JEE Practice Questions (Single Choice Type)

Hint

(b) \(f: N \rightarrow N, f(n)=2 n+3\)
Here, the range of the function is \(\{5,6,7, \ldots\}\) or \(N-\{1,2\), \(3,4\}\)
which is a subset of \(N\) (co-domain).
Hence, function is into.
Also, it is clear that \(f(n)\) is one-one or injective.
Hence. \(f(n)\) is iniective only.

Hint

(b)
\(
\begin{aligned}
& f(x)=\sin \left(\log \left(x+\sqrt{1+x^2}\right)\right) \\
\Rightarrow & f(-x)=\sin \left[\log \left(-x+\sqrt{1+x^2}\right)\right] \\
\Rightarrow & f(-x)=\sin \log \left(\left(\sqrt{1+x^2}-x\right) \frac{\left(\sqrt{1+x^2}+x\right)}{\left(\sqrt{1+x^2}+x\right)}\right) \\
\Rightarrow & f(-x)=\sin \log \left[\frac{1}{\left(x+\sqrt{1+x^2}\right)}\right]
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow f(-x)=\sin \left[-\log \left(x+\sqrt{1+x^2}\right)\right] \\
& \Rightarrow f(-x)=-\sin \left[\log \left(x+\sqrt{1+x^2}\right)\right] \\
& \Rightarrow f(-x)=-f(x) \\
& \Rightarrow f(x) \text { is an odd function. }
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
&\begin{aligned}
& \frac{x^2+14 x+9}{x^2+2 x+3}=y \\
& \Rightarrow x^2+14 x+9=x^2 y+2 x y+3 y \\
& \Rightarrow x^2(y-1)+2 x(y-7)+(3 y-9)=0
\end{aligned}\\
&\text { Since } x \text { is real, }\\
&\begin{aligned}
& \therefore \quad 4(y-7)^2-4(3 y-9)(y-1)>0 \\
& \Rightarrow \quad 4\left(y^2+49-14 y\right)-4\left(3 y^2+9-12 y\right)>0 \\
& \Rightarrow \quad(y+5)(y-4)<0 ; \\
& \therefore \quad y \text { lies between }-5 \text { and } 4 .
\end{aligned}
\end{aligned}
\)

Hint

(c)
\(
\begin{aligned}
& y=f(x)=\cos ^2 x+\sin ^4 x \\
\Rightarrow \quad & y=f(x)=\cos ^2 x+\sin ^2 x\left(1-\cos ^2 x\right) \\
\Rightarrow \quad & y=\cos ^2 x+\sin ^2 x-\sin ^2 x \cos ^2 x \\
\Rightarrow \quad & y=1-\sin ^2 x \cos ^2 x \\
\Rightarrow \quad & y=1-\frac{1}{4} \sin ^2 2 x
\end{aligned}
\)
\(
\therefore \frac{3}{4} \leq f(x) \leq 1 \quad\left(\because 0 \leq \sin ^2 2 x \leq 1\right)
\)
\(
\Rightarrow f(x) \in[3 / 4,1]
\)

Hint

(c)

\(
\begin{aligned}
&f(x) \text { is to be defined when } x^2-1>0 \text { and } 3+x>0 \text { and }\\
&\begin{aligned}
& 3+x \neq 1 \\
& \Rightarrow \quad x^2>1 \text { and } x>-3 \text { and } x \neq-2 \\
& \Rightarrow \quad x<-1 \text { or } x>1 \text { and } x>-3 \text { and } x \neq-2 \\
& \therefore \quad D_f=(-3,-2) \cup(-2,-1) \cup(1, \infty)
\end{aligned}
\end{aligned}
\)

Hint

(b) We have \(f(x)=\left[\log _{10}\left(\frac{5 x-x^2}{4}\right)\right]^{1 / 2} \dots(1)[latex]
From (1), clearly [latex]f(x)\) is defined for those values of \(x\) for
\(
\begin{aligned}
& \text { which } \log _{10}\left[\frac{5 x-x^2}{4}\right] \geq 0 \\
& \Rightarrow\left(\frac{5 x-x^2}{4}\right) \geq 10^0 \\
& \Rightarrow\left(\frac{5 x-x^2}{4}\right) \geq 1 \\
& \Rightarrow x^2-5 x+4 \leq 0 \\
& \Rightarrow(x-1)(x-4) \leq 0
\end{aligned}
\)
Hence, the domain of the function is \([1,4]\).

Hint

(b) \(f(x)=\frac{\sin ^{-1}(3-x)}{\log (|x|-2)}\)
Let \(g(x)=\sin ^{-1}(3-x)\)
\(
\Rightarrow-1 \leq 3-x \leq 1
\)
The domain of \(g(x)\) is \([2,4]\)
and let \(h(x)=\log (|x|-2)\)
\(
\begin{aligned}
& \Rightarrow \quad|x|-2>0 \text { or }|x|>2 \\
& \Rightarrow \quad x<-2 \text { or } x>2 \\
& \Rightarrow \quad(-\infty,-2) \cup(2, \infty)
\end{aligned}
\)
We know that
\(
(f / g)(x)=\frac{f(x)}{g(x)} \forall x \in D_1 \cap D_2-\{x \in R: g(x)=0\}
\)
\(\therefore\) the domain of \(f(x)=(2,4]-\{3\}=(2,3) \cup(3,4]\).

Hint

(c)

\(
\begin{aligned}
& f(x)=\log |\log x|, f(x) \text { is defined if }|\log x|>0 \text { and } x>0, \text { i.e., } \\
& \text { if } x>0 \text { and } x \neq 1 \\
& \Rightarrow x \in(0,1) \cup(1, \infty) .
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
& \text { Here } x+3>0 \text { and } x^2+3 x+2 \neq 0 \\
& \therefore \quad x>-3 \text { and }(x+1)(x+2) \neq 0, \text { i.e., } x \neq-1,-2 \\
& \therefore \quad \text { The domain }=(-3, \infty)-\{-1,-2\} .
\end{aligned}
\)

Hint

(b) \(y=f(x)=\sqrt{3} \sin x-\cos x+2=2 \sin \left(x-\frac{\pi}{6}\right)+2 \dots(1)\)
Since \(f(x)\) is one-one and onto, \(f\) is invertible.
From (1) \(\sin \left(x-\frac{\pi}{6}\right)=\frac{y-2}{2}\)
\(
\Rightarrow \quad x=\sin ^{-1} \frac{y-2}{2}+\frac{\pi}{6}
\)
\(
\Rightarrow f^{-1}(x)=\sin ^{-1}\left(\frac{x-2}{2}\right)+\frac{\pi}{6}
\)

Hint

(a)

\(
\begin{aligned}
&F(n+1)=\frac{2 F(n)+1}{2} \Rightarrow F(n+1)-F(n)=\frac{1}{2}\\
&\text { Put } n=1,2,3, \ldots, 100 \text { and add, we get }\\
&\begin{aligned}
& F(101)-F(1)=100 \times \frac{1}{2} \\
& \Rightarrow \quad F(101)=52
\end{aligned} \quad[\because F(1)=2]
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
&\text { Given function is defined if }{ }^{10} C_{x-1}>3{ }^{10} C_x\\
&\begin{aligned}
& \Rightarrow \frac{1}{11-x}>\frac{3}{x} \Rightarrow 4 x>33 \\
& \Rightarrow x \geq 9 \text { but } x \leq 10 \Rightarrow x=9,10 .
\end{aligned}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\text { For the domain } \sin (\ln x)>\cos (\ln x) \text { and } x>0\\
&2 n \pi+\frac{\pi}{4}<\ln x<2 n \pi+\frac{5 \pi}{4}, n \in N \cup\{0\}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& \text { Put } x=0 \Rightarrow f(2)=2 f(0)-f(1)=2 \times 2-3=1 \\
& \text { Put } x=1 \Rightarrow f(3)=6-1=5 \\
& \text { Put } x=2 \Rightarrow f(4)=2 f(2)-f(3)=2 \times 1-5=-3 \\
& \text { Put } x=3 \Rightarrow f(5)=2 f(3)-f(4)=2(5)-(-3)=13 \text {. }
\end{aligned}
\)

Hint

(c)
\(
\text { Here, } \frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}
\)
\(
\begin{aligned}
&\text { Now, } 2 \leq x^2+2<\infty \text { for all } x \in R\\
&\begin{aligned}
& \Rightarrow \frac{1}{2} \geq \frac{1}{x^2+2}>0 \\
& \Rightarrow-\frac{1}{2} \leq \frac{-1}{x^2+2}<0 \\
& \Rightarrow \frac{1}{2} \leq 1-\frac{1}{x^2+2}<1 \\
& \Rightarrow \frac{\pi}{6} \leq \sin ^{-1}\left(1-\frac{1}{x^2+2}\right)<\frac{\pi}{2}
\end{aligned}
\end{aligned}
\)

Hint

(b) 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 \(x \in R-\{(-1,1) \cup\{n \mid n \in Z\}\}\).

