} + c\,(x + 1) + d] - (b{x^2} + cx + d) = 8x + 3$

$ \Rightarrow \,\,(2b)\,x + (b + c) = 8x + 3$

$ \Rightarrow \,\,2b = 8,\,\,b + c = 3\,\,\, \Rightarrow \,\,b = 4,\,\,c = - 1.$

$ = \frac{1}{2}\,\left[ {{a^{x + y}} + {a^{ - x - y}} + {a^{x - y}} + {a^{ - x + y}}} \right]$

$ = \frac{1}{2}\,\left[ {{a^x}({a^y} + {a^{ - y}}) + {a^{ - x}}({a^y} + {a^{ - y}})} \right]$

$ = \frac{1}{2}\,({a^x} + {a^{ - x}})\,\,({a^y} + {a^{ - y}}) = 2f(x)\,f(y)$.

$\cos \,(\log {x^2})\cos \,(\log {y^2}) - \frac{1}{2}\,\left[ {\cos \log \frac{{{x^2}}}{2} + \cos \log \frac{{{x^2}}}{{{y^2}}}} \right]$

$ = \frac{1}{2}\left[ {\cos \,(\log {x^2} + \log {y^2}) + \cos \,(\log {x^2} - \log {y^2})} \right]$

$ - \frac{1}{2}\left[ {\cos \,\log \frac{{{x^2}}}{2} + \cos \,(\log {x^2} - \log {y^2})} \right]$

$ = \frac{1}{2}\,\left[ {\cos \log {x^2}{y^2} - \cos \log \frac{{{x^2}}}{2}} \right]$.

Now let $y = f(x)\,\,.\,\,f(4) - \frac{1}{2}\,\left[ {f\left( {\frac{x}{4}} \right) + f(4x)} \right]$

==> $y = \cos \,(\log x).\cos \,(\log 4) $

$- \frac{1}{2}\,\left[ {\cos \,\log \,\left( {\frac{x}{4}} \right) + \cos \,(\log 4x)} \right]$

==> $y = \cos \,(\log x)\,\cos \,(\log 4)$

$ - \frac{1}{2}\,\left[ {\cos \,(\log x - \log 4) + \cos \,(\log x + \log 4)} \right]$

==> $y = \cos \,(\log x)\,\cos \,(\log 4) - \frac{1}{2}\,\left[ {2\,\cos \,(\log x)\,\cos \,(\log 4)} \right]$

==> $y = 0$.

$= \frac{{{x^2} + 1 - 2}}{{{x^2} + 1}} = 1 - \frac{2}{{{x^2} + 1}}$

$\because {{x^2} + 1} > 1;   \,\, \therefore  \frac{2}{{{x^2} + 1}} \le 2$

$\therefore  1 - \frac{2}{{{x^2} + 1}} \ge 1 - 2$; 

$\therefore - 1 \le f(x) < 1$

Thus $f(x)$ has the minimum value equal to $-1$.

Let $g(x) = {\sin ^{ - 1}}(3 - x)$ ==> $ - 1 \le 3 - x \le 1$

Domain of $g(x)$ is $[2, 4]$

and let $h(x) = \log \left[ {|x| - 2} \right]$ ==> $|x| - 2 > 0$

==> $|x|\, > 2$ ==> $x < - 2$ or $x > 2$

==> $( - \infty ,\, - 2) \cup (2,\,\infty )$

we know that

$(f/g)(x) = $ $\frac{{f(x)}}{{g(x)}}\forall x \in {D_1} \cap {D_2} - \left\{ {x \in R:g(x) = 0} \right\}$

$\therefore$ Domain of $f(x) = (2,\,4] - \{ 3\} $$ = (2,\,3) \cup (3,\,4]$.

So the given function is defined for all $x \in R - \{ ( - 1,\,\,1)\,\, \cup \,\,(n\,\,|\,\,n \in Z)\} .$

So, it is an odd function.

In $(b)$, $f( - x) = ( - x)\frac{{{a^{ - x}} - 1}}{{{a^{ - x}} + 1}} = - x\frac{{1 - {a^x}}}{{1 + {a^x}}} = x\frac{{{a^x} - 1}}{{{a^x} + 1}} = f(x)$

So, it is an even function.

