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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Easy

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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421 Easy
421 of 478
Understanding +4 | -1
The function f(x)=log⁡(x+x2+1)f(x)=\log \left(x+\sqrt{x^2+1}\right), is
View answer and explanation
Correct option: C
an odd function
Option C (an odd function) is correct because f(-x) = -f(x), satisfying the definition of an odd function.
422 Easy
422 of 478
Application +4 | -1
In above problem, the range of f(x)∀x∈[−1,1]f(x) \forall x \in[-1,1] is
View answer and explanation
Correct option: D
[−1,1][-1,1]
Option D ([-1,1]) is correct because the range is determined by the minimum and maximum values of the function within the given interval.
423 Easy
423 of 478
Understanding +4 | -1
Function sin⁡−1x\sin ^{-1} \sqrt{x} is defined in the interval
View answer and explanation
Correct option: B
[0,1][0,1]
Option B ([0,1]) is correct because the domain of sin⁡−1x\sin^{-1} \sqrt{x} requires 0≤x≤10 \le x \le 1 to ensure a real and defined value.
424 Easy
424 of 478
Understanding +4 | -1
Function f:N→N,f(x)=2x+3f: N \rightarrow N, f(x)=2 x+3 is:
View answer and explanation
Correct option: B
One-one into
Option B (One-one into) is correct because the function is injective (one-to-one) but not surjective (onto) as the range is a subset of the codomain.
425 Easy
425 of 478
Application +4 | -1
The value of bb and cc for which the identit f(x+1)−f(x)=8x+3f(x+1)-f(x)=8 x+3 is satisfied, wher f(x)=bx2+cx+df(x)=b x^{2}+c x+d, are
View answer and explanation
Correct option: B
b=4,c=−1b=4, c=-1
Option B (b=4,c=−1b=4, c=-1) is correct because substituting f(x)=bx2+cx+df(x)=b x^{2}+c x+d into f(x+1)−f(x)=8x+3f(x+1)-f(x)=8 x+3 and solving for bb and cc gives b=4b=4 and c=−1c=-1.
426 Easy
426 of 478
Analyzing +4 | -1
If f(x)=ax+bf(x)=a x+b and g(x)=cx+dg(x)=c x+d, then f(g(x))=g(f(x))f(g(x))=g(f(x)) ⇔\Leftrightarrow
View answer and explanation
Correct option: C
f(d)=g(b)f(d)=g(b)
Option C (f(d)=g(b)f(d)=g(b)) is correct because it is the simplified condition derived from equating f(g(x))f(g(x)) and g(f(x))g(f(x)).
427 Easy
427 of 478
Analyzing +4 | -1
The range of the function f(x)=cos⁡[x]f(x)=\cos [x] for −π/2-\pi / 2 <x<π/2< x < \pi / 2 contains
View answer and explanation
Correct option: B
{cos⁡1,1,cos⁡2}\{\cos 1,1, \cos 2\}
Option B ({cos 1, 1, cos 2}) is correct because the range of the function is determined by evaluating cos[x] for integer values of [x] within the given interval.
428 Easy
428 of 478
Application +4 | -1
If a function ff satisfies f{f(x)}=x+1f\left\{ {f\left( x \right)} \right\} = x + 1 for all real values of xx and if f(0)=12f\left( 0 \right) = \frac{1}{2}, then f(1)f\left( 1 \right) is equal to
View answer and explanation
Correct option: C
32\frac{3}{2}
Option C (3/2) is correct because by using the given condition f{f(x)} = x + 1 and f(0) = 1/2, we can find f(1) by substituting appropriate values.
429 Easy
429 of 478
Analyzing +4 | -1
If f(x)=xb+3,g(x)=ax+cf(x)=x^{b}+3, g(x)=a x+c. If (g(f(x)))−1=(x−72)13(g(f(x)))^{-1}=\left(\frac{x-7}{2}\right)^{\frac{1}{3}}, then fog(ac)+gof(b)f o g(a c)+g o f(b) is:
View answer and explanation
Correct option: A
189
Option A (189) is correct because after finding the values of a, b, and c, we can calculate f o g(ac) + g o f(b) which equals 189.