Hint

(b)
\(
\begin{aligned}
& \cos ^{-1}\left(\frac{2-|x|}{4}\right) \text { exists if }-1 \leq \frac{2-|x|}{4} \leq 1 \\
& \Rightarrow-6 \leq-|x| \leq 2 \\
& \Rightarrow-2 \leq|x| \leq 6 \\
& \Rightarrow|x| \leq 6 \\
& \Rightarrow-6 \leq x \leq 6
\end{aligned}
\)
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
\(
\{x \mid-6 \leq x \leq 6\} \cap\{x \mid x \neq 2, x<3\}=[-6,2) \cup(2,3) .
\)

Hint

(b)
\(
\begin{aligned}
& f(x) \text { is defined for } \log \left(\frac{1}{|\sin x|}\right) \geq 0 \\
& \Rightarrow \frac{1}{|\sin x|} \geq 1 \text { and }|\sin x| \neq 0
\end{aligned}
\)
\(
\Rightarrow \quad|\sin x| \neq 0 \quad\left[\because \frac{1}{|\sin x|} \geq 1 \text { for all } x\right]
\)
\(
\begin{aligned}
&\Rightarrow \quad x \neq n \pi, n \in Z\\
&\text { Hence, the domain of } f(x)=R-\{n \pi: n \in Z\} \text {. }
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& f(x) \text { is defined if }-\log _{1 / 2}\left(1+\frac{1}{x^{1 / 4}}\right)-1>0 \\
& \Rightarrow \quad \log _{1 / 2}\left(1+\frac{1}{x^{1 / 4}}\right)<-1 \\
& \Rightarrow \quad 1+\frac{1}{x^{1 / 4}}>\left(\frac{1}{2}\right)^{-1} \\
& \Rightarrow \quad \frac{1}{x^{1 / 4}}>1 \\
& \Rightarrow \quad 0<x<1
\end{aligned}
\)

Hint

(c)
\(
\begin{aligned}
&\text { For the function to get defined } 0 \leq x^2+x+1 \leq 1 \text {, }\\
&\text { but } x^2+x+1 \geq \frac{3}{4} \Rightarrow \frac{\sqrt{3}}{2} \leq \sqrt{x^2+x+1} \leq 1
\end{aligned}
\)
\(
\Rightarrow \quad \frac{\pi}{3} \leq \sin ^{-1}\left(\sqrt{x^2+x+1}\right) \leq \frac{\pi}{2} .
\)

Hint

(c)

\(
\text { Clearly, from the graph } f(x)=\left\{\begin{array}{cc}
\frac{1}{64}, & 0 \leq x \leq \frac{1}{8} \\
x^2, & \frac{1}{8}<x \leq 1 \\
x^3, & x>1
\end{array}\right.
\)

Hint

(c) The period of \(\cos (\sin n x)\) is \(\frac{\pi}{n}\) and the period of \(\tan \left(\frac{x}{n}\right)\) is \(\pi n\).
Thus, \(6 \pi=\operatorname{LCM}\left(\frac{\pi}{n}, \pi n\right)\).
By checking for the different values of \(n, n=6\).

Hint

(b) \(\text { Draw the graph of } y=\log _{0.5}|x| \text { and } y=2|x|\)

Clearly, from the graph, there are two solutions.

Hint

(b) \(f(x)=\left|\sin ^3 \frac{x}{2}\right|+\left|\cos ^5 \frac{x}{5}\right|\)
The period of \(\sin ^3 x\) is \(2 \pi\)
\(\Rightarrow\) The period of \(\sin ^3 \frac{x}{2}\) is \(\frac{2 \pi}{1 / 2}=4 \pi\)
\(\Rightarrow\) The period of \(\left|\sin ^3 \frac{x}{2}\right|\) is \(2 \pi\)
The period of \(\cos ^5 x\) is \(2 \pi\)
\(\Rightarrow\) The period of \(\cos ^5 \frac{x}{5}\) is \(\frac{2 \pi}{\left(\frac{1}{5}\right)}=10 \pi\)
\(\Rightarrow\) The period of \(\left|\cos ^5 \frac{x}{2}\right|\) is \(5 \pi\)
\(
\text { Now the period of } f(x)=\operatorname{LCM} \text { of }\{2 \pi, 10 \pi\}=10 \pi \text {. }
\)

Hint

(a)

\(
\begin{aligned}
&\text { Given } f(x)=\sqrt[n]{x^m}, n \in N \text { is an even function where } m \in I .\\
&\begin{aligned}
& \Rightarrow f(x)=f(-x) \\
& \Rightarrow \sqrt[n]{x^m}=\sqrt[n]{(-x)^m} \\
& \Rightarrow x^m=(-x)^m \\
& \Rightarrow m \text { is an even integer } \\
& \Rightarrow m=2 k, k \in I
\end{aligned}
\end{aligned}
\)

Hint

(c) From the given data \(g(x)\) must be linear function
Hence, \(g(x)=a x+b\)
Also \(g(2)=2 a+b=3\) and \(g(4)=4 a+b=7\)
Solving, we get \(a=2\) and \(b=-1\)
Hence, \(g(x)=2 x-1\)
Then, \(g(6)=11\).

Hint

(b) The period of \(\sin \pi x\) and \(\cos 2 \pi x\) is 2 and 1 , respectively
The period of \(2^{\{x\}}\) is 1
The period of \(3^{\{x / 2\}}\) is 2
Hence, the period of \(f(x)\) is LCM of 1 and \(2=2\).

Hint

(a)

\(
\begin{aligned}
&\begin{aligned}
& |x-2|+a= \pm 4 \\
& \Rightarrow \quad|x-2|= \pm 4-a
\end{aligned}\\
&\text { for } 4 \text { real roots, } 4-a>0 \text { and }-4-a>0\\
&\Rightarrow \quad a \in(-\infty,-4)
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\text { We have } f(x+y)+f(x-y)\\
&\begin{aligned}
& =\frac{1}{2}\left[a^{x+y}+a^{-x-y}+a^{x-y}+a^{-x+y}\right] \\
& =\frac{1}{2}\left[a^x\left(a^y+a^{-y}\right)+a^{-x}\left(a^y+a^{-y}\right)\right] \\
& =\frac{1}{2}\left(a^x+a^{-x}\right)\left(a^y+a^{-y}\right)=2 f(x) f(y)
\end{aligned}
\end{aligned}
\)

Hint

(d)

\begin{aligned}
& \log _3\left(x^2-6 x+11\right) \leq 1 \\
& \Rightarrow \quad 0<x^2-6 x+11 \leq 3 \\
& \Rightarrow \quad x \in[2,4]
\end{aligned}

Hint

(d) \(x^2-[x]^2 \geq 0 \Rightarrow x^2 \geq[x]^2\)
This is true for all positive values of \(x\) and all negative integer \(x\).

Hint

(b)

Clearly, from the graph, the range is \([1, f(-1)] \equiv[1,5]\) If \(x<1, f(x)=-(x-1)-(x-2)=-2 x+3\).
In this interval, \(f(x)\) is decreasing.
If \(1 \leq x<2, f(x)=x-1-(x-2)=1\)
In this interval, \(f(x)\) is constant.
\(
\text { If } 2 \leq x \leq 3, f(x)=x-1+x-2=2 x-3
\)
In this interval, \(f(x)\) is increasing.
\(\therefore \max f(x)=\) the greatest among \(f(-1)\) and \(f(3)=5, \min f(x)\)
\(
=f(1)=1
\)
So, the range \(=[1,5]\).