In $(c)$, $f( - x) = - \sin \left[ {\log (x + \sqrt {1 + {x^2}} )} \right]$

So, it is an odd function.

In $(d)$, $f( - x) = \sin ( - x) = - \sin x = - f(x)$

So, it is an odd function.

$\therefore$ ${e^x} - y = \sqrt {1 + {y^2}} $

Squaring both the sides, ${({e^x} - y)^2} = (1 + {y^2})$

${e^{2x}} + {y^2} - 2y{e^x} = 1 + {y^2} \Rightarrow {e^{2x}} - 1 = 2y{e^x}$

==> $2y = \frac{{{e^{2x}} - 1}}{{{e^x}}} \Rightarrow 2y = {e^x} - {e^{ - x}}$

Hence, $y = \frac{{{e^x} - {e^{ - x}}}}{2}$.

==> $f(x) = \frac{{{{\cos }^2}x + {{\sin }^2}x(1 - {{\cos }^2}x)}}{{{{\sin }^2}x + {{\cos }^2}x(1 - {{\sin }^2}x)}}$

==> $f(x) = \frac{{{{\sin }^2}x + {{\cos }^2}x - {{\sin }^2}x{{\cos }^2}x}}{{{{\sin }^2}x + {{\cos }^2}x - {{\sin }^2}x{{\cos }^2}x}}$

==> $f(x) = 1$ ==> $f(2002) = 1$.

This is true for $x \in R.$.

$f(x) = f(y) \Rightarrow \frac{{x - m}}{{x - n}} = \frac{{y - m}}{{y - n}} \Rightarrow x = y$

$\therefore$ $f$ is one-one.

Let $\alpha$ $\in$ $R$ such that $f(x) = \alpha \Rightarrow \frac{{x - m}}{{x - n}} = \alpha $

==> $x = \frac{{m - n\alpha }}{{1 - \alpha }}$

Clearly $x \notin R$ for $\alpha = 1$. So, $f$ is not onto.

==> $f( - x) = \sin \,[\log \,( - x + \sqrt {1 + {x^2}} )]$

==> $f( - x) = \sin \,\log \left( {(\sqrt {1 + {x^2}} - x)\frac{{(\sqrt {1 + {x^2}} + x)}}{{(\sqrt {1 + {x^2}} + x)}}} \right)$

==> $f( - x) = \sin \,\log \,\left[ {\frac{1}{{(x + \sqrt {1 + {x^2}} )}}} \right]$

==> $f( - x) = \sin \left[ {\log {{(x + \sqrt {1 + {x^2}} )}^{ - 1}}} \right]$

==> $f( - x) = \sin \left[ { - \log (x + \sqrt {1 + {x^2}} )} \right]$

==> $f( - x) = - \sin \left[ {\log (x + \sqrt {1 + {x^2}} )} \right]$==> $f( - x) = - f(x)$

$f(x)$ is odd function.

$ = \frac{{\left( {\frac{{x - 3}}{{x + 1}}} \right) - 3}}{{\left( {\frac{{x - 3}}{{x + 1}}} \right) + 1}} = \frac{{x - 3 - 3x - 3}}{{x - 3 + x + 1}} = \frac{{3 + x}}{{1 - x}}$

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

$ = \frac{{\left( {\frac{{x - 3}}{{x + 1}}} \right) - 3}}{{\left( {\frac{{x - 3}}{{x + 1}}} \right) + 1}} = \frac{{x - 3 - 3x - 3}}{{x - 3 + x + 1}} = x$

$⇒$ $f\left( {\frac{{200x}}{{100 + {x^2}}}} \right) = \log \left[ {\frac{{10 + \frac{{200x}}{{100 + {x^2}}}}}{{10 - \frac{{200x}}{{100 + {x^2}}}}}} \right] = \log {\left[ {\frac{{10(10 + x)}}{{10(10 - x)}}} \right]^2}$

$ = 2\log \left( {\frac{{10 + x}}{{10 - x}}} \right) = 2f(x)$

$\therefore$ $f(x) = \frac{1}{2}f\left( {\frac{{200x}}{{100 + {x^2}}}} \right) $

$\Rightarrow k = \frac{1}{2} = 0.5.$

==> $f(cx + d) = g(ax + b)$ 

==> $a[cx + d] + b = c[ax + b] + d$

==> $ad + b = cb + d$==> $f(d) = g(b)$.

