Chapter 1

(1) \(|a-a|=0<1 \therefore a R a \forall a \in R\) Therefore, \(R\) is reflexive. Again \(a R b \Rightarrow|a-b| \leq \mid\) and \(|b-a| \leq \mid \Rightarrow b R a\) Therefore, \(R\) is symmetric. Therefore, \(R\) is not anti-symmetric. Further, \(1 R 2\) and \(2 R 3\) but \(1 \not R 3,[\because|1-3|=2>1]\) Therefore, \(R\) is not transitive.

$$ \begin{aligned} &f(x)=e^{x^{3}-3 x+2} \\ &\text { Let } g(x)=x^{3}-3 x+2 \\ &\quad g^{\prime}(x)=3 x^{2}-3=3\left(x^{2}-1\right) \\ &\geq 0 \text { for } x \in(-\infty,-1] \end{aligned} $$ Now, the range of \(f(x)\) is \(\left(0, e^{4}\right]\) But co-domain is \(\left(0, e^{5}\right]\) Hence, \(f(x)\) is an into function.

\(f(x) \frac{2^{x}+1}{4^{x}-1}=\frac{1}{2^{x}-1}\) is one-one function.

Given function is defined if \({ }^{10} C_{x-1}>3{ }^{10} C_{x}\) or \(\frac{1}{11-x}>\frac{3}{x} \quad\) or \(4 x>33\) or \(\quad x \geq 9\) But \(x \leq 10\) \(\therefore \quad x=9,10\)

$$ \begin{aligned} &f(x)=\frac{\sin ^{-1}(3-x)}{\log (|x|-2)} \\ &\text { Let } g(x)=\sin ^{-1}(3-x) \\ &\text { or }-1 \leq 3-x \leq 1 \\ &\text { The domain of } g(x) \text { is }[2,4] \text {. } \\ &\text { Let } h(x)=\log (|x|-2) \\ &\text { i.e., }|x|-2>0 \text { or }|x|>2 \\ &\text { i.e., } x<-2 \text { or } x>2 \\ &\therefore \quad \text { Domain }(-\infty,-2) \cup(2, \infty) \\ &\text { We know that } \\ &\qquad(f / g)(x)=\frac{f(x)}{g(x)} \forall x \in D_{1} \cap D_{2}-\\{x \in R: g(x)=0\\} \end{aligned} $$ Therefore, the domain of \(f(x)\) is \((2,4]-\\{3\\}=(2,3) \cup(3,4]\)

Here, \(x+3>0\) and \(x^{2}+3 x+2 \neq 0\) Therefore, \(x>-3\) and \((x+1)(x+2) \neq 0\), i.e., \(x \neq-1,-2\). Therefore, the domain is \((-3, \infty)-\\{-1,-2\\}\).

$$ x^{2}-[x]^{2} \geq 0 \text { or } x^{2} \geq[x]^{2} $$ This is true for all non-negative values of \(x\) and all negative integers \(x\).

\(f(x)\) is to be defined when \(x^{2}-1>0\) and \(3+x>0\) and \(3+x \neq 1\), i.e., \(x^{2}>1\) and \(x>-3\) and \(x \neq-2\) i.e., \(x<-1\) or \(x>1\) and \(x>-3\) and \(x \neq-2\) \(\therefore \quad D_{f}=(-3,-2) \cup(-2,-1) \cup(1, \infty)\)

We have \(f(x)=\left[\log _{10}\left(\frac{5 x-x^{2}}{4}\right)\right]^{1 / 2}\) From (1), clearly, \(f(x)\) is defined for those values of \(x\) for which $$ \log _{10}\left[\frac{5 x-x^{2}}{4}\right] \geq 0 $$

The function \(\sec ^{-1} x\) is defined for all \(x \in R-(-1,1)\) and the function \(\frac{1}{\sqrt{x-[x]}}\) is defined for all \(x \in R-Z\). So, the given function is defined for all \(x \in R-\\{(-1,1) \cup\) \(\\{n \mid n \in Z\\}\\} .\)

It is given that \(2^{x}+2^{y}=2 \forall x, y \in R\) or \(\quad 2^{y}=2-2^{x}\) or \(y=\log _{2}\left(2-2^{x}\right)\) Therefore, function is defined only when \(2-2^{x}>0\) or \(2^{x}<2\) or \(x<1\)

\(\cos ^{-1}\left(\frac{2-|x|}{4}\right)\) exists if $$ -1 \leq \frac{2-|x|}{4} \leq 1 $$ or \(\quad-6 \leq-|x| \leq 2\) or \(\quad-2 \leq|x| \leq 6\) or \(\quad|x| \leq 6\) or \(\quad-6 \leq x \leq 6\) The function \([\log (3-x)]^{-1}=\frac{1}{\log (3-x)}\) is defined if \(3-x>0\) and \(x \neq 2\), i.e., if \(x \neq 2\) and \(x<3\). Thus, the domain of the given function is $$ \\{x \mid-6 \leq x \leq 6\\} \cap\\{x \mid x \neq 2, x<3\\}=[-6,2) \cup(2,3) $$