Hint

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

Hint

(b) \(x^2 F(x)+F(1-x)=2 x-x^4 \dots(1)\)
\(
\begin{aligned}
&\text { Replacing } x \text { by } 1-x \text {, we get }\\
&\Rightarrow \quad(1-x)^2 F(1-x)+F(x)=2(1-x)-(1-x)^4 \dots(2)
\end{aligned}
\)
\(
\text { Eliminating } F(1-x) \text { from (1) and (2), we get } F(x)=1-x^2 \text {. }
\)

Hint

(b)
\(
f(-x)=\left\{\begin{array}{cc}
(-x)^2 \sin \frac{\pi(-x)}{2}, & |-x|<1 \\
(-x)|-x|, & |-x| \geq 1
\end{array}\right.
\)
\(
=\left\{\begin{array}{cc}
-x^2 \sin \frac{\pi x}{2}, & |x|<1 \\
-x|x|, & |x| \geq 1
\end{array}\right.
\)
\(
=-f(x)
\)

Hint

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

Hint

(c)

\(
\begin{aligned}
f(x) & =\frac{1}{x}, g(x)=\frac{1}{x^2} \text { and } h(x)=x^2 \\
f(g(x)) & =x^2 ; x \neq 0 \\
h(g(x)) & =\frac{1}{x^4}=(g(x))^2, x \neq 0
\end{aligned}
\)

Hint

(c)

\(
\sum_{r=1}^{2000} \frac{\{x+r\}}{2000}=\sum_{r=1}^{2000} \frac{\{x\}}{2000}=2000 \frac{\{x\}}{2000}=\{x\}
\)

Hint

(b)

\(
\begin{aligned}
f(x) & =x^n+1 \\
f(3) & =3^n+1=28 \\
3^n & =27 \\
n & =3 \\
f(4) & =4^3+1=65 .
\end{aligned}
\)

Hint

(b)
\(
\begin{aligned}
& \because f(x+1)-f(x)=8 x+3 \\
& \Rightarrow\left\{b(x+1)^2+c(x+1)+d\right\}-\left\{b x^2+c x+d\right\}=8 x+3 \\
& \Rightarrow b\left\{(x+1)^2-x^2\right\}+c=8 x+3 \\
& \Rightarrow b(2 x+1)+c=8 x+3
\end{aligned}
\)
On comparing co-efficient of \(x\) and constant term, we get \(2 b=8\) and \(b+c=3\) then \(b=4\) and \(c=-1\).

Hint

(d) If \(f\) is injective and \(g\) is surjective
\(\Rightarrow f o g\) is injective
\(\Rightarrow f o f\) is injective.

Hint

(c) \(f(x)=\left\{\begin{array}{ll}x-1, & x \text { is even } \\ x+1, & x \text { is odd }\end{array}\right.\), which is clearly are one-one and onto.

Hint

(c)

\(
\begin{aligned}
& \frac{1}{2}(g \circ f)(x)=2 x^2-5 x+2 \text { or } \frac{1}{2} g[f(x)]=2 x^2-5 x+2 \\
& \therefore\left[\{f(x)\}^2+\{f(x)\}-2\right]=2\left[2 x^2-5 x+2\right] \\
& \Rightarrow f(x)^2+f(x)-\left(4 x^2-10 x+6\right)=0 \\
& \therefore f(x)=\frac{-1 \pm \sqrt{1+4\left(4 x^2-10 x+6\right)}}{2} \\
& =\frac{-1 \pm \sqrt{\left(16 x^2-40 x+25\right.}}{2}=\frac{-1 \pm(4 x-5)}{2}=2 x-3 \text { or }-2 x+2
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& \text { Since } f(x) \text { and } f^{-1}(x) \text { are symmetric about the line } y=-x \text {. } \\
& \text { If }(\alpha, \beta) \text { lies on } y=f(x) \text { then }(-\beta,-\alpha) \text { on } y=f^{-1}(x) \\
& \Rightarrow \quad(-\alpha,-\beta) \text { lies on } y=f(x) \\
& \Rightarrow \quad y=f(x) \text { is odd. }
\end{aligned}
\)

Hint

(c) Let \(x, y \in N\) such that \(f(x)=f(y)\)
\(
\begin{aligned}
& \text { Then } f(x)=f(y) \\
& \Rightarrow \quad x^2+x+1=y^2+y+1 \\
& \Rightarrow \quad(x-y)(x+y+1)=0 \\
& \Rightarrow \quad x=y \text { or } x=(-y-1) \notin N
\end{aligned}
\)
\(\therefore \quad f\) is one-one.
Also, \(f(x)\) does not take all positive integral values. Hence \(f\) is into.

Hint

(c)

\(
\begin{aligned}
f(x) & =\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)+2 \sqrt{2} \\
f(x) & =\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)+2 \sqrt{2} \\
Y & =[\sqrt{2}, 3 \sqrt{2}] \text { and } X=\left[-\frac{3 \pi}{4}, \frac{\pi}{4}\right] \text { or }\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& f(7)+f(-7)=-10 \\
& \Rightarrow f(7)=-17 \\
& \Rightarrow f(7)+17 \cos x=-17+17 \cos x \text { which has the range } \\
& {[-34,0] .}
\end{aligned}
\)

Hint

(c) \(f(x)=\frac{\sin [x] \pi}{x^2+x+1}\)
Let \([x]=n \in\) integer
\(
\begin{aligned}
& \Rightarrow \quad \sin [x] \pi=0 \\
& \Rightarrow \quad f(x)=0
\end{aligned}
\)
\(\Rightarrow f(x)\) is constant function.

Hint

(b) Two triangles may have equal areas
\(\therefore f\) is not one-one.
Since each positive real number can represent area of a triangle.
\(\therefore f\) is onto.

Hint

(c)

\(
\begin{aligned}
& \frac{y-x}{y+x}=k(k>1) ; \quad y-x=k(y+x) \\
& \Rightarrow \quad y(1+k)=x(1+k) \\
& \Rightarrow \quad y=\left(\frac{1+k}{1-k}\right) x, \text { where } \frac{1+k}{1-k}<-1
\end{aligned}
\)

Hint

(c)
\(
\begin{aligned}
& g(x)=x^3+\tan x+\left[\frac{x^2+1}{P}\right] \\
& \Rightarrow g(-x)=(-x)^3+\tan (-x)+\left[\frac{(-x)^2+1}{P}\right] \\
& \Rightarrow g(-x)=-x^3-\tan x+\left[\frac{x^2+1}{P}\right] \\
& \Rightarrow g(x)+g(-x)=0
\end{aligned}
\)
because \(g(x)\) is a odd function
\(
\begin{aligned}
& \therefore\left(-x^3-\tan x+\left[\frac{x^2+1}{P}\right]\right)+\left(-x^3-\tan x\right) \\
& \left.\Rightarrow 2\left[\frac{x^2+1}{P}\right]\right)=0 \\
& \Rightarrow\left[\frac{\left(x^2+1\right)}{P}\right]=0 \Rightarrow 0 \leq \frac{x^2+1}{P}<1
\end{aligned}
\)
Now \(x \in[-2,2]\)
\(
\Rightarrow 0 \leq \frac{5}{P}<1 \Rightarrow P>5
\)

Hint

(d) Let \(2 x+\frac{y}{8}=\alpha\) and \(2 x-\frac{y}{8}=\beta\), then \(x=\frac{\alpha+\beta}{4}\) and \(y=4(\alpha-\beta)\).
Given, \(f\left(2 x+\frac{y}{8}, 2 x-\frac{y}{8}\right)=x y\)
\(
\begin{aligned}
& \Rightarrow f(\alpha, \beta)=\alpha^2-\beta^2 \\
& \Rightarrow f(m, n)+f(n, m)=m^2-n^2+n^2-m^2=0 \text { for all } m, n
\end{aligned}
\)

