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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 1-30 of 650 questions
1 Medium
1 of 650
Application +4 | -1
Let f:(2,  3)→(0,  1)f:\left( {2,\;3} \right) \to \left( {0,\;1} \right) be defined by f(x)=x−[x]f\left( x \right) = x - \left[ x \right], then f−1(x){f^{ - 1}}\left( x \right) equals
View answer and explanation
Correct option: D
x+2x + 2
Option D (x+2x + 2) is correct because the inverse function is found by solving for x in terms of y.
2 Easy
2 of 650
Understanding +4 | -1
Let f:R→Rf: R \rightarrow R be defined as f(x)=x4f(x)=x^4, Choose the correct answer.
View answer and explanation
Correct option: D
ff is neither one-one nor onto
Option D (ff is neither one-one nor onto) is correct because the function f(x)=x4f(x) = x^4 does not satisfy the conditions for being one-to-one (injective) or onto (surjective) over the real numbers.
3 Easy
3 of 650
Analyzing +4 | -1
Which of the following functions are odd functions
View answer and explanation
Correct option: C
f(x)=ax−1ax+1f(x)=\frac{a^x-1}{a^x+1}
Option C, f(x)=ax−1ax+1f(x)=\frac{{a^x-1}}{{a^x+1}}, is the correct answer because it satisfies the condition f(−x)=−f(x)f(-x) = -f(x) for odd functions.
4 Easy
4 of 650
Analyzing +4 | -1
Let f(x)=(x+1)2−1(x≥−1)f(x)=(x+1)^2-1(x \geq-1). Then the set S={x:f(x)=f−1(x)}S=\left\{x: f(x)=f^{-1}(x)\right\} contains
View answer and explanation
Correct option: C
{0,−1}\{0,-1\}
The correct answer is C ({0,-1}) because the set S={x:f(x)=f−1(x)}S = \{x: f(x) = f^{-1}(x)\} contains 0 and -1, found by solving f(f(x))=xf(f(x)) = x for x≥−1x \geq -1.
5 Easy
5 of 650
Analyzing +4 | -1
If f(x+y)=f(x)f(y)f(x + y) = f(x)f(y) and ∑x=1∞f(x)=2,x,y∈N,\sum\limits_{x = 1}^\infty {f(x) = 2,x,y \in N}, where N is the set of all natural numbers, then the value of f(4)f(2)\frac{{f(4)}}{{f(2)}} is
View answer and explanation
Correct option: D
49\frac{4}{9}
Option D is correct because by using the given functional relation and the sum, we find f(x) = (2/3)^x, and thus f(4)/f(2) = 4/9.
6 Easy
6 of 650
Application +4 | -1
If the function f:[1,∞)→[1,∞)\mathrm{f}:[1, \infty) \rightarrow[1, \infty) is defined by f(x)=2x(x−1)f(x)=2^{x(x-1)}, then f−1(x)f^{-1}(x) is
View answer and explanation
Correct option: B
12(1+1+4log⁡2x)\frac{1}{2}\left(1+\sqrt{1+4 \log _2 x}\right)
Option B is the correct answer because it represents the inverse function derived by solving for x in terms of y and swapping variables, considering the domain and range.
7 Medium
7 of 650
Application +4 | -1
Domain of the function f(x)=[log⁡10(5x−x24)]1/2f(x)=\left[\log _{10}\left(\frac{5 x-x^{2}}{4}\right)\right]^{1 / 2} is
View answer and explanation
Correct option: B
1≤x≤41 \leq x \leq 4
Option B (1 ≤ x ≤ 4) is correct because the domain requires the argument of the logarithm to be positive and the logarithm itself to be non-negative.