\(f(x)\) is defined if $$ -\log _{1 / 2}\left(1+\frac{1}{x^{1 / 4}}\right)-1>0 $$ or \(\quad \log _{1 / 2}\left(1+\frac{1}{x^{1 / 4}}\right)<-1\) or \(1+\frac{1}{x^{1 / 4}}>\left(\frac{1}{2}\right)^{-1}\) or \(\quad \frac{1}{x^{1 / 4}}>1\) or \(\quad 0

Draw the graph of \(y=\log _{0.5}|x|\) and \(y=2|x|\). Clearly, from the graph, there are two solutions.

$$ f(x)=\frac{1}{\sqrt{4 x-\left|x^{2}-10 x+9\right|}} $$ For \(f(x)\) to be defined, \(\left|x^{2}-10 x+9\right|<4 x\) or \(x^{2}-10 x+9<4 x\) and \(x^{2}-10 x+9>-4 x\) or \(x^{2}-14 x+9<0\) and \(x^{2}-6 x+9>0\) or \(x \in(7-\sqrt{40}, 7+\sqrt{40})\) and \(x \in R-\\{3\\}\) or \(x \in(7-\sqrt{40}, 3) \cup(3,7+\sqrt{40})\)

We have \(f(x)=\sqrt{\sin x+\cos x}+\sqrt{7 x-x^{2}-6}\) $$ =\sqrt{\sqrt{2} \sin \left(\frac{\pi}{4}+x\right)}+\sqrt{7 x-x^{2}-6} $$ \(f(x)\) is defined if $$ \text { (i) } \begin{aligned} 7 x-x^{2}-6 \geq 0 \\ \Rightarrow \quad 1 \leq x \leq 6 \end{aligned} $$ (ii) \(\sin \left(\frac{\pi}{4}+x\right) \geq 0\) $$ \begin{aligned} & x+\frac{\pi}{4} \in \ldots[-2 \pi,-\pi] \cup[0, \pi] \cup[2 \pi, 3 \pi] \ldots \\ \Rightarrow & x \in \ldots\left[-\frac{\pi}{4}, \frac{3 \pi}{4}\right] \cup\left[\frac{7 \pi}{4}, \frac{11 \pi}{4}\right] \ldots \end{aligned} $$ From (1) and (2), we get $$ x \in\left[1, \frac{3 \pi}{4}\right] \cup\left[\frac{7 \pi}{4}, 6\right] $$

Given, \(y=x^{\log _{x} \pi}=\pi\) Domain is \(x \in(0,1) \cup(1, \infty)\) Range is \(\\{\pi\\}\)

\(\begin{aligned} y=f(x) &=\cos ^{2} x+\sin ^{4} x \\ &=\cos ^{2} x+\sin ^{2} x\left(1-\cos ^{2} x\right) \\ &=\cos ^{2} x+\sin ^{2} x-\sin ^{2} x \cos ^{2} x \\ &=1-\sin ^{2} x \cos ^{2} x \\ &=1-\frac{1}{4} \sin ^{2} 2 x \\\ \therefore \quad & \frac{3}{4} \leq f(x) \leq 1 \\ \therefore \quad f(x) & \in[3 / 4,1] \end{aligned}\)

We have \(f(x)={ }^{7-x} P_{x-3}=\frac{(7-x) !}{(10-2 x) !}\) We must have \(7-x>0, x \geq 3\), and \(7-x \geq x-3\). Therefore, \(x<7, x \geq 3\), and \(x \leq 5\) or \(\quad 3 \leq x \leq 5\) or \(x=3,4,5\) Now, \(f(3)=\frac{4 !}{4 !}=1, f(4)=\frac{3 !}{2 !}=3, f(5)=\frac{2 !}{0 !}=2\) Hence, \(R_{f}=\\{1,2,3\\}\)

Since \(\\{x\\} \in[0,1), \sin \\{x\\} \in(0, \sin 1)\) as \(f(x)\) is defined if \(\sin \\{x\\} \neq 0\), i.e. $$ \frac{1}{\sin \\{x\\}} \in\left(\frac{1}{\sin 1}, \infty\right) \text { or }\left[\frac{1}{\sin \\{x\\}}\right] \in\\{1,2,3, \ldots\\} $$ Note that \(1<\frac{\pi}{3}\) or \(\sin 1<\sin \frac{\pi}{3}=0.866\) or \(\frac{1}{\sin 1}>1.155\)