Hint

(b)
\(
\begin{aligned}
& \text { Given } f(x+y)=f(x)+f(y)-x y-1 \forall x, y \in R \\
& f(1)=1 \\
& f(2)=f(1+1)=f(1)+f(1)-1-1=0 \\
& f(3)=f(2+1)=f(2)+f(1)-2 \cdot 1-1=-2 \\
& f(n+1)=f(n)+f(1)-n-1=f(n)-n<f(n)
\end{aligned}
\)
\(
\begin{aligned}
& \text { Thus, } f(1)>f(2)>f(3)>\ldots \text { and } f(1)=1 . \\
& \therefore \quad f(1)=1 \text { and } f(n)<1, \text { for } n>1 . \\
& \text { Hence, } f(n)=n, n \in N \text { has only one solution } n=1 .
\end{aligned}
\)

Hint

(c)
\(
f(x)=\frac{e^x-e^{|x|}}{e^x+e^{|x|}}=\left\{\begin{array}{cc}
0, & x \geq 0 \\
\frac{e^x-e^{-x}}{e^x+e^{-x}}, & x<0
\end{array}\right.
\)
Clearly, \(f(x)\) is identically zero if \(x \geq 0 \dots(1)\)
\(
\begin{aligned}
& \text { If } x<0, \text { let } y=f(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}} \Rightarrow e^{2 x}=\frac{1+y}{1-y} \\
& \because \quad x<0 \Rightarrow e^{2 x}<1 \Rightarrow 0<e^{2 x}<1 \\
& \therefore \quad 0<\frac{1+y}{1-y}<1
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \frac{1+y}{1-y}>0 \text { and } \frac{1+y}{1-y}<1 \\
& \Rightarrow(y+1)(y-1)<0 \text { and } \frac{2 y}{1-y}<0 \\
& \Rightarrow-1<y<1 \text { and } y<0 \text { or } y>1 \\
& \Rightarrow-1<y<0 \dots(2)
\end{aligned}
\)
Combining (1) and (2), we get \(-1<y \leq 0 \Rightarrow\) Range \(=(-1,0]\).

Hint

(c)
\(
\begin{aligned}
& f(2 x+3)+f(2 x+7)=2 \dots(1) \\
& \text { Replace } x \text { by } x+2, f(2 x+7)+f(2 x+11)=2 \dots(2) \\
& \text { from }(1)-(2) \text { we get } f(2 x+3)-f(2 x+11)=0 \\
& \Rightarrow f(2 x+3)=f(2 x+11) \\
& \Rightarrow f(2 x+3)=f(2(x+4)+3) \\
& \Rightarrow \text { Period of } f(x) \text { is } 8
\end{aligned}
\)

Hint

(c) \(\text { Since co-domain }=\left[0, \frac{\pi}{2}\right)\)
\(\therefore\) for \(f\) to be onto, the range \(=\left[0, \frac{\pi}{2}\right)\)
This is possible only when \(x^2+x+a \geq 0 \quad \forall x \in R\)
\(
\therefore 1^2-4 a \leq 0 \Rightarrow a \geq \frac{1}{4}
\)

Hint

(d)
\(
\begin{aligned}
& f(x)=\frac{1}{\sqrt{\{\sin x\}+\{\sin (\pi+x)\}}}=\frac{1}{\sqrt{\{\sin x\}+\{-\sin x\}}} \\
& \text { Now }\{\sin x\}+\{-\sin x\}=\left\{\begin{array}{lc}
0, & \sin x \text { is an integer } \\
1, & \sin x \text { is not an integer }
\end{array}\right.
\end{aligned}
\)
For \(f(x)\) to get defined \(\{\sin x\}+\{-\sin x\} \neq 0\)
\(
\begin{aligned}
& \Rightarrow \quad \sin x \neq \text { integer } \\
& \Rightarrow \quad \sin x \neq \pm 1,0 \\
& \Rightarrow \quad x \neq \frac{n \pi}{2}, n \in I
\end{aligned}
\)
Hence, the domain is \(R-\left\{\frac{n \pi}{2} / n \in I\right\}\).

Hint

(a)
\(
\begin{aligned}
&f(-x)=\frac{\cos (-x)}{\left[-\frac{2 x}{\pi}\right]+\frac{1}{2}}=\frac{\cos x}{-1-\left[\frac{2 x}{\pi}\right]+\frac{1}{2}}\\
&\text { (as } x \text { is not an integral multiple of } \pi \text { ) }
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow f(-x)=-\frac{\cos x}{\left[\frac{2 x}{\pi}\right]+\frac{1}{2}}=-f(x)\\
&\Rightarrow f(x) \text { is an odd function. }
\end{aligned}
\)

Hint

(d)
\(
\begin{aligned}
& f(x)=\alpha x^3-\beta x-(\tan x) \operatorname{sgn} x \\
& f(-x)=f(x) \\
& \Rightarrow-\alpha x^3+\beta x-\tan x \operatorname{sgn} x=\alpha x^3-\beta x-(\tan x)(\operatorname{sgn} x) \\
& \Rightarrow 2\left(-\alpha x^2-\beta\right) x=0 \forall x \in R \\
& \Rightarrow \alpha=0 \text { and } \beta=0 \\
& \therefore[a]^2-5[a]+4=0 \text { and } 6\{a\}^2-5\{a\}+1=0 \\
& \Rightarrow(3\{x\}-1)(2\{x\}-1)=0 \\
& \therefore a=1+\frac{1}{3}, 1+\frac{1}{2}, 4+\frac{1}{3}, 4+\frac{1}{2}
\end{aligned}
\)
\(
\text { Sum of values of } a=\frac{35}{3}
\)

Hint

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

Hint

(b)
\(
\begin{aligned}
& 3 f(x)+2 f\left(\frac{x+59}{x-1}\right)=10 x+30 \\
& \text { For } x=7,3 f(7)+2 f(11)=70+30=100 \text {. } \\
& \text { For } x=11,3 f(11)+2 f(7)=140 \text {. }
\end{aligned}
\)
\(
\frac{f(7)}{-20}=\frac{f(11)}{-220}=\frac{-1}{9-4} \Rightarrow f(7)=4
\)

Hint

(c)
\(
\begin{aligned}
f(x) & =[6 x+7]+\cos \pi x-6 x \\
& =[6 x]+7+\cos \pi x-6 x \\
& =7+\cos \pi x-\{6 x\}
\end{aligned}
\)
\(\{6 x\}\) has the period \(1 / 6\) and \(\cos \pi x\) has the period 2, then the period of \(f(x)=\mathrm{LCM}\) of 2 and \(1 / 6\) which is 2. Hence, the period is 2.

Hint

(d) \(f(x)=\frac{a^x-1}{x^n\left(a^x+1\right)}\)
\(f(x)\) is symmetrical about \(y\)-axis
\(
\begin{aligned}
& \Rightarrow f(x)=f(-x) \\
& \Rightarrow \frac{a^x-1}{x^n\left(a^x+1\right)}=\frac{a^{-x}-1}{(-x)^n\left(a^{-x}+1\right)} \\
& \Rightarrow \frac{a^x-1}{x^n\left(a^x+1\right)}=\frac{1-a^x}{(-x)^n\left(1+a^x\right)} \Rightarrow x^n=-(-x)^n
\end{aligned}
\)
\(\Rightarrow\) the value of \(n\) which satisfy this relation is \(-\frac{1}{3}\).

Hint

(a)
\(
\begin{aligned}
& x^2 f(x)-2 f\left(\frac{1}{x}\right)=g(x) \text { and } 2 f\left(\frac{1}{x}\right)-4 x^2 f(x)=2 x^2 g\left(\frac{1}{x}\right) \\
& \text { (Replacing } x \text { by } \frac{1}{x} \text { ) } \\
& \Rightarrow-3 x^2 f(x)=g(x)+2 x^2 g\left(\frac{1}{x}\right) \\
& \text { (Eliminating } f\left(\frac{1}{x}\right) \text { ) } \\
& \Rightarrow f(x)=-\left(\frac{g(x)+2 x^2 g\left(\frac{1}{x}\right)}{3 x^2}\right)
\end{aligned}
\)
\(\because g(x)\) and \(x^2\) are odd and even functions, respectively.
So, \(f(x)\) is an odd function. But \(f(x)\) is given even
\(\Rightarrow f(x)=0 \forall x\). Hence, \(f(5)=0\).