430 Easy
430 of 478
Application +4 | -1
Set AA has 3 elements and set BB has 4 elements. The number of injection that can be defined from AA to BB is
View answer and explanation
Correct option: C
24
Option C (24) is correct because the number of injections from a set A with 3 elements to a set B with 4 elements is given by 4P3 = 24.
431 Easy
431 of 478
Application +4 | -1
The domain of the function f(x)=x−x2+4+x+4−xf(x)=\sqrt{x-x^{2}}+\sqrt{4+x}+\sqrt{4-x} is
View answer and explanation
Correct option: D
[0,1][0,1]
Option D ([0,1]) is correct because the domain is the intersection of the domains of each square root term.
432 Easy
432 of 478
Application +4 | -1
Given f(x)=1(1−x),g(x)=f{f(x)}f(x)=\frac{1}{(1-x)}, g(x)=f\{f(x)\} and h(x)=f[f{f(x)}]h(x)=f[f\{f(x)\}]. Then the value of f(x)⋅g(x).h(x)f(x) \cdot g(x) . h(x) is
View answer and explanation
Correct option: B
-1
Option B (-1) is correct because, after simplifying the composite functions, the product f(x) * g(x) * h(x) results in -1.
433 Easy
433 of 478
Analyzing +4 | -1
Let a function f:N→Nf:N \to N be defied by f(n)=[2n,n=2,4,6,8....n−1,n=3,7,11,15....n+12,n=1,5,9,13....f\left( n \right) = \left[ {\begin{array}{cccccccccccccccccccc} {2n,}&{n = 2,4,6,8....} \\ {n - 1,}&{n = 3,7,11,15....} \\ {\frac{{n + 1}}{2},}&{n = 1,5,9,13....} \end{array}} \right. then ff is
View answer and explanation
Correct option: D
One-one and onto
Option D (One-one and onto) is correct because the function covers all natural numbers in its range and each element in the domain maps to a unique element in the range.
434 Easy
434 of 478
Application +4 | -1
If f(x)=[x−2]f\left( x \right) = \left[ {x - 2} \right], where [x]\left[ x \right] denotes the greatest integer less than or equal to xx, then f(2,  5)f\left( {2,\;5} \right) is equal to
View answer and explanation
Correct option: B
1
Option C (0) is correct because the greatest integer function returns the largest integer less than or equal to the input, and [0.5] = 0.
435 Easy
435 of 478
Analyzing +4 | -1
The domain of definition of the function

f(x)=log⁡3{−log⁡4(6x−46x+5)},  isf\left( x \right) = {\log _3}\left\{ { - {{\log }_4}\left( {\frac{{6x - 4}}{{6x + 5}}} \right)} \right\},\;\mathrm{is}
View answer and explanation
Correct option: A
(2/3,∞)\left( {2/3,\infty } \right)
Option A (2/3,∞)\left( {2/3,\infty } \right) is correct because it satisfies the nested logarithmic domain restrictions.
436 Easy
436 of 478
Analyzing +4 | -1
Let f(x)=ax(a>0)f(x) = {a^x}(a > 0) be written as f(x)=f1(x)+f2(x),f(x) = {f_1}(x) + {f_2}(x), where f1(x){f_1}(x) is an even function and f2(x){f_2}(x) is an odd function. Then f1(x+y)+f1(x−y){f_1}(x + y) + {f_1}(x - y) equals
View answer and explanation
Correct option: D
2f1(x)⋅f1(y)2{f_1}(x) \cdot {f_1}(y)
Option D (2f1(x)⋅f1(y)2{f_1}(x) \cdot {f_1}(y)) is the correct answer because it results from correctly decomposing the exponential function into even and odd parts and simplifying the expression.