\(f(x)=\sqrt{|x|-\\{x\\}}\) is defined if \(|x| \geq\\{x\\}\) or \(\quad x \in\left(-\infty-\frac{1}{2}\right] \cup[0, \infty)\) or \(Y \in[0, \infty)\)

Let \(g(x)=(x+1)(x+2)(x+3)(x+4)\) The rough graph of \(g(x)\) is given as follows: Now, \(g_{\min }=-1\), for which \(x^{2}+5 x=-1\) has real roots in \([-6,6]\). Also, \(g(6)=7 \times 8 \times 9 \times 10=5040\). Hence, the range of \(g(x)\) is \([-1,5040]\) for \(x \in[-6,6]\), Then. the range of \(f(x)\) is \([4,5045]\).

\(\left[\sqrt{n^{2}+1}\right]=n\) or \(\quad\left[\sqrt{n^{2}+\lambda}\right]=n+2\) or \(\quad n+2 \leq \sqrt{n^{2}+\lambda}

We must have \(-1 \leq\left[2 x^{2}-3\right] \leq 1\) or \(-1 \leq 2 x^{2}-3<2\) or \(1 \leq x^{2}<\frac{5}{2}\) or \(x \in\left(-\sqrt{\frac{5}{2}},-1\right] \cup\left[1, \sqrt{\frac{5}{2}}\right)\)

$$ \begin{aligned} &f(x)=\sin \left(\log \left(x+\sqrt{1+x^{2}}\right)\right) \\ &\text { or } \begin{aligned} & f(-x) &=\sin \left[\log \left(-x+\sqrt{1+x^{2}}\right)\right] \\ &=\sin \log \left(\left(\sqrt{1+x^{2}}-x\right) \frac{\left(\sqrt{1+x^{2}}+x\right)}{\left(\sqrt{1+x^{2}}+x\right)}\right) \\ &=\sin \log \left[\frac{1}{\left(x+\sqrt{1+x^{2}}\right)}\right] \\ &=\sin \left[-\log \left(x+\sqrt{1+x^{2}}\right)\right] \\ &=-\sin \left[\log \left(x+\sqrt{1+x^{2}}\right)\right] \end{aligned} \\ &\therefore \quad f(-x)=-f(x) \end{aligned} $$ Therefore, \(f(x)\) is an odd function.

Since \(f(x)\) is an odd function, $$ \left[\frac{x^{2}}{a}\right]=0 \text { for all } x \in[-10,10] $$ or \(0 \leq \frac{x^{2}}{a}<1\) for all \(x \in[-10,10]\) or \(a>100\)

$$ \begin{aligned} f(-x)=\frac{\cos (-x)}{\left[-\frac{2 x}{\pi}\right]+\frac{1}{2}} &=\frac{\cos x}{-1-\left[\frac{2 x}{\pi}\right]+\frac{1}{2}} \quad \begin{array}{r} \text { (As } x \text { is not an integral } \\ \text { multiple of } \pi) \end{array} \\ &=-\frac{\cos x}{\left[\frac{2 x}{\pi}\right]+\frac{1}{2}}=-f(x) \end{aligned} $$ Therefore, \(f(x)\) is an odd function.

For odd function, $$ \begin{aligned} f(x) &=-f(-x) \\ &=- \begin{cases}\sin (-x)+\cos (-x) & 0<-x<\pi / 2 \\ a, & \quad-x=\pi / 2 \\ \tan ^{2}(-x)+\operatorname{cosec}(-x), & \pi / 2<-x<\pi\end{cases} \\ &= \begin{cases}\sin x-\cos x, & -\pi / 2

$$ \begin{aligned} f(x) &=[6 x+7]+\cos \pi x-6 x \\ &=[6 x]+7+\cos \pi x-6 x \\ &=7+\cos \pi x-\\{6 x\\} \end{aligned} $$ \(\\{6 x\\}\) has period \(1 / 6\) and \(\cos \pi x\) has period 2. Then, Period of \(f(x)=L C M\) of 2 and \(1 / 6=2\) Hence, the period is 2 .

\(f(x)\) is defined for \(x \in(0,1)\) So, \(f\left(e^{x}\right)+f(\ln |x|)\) is defined for \(0

\(y=2^{x(x-1)}\) or \(x^{2}-x-\log _{2} y=0\) $$ \text { or } \quad x=\frac{1}{2}\left(1 \pm \sqrt{1+4 \log _{2} y}\right) $$ Since \(x \in[1, \infty)\), we choose \(x=\frac{1}{2}\left(1+\sqrt{1+4 \log _{2} y}\right)\) or \(\quad f^{-1}(x)=\frac{1}{2}\left(1+\sqrt{1+4 \log _{2} x}\right) .\)