Hint

(a) Given \(f(x+y)=f(x) f(y)\). Put \(x=y=0\), then \(f(0)=1\).
Put \(y=-x\), then \(f(0)=f(x) f(-x) \Rightarrow f(-x)=\frac{1}{f(x)}\)
Now, \(g(x)=\frac{f(x)}{1+\{f(x)\}^2}\)
\(
\begin{aligned}
g(-x) & =\frac{f(-x)}{1+\{f(-x)\}^2}=\frac{\frac{1}{f(x)}}{1+\frac{1}{\{f(x)\}^2}} \\
& =\frac{f(x)}{1+\{f(x)\}^2}=g(x)
\end{aligned}
\)

Hint

(d) The equation is \(x^2+2 a x+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}}\), \(\Rightarrow f(x)=f^{-1}(x)\) which has the solution if \(x^2+2 a x+\frac{1}{16}=x\)
\(
\Rightarrow \quad x^2+(2 a-1) x+\frac{1}{16}=0
\)
\(
\begin{aligned}
& \text { For real and distinct roots }(2 a-1)^2-4 \frac{1}{16} \geq 0 \\
& \Rightarrow 2 a-1 \leq \frac{-1}{2} \text { or } 2 a-1 \geq \frac{1}{2} \Rightarrow a \leq \frac{1}{4} \text { or } a \geq \frac{3}{4} .
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
&\begin{aligned}
& f(x)-1+f(1-x)-1=0 ; \text { so } g(x)+g(1-x)=0 \\
& \text { Replacing } x \text { by } x+\frac{1}{2}, \text { we get } g\left(\frac{1}{2}+x\right)+g\left(\frac{1}{2}-x\right)=0 .
\end{aligned}\\
&\text { So, it is symmetrical about }\left(\frac{1}{2}, 0\right) \text {. }
\end{aligned}
\)

Hint

(a) When \([x]=0\) we have \(\sin ^{-1}\left(\cos ^{-1} 0\right)=\sin ^{-1}(\pi / 2)\), not defined.
When \([x]=-1\) we have \(\sin ^{-1}\left(\cos ^{-1}-1\right)=\sin ^{-1}(\pi)\), not defined.
When \([x]=1\) we have \(\sin ^{-1}\left(\cos ^{-1} 1\right)=\sin ^{-1}(0)=0\).
Hence, \(x \in[1,2)\) and the range of function is \(\{0\}\).

Hint

(a)

\(
\begin{aligned}
& \text { Putting } x=1, f(2)+f(0)=2 f(1) \Rightarrow f(2)=2 f(1) \\
& \text { Putting } x=2, f(3)+f(1)=2 f(2) \\
& \Rightarrow \quad f(3)=2 \times 2 f(1)-f(1)=3 f(1), \text { and so on. } \\
& \therefore \quad f(n)=n f(1), \text { for } n=1,2, \ldots, n \\
& \quad f(n+1)+f(n-1)=2 f(n) \\
& \Rightarrow \quad f(n+1)+(n-1) f(1)=2 n f(1) \\
& \Rightarrow \quad f(n+1)=(n+1) f(1)
\end{aligned}
\)

Hint

(d) \(\because\{x\} \in[0,1)\)
\(\sin \{x\} \in(0, \sin 1)\) as \(f(x)\) is defined if \(\sin \{x\} \neq 0\)
\(
\Rightarrow \frac{1}{\sin \{x\}} \in\left(\frac{1}{\sin 1}, \infty\right) \Rightarrow\left[\frac{1}{\sin \{x\}}\right] \in\{1,2,3, \ldots\}
\)
Note that \(1<\frac{\pi}{3} \Rightarrow \sin 1<\sin \frac{\pi}{3}=0.866 \Rightarrow \frac{1}{\sin 1}>1.155\).

Hint

(c)
\(
\begin{aligned}
& \text { We have }\left[\cos ^{-1} x\right] \geq 0 \forall x \in[-1,1] \\
& \text { and }\left[\cot ^{-1} x\right] \geq 0 \forall x \in R \\
& \text { Hence, }\left[\cot ^{-1} x\right]+\left[\cot ^{-1} x\right]=0 \\
& \Rightarrow \quad\left[\cot ^{-1} x\right]=\left[\cot ^{-1} x\right]=0 \\
& \text { If }\left[\cos ^{-1} x\right]=0 \Rightarrow x \in(\cos 1,1]
\end{aligned}
\)
\(
\begin{aligned}
& \text { If }\left[\cot ^{-1} x\right]=0 \Rightarrow x \in(\cot 1, \infty) \\
& \Rightarrow x \in(\cot 1,1]
\end{aligned}
\)

Hint

(d) The period of \(f(x)\) is \(7 \Rightarrow\) The period of \(f\left(\frac{x}{3}\right)\) is \(\frac{7}{1 / 3}=21\)
The period of \(g(x)\) is \(11 \Rightarrow\) The period of \(g\left(\frac{x}{5}\right)\) is \(\frac{11}{1 / 5}=55\)
Hence, \(T_1=\) period of \(f(x) g\left(\frac{x}{5}\right)=7 \times 55=385\) and
\(
\begin{aligned}
& T_2=\text { period of } g(x) f\left(\frac{x}{3}\right)=11 \times 21=231 \\
& \begin{aligned}
\therefore \text { Period of } F(x) & =\operatorname{LCM}\left\{T_1, T_2\right\} \\
& =\operatorname{LCM}\{385,231\} \\
& =7 \times 11 \times 3 \times 5 \\
& =1155
\end{aligned}
\end{aligned}
\)

Hint

(d)
\(
\begin{aligned}
& \sin ^2 x+\sin ^2\left(x+\frac{\pi}{3}\right)+\cos x \cos \left(x+\frac{\pi}{3}\right) \\
& =\sin ^2 x+\left(\frac{\sin x}{2}+\frac{\sqrt{3} \cos x}{2}\right)^2+\cos x\left(\frac{\cos x}{2}-\frac{\sqrt{3} \sin x}{2}\right) \\
& =\sin ^2 x+\frac{\sin ^2 x}{4}+\frac{3 \cos ^2 x}{4}+\frac{\cos ^2 x}{2} \\
& =\frac{5 \sin ^2 x}{4}+\frac{5 \cos ^2 x}{4}=5 / 4
\end{aligned}
\)
Hence, \(f(x)=c^{5 / 4}=\) constant, which is periodic whose period cannot be determined.

Hint

(a)

\(
\begin{aligned}
&\begin{aligned}
& f(x+f(y))=f(x)+y, f(0)=1 \\
& \text { Putting } y=0 \text {, we get } f(x+f(0))=f(x)+0 \\
& \Rightarrow f(x+1)=f(x) \forall x \in R
\end{aligned}\\
&\text { Thus, } f(x) \text { is the period with } 1 \text { as one of its period. }\\
&\Rightarrow \quad f(7)=f(6)=f(5)=\cdots=f(1)=(0)=1
\end{aligned}
\)

Hint

(c)

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

Hint

(a)

\(
\begin{aligned}
f(x y) & =\frac{f(x)}{y} \\
f(y) & =\frac{f(1)}{y} \quad \text { (putting } x=1 \text { ) }
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow \quad f(30)=\frac{f(1)}{30} \text { or } f(1)=30 \times f(30)=30 \times 20=600 .\\
&\text { Now, } f(40)=\frac{f(1)}{40}=\frac{600}{40}=15 \text {. }
\end{aligned}
\)

Hint

(c) Case I:
\(
\begin{aligned}
& 0<|x|-1<1 \Rightarrow 1<|x|<2, \text { then } \\
& x^2+4 x+4 \leq 1 \\
& \Rightarrow x^2+4 x+3 \leq 0 \\
& \Rightarrow-3 \leq x \leq-1 \dots(1)
\end{aligned}
\)
So \(x \in(-2,-1)\)
Case II:
\(
\begin{aligned}
& |x|-1>1 \Rightarrow|x|>2, \text { then } x^2+4 x+4 \geq 1 \\
& \Rightarrow x^2+4 x+3 \geq 0 \\
& \Rightarrow x \geq-1 \text { or } x \leq-3
\end{aligned}
\)
So, \(x \in(-\infty,-3] \cup(2, \infty) \dots(2)\)
\(
\text { From (1) and (2), } x \in(-\infty,-3] \cup(-2,-1) \cup(2, \infty) \text {. }
\)

Hint

(d) \(f(x)=\frac{n(n+1)}{2}+[\sin x]+\left[\sin \frac{x}{2}\right]+\cdots+\left[\sin \frac{x}{n}\right]\)
Thus, the range of \(f(x)=\left\{\frac{n(n+1)}{2}, \frac{n(n+1)}{2}+1\right\}\) as \(x \in[0, \pi]\).