437 Easy
437 of 478
Application +4 | -1
Let g(x)=1+x−[x]g\left( x \right) = 1 + x - \left[ x \right] and f(x)={−1,                    x<0            0,                          x=0,  1                        x>0        then    forf\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\begin{array}{cccccccccccccccccccc} { - 1,\;\;\;\;\;\;\;\;\;\;x < 0\;\;\;\;\;\;}\\ {0,\;\;\;\;\;\;\;\;\;\;\;\;\;x = 0,\;} \end{array}}\\ {1\;\;\;\;\;\;\;\;\;\;\;\;x > 0\;\;\;} \end{array}} \right.\;\mathrm{then}\;\mathrm{\;for} for x,f[g(x)]x,f\left[ {g\left( x \right)} \right] is equal to
View answer and explanation
Correct option: B
11
Option B (1) is correct because g(x) is always greater than 1, and f(x) is 1 for any input greater than 0.
438 Easy
438 of 478
Application +4 | -1
Domain of f(x)=log⁡∣log⁡x∣f(x)=\log |\log x| is
View answer and explanation
Correct option: C
(0,1)∪(1,∞)(0,1) \cup(1, \infty)
Option C ((0,1) ∪(1, ∞)) is correct because the argument of the outer log must be positive, and the argument of the inner log must be positive and not equal to 1.
439 Easy
439 of 478
Application +4 | -1
If f(x)=xx−1=1yf(x)=\frac{x}{x-1}=\frac{1}{y}, then f(y)=f(y)=
View answer and explanation
Correct option: D
1−x1-x
Option D (1−x1-x) is correct because solving for y and then substituting into f(y) yields 1-x.
440 Easy
440 of 478
Understanding +4 | -1
Let A\mathrm{A} and B\mathrm{B} be non-empty sets. Let f:A→Bf: \mathrm{A} \rightarrow \mathrm{B} and g:B→Ag: \mathrm{B} \rightarrow \mathrm{A} be two functions such that f(g(x))=x,∀x∈Bf(g(x))=x, \forall x \in \mathbf{B}. Which of the following statements are true (T) or false (F) ?
(i) The function ff is one-one function
(ii) The function ff is onto
(iii) The function gg is one-one
(iv) The function gg is onto
View answer and explanation
Correct option: A
FTTF
Option C (TTFF) is correct because the function f is onto, and the function g is one-one, given the condition f(g(x)) = x.
441 Easy
441 of 478
Analyzing +4 | -1
If f:R→R,f(x)={x∣x∣−4,x∈Qx∣x∣−3x∉Q\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\mathrm{x}|\mathrm{x}|-4, & x \in Q \\ \mathrm{x}|\mathrm{x}|-\sqrt{3} & x \notin Q\end{array}\right., then f(x)\mathrm{f}(\mathrm{x}) is
View answer and explanation
Correct option: D
many to one and into
Option D (many to one and into) is correct because multiple inputs can map to the same output, and not all real numbers are in the range.
442 Easy
442 of 478
Analyzing +4 | -1
Let A={1,  2,  3,  ...,  10}A = \{ 1,\,\,2,\,\,3,\,\,...,\,\,10\} and f:A→Af:A \to A be defined as f(k)={k+1,if  k  is  oddk,if  k  is  evenf(k) = \left\{ \begin{array}{l} k + 1,\quad if\,\,k\,\,is\,\,odd\\ k,\quad {\kern 1pt} {\kern 1pt} \quad if\,\,k\,\,is\,\,even \end{array} \right.
Then, the number of possible functions g:A→A,g:A \to A, such that gof=fgof = f is
View answer and explanation
Correct option: A
105{10^5}
Option A is correct. The number of possible functions gg is found by recognizing that g(x)=xg(x) = x for even xx, and for odd xx, g(x)g(x) can take any value in AA, leading to 10510^5 possibilities.