8 Easy
8 of 650
Understanding +4 | -1
The domain of f(x)=log⁡2(x+3)x2+3x+2f(x)=\frac{\log _{2}(x+3)}{x^{2}+3 x+2} is
View answer and explanation
Correct option: D
(−3,+∞)−{−1,−2}(-3,+\infty)-\{-1,-2\}
Option D ((-3,+\infty)-\{-1,-2\\}(-3,+\infty)-\{-1,-2\\}) is correct because the argument of the logarithm must be greater than 0 (x > -3) and the denominator cannot be zero (x≠−1,−2x \neq -1, -2).
9 Medium
9 of 650
Application +4 | -1
If f(x)=2x−1x+5(x≠−5)f\left( x \right) = \frac{{2x - 1}}{{x + 5}}\left( {x \ne - 5} \right), then f−1(x){f^{ - 1}}\left( x \right) is equal to
View answer and explanation
Correct option: B
5x+12−x,  x≠2\frac{{5x + 1}}{{2 - x}},\;x \ne 2
Option B (5x+12−x,  x≠2\frac{{5x + 1}}{{2 - x}},\;x \ne 2) is correct because it represents the inverse function obtained by swapping x and y and solving for y.
10 Medium
10 of 650
Analyzing +4 | -1
Let f:(e,∞)→Rf:\left( {e,\infty } \right) \to R be defined by f(x)=log⁡[log⁡(log⁡x)],f\left( x \right) = \log \left[ {\log \left( {\log x} \right)} \right], then
View answer and explanation
Correct option: C
f is both one-one and onto
Option C (f is both one-one and onto) is correct because the function is strictly increasing and its range covers all real numbers.
11 Easy
11 of 650
Application +4 | -1
If g(x)=x2+x−2g(x)=x^2+x-2 and 12g∘f(x)=2x2−5x+2\frac{1}{2} g \circ f(x)=2 x^2-5 x+2, then f(x)f(x) is
View answer and explanation
Correct option: A
2x−32 x-3
Option A (2x−32x - 3) is correct because solving for f(x)f(x) in the equation 12g(f(x))=2x2−5x+2\frac{1}{2}g(f(x)) = 2x^2 - 5x + 2 yields f(x)=2x−3f(x) = 2x - 3.
12 Easy
12 of 650
Understanding +4 | -1
Let f(x)f\left( x \right) be an even function. Then f′(x)f'\left( x \right)
View answer and explanation
Correct option: B
Is an odd function
The correct answer is B (Is an odd function) because the derivative of an even function is always an odd function.
13 Easy
13 of 650
Application +4 | -1
If f(x)⇒cos⁡x∣f(x) \Rightarrow \cos x \mid and g(x)=[x]g(x)=[x], then gg of (x)(x) is equal to
View answer and explanation
Correct option: C
[∣cos⁡x∣][|\cos x|]
Option C ([∣cos⁡x∣][|\cos x|]) is correct because g(f(x))=g(∣cos⁡x∣)=[∣cos⁡x∣]g(f(x)) = g(|\cos x|) = [|\cos x|].
14 Easy
14 of 650
Understanding +4 | -1
The graph of the function of y=f(x)y = f\left( x \right) is symmetrical about the line x=2x = 2, then
View answer and explanation
Correct option: B
f(2+x)=f(2−x)f\left( {2 + x} \right) = f\left( {2 - x} \right)
Option B (f(2+x)=f(2−x)f(2 + x) = f(2 - x)) is correct because it represents the symmetry of the function f(x)f(x) about the line x=2x = 2.
15 Easy
15 of 650
Analyzing +4 | -1
If f:[0,  ∞]→[0,  ∞]  and  f(x)=x1+xf:\left[ {0,\;\infty } \right] \to \left[ {0,\;\infty } \right]\;and\;f\left( x \right) = \frac{x}{{1 + x}}, then f is
View answer and explanation
Correct option: B
One-one but not onto
Option B (One-one but not onto) is correct because the function is strictly increasing, making it one-to-one, but its range is not equal to its codomain, making it not onto.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
16 Easy
16 of 650
Understanding +4 | -1
Assertion: If f(x)=1x−2,x≠2f(x)=\frac{1}{x-2}, x \neq 2 and g(x)=(x−2)2g(x)=(x-2)^2, then (f+g)(x)=1+(x−2)3x−2,x≠2({f}+{g})({x})=\frac{1+({x}-2)^3}{{x}-2}, {x} \neq 2.

