JEE Functions

This page will teach you how to master JEE Functions. We highlight the main concepts, provide a list of examples with solutions, and include problems for you to try. Once you are confident, you can take the quiz to establish your mastery.

A function $f$ from a non-empty set $A$ to another non-empty set $B$ is a correspondence that assigns to each element of set $A$, one and only one element of set $B$. Mathematically, we write

$f:A\to B, \text{ where } y=f(x), x \in A \text{ and } y \in B.$

As per JEE syllabus, the main concepts are definition of function, mappings, even and odd functions, composition of functions and periodic functions.

Definition of function

• Function as a set of ordered pairs
• Domain, co-domain and range

Mappings

• One-one and many-one functions
• Onto and into functions
• Bijective fuctions

Even and odd functions

• Definition and properties
• Even and odd extensions

Composition of functions

• Definition and properties

• Determining the composite function

Periodic functions

• Properties
• Period and fundamental period

Find the domain of the real-valued function $f(x)=\dfrac{1}{\sqrt{x^2-\lfloor x \rfloor ^2}}$, where $\lfloor \cdot \rfloor$ represents the greatest integer function.

$\begin{array}{ll} (a)~\mathbb R - \mathbb Z &&&&&(b)~\mathbb R^+ - \mathbb Z\\ (c)~\mathbb Z &&&&&(d)~\mathbb R^+ \end{array}$

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Concepts tested: domain

Answer: $(b)~\mathbb R^+ - \mathbb Z$

Solution:
For $f(x)$ to be real-valued, $x^2-\lfloor x \rfloor ^2$ has to be positive:
$\begin{aligned} x^2-\lfloor x \rfloor ^2&>0 \\ (x-\lfloor x \rfloor )(x+\lfloor x \rfloor )&>0 \\ \{x\} (x+\lfloor x\rfloor )&>0 \\ x+ \lfloor x\rfloor &>0 \qquad (\text{when } \{x\} \neq 0) \\ x &\in \mathbb R^+. \end{aligned}$ As $\{ x \} \geq 0$, excluding the values at which $\{ x \}=0,$ i.e. $x \in \mathbb Z,$ the domain of the real-valued function $f(x)$ is $\mathbb R^+ - \mathbb Z. \ _\square$

Common mistakes: It is wrong to consider that $x^2-\lfloor x\rfloor ^2=\{ x^2\}$ because the definition of fractional part function says that $x^2-\lfloor x^2\rfloor =\{ x^2\}$.

If $f(x)=\left( \sqrt 2 \right)^{\sin^2x }$, then what is the smallest positive value of $P$ such that $f(x)=f(x+P)$ for all real values of $x$ ?

$\begin{array}{ll} (a)~\frac{\pi}{2} &&&&&(b)~\pi \\ (c)~2\pi &&&&&(d)~\text{ No such value of } P \text{ exists.} \end{array}$

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Concepts tested: periodic functions

Answer: $(b)~\pi$

Solution:
As per the definition of the fundamental period, we are asked to find the fundamental period of the function $f(x)$ through the statement of the problem. Let $q(x)=\sin^2x$. Then since the fundamental period of $q(x)$ is $\pi,$ the fundamental period of $\left( \sqrt 2\right)^{q(x)}$ is also $\pi$. $_\square$

Common mistakes: Thought that $f(x)$ was exponential and thus couldn't be periodic? $e ^ { \{ x \} }$ is a counter-example.

Which of following statements is true about the function $f(x)=\dfrac{x(\sin x+\tan x)}{\left\lfloor \dfrac{x+\pi}{\pi} \right\rfloor -0.5}$ , where $x$ is not an integer multiple of $\pi$ and $\lfloor \cdot \rfloor$ represents the greatest integer function ?

$$ $\begin{array} { l l } (a) ~ f(x) \text{ is an odd function} &&&&& (b)~ f(x) \text{ is an even function} \\ (c) ~ f(x) \text{ is a constant function} &&&&& (d)~ f(x) \text{ is neither an odd nor an even function} \\ \end{array}$

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Concepts tested: even and odd functions

Answer: $(a)~f(x)\text{ is an odd function}$

Solution:

We have

$\begin{aligned} f(x)&=\dfrac{x(\sin x+\tan x)}{\left\lfloor \frac{x+\pi}{\pi} \right\rfloor -0.5}\\ &=\dfrac{x(\sin x+\tan x)}{\left\lfloor \frac{x}{\pi} \right\rfloor+1 -0.5}\\ &=\dfrac{x(\sin x+\tan x)}{\left\lfloor \frac{x}{\pi} \right\rfloor +0.5}.\\ \Rightarrow f(-x)&=\dfrac{-x(\sin(- x )+\tan (-x) )}{\left\lfloor -\frac{x+\pi}{\pi} \right\rfloor -0.5}\\ &=\dfrac{-x(-\sin x-\tan x )}{\left\lfloor -\frac{x}{\pi} +1\right\rfloor -0.5}. \end{aligned}$

Since we can say that $\lfloor -\lambda\rfloor =-1-\lfloor \lambda\rfloor$ if $\lambda$ is not an integer, it follows that

$f(-x)=\dfrac{x(\sin x+\tan x)}{-1-\left[ \frac{x}{\pi} \right] +0.5} =\displaystyle -\dfrac{x(\sin x+\tan x)}{\left[ \frac{x}{\pi} \right] +0.5}=-f(x).$

Hence $f(x)$ is an odd function, provided $x$ is not an integer multiple of $\pi$. $_\square$

Common mistakes: If you thought that $\lfloor -\lambda \rfloor = -\lfloor \lambda\rfloor$, you would get that $f(x)$ is neither odd nor even.