$ = f\,\left[ {\frac{{\left( {\frac{x}{{\sqrt {1 + {x^2}} }}} \right)}}{{\sqrt {1 + \frac{{{x^2}}}{{1 + {x^2}}}} }}} \right] $

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

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

$= \frac{{\frac{x}{{\sqrt {1 + 2{x^2}} }}}}{{\sqrt {\left[ {1 + \frac{{{x^2}}}{{1 + 2{x^2}}}} \right]} }} $

$= \frac{x}{{\sqrt {1 + 3{x^2}} }}.$

and $\psi \,\left\{ {\phi \,(x)\,} \right\} = \psi \,({x^2} + 1) = {3^{{x^2} + 1}}$.

==> $f(1 + \sqrt x ) = 3 + 2\sqrt x + x$

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

then, $f(y) = 3 + 2(y - 1) + {(y - 1)^2}$$ = 2 + {y^2}$

therefore, $f(x) = 2 + {x^2}$.

and $(gof)(x) = g(f(x)) = c(ax + b) + d$

Given that, $(fog)(x) = (gof)(x)$ and at $a = 1,\,b = 2$

==> $cx+d+2 = cx +2c +d $ ==> $c = 1$ and $d$ is arbitrary.

==> $f\,o\,g = f(g(x)) = 3(g(x)) + 10$

$= 3({x^2} - 1) + 10$=$3{x^2} + 7$.....(i)

Let $3{x^2} + 7 = y$ ==> $3{x^2} = y - 7$

==> ${x^2} = \frac{{y - 7}}{3} \Rightarrow x = {\left( {\frac{{y - 7}}{3}} \right)^{1/2}}$

We know that $f(x) = y$, then $x = {f^{ - 1}}(y)$

so ${(fog)^{ - 1}} = {\left( {\frac{{x - 7}}{3}} \right)^{1/2}}$.

==> ${(2x + 3)^2} + 7 = 8$ ==> $2x + 3 = \pm 1$ ==> $x = - 1,\, - 2$.

So, ${R^{ - 1}}$ $= \{(1, 2), (2, 4), (3, 6),.....\}.$

$\Rightarrow x = \frac{1}{2}{\log _{10}}\left( {\frac{{1 + y}}{{1 - y}}} \right)$

Let $y = f(x)$ ==> $x = \pi ,\,\,f(\pi ) = - \tan \frac{\pi }{4} = - 1$

==> ${f^{ - 1}}(y) = \frac{1}{2}{\log _{10}}\left( {\frac{{1 + y}}{{1 - y}}} \right)$ 

==>${f^{ - 1}}(x) = \frac{1}{2}{\log _{10}}\left( {\frac{{1 + x}}{{1 - x}}} \right)$

Since, $A \times A$ contains $100$ ordered pairs $(a, b)$ out of which $10$ ordered pairs are such that $a=b$.

For a reflexive relation $(a, a)$ must be present and others have a choice of to be present or not.

So, number of reflexive relations $=2^{90}$.

For a symmetric relation if $(a, b)$ is present then $(b, a)$ is also present

(where $a \neq b$ ). There are $45$ such pairs of ordered pairs.

So, number of reflexive relations which are also symmetric $=2^{45}$

$\therefore$ Required number of relations $=2^{90}-2^{45}$.

Given, $\ln f(x)=10 \ln \sin x+$ $\ln \left(1+\cos ^2 x+\cos ^4 x+\cos ^8 x\right)$

$\frac{f^{\prime}(x)}{f(x)}=10 \cot x-\sin 2 x$

${\left[\frac{1+2 \cos ^2 x+4 \cos ^6 x}{1+\cos ^2 x+\cos ^4 x+\cos ^8 x}\right] }$

$\left[\frac{1+2 \cos ^2 x+4 \cos ^6 x}{1+\cos ^2 x+\cos ^4 x+\cos ^8 x}\right]$

$\because f(x)$ is periodic with period $\pi$ and $\frac{f^{\prime}(x)}{f(x)}$ is continuous in $(0, \pi) .$

$\frac{f^{\prime}(x)}{f(x)}=\underbrace{10 \cot x}_{\begin{array}{c}\text { Range is }(-\infty, \infty) \\ \text { for } x \in(0, \pi)\end{array}}$

$\underbrace{-\sin 2 x\left[\frac{1+2 \cos ^2 x+4 \cos ^6 x}{1+\cos ^2 x+\cos ^4 x+\cos ^8 x}\right]}_{\text {finite number for } x \in(0, \pi)}$

So, $\frac{f^{\prime}(x)}{f(x)}$ has range $(-\infty, \infty)$ Hence, $\lambda \in R$.