Reason : If ff and gg are two functions, then their sum is defined by (f+g)(x)=f(x)+g(x) ∀ x∈D1∩D2({f}+{g})({x})={f}({x})+{g}({x}) \, \forall \, {x} \in {D}_1 \cap {D}_2, where D1D_1 and D2D_2 are domains of ff and gg, respectively.
View answer and explanation
Correct option: A
Assertion is correct, Reason is correct; Reason is a correct explanation for assertion
Assertion is correct, Reason is correct; Reason is a correct explanation for assertion because the sum of two functions is defined as the sum of their individual values at each point in the intersection of their domains.
17 Easy
17 of 650
Application +4 | -1
Let f\mathrm{f} be an odd function defined on the set of real numbers such that for x≥0,f(x)=3sin⁡x+4cos⁡xx \geq 0, f(x)=3 \sin x+4 \cos x.
Then f(x)\mathrm{f}(\mathrm{x}) at x=−11π6\mathrm{x}=-\frac{11 \pi}{6} is equal to:
View answer and explanation
Correct option: C
32−23\frac{3}{2}-2 \sqrt{3}
Option C (32−23\frac{{3}}{{2}} - 2 \sqrt{{3}}) is correct because it is the value of the odd function f(x) at x = -11π/6, calculated using the properties of sine and cosine.
18 Easy
18 of 650
Application +4 | -1
Let f(x)=x∣x∣f(x)=x|x| and g(x)=∣x∣g(x)=\sqrt{|x|} then the number of elements in the set {x∈R:f(x)=g(x)}\{x \in \mathbf{R}: f(x)=g(x)\} is
View answer and explanation
Correct option: B
2
Option B (2) is correct because there are two real solutions to the equation f(x) = g(x), which are x = 0 and x = 1.
19 Medium
19 of 650
Application +4 | -1
If [x]2−5[x]+6=0[x]^2-5[x]+6=0, where [.] denotes the greatest integer function, then
View answer and explanation
Correct option: D
x∈[2,4)x \in[2,4)
Option D (x∈[2,4)x \in [2,4)) is correct because it encompasses both possible solutions for xx derived from the given equation involving the greatest integer function.
20 Hard
20 of 650
Application +4 | -1
Let the functionf(x)=x2+x+sin⁡x−cos⁡x+log⁡(1+∣x∣)f\left( x \right) = {x^2} + x + \sin x - \cos x + \log \left( {1 + \left| x \right|} \right) be defined on the interval [0,  1].\left[ {0,\;1} \right]. The odd extension of f(x)f\left( x \right) to the interval [−1,  1]\left[ { - 1,\;1} \right] is
View answer and explanation
Correct option: B
−x2+x+sin⁡x+cos⁡x−log⁡(1+∣x∣)- {x^2} + x + \sin x + \cos x - \log \left( {1 + \left| x \right|} \right)
Option B is correct because the odd extension of f(x) involves changing the even terms' signs when x is negative.
21 Medium
21 of 650
Analyzing +4 | -1
The domain of the function f(x)f\left( x \right) given by
f(x)=−log⁡0.3(x−1)−x2+3x+18,    isf\left( x \right) = \sqrt {\frac{{ - {{\log }_{0.3}}\left( {x - 1} \right)}}{{ - {x^2} + 3x + 18}}} ,\;\;\mathrm{is}, is
View answer and explanation
Correct option: B
(2,  6)\left( {2,\;6} \right)
Option B (2,  6)\left( {2,\;6} \right) is correct because it satisfies the logarithmic and rational function domain restrictions.
