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
\[f:A\to B, \text{ where } y=f(x), x \in A \text{ and } y \in B.\]
JEE Conceptual Theory
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
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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
JEE Main Problems
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.
\[\begin{array}{ll} (a)~\mathbb R - \mathbb Z &&&&&(b)~\mathbb R^+ - \mathbb Z\\ (c)~\mathbb Z &&&&&(d)~\mathbb R^+ \end{array} \]
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:
\[\begin{align} x^2-\lfloor x \rfloor ^2&>0 \\ (x-\lfloor x \rfloor )(x+\lfloor x \rfloor )&>0 \\ \{x\} (x+\lfloor x\rfloor )&>0 \\ x+ \lfloor x\rfloor &>0 \qquad (\text{when } \{x\} \neq 0) \\ x &\in \mathbb R^+. \end{align}\] 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 \(x\) ?
\[ \begin{array}{ll} (a)~\frac{\pi}{2} &&&&&(b)~\pi \\ (c)~2\pi &&&&&(d)~\text{ No such value of } P \text{ exists.} \end{array}\]
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 ?
\[ \] \[ \begin{array} { l l } (a) ~ f(x) \text{ is an odd function} &&&&& (b)~ f(x) \text{ is an even function} \\ (c) ~ f(x) \text{ is a constant function} &&&&& (d)~ f(x) \text{ is neither an odd nor an even function} \\ \end{array} \]
Concepts tested: even and odd functions
Answer: \((a)~f(x)\text{ is an odd function}\)
Solution:
We have
\[\begin{align} f(x)&=\dfrac{x(\sin x+\tan x)}{\left\lfloor \frac{x+\pi}{\pi} \right\rfloor -0.5}\\ &=\dfrac{x(\sin x+\tan x)}{\left\lfloor \frac{x}{\pi} \right\rfloor+1 -0.5}\\ &=\dfrac{x(\sin x+\tan x)}{\left\lfloor \frac{x}{\pi} \right\rfloor +0.5}.\\ \Rightarrow f(-x)&=\dfrac{-x(\sin(- x )+\tan (-x) )}{\left\lfloor -\frac{x+\pi}{\pi} \right\rfloor -0.5}\\ &=\dfrac{-x(-\sin x-\tan x )}{\left\lfloor -\frac{x}{\pi} +1\right\rfloor -0.5}. \end{align}\]
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.
\[ \large f(x) = \frac{\alpha x}{x+1} \]
For \(x\ne-1\), consider a function \(f\) as described above for some constant \(\alpha\). What is the value of \( \alpha\) for which \(f(f(x)) \) is an identity function?
Clarification: \(g(x) \) is an identity function if \(g(x) = x\).
Image Credit: Flickr Eliana Lúcio.
JEE Advanced Problems
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 \(f(x)\)?
\[ \] \[ \begin{array} { l l } (a) \, f(x) \text{ is one-one on } [R,2R] &&&&&(b) \, f(x) \text{ is one-one on } [R,3R] \\ (c) \, f(x) \text{ is one-one on } [0,4R] &&&&&(d) \, f(x) \text{ is one-one on } [0,R] \cup [3R,4R] \\ \end{array} \]
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 \(f(g(x))\).
\[ \begin{array} { l l } A) \, [0,1) & \quad \quad \quad \quad \quad \quad \quad \quad \quad & B) \, (-\infty,1) \\ C) \, [1,\infty) & & D) \, \mathbb R \\ \end{array} \]
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.
Common mistakes:
- Not having deep understanding of how to figure out the exact composition function \(f(g(x))\).
\[f(x)=\left( \lfloor a \rfloor^2-5\lfloor a \rfloor +4 \right)x^3-\left( 6\{a\}^2-5\{a\}+1 \right)x -(\tan (x)) \text{sgn}(x)\]
Provided that \(f(x)\) be an even function for all \(x \in \mathbb{R}\). If sum of all possible values of \(a\) is \(\dfrac{P}{Q}\) for coprime positive integers \(P,Q\), then find the value of \((P+Q)\).
Details And Assumptions:
\(\lfloor ..\rfloor\) is floor function or greatest integer function.
\(\{ .. \}\) is fractional part function.
\(\text{sgn}(x)\) is signum function defined as : \(\text{sgn}(x)=\begin{cases} 1 \quad , x>0 \\ 0 \quad , x=0 \\ -1 \quad , x<0 \end{cases}\).
A function \(f:\mathbb{R} -\{a_1,a_2\} \to \mathbb{R},\) where \(\mathbb{R}\) is the set of Real numbers, is defined by \[f(x)=\frac{Ax^2+6x-8}{A+6x-8x^2}\] How many integral values of \(A\) exist for which \(f(x)\) is onto?
Details : \(a_1,a_2\) are the roots of the quadratic equation \(A+6x-8x^2=0\).
Once you are confident of Functions, move on to JEE Function Graphs or JEE Limits,Continuity and Differentiability.