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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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Showing 1-30 of 478 questions
1 Easy
1 of 478
Application +4 | -1
If f:(−∞,2]⟶(−∞,4]f:(-\infty, 2] \longrightarrow(-\infty, 4], where f(x)=x(4−x)f(x)=x(4-x), then f−1(x)f^{-1}(x) is given by:
View answer and explanation
Correct option: A
2−4−x2-\sqrt{4-x}
The correct answer is A (2−4−x2-\sqrt{4-x}) because it represents the inverse function of f(x) with the given domain and range.
2 Easy
2 of 478
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 478
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 478
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 478
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 478
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 Easy
7 of 478
Analyzing +4 | -1
The function f:N→Nf: \mathrm{N} \rightarrow \mathrm{N} defined by f(x)=x−5[x5]\mathrm{f}(\mathrm{x})=\mathrm{x}-5\left[\frac{\mathrm{x}}{5}\right], where N\mathrm{N} is set of natural numbers and [x][x] denotes the greatest integer less than or equal to xx, is :
View answer and explanation
Correct option: D
neither one-one nor onto.
Option D (neither one-one nor onto) is correct because the function does not satisfy the conditions for being one-to-one or onto.
8 Easy
8 of 478
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 Easy
9 of 478
Application +4 | -1
If A={3,4,5,6}B={1,2,3,−−−−10}A=\{3,4,5,6\} B=\{1,2,3,----10\} be two sets. A function f{f} is defined from AA to B{B} such that f(i)−f(j)≥2 ∀ i>jf(i)-f(j) \geq 2 \,\forall \,i > j where i,j∈Ai, j \in A and number of such functions is N{N} then N5\frac{N}{5} is equal to
View answer and explanation
Correct option: D
7
Option D (7) is correct because the number of such functions N is 35, and N/5 equals 7.
10 Easy
10 of 478
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.
11 Easy
11 of 478
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.
12 Easy
12 of 478
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|].
13 Easy
13 of 478
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.
14 Easy
14 of 478
Analyzing +4 | -1
The range of the function f(x)=2−x+1+xf(x)=\sqrt{2-x}+\sqrt{1+x} is
View answer and explanation
Correct option: B
[3,6][\sqrt{3}, \sqrt{6}]
Option B ([3,6][\sqrt{3}, \sqrt{6}]) is correct because the range is determined by analyzing the function's behavior within its domain, considering its minimum and maximum values.
15 Easy
15 of 478
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 478
Analyzing +4 | -1
Let the sets AA and BB denote the domain and range respectively of the function f(x)=1⌈x⌉−xf(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}, where ⌈x⌉\lceil x\rceil denotes the smallest integer greater than or equal to xx. Then among the statements

(S1):A∩B=(1,∞)−N(\mathrm{S} 1): A \cap B=(1, \infty)-\mathbb{N} and

(S2):A∪B=(1,∞)(\mathrm{S} 2): A \cup B=(1, \infty)
View answer and explanation
Correct option: C
Neither (S1) nor (S2) is true
The correct answer is C (