Given, $x^3-[x]^3=(x-[x])^3$

$(x-\mid x])\left(x^2+[x]^2+x[x]\right)=(x-[x])^3$

$x-[x]=0$ or $x[x]=0$

$\{x\}=0$ or $x[x]=0$

$x \in Z$ or $x \in[0,1)$

Hence, solution in $x \in(0, 1 ) \cup Z$.

Herc, $f(x)$ cannot be many one.

Since, If $f(x)$ is many one, then there exist $x_1$ and $x_2$ such that $f\left(x_1\right)=f\left(x_2\right)$ but $x_1 \neq x_2$. Then $0 \geq \left|x_1-x_2\right|$, which is a contradiction $\therefore f(x)$ is one-one.

Given, $|f(x)-f(y)| \geq |x-y|$

At anv point $x=a$

$\lim _{h \rightarrow 0}|f(a+h)-f(a)|_{\geq 1}|(a+h)-a|$

$\Rightarrow\left|f^{\prime}(a)\right| \geq 1 \Rightarrow f^{\prime}(a) \geq 1 \text { or } f^{\prime}(a) \leq-1$

But both can not hold simultancously

( $\because f(x)$ is one-one, proved above)

Also $|x-y|$ can assume very large value and $|f(x)|-f(y) \geq |x-y|$

$\therefore f(x)$ is onto.

Case $I$

$f^{\prime}(a) \geq 1$

Let $y=k=$ finite

$|(f(x)-f(y))| \geq |x-y|$

$-(f(x)-f(y)) \geq -(x-y)$

$f(y)-f(x) \geq y-x$

$f(k)-f(x) \geq k-x$

$f(x) \leq f(k)-k+x$

When, $x \rightarrow-\infty$

Then, $f(x) \leq f(x)-k-\infty$

$f(x) \rightarrow-\infty$

Case $II$

$f^{\prime}(a) \leq-1$

Let $y=k=$ finite and $x > k, f(x)$

$|f(x)-f(k)| \geq|x-k|$

$-(f(x)-f(k)) \geq x-k$

$f(x) \leq f(k)+k-x$

When, $x \rightarrow \infty$ then $f(x) \leq f(k)+k-\infty$

Then, $f(x) \rightarrow-\infty$

Given function

$f(x)=\left[\begin{array}{cl} x \sin \left(\frac{1}{x}\right) & , \text { when } x \neq 0 \\ 1 & , \text { when } x=0 \end{array}\right.$

Now, for $x=0, f(x)=1$ and for $x \neq 0, f(x)=1 \Rightarrow \sin \frac{1}{x}=\frac{1}{x}$ has no solution.

$\therefore$ The $\operatorname{set} A=\{x \in R: f(x)=1\}$ has exactly one element.

Given function $f: R \longrightarrow R$ be a continuous function such that $f\left(x^2\right)=f\left(x^3\right) \forall x \in R$

then $f(x)=f\left(x^{23}\right) \quad$ [on replacing $x$ by $x^{1 / 3}$ ]

Similarly,

$f(x)=f\left(x^{23}\right)=f\left(x^{4 / 9}\right)=f\left(x^{2 / 27}\right)=$

$\quad \ldots=f\left(x^{(23)^n}\right)$

$=f\left(x^0\right) \text { [as } x \text { tends to infinity] }=f(1)$

$\therefore f(x)=f(1)=\text { constant }$

The function $f(x)=$ constant is even and differentiable everywhere.