443 Easy
443 of 478
Analyzing +4 | -1
The domain of the function f(x)=sin⁡−1[2x2−3]+log⁡2(log⁡12(x2−5x+5)),f(x)=\sin ^{-1}\left[2 x^2-3\right]+\log _2\left(\log _{\frac{1}{2}}\left(x^2-5 x+5\right)\right), where [t][t] is the greatest integer function, is
View answer and explanation
Correct option: C
(1,5−52)\left(1, \frac{5-\sqrt{5}}{2}\right)
Option C (1,5−52)\left(1, \frac{{5 - \sqrt 5 }}{{2}}\right) is correct because it represents the intersection of the domains of the inverse sine and logarithmic functions, considering the greatest integer function.
444 Easy
444 of 478
Analyzing +4 | -1
Let f(x)=9x/(9x+3)f(x)=9^x /\left(9^x+3\right) for all x∈Rx \in \mathbb{R}. Then the value of ∑r=12008f(r/2009)\sum_{r=1}^{2008} f(r / 2009) is
View answer and explanation
Correct option: A
1004
The correct answer is A (1004) because the sum can be simplified by pairing terms f(x)f(x) and f(1−x)f(1-x), which sum to 1, and then summing these pairs.
445 Easy
445 of 478
Analyzing +4 | -1
If f(x)=log⁡e(x+3)x2+3x+2f(x)=\frac{\log _e(x+3)}{x^2+3 x+2}, then the domain of f′(x)f^{\prime}(x) is
View answer and explanation
Correct option: D
(−3,∞)−{−1,−2(-3, \infty)-\{-1,-2,}.\} .
Option D ((-3, ∞)-\{-1,-2,\},$.) is correct because it considers the domain of the logarithm and excludes points where the denominator is zero.
446 Easy
446 of 478
Understanding +4 | -1
Range of the function f(x)=sin⁡2(x4)+cos⁡2(x4)f(x)=\sin ^{2}\left(x^{4}\right)+\cos ^{2}\left(x^{4}\right) is
View answer and explanation
Correct option: B
{1}\{1\}
Option B (\left\\{1\\}\right.\left\\{1\\}\right.) is correct because sin⁡2(x4)+cos⁡2(x4)\sin^2(x^4) + \cos^2(x^4) is always equal to 1 due to the trigonometric identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1.
447 Easy
447 of 478
Application +4 | -1
If f(x)=x2+λx+1f(x)=x^2+\lambda x+1 for all real xx and ff is an even function, then the value of λ\lambda is
View answer and explanation
Correct option: D
0
Option D (0) is correct because for a function to be even, f(x)=f(−x)f(x) = f(-x) for all x, which implies that the coefficient of the linear term must be zero.
448 Easy
448 of 478
Application +4 | -1
If f(x)=(25−x4)1/4  for  0<x<5,f\left( x \right) = {\left( {25 - {x^4}} \right)^{1/4}}\;for\;0 < x < \sqrt 5 , then (f(12))=\left( {f\left( {\frac{1}{2}} \right)} \right) =
View answer and explanation
Correct option: D
2−1{2^{ - 1}}
Option D (2−1{2^{ - 1}}) is correct because f(f(x)) = x, so f(f(1/2)) = 1/2, which is 2(-1).
449 Easy
449 of 478
Understanding +4 | -1
The mapping f:N→Nf:N \to N given f(n)=1+n2,  n∈Nf\left( n \right) = 1 + {n^2},\;n \in N where NN is the set of natural number, is
View answer and explanation
Correct option: C
One-to-one but not onto
Option C (One-to-one but not onto) is correct because each natural number maps to a unique natural number, but not all natural numbers are in the range.
450 Easy
450 of 478
Analyzing +4 | -1
The function f(x)=sec[log⁡(x+1+x2)]f\left( x \right) = \mathrm{sec}\left[ {\log (x + \sqrt {1 + {x^2})} } \right] is
View answer and explanation
Correct option: B
Even
Option B (Even) is correct because the function can be shown to satisfy the condition f(x) = f(-x), indicating even symmetry.
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