Hint

(b)
\(
\begin{aligned}
& {[x]^2=x+2\{x\} } \\
\Rightarrow & {[x]^2=[x]+3\{x\} } \\
\Rightarrow & \{x\}=\frac{[x]^2-[x]}{3} \\
\Rightarrow & 0 \leq \frac{[x]^2-[x]}{3}<1 \\
\Rightarrow & 0 \leq[x]^2-[x]<3
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad[x] \in\left(\frac{1-\sqrt{3}}{2}, 0\right] \cup\left[1, \frac{1+\sqrt{3}}{2}\right) \\
& \Rightarrow \quad[x]=-1,0,1,2 \\
& \Rightarrow \quad\{x\}=\frac{2}{3}, 0,0, \frac{2}{3}, \text { (respectively) } \\
& \Rightarrow \quad x=-\frac{1}{3}, 0,1, \frac{8}{3}
\end{aligned}
\)

Hint

(b) We must have
\(
2\{x\}^2-3\{x\}+1 \geq 0 \Rightarrow\{x\} \geq 1 \text { or }\{x\} \leq 1 / 2
\)
Thus, we have \(0 \leq\{x\} \leq 1 / 2 \Rightarrow x \in\left[n, n+\frac{1}{2}\right], n \in I\).

Hint

(b)
\(
\begin{aligned}
&\left[x^2+\frac{1}{2}\right]=\left[x^2-\frac{1}{2}+1\right]=1+\left[x^2-\frac{1}{2}\right] .\\
&\text { Thus, from domain point of view, }
\end{aligned}
\)
\(
\begin{aligned}
& {\left[x^2-\frac{1}{2}\right]=0,-1 \Rightarrow\left[x^2+\frac{1}{2}\right]=1,0} \\
& \Rightarrow f(x)=\sin ^{-1}(1)+\cos ^{-1}(0) \text { or } \sin ^{-1}(0)+\cos ^{-1}(-1) \\
& \Rightarrow f(x)=\{\pi\}
\end{aligned}
\)

Hint

(c) The period of \(\cos (\sin n x)\) is \(\frac{\pi}{n}\) and the period of \(\tan \left(\frac{x}{n}\right)\) is \(\pi n\).
Thus, \(6 \pi=\operatorname{LCM}\left(\frac{\pi}{n}, \pi n\right)\)
\(
\Rightarrow 6 \pi=\frac{\pi}{n} \lambda_1 \Rightarrow n=\frac{\lambda_1}{6}, \text { and } 6 \pi=\lambda_2 \pi n \Rightarrow n=\frac{6}{\lambda_2}, \lambda_1,
\)
\(\lambda_2 \in I^{+}\)
\(
\begin{aligned}
&\text { From } n=\frac{6}{\lambda_2} \Rightarrow n=6,3,2,1 .\\
&\text { Clearly, for } n=6 \text {, we get the period of } f(x) \text { to be } 6 \pi \text {. }
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\text { We must have } a x^3+(a+b) x^2+(b+c) x+c>0\\
&\begin{aligned}
& \Rightarrow a x^2(x+1)+b x(x+1)+c(x+1)>0 \\
& \Rightarrow(x+1)\left(a x^2+b x+c\right)>0 \\
& \Rightarrow a(x+1)\left(x+\frac{b}{2 a}\right)^2>0 \text { as } b^2=4 a c \\
& \Rightarrow x>-1 \text { and } \neq-\frac{b}{2 a}
\end{aligned}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\begin{aligned}
f(x) & =[x]+[2 x]+[3 x]+\cdots+[n x]-(x+2 x+3 x+\cdots+n x) \\
& =-(\{x\}+\{2 x\}+\{3 x\}+\cdots+\{n x\})
\end{aligned}\\
&\text { The period of } f(x)=\operatorname{LCM}\left(1, \frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{n}\right)=1 \text {. }
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
&\begin{aligned}
& f\left(x+\frac{1}{2}\right)+f\left(x-\frac{1}{2}\right)=f(x) \\
& \Rightarrow f(x+1)+f(x)=f\left(x+\frac{1}{2}\right) \\
& \Rightarrow f(x+1)+f\left(x-\frac{1}{2}\right)=0 \\
& \Rightarrow f\left(x+\frac{3}{2}\right)=-f(x) \\
& \Rightarrow f(x+3)=-f\left(x+\frac{3}{2}\right)=f(x)
\end{aligned}\\
&\therefore f(x) \text { is periodic with period } 3 .
\end{aligned}
\)

Hint

(c) \(f(x)+3 x f\left(\frac{1}{x}\right)=2(x+1) \dots(1)\)
Replacing \(x\) by \(\frac{1}{x}\), we get
\(
\begin{aligned}
& f\left(\frac{1}{x}\right)+3 \frac{1}{x} f(x)=2\left(\frac{1}{x}+1\right) \\
& \Rightarrow x f\left(\frac{1}{x}\right)+3 f(x)=2(x+1) \dots(2)
\end{aligned}
\)
From (1) and (2), we have \(f(x)=\frac{x+1}{2} \Rightarrow f(99)=50\)

Hint

(a) Let \(y=\frac{x+5}{x+2}=1+\frac{3}{x+2} \Rightarrow x=1\)
Also, \(y-1=\frac{3}{x+2} \Rightarrow x+2=\frac{3}{y-1}\)
\(
\Rightarrow \quad x=-2+\frac{3}{y-1}
\)
\(\Rightarrow y=2\) only as \(x\) and \(y\) are natural numbers.

Hint

(d)
\(
f(f(x))= \begin{cases}(f(x))^2, & \text { for } f(x) \geq 0 \\ f(x), & \text { for } f(x)<0\end{cases}
\)
\(
=\left\{\begin{array}{l}
\left(x^2\right)^2, x^2 \geq 0, x \geq 0 \\
x^2, x \geq 0, x<0 \\
x^2, x^2<0, x \geq 0 \\
x, x<0, x<0
\end{array}\right.
\)
\(
=\left\{\begin{array}{l}
x^4, x \geq 0 \\
x, x<0
\end{array}\right.
\)

Hint

(c) From the given data
\(
\begin{aligned}
& f(1-x)=f(1+x) \dots(1) \\
& \text { and } f(2-x)=f(2+x) \dots(2)
\end{aligned}
\)
In (2) replacing \(x\) by \(1+x\), we have
\(
\begin{aligned}
& & f(1-x) & =f(3+x) \\
\Rightarrow & & f(1+x) & =f(3+x) [\text { from (1) }]\\
\Rightarrow & & f(x) & =f(2+x)
\end{aligned}
\)

Hint

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

Hint

(d)

\(
\begin{aligned}
&\text { We must have }-1 \leq\left[2 x^2-3\right] \leq 1\\
&\begin{aligned}
& \Rightarrow \quad-1 \leq 2 x^2-3<2 \Rightarrow 1 \leq x^2<\frac{5}{2} \\
& \Rightarrow \quad x \in\left(-\sqrt{\frac{5}{2}},-1\right] \cup\left[1, \sqrt{\frac{5}{2}}\right)^2
\end{aligned}
\end{aligned}
\)

Hint

(c)
\(
\cos ^{-1}\left(\frac{1+x^2}{2 x}\right) \text { is defined if }\left|\frac{1+x^2}{2 x}\right| \leq 1 \text { and } x \neq 0
\)
\(
\begin{aligned}
& \Rightarrow \quad 1+x^2-2|x| \leq 0 \\
& \Rightarrow \quad(|x|-1)^2 \leq 0 \\
& \Rightarrow \quad x=1,-1
\end{aligned}
\)
Thus, the domain of \(f(x)\) is \(\{1,-1\}\). Hence, the range is \(\{1,1+\pi\}\).