Given function

$f(x)=\left[\begin{array}{cc} \frac{\sin \left(x^2\right)}{x} & , x \neq 0 \\ 0 & , \text { if } x=0\end{array}\right.$

then $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sin x^2}{x}=\lim _{x \rightarrow 0}$

$x \frac{\sin x^2}{x^2}=0=f(0)$

Hence, $f(x)$ is continuous at $x=0$

Now, for differentiability

$RHD ($ at $x=0)$

$=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{\sin h^2}{h^2}=1$

and $LHD$ (at $x=0)$

$=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)-\lim _{h \rightarrow 0} \sin h^2}{-h}=1$

So, $f(x)$ is differentiable at $x=0$

$\therefore \quad f^{\prime}(x)=\left[\begin{array}{cc}2 \cos \left(x^2\right)-\frac{\sin x^2}{x^2} & , \text { if } x \neq 0 \\ 1 & , \text { if } x=0\end{array}\right]$ $\because \lim _{x \rightarrow 0} f^{\prime}(x)=\lim _{x \rightarrow 0}\left[2 \cos \left(x^2\right)-\frac{\sin x^2}{x^2}\right]$ $=2-1=1$ $\therefore \lim _{x \rightarrow 0} f^{\prime}(x)=f^{\prime}(0)$ So,f $f(x)$ is differentiable and the

Given,

$f(x)=x^6-2 x^5+x^3+x^2-x-1$

$=x^2\left(x^4-x^3-x^2-1\right)$

$-x\left(x^4-x^3-x^2-1\right)+2 x^2-2 x-1$

$\Rightarrow f(x)=\left(x^2-x\right) g(x)+2 x^2-2 x-1$

$\therefore f(a)=2 a^2-2 a-1$

$\therefore \Sigma f(a)=f(a)+f(b)+(c)+f(d)$

$=2 \Sigma a^2-2 \Sigma a-\Sigma 1$

$=2\left((\Sigma a)^2-2 \Sigma a b\right)-2 \Sigma a-4$

$\because a, b, c, d$ are roots of equation

$g(x)=x^4-x^3-x^2-1=0$

then $\Sigma a=1, \Sigma a b=-1$

$\therefore \Sigma f(a)=2\left[(1)^2-2(-1)\right]-2(1)-4$

$=6-2-4=0$

It is given that for $n \in N$

$f_n=(n+1)^{1 / 3}-n^{1 / 3}\,(n+1)-n$

$(n+1)^{2/3}+(n+1)^{23} n^{2/3}+n^{2/3}$

$3 n^{2 / 3} < (n+1)^{2 / 3}+(n+1)^{2/3} n^{2/3}+n^{2/3} < 3(n+1)^{2 / 3}$

$\Rightarrow \frac{1}{3(n+1)^{2/3}} <  \frac{1}{(n+1)^{2 / 3}+(n+1)^{2 / 3} n^{2/3}+n^{2/3}} < \frac{1}{3 n^{2/3}}$

$\frac{1}{3(n+1)^{2/3}} < f_n < \frac{1}{3 n^{2/3}}$

Similarly,

$\frac{1}{3(n+2)^{23}} < f_{n+1} < \frac{1}{3(n+1)^{23}}$

$\therefore \quad f_{n+1} < \frac{1}{3(n+1)^{23}} < f_{n+1}, \forall n \in N$

So, $\operatorname{set} A=N$.

Let $h(x)=g(f(x))$ where, $g(x)$ is injective and $h(x)$ is surjective. $h(x)$ is surjective.

$\therefore$ Codomain of $h(x)=$ Range of $h(x)$.

Range of $h(x)=[0,2]$ which is also

codomain of $g$. So, must be surjective.

Now, domain of $g=[-1,1]$ which must

be range of $f$. But codomain of $f=[-1,1]$

So, it must be surjective.

We have,,

$-1 < f(0) < f(1) < 1$

$g=[-1,1] \rightarrow[0,1]$

$gof(x)=x, \forall x \in[0,1]$

Only condition that $g(x)$ should satisfy for $g o f(x)=x, \forall x \in[0,1]$ is that $g(x)$ should attain all values in $[0,1]$ when range of $f(x)$ a subset of $(-1,1)$ is used as image for $g(x)$. Thus, there can be infinite such functions $g(x)$ with domain $[-1,1]$ and $\operatorname{range}[0,1]$