The base of a tower has the shape of a truncated right circular cone which is surmounted by a right circular cylinder which is again surmounted by a hemisphere of radius $R$. The truncated right circular cone is having a lower base of radius $2R$, upper base of radius $R$ and height $R$. The cylinder is of radius $R$ and height $2R$. Suppose that the cross-sectional area of the tower is given by $f(x)$, where $x$ is the distance of the cross-section from the lower base of the cone. Then which of the followings is true about $f(x)$?

$$ $\begin{array} { l l } (a) \, f(x) \text{ is one-one on } [R,2R] &&&&&(b) \, f(x) \text{ is one-one on } [R,3R] \\ (c) \, f(x) \text{ is one-one on } [0,4R] &&&&&(d) \, f(x) \text{ is one-one on } [0,R] \cup [3R,4R] \\ \end{array}$

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Concepts tested: one-one functions

Answer: $(d)$ $f(x)$ is one-one on $[0,R] \cup [3R,4R]$

Solution:

Step 1: Figure out the function
The function $f(x)$ is defined in $[0,4R]$ and is given by $f(x)=\begin{cases} \pi(2R-x)^2 &&0 \leq x \leq R \\ \pi R^2 &&R < x < 3R \\ \pi(6Rx-x^2-8R^2) &&3R \leq x \leq 4R. \end{cases}$

Step 2: Analyse the function in all the relevant intervals

• $0 \leq x \leq R$ : $f$ decreases on $[0,R]$ and takes values in $[\pi R^2,4\pi R^2]$.

• $R < x < 3R$ : $f$ is constant on $(R,3R)$ and equals to $\pi R^2$.

• $3R \leq x \leq 4R$ : $f$ decreases on $[3r,4R]$ and takes values in $[0,4 \pi R^2]$.

Step 3: Check condition of one-one on all intervals
$f$ is decreasing function in the interval $[0,R] \cup [3R,4R]$ and constant function in the interval $(R,3R)$. Hence, one-one in the interval $[0,R] \cup [3R,4R]$, but many-one in the interval $(R,3R)$.

Common mistakes: If you didn't figure out the function with respect to the given variable $x$ exactly, then you'll come out with inappropriate results.

Let $f(x)=\begin{cases} x+1 \quad , x<0 \\ x^2 \quad \quad \ , x \geq 0 \end{cases}$ and $g(x)= \begin{cases} x^3 \quad \quad \ \ , x<1 \\ 2x-1 , \quad x \geq 1 \end{cases}$. Find the range of the function $f(g(x))$.

$\begin{array} { l l } A) \, [0,1) & \quad \quad \quad \quad \quad \quad \quad \quad \quad & B) \, (-\infty,1) \\ C) \, [1,\infty) & & D) \, \mathbb R \\ \end{array}$

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Concepts tested: Composition of functions

Answer: D) $\mathbb R$

Solution:
Step 1: Writing $f(g(x))$ in terms of g(x)
$f(g(x))=\begin{cases} g(x)+1 \quad , g(x)<0 \\ g(x)^2 \quad \quad, g(x) \geq 0 \end{cases}$
Step 2: Breaking $f(g(x))$ into intervals of $x$
$f(g(x))=\begin{cases} x^3+1 \quad \quad \quad , x^3<0 \text{ and } x<1 \\ 2x-1+1 \quad , 2x-1<0 \text{ and } x \geq 1 \\ (x^3)^2 \quad \quad \quad , x^3 \geq 0 \text{ and } x<1 \\ (2x-1)^2 \quad , 2x-1 \geq 0 \text{ and } x \geq 1\end{cases} =\begin{cases} x^3+1 \quad \quad \quad \ , x<0 \\ x^6 \quad \quad \quad \quad, 0 \leq x <1 \\ (2x-1)^2 \quad \quad , x \geq 1 \end{cases}$
Step 3: Find range for the separate intervals

• For $x<0$ : $f(g(x))=x^3+1$ and range of $x^3+1$ is $(-\infty,1)$.

• For $0 \leq x <1$ : $f(g(x))=x^6$ and range of $x^6$ is $[0,1)$.

• For $x \geq 1$ : $f(g(x))=(2x-1)^2$ and range of $(2x-1)^2$ is $[1,\infty)$.

Step 4: Taking union of all the above intervals of range
Since $(-\infty,1) \cup [0,1) \cup [1,\infty)=\mathbb R$, hence the range of $f(g(x))$ is all real numbers.

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Common mistakes:

• Not having deep understanding of how to figure out the exact composition function $f(g(x))$.

Once you are confident of Functions, move on to JEE Function Graphs or JEE Limits,Continuity and Differentiability.