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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Mixed difficulty

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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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Showing 571-600 of 650 questions
571 Easy
571 of 650
Understanding +4 | -1
f(x)=∣sin⁡x∣f\left( x \right) = \left| {\sin x} \right| has an inverse if its domain is
View answer and explanation
Correct option: B
[0,  π/2]\left[ {0,\;\pi /2} \right]
Option B ([0,;π/2{0,\\;\pi /2}]) is correct because the absolute value of sin(x) is one-to-one on this interval, allowing for an inverse function to exist.
572 Easy
572 of 650
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.
573 Medium
573 of 650
Understanding +4 | -1
The function f:R→Rf: R \rightarrow R defined by f(x)=exf(x)=e^x is:
View answer and explanation
Correct option: C
One-one and into
Option C (One-one and into) is correct because the exponential function f(x)=exf(x) = e^x is injective (one-to-one) but not surjective (onto) when mapping from R to R.
574 Easy
574 of 650
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.
575 Easy
575 of 650
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.
576 Easy
576 of 650
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.
577 Medium
577 of 650
Application +4 | -1
If f(x)=ex  and  g(x)=log⁡ex,f\left( x \right) = {e^x}\;and\;g\left( x \right) = {\log _e}x, then which of the following is true?
View answer and explanation
Correct option: B
f{g(x)}=g{f(x)}f\left\{ {g\left( x \right)} \right\} = g\left\{ {f\left( x \right)} \right\}
Option B is correct because f{g(x)} = e^(log_e(x)) = x and g{f(x)} = log_e(e^x) = x, so f{g(x)} = g{f(x)}.
578 Medium
578 of 650
Understanding +4 | -1
Which of the following relation is NOT a function?
View answer and explanation
Correct option: C
h={(n,1n)∣n∈I}h=\left\{\left(n, \frac{1}{n}\right) \mid n \in \mathrm{I}\right\}
Option C is correct because the function h(n) is not defined when n = 0, as it would involve division by zero.
579 Easy
579 of 650
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.
580 Easy
580 of 650
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.
581 Hard
581 of 650
Understanding +4 | -1
The function f:N→Nf: N \rightarrow N (NN is the set of natural numbers) defined by f(n)=2n+3f(n)=2 n+3, is
View answer and explanation
Correct option: B
Injective
Option B (Injective) is correct because for every distinct n in N, f(n) produces a distinct value, meaning the function is one-to-one.
582 Medium
582 of 650
Application +4 | -1
If f:R→Sf:R \to S, defined by f(x)=sin⁡x−3cos⁡x−1f\left( x \right) = \sin x - \sqrt 3 \cos x - 1 is onto, then the intervel of s is
View answer and explanation
Correct option: D
[−1,3]\left[ { - 1,\mathrm{ }3} \right]
Option D is correct because the range of sin(x) - sqrt(3)cos(x) - 1 is [-1, 3] when the function is onto.
583 Medium
583 of 650
Understanding +4 | -1
Assertion: Let lim⁡x→af(x)=l\lim \limits_ {{x} \rightarrow {a}} {f}({x})=l and lim⁡x→ag(x)=m\lim \limits_ {{x} \rightarrow {a}} {g}({x})={m}. If ll and m{m} both exist, then lim⁡x→a(fg)(x)=lim⁡x→af(x).lim⁡x→ag(x)=lm.\lim \limits_{{x \rightarrow a}}(f g)(x)=\lim \limits_{{x \rightarrow a}} f(x). \lim \limits_{{x \rightarrow a}} g(x)=l m.

Reason: Let f{f} be a real valued function defined by f(x)=x2+1f(x)=x^2+1, then f′(2)=4f^{\prime}(2)=4.
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a
correct explanation for assertion
Option B (Assertion is correct, reason is correct; reason is not a correct explanation for assertion) is correct because both the assertion and reason are true statements, but the reason does not explain the assertion.
584 Easy
584 of 650
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.
585 Easy
585 of 650
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.
586 Medium
586 of 650
Analyzing +4 | -1
The domain of definition of the function