Hint

(a)

\(
\begin{aligned}
& f(f(x))=\left\{\begin{array}{cc}
f(x), & f(x) \text { is rational } \\
1-f(x), & f(x) \text { is irrational }
\end{array}\right. \\
& f(f(x))=\left\{\begin{array}{cc}
x, & x \text { is rational } \\
1-(1-x)=x, & x \text { is irrational }
\end{array}\right.
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
& y=|\sin x|+|\cos x| \\
& \Rightarrow y^2=1+|\sin 2 x| \\
& \Rightarrow 1 \leq y^2 \leq 2 \\
& \Rightarrow y \in[1, \sqrt{2}] \\
& \Rightarrow f(x)=1 \forall x \in R
\end{aligned}
\)

Hint

(d) \(f(x)=\ln \left(\frac{x^2+e}{x^2+1}\right)=\ln \left(\frac{x^2+1+e-1}{x^2+1}\right)=\ln \left(1+\frac{e-1}{x^2+1}\right)\)
Now, \(1 \leq x^2+1<\infty\)
\(
\begin{aligned}
& \Rightarrow 0<\frac{1}{x^2+1} \leq 1 \Rightarrow 0<\frac{e-1}{x^2+1} \leq e-1 \\
& \Rightarrow 1<1+\frac{e-1}{x^2+1} \leq e \Rightarrow 0<\ln \left(1+\frac{e-1}{x^2+1}\right) \leq 1
\end{aligned}
\)
Hence, the range is \((0,1]\).

Hint

(d)

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

Hint

(b)

\(
\begin{aligned}
&\text { Given } y=2^{x(x-1)}\\
&\begin{aligned}
& \Rightarrow x(x-1)=\log _2 y \\
& \Rightarrow x^2-x-\log _2 y=0 \\
& \Rightarrow x=\frac{1 \pm \sqrt{1+4 \log _2 y}}{2}
\end{aligned}\\
&\text { Only } x=\frac{1+\sqrt{1+4 \log _2 y}}{2} \text { lies in the domain. }\\
&\Rightarrow f^{-1}(x)=\frac{1}{2}\left[1+\sqrt{1+4 \log _2 x}\right]
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
& x \sin x=1 \dots(1) \\
& \Rightarrow \quad y=\sin x=\frac{1}{x}
\end{aligned}
\)
Root of equation (1) will be given by the point(s) of intersection of the graphs \(y=\sin x\) and \(y=\frac{1}{x}\). Graphically, it is clear that we get four roots.

Hint

(c) See the graph of \(y=2 \cos x\) and \(y=|\sin x|\). Their points of intersection represent the solution of the given equation.

Hint

(a)
\(
\begin{aligned}
&a f(x+1)+b f\left(\frac{1}{x+1}\right)=(x+1)-1 \dots(1)\\
&\text { Replacing } x+1 \text { by } \frac{1}{x+1} \text {, we get }\\
&\therefore a f\left(\frac{1}{x+1}\right)+b f(x+1)=\frac{1}{x+1}-1 \dots(2)
\end{aligned}
\)
\(
\begin{aligned}
& \text { (1) } \times a-(2) \times b \Rightarrow\left(a^2-b^2\right) f(x+1)=a(x+1) \\
& -a-\frac{b}{x+1}+b \\
& \text { Putting } x=1,\left(a^2-b^2\right) f(2)=2 a-a-\frac{b}{2}+b=a+\frac{b}{2} \\
& \qquad=\frac{2 a+b}{2}
\end{aligned}
\)

Hint

(c)

In \(\left(-\frac{\pi}{2}, 0\right)\), the graph of \(y=\tan x\) lies below the line \(y=x\) which is the tangent at \(x=0\) and in \(\left(0, \frac{\pi}{2}\right)\) it lies above the lies \(y=x\).
For \(m>1\), the line \(y=m x\) lies below \(y=x\) in \(\left(-\frac{\pi}{2}, 0\right)\) and above \(y=x\) in \(\left(0, \frac{\pi}{2}\right)\). Thus graphs of \(y=\tan x\) and \(y=m x, m>1\), meet at three points including \(x=0\) in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) independent of \(m\).

Hint

(c)
\(
\text { Given } \begin{aligned}
f(x) & =[\sin x+[\cos x+[\tan x+[\sec x]]]] \\
& =[\sin +p], \text { where } p=[\cos x+[\tan x+[\sec x]]] \\
& =[\sin x]+p,(\operatorname{as} p \text { is an integer }) \\
& =[\sin x]+[\cos x+[\tan x+[\sec x]]] \\
& =[\sin x]+[\cos x]+[\tan x]+[\sec x]
\end{aligned}
\)
\(
\begin{aligned}
&\text { Now, for } x \in(0, \pi / 4), \sin x \in\left(0, \frac{1}{\sqrt{2}}\right), \cos x \in\left(\frac{1}{\sqrt{2}}, 1\right) \text {, }\\
&\begin{aligned}
& \tan x \in(0,1), \sec x \in(1, \sqrt{2}) \\
& \Rightarrow \quad[\sin x]=0,[\cos x]=0,[\tan x]=0 \text { and }[\sec x]=1 \\
& \Rightarrow \quad \text { The range of } f(x) \text { is } 1 .
\end{aligned}
\end{aligned}
\)

Hint

(d)

\(
f(3 x+2)+f(3 x+29)=0 \dots(1)
\)
Replacing \(x\) by \(x+9\), we get
\(
\begin{aligned}
& f(3(x+9)+2)+f(3(x+9)+29)=0 \\
& \Rightarrow f(3 x+29)+f(3 x+56)=0 \dots(2)
\end{aligned}
\)
From (1) and (2), we get
\(
\begin{aligned}
& f(3 x+2)=f(3 x+56) \\
& \Rightarrow \quad f(3 x+2)=f(3(x+18)+2)
\end{aligned}
\)
\(\Rightarrow f(x)\) is periodic with period 54.

Hint

(c)
(a) \(f(x)=\sin x[latex] and [latex]g(x)=\cos x, x \in[0, \pi / 2]\)
Here, both \(f(x)\) and \(g(x)\) are one-one functions, but \(h(x)=f(x)+g(x)=\sin x+\cos x\) is many-one as \(h(0)=h(\pi / 2)=1\).
(b) \(h(x)=f(x) g(x)=\sin x \cos x=\frac{\sin 2 x}{2}\) is many-one, as \(h(0)=h(\pi / 2)=0\).
(c) It is a fundamental property.

Hint

(c)

\(
\begin{aligned}
& f(x) \text { is definea for } x \in(0,1) \\
& \Rightarrow \quad f\left(e^x\right)+f(\ln |x|) \text { is defined for, } \\
& \quad 0<e^x<1 \text { and } 0<\ln |x|<1 \\
& \Rightarrow \quad-\infty<x<0 \text { and } 1<|x|<e \\
& \Rightarrow \quad x \in(-\infty, 0) \text { and } x \in(-e,-1) \cup(1, e) \\
& \Rightarrow \quad x \in(-e,-1)
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
&|\cos x|+\cos x= \begin{cases}0, & \cos x \leq 0 \\ 2 \cos x & , \cos x>0\end{cases}\\
&\text { For } f(x) \text { to defined } \cos x>0\\
&\Rightarrow x \in\left(\frac{(4 n-1) \pi}{2}, \frac{(4 n+1) \pi}{2}\right) n \in Z(1 \text { st and } 4 \text { th quadrant }) .
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\text { Let } 2 x+3 y=m \text { and } 2 x-7 y=n\\
&\begin{aligned}
& \Rightarrow y=\frac{m-n}{10} \text { and } x=\frac{7 m+3 n}{20} \\
& \Rightarrow f(m, n)=7 m+3 n \\
& \Rightarrow f(x, y)=7 x+3 y
\end{aligned}
\end{aligned}
\)

Hint

(d) Image \(b_1\) is assigned to any three of the six pre-images in \({ }^6 C_3\) ways.
Rest two images can be assigned to remaining three preimages in \(2^3-2\) ways (as function is onto).
Hence number of functions are \({ }^6 C_3 \times\left(2^3-2\right)=20 \times 6=120\)

Hint

(d) \(y=f(x)\) and \(y=g(x)\) are mirror image of each other about line \(y=a\)

\(
\begin{aligned}
& \Rightarrow \text { for some } x=b, g(b)-a=a-f(b) \\
& \Rightarrow f(b)+g(b)=2 a \\
& \Rightarrow h(b)=f(b)+g(b)=2 a \text { ( constant) }
\end{aligned}
\)
Hence \(h(x)\) is constant function. Thus it is neither oneone nor onto.