Given that \(f(x)=(x+1)^{2}, x \geq-1\) Now, if \(g(x)\) is the reflection of \(f(x)\) in the line \(y=x\), then \(g(x)\) is an inverse function of \(y=f(x)\) Given \(y=(x+1)^{2} \quad(x \geq-1\) and \(y \geq 0)\) or \(\quad x=\pm \sqrt{y}-1\) or \(g(x)=f^{-1}(x)=\sqrt{x}-1, x \geq 0\)

By checking for different functions, we find that for $$ f(x)=\frac{1-x}{1+x}, f^{-1}(x)=f(x) $$

\(\frac{1}{2}(\) gof \()(x)=2 x^{2}-5 x+2\) or \(\quad \frac{1}{2} g[f(x)]=2 x^{2}-5 x+2\) \(\therefore \quad\left[\\{f(x)\\}^{2}+\\{f(x)\\}-2\right]=2\left[2 x^{2}-5 x+2\right]\) or \(f(x)^{2}+f(x)-\left(4 x^{2}-10 x+6\right)=0\) \(\therefore \quad f(x)=\frac{-1 \pm \sqrt{1+4\left(4 x^{2}-10 x+6\right)}}{2}\) \(=\frac{-1 \pm \sqrt{\left(16 x^{2}-40 x+25\right)}}{2}=\frac{-1 \pm(4 x-5)}{2}\) \(=2 x-3\) or \(-2 x+2\)

\(f(x)=x^{n}+1\) or \(\quad f(3)=3^{n}+1=28\) or \(\quad 3^{n}=27\) \(\therefore \quad n=3\) or \(f(4)=4^{3}+1=65\)

$$ \left(x+\frac{1}{2}\right)+f\left(x-\frac{1}{2}\right)=f(x) $$ Replacing \(x\) by \(\left(x+\frac{1}{2}\right)\), we get $$ f(x+1)+f(x)=f\left(x+\frac{1}{2}\right) $$ of \(f(x+1)+f\left(x-\frac{1}{2}\right)=0\) (from (1) and (2)) or \(f\left(x+\frac{3}{2}\right)=-f(x)\) (Replacing \(x\) by \(\left(x+\frac{1}{2}\right)\) ) of \(f(x+3)=-f\left(x+\frac{3}{2}\right)=f(x) \quad\) (Replacing \(x\) by \(\left.\left(x+\frac{3}{2}\right)\right)\) Therefore, \(f(x)\) is periodic with period 3 .

$$ f(3 x+2)+f(3 x+29)=0 $$ Replacing \(x\) by \(x+9\), we get $$ f(3(x+9)+2)+f(3(x+9)+29)=0 $$ or \(f(3 x+29)+f(3 x+56)=0\) From (1) and (2), we get $$ f(3 x+2)=f(3 x+56) $$ or \(f(3 x+2)=f(3(x+18)+2)\) Therefore, \(f(x)\) is periodic with period 54 .

From the given data, $$ f(1-x)=f(1+x) $$ and \(f(2-x)=f(2+x)\) \(\ln (2)\), replacing \(x\) by \(1+x\), we have $$ f(1-x)=f(3+x) $$ or \(f(1+x)=f(3+x)\) or \(f(x)=f(2+x)\)

\(|x-2|+a=\pm 4\) or \(|x-2|=\pm 4-a\) For four real roots, \(4-a>0\) and \(-4-a>0\) or \(a \in(-\infty,-4)\)

Let \(\begin{aligned} f(x)=& x+2|x+1|+2|x-1| \\ &= \begin{cases}x-2(x+1)-2(x-1), & x<-1 \\ x+2(x+1)-2(x-1), & -1 \leq x \leq 1 \\\ x+2(x+1)+2(x-1), & x>1\end{cases} \\ &= \begin{cases}-3 x, & x<-1 \\\ x+4, & -1 \leq x \leq 1 \\ 5 x, & x>1\end{cases} \end{aligned}\)

$$ \begin{aligned} &\log _{4}\left(\frac{2 f(x)}{1-f(x)}\right)=x \\ &\Rightarrow \quad \frac{2 f(x)}{1-f(x)}=4^{x} \end{aligned} $$ Replacing \(x\) by \(1-x\), we get $$ \frac{2 f(1-x)}{1-f(1-x)}=4^{1-x} $$ Multiplying (1) and (2), we get $$ \begin{aligned} & \frac{2 f(x)}{1-f(x)}\left(\frac{2 f(1-x)}{1-f(1-x)}\right)=4^{1-x} \\ \Rightarrow & f(x) f(1-x)=1-f(x)-f(1-x)+f(x) f(1-x) \\ \Rightarrow & f(x)+f(1-x)=1 \end{aligned} $$ Putting \(x=2010\), we get $$ f(2010)+f(-2009)=1 $$