22 Easy
22 of 650
Application +4 | -1
Numerical value of the expression ∣3x3+12x2+2∣\left|\frac{3 x^{3}+1}{2 x^{2}+2}\right| for x=−3x=-3 is
View answer and explanation
Correct option: A
4
Option A (4) is correct because substituting x = -3 into the expression and taking the absolute value results in 4.
23 Easy
23 of 650
Application +4 | -1
The range of the function
f(x)=1+sin⁡x+sin⁡3x+sin⁡5x+…f\left( x \right) = 1 + \sin x + {\sin ^3}x + {\sin ^5}x + \ldots when x∈(−π/2,π/2)x \in \left( { - \pi /2,\pi /2} \right), is
View answer and explanation
Correct option: B
RR
Option B (R) is correct because the function can be simplified to 1 + tan(x)sec(x), which has a range of all real numbers over the given interval.
24 Medium
24 of 650
Application +4 | -1
If f:R→Rf: R \rightarrow R satisfies f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y), for all xx, y∈Ry \in R and f(1)=7f(1)=7, then ∑r=1nf(r)\sum_{r=1}^n f(r) is
View answer and explanation
Correct option: A
7n(n+1)2\frac{7 n(n+1)}{2}
Option A (7n(n+1)2\frac{{7 n(n+1)}}{{2}}) is correct because the sum of the function values from 1 to n is a multiple of 7 times the sum of the first n natural numbers.
25 Easy
25 of 650
Application +4 | -1
Let f:R→Rf: R \rightarrow R be defined by f(x)=3x−4f(x)=3 x-4, then f−1(x)f^{-1}(x) is
View answer and explanation
Correct option: A
13(x+4)\frac{1}{3}(x+4)
Option A (13(x+4)\frac{1}{3}(x+4)) is correct because it represents the inverse function obtained by swapping x and y and solving for y.
26 Medium
26 of 650
Application +4 | -1
If f(x)=x2+1f(x)=x^2+1, then f−1(17)f^{-1}(17) and f−1(−3)f^{-1}(-3) will be
View answer and explanation
Correct option: D
None of these
Option D is correct because f−1(17)=±4f^{-1}(17) = \pm 4 and f−1(−3)f^{-1}(-3) is not a real number.
27 Easy
27 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=24^4P_3 = 24.
28 Easy
28 of 650
Understanding +4 | -1
{x∈R:2x−1x3+4x2+3x∈R}\left\{ {x \in R:\frac{{2x - 1}}{{{x^3} + 4{x^2} + 3x}} \in R} \right\} equals
View answer and explanation
Correct option: C
R−{0,  −1,  −3}R - \left\{ {0,\; - 1,\; - 3} \right\}
Option C (R−{0,  −1,  −3}R - \left\{ {0,\; - 1,\; - 3} \right\}) is correct because the expression is real for all real numbers except where the denominator is zero, which occurs at x = 0, -1, and -3.
29 Medium
29 of 650
Understanding +4 | -1
Consider the following statements.

Statement-I: An onto function f:{1,2,3}→{1,2,3}{f}:\{1,2,3\} \rightarrow\{1,2,3\} is always one-one.

Statement-II: A one-one function f:{1,2,3}→{1,2,3}{f}:\{1,2,3\} \rightarrow\{1,2,3\} must be onto.
View answer and explanation
Correct option: C
Both I and II are true
Option C (Both I and II are true) is correct because for a function from a finite set to itself, being onto implies being one-to-one, and vice versa.
30 Medium
30 of 650
Application +4 | -1
Suppose that g(x)=1+xg(x)=1+\sqrt{x} and f(g(x))=3+2x+xf(g(x))=3+2 \sqrt{x}+x, then f(x)f(x) is
View answer and explanation
Correct option: B
2+x22+x^2
Option B (2+x22+x^2) is correct because substituting g(x)g(x) into f(g(x))f(g(x)) and solving for f(x)f(x) yields 2+x22+x^2.
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  • ✓ This set contains 650 single-correct-answer MCQs on FUNCTIONS.
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