Hint

(c) Clearly \(f(x+\pi)=f(x), g(x+\pi)=g(x)\) and \(\phi\left(x+\frac{\pi}{2}\right) =\{(-1) f(x)\}\{(-1) g(x)\}=\phi(x)\).

Hint

(b) In the sum, \(f(1)+f(2)+f(3)+\cdots+f(n), 1\) occurs \(n\) times, \(\frac{1}{2}\) occurs \((n-1)\) times, \(\frac{1}{3}\) occurs \((n-2)\) times and so on
\(
\begin{aligned}
& \therefore f(1)+f(2)+f(3)+\cdots+f(n) \\
& =n \cdot 1+(n-1) \cdot \frac{1}{2}+(n-2) \cdot \frac{1}{3}+\cdots+1 \cdot \frac{1}{n} \\
& =n\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots+\frac{n-1}{n}\right) \\
& =n f(n)-\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+\cdots+\left(1-\frac{1}{n}\right)\right]
\end{aligned}
\)
\(
\begin{aligned}
& =n f(n)-[n-f(n)] \\
& =(n+1) f(n)-n
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& h(x)=\log (f(x) \cdot g(x))=\log e^{\{y\}+[y]}=\{y\}+[y]=e^{|x|} \operatorname{sgn} x \\
& \therefore h(x)=e^{|x|} \operatorname{sgn} x= \begin{cases}e^x, & x>0 \\
0, & x=0 \\
-e^{-x}, & x<0\end{cases} \\
& \Rightarrow h(-x)=\left\{\begin{array}{ll}
e^{-x}, & x<0 \\
0, & x=0 \\
-e^x, & x>0
\end{array} \Rightarrow h(x)+\dot{h}(-x)=0 \text { for all } x .\right.
\end{aligned}
\)

Hint

(b)

Hint

(d)
\(
\begin{aligned}
& \text { d. }[y+[y]]=2 \cos x \\
& \quad \Rightarrow \quad[y]+[y]=2 \cos x \quad(\because[x+n]=[x]+n \text { if } n \in I) \\
& \Rightarrow \quad 2[y]=2 \cos x \Rightarrow[y]=\cos x \dots(1)
\end{aligned}
\)
\(
\begin{aligned}
&\text { Also } \begin{aligned}
y & =\frac{1}{3}[\sin x+[\sin x+[\sin x]]] \\
& =\frac{1}{3}(3[\sin x]) \\
& =[\sin x] \dots(2)
\end{aligned}\\
&\text { From (1) and (2) }\\
&\begin{aligned}
& {[[\sin x]]=\cos x} \\
& \Rightarrow[\sin x]=\cos x
\end{aligned}
\end{aligned}
\)

The number of solutions is 0.

Hint

(a) \(\cos ^{-1}(\cos x)=[x]\)

The solutions are clearly \(0,1,2,3\) and \(3=2 \pi-x\) or \(x=2 \pi-3\).

Hint

(c)

\(
\begin{aligned}
& \text { Given } f(x)=\sqrt{(1-\cos x) \sqrt{(1-\cos x) \sqrt{(1-\cos x) \sqrt{\ldots \infty}}}} \\
& \Rightarrow f(x)=(1-\cos x)^{\frac{1}{2}}(1-\cos x)^{\frac{1}{4}}(1-\cos x)^{\frac{1}{8}} \ldots \infty \\
& \Rightarrow f(x)=(1-\cos x)^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots \infty} \\
& \Rightarrow f(x)=(1-\cos x)^{\frac{\frac{1}{2}}{1-\frac{1}{2}}} \\
& \Rightarrow f(x)=1-\cos x \\
& \Rightarrow \text { The range of } f(x) \text { is }[0,2)
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& -5 \leq|k x+5| \leq 7 \\
& \Rightarrow \quad-12 \leq k x \leq 2 \text { where }-6 \leq x \leq 1 \\
& \Rightarrow \quad-6 \leq \frac{k}{2} x \leq 1 \text { where }-6 \leq x \leq 1 \\
& \therefore k=2 . \quad[\because \text { the range of } h(x)=\text { the domain of } f(x)]
\end{aligned}
\)

Hint

(a) Let \(g(x)=(x+1)(x+2)(x+3)(x+4)\)
The rough graph of \(g(x)\) is given as

\(
\begin{aligned}
\therefore g(x) & =(x+1)(x+2)(x+3)(x+4) \\
& =(x+1)(x+4)(x+2)(x+3) \\
& =\left(x^2+5 x+4\right)\left(x^2+5 x+6\right) \\
& =t(t+2)=(t+1)^2-1,
\end{aligned}
\)
where \(t=x^2+5 x\)
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 \([4,5045]\).

Hint

(d) \(f(x)=\sqrt{x^{12}-x^9+x^4-x+1}\)
We must have \(x^{12}-x^9+x^4-x+1 \geq 0 \dots(1)\)
Obviously (1) is satisfied by \(x \in(-\infty, 0]\)
Also, \(x^9\left(x^3-1\right)+x\left(x^3-1\right)+1 \geq 0 \forall x \in[1, \infty)\)
Further, \(x^{12}-x^9+x^4-x+1=(1-x)+x^4\left(1-x^5\right)+x^{12}\) is also satisfied by \(x \in(0,1)\).
Hence, the domain is \(R\).

Hint

(a)
\(
\begin{aligned}
& f(x)=\sec ^{-1}\left(\log _3 \tan x+\log _{\tan x} 3\right) . \\
& f(x)=\sec ^{-1}\left(\log _3 \tan x+\frac{1}{\log _3 \tan x}\right)
\end{aligned}
\)
Now for \(\log _3 \tan x\) to get defined, \(\tan x \in(0, \infty)\)
\(\Rightarrow \log _3 \tan x \in(-\infty, \infty)\) or \(\log _3 \tan x \in R\)
Also \(x+\frac{1}{x} \leq-2\) or \(x+\frac{1}{x} \geq 2\)
\(
\begin{aligned}
\Rightarrow & \log _3 \tan x+\frac{1}{\log _3 \tan x} \leq-2 \text { or } \\
& \log _3 \tan x+\frac{1}{\log _3 \tan x} \geq 2 \\
\Rightarrow & \sec ^{-1}\left(\log _3 \tan x+\frac{1}{\log _3 \tan x}\right) \leq \sec ^{-1}(-2) \text { or } \\
& \sec ^{-1}\left(\log _3 \tan x+\frac{1}{\log _3 \tan x}\right) \geq \sec ^{-1} 2 \\
\Rightarrow & f(x) \leq \frac{2 \pi}{3} \text { or } f(x) \geq \frac{\pi}{3} \\
\Rightarrow & f(x) \in\left[\frac{\pi}{3}, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \frac{2 \pi}{3}\right]
\end{aligned}
\)

Hint

(a)

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 \(7-x \geq x-3\)
\(
\begin{aligned}
& \Rightarrow \quad x<7, x \geq 3 \text { and } x \leq 5 \\
& \Rightarrow \quad 3 \leq x \leq 5 \\
& \Rightarrow \quad x=3,4,5 \\
& \text { Now, } f(3)=\frac{4!}{4!}=1, f(4)=\frac{3!}{2!}=3, f(5)=\frac{2!}{0!}=2 . \\
& \text { Hence, } R_f=\{1,2,3\} .
\end{aligned}
\)

Hint

(b) We have \(f(x-y)=f(x) f(y)-f(a-x) f(a+y)\)
Putting \(x=a\) and \(y=a-x\), we get
\(
\begin{aligned}
& \quad f(a-(x-a))=f(a) f(x-a)-f(0) f(x) \\
& \text { Putting } x=0, y=0, \text { we get } \\
& \quad f(0)=f(0)(f(0))-f(a) f(a) \\
& \Rightarrow \quad f(0)=(f(0))^2-(f(a))^2 \\
& \Rightarrow \quad 1=(1)^2-(f(a))^2 \\
& \Rightarrow \quad f(a)=0 \\
& \Rightarrow \quad f(2 a-x)=-f(x)
\end{aligned}
\)