f(x)=x⋅1+2(x+4)−0.52−(x+4)0.5+(x+4)0.5+4(x+4)0.5isf\left( x \right) = x \cdot \frac{{1 + 2{{\left( {x + 4} \right)}^{ - 0.5}}}}{{2 - {{\left( {x + 4} \right)}^{0.5}}}} + {\left( {x + 4} \right)^{0.5}} + 4{\left( {x + 4} \right)^{0.5}}\mathrm{is}
View answer and explanation
Correct option: D
(−4,  0)∪(0,  ∞)\left( { - 4,\;0} \right) \cup \left( {0,\;\infty } \right)
The correct answer is D ((-4, 0) ∪ (0, ∞)) because it represents the values of x for which the square root is defined (x + 4 > 0) and the denominator is non-zero (x ≠ 0).
587 Easy
587 of 650
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.
588 Easy
588 of 650
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.
589 Easy
589 of 650
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.
590 Medium
590 of 650
Application +4 | -1
Let f(x)=x+1  and  ϕ(x)=x−2.f\left( x \right) = x + 1\;and\;\phi \left( x \right) = x - 2.. Then the values of x satisfying ∣f(x)+ϕ(x)∣=∣f(x)∣+∣ϕ(x)∣\left| {f\left( x \right) + \phi \left( x \right)} \right| = \left| {f\left( x \right)} \right| + \left| {\phi \left( x \right)} \right| are :
View answer and explanation
Correct option: B
[2,∞)\left[ {2,\infty } \right)
Option B ([2, ∞)) is correct because the condition |f(x) + φ(x)| = |f(x)| + |φ(x)| is satisfied when both f(x) and φ(x) are non-negative.
591 Easy
591 of 650
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.
592 Medium
592 of 650
Analyzing +4 | -1
Which of the following functions is not an are not an injective map(s)?
View answer and explanation
Correct option: B
g(x)=x+1x,x∈(0,∞)g\left( x \right) = x + \frac{1}{x},x \in \left( {0,\infty } \right)
Option B (g(x)=x+1x,x∈(0,∞)g\left( x \right) = x + \frac{1}{x},x \in \left( {0,\infty } \right)) is correct because it is not injective, as different values of x can produce the same output.
593 Easy
593 of 650
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.
594 Medium
594 of 650
Understanding +4 | -1
The domain of the function f(x)=(log⁡x)32f(x)=\left(\log ^x\right)^{32} is
View answer and explanation
Correct option: D
[1,∞][1, \infty]
Option D ([1,∞][1, \infty]) is correct because the base of a logarithm must be positive and not equal to 1, and the argument must be positive.
595 Medium
595 of 650
Application +4 | -1
Let f:N→Nf: N \rightarrow N where NN is set of natural numbers be a function such that f(x+y)=f(xy) ∀ x≥4f(x+y)=f(x y) \, \forall \, x \geq 4 and y≥4y \geq 4, then
View answer and explanation
Correct option: B
f(8)=f(9)f(8)=f(9)
Option B (f(8)=f(9)f(8)=f(9)) is correct because it is derived from the given functional equation by substituting appropriate values for x and y.
596 Easy
596 of 650
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.
597 Easy
597 of 650
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.
598 Easy
598 of 650
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.
599 Medium
599 of 650
Application +4 | -1
Let f(x)=210⋅x+1f(x)=2^{10} \cdot x+1 and g(x)=310⋅x−1g(x)=3^{10} \cdot x-1. If (f∘g)(x)=x(f \circ g)(x)=x, then x\mathrm{x} is equal to :
View answer and explanation
Correct option: D
1−2−10310−2−10\frac{1-2^{-10}}{3^{10}-2^{-10}}
Option D (1−2−10310−2−10\frac{1-2^{-10}}{3^{10}-2^{-10}}) is correct because it is the solution to the equation f(g(x))=xf(g(x)) = x after substituting the given functions.
600 Hard
600 of 650
Analyzing +4 | -1
Suppose a,ba,b and cc are real numbers such that ab>1\frac{a}{b} > 1 and ac<0\frac{a}{c} < 0. Which one of the following is true?
View answer and explanation
Correct option: C
(a−c)(b−c)>0\left( {a - c} \right)\left( {b - c} \right) > 0
The correct answer is C. Given the conditions, we can deduce the signs of a, b, and c, and thus determine that (a-c) and (b-c) have the same sign, making their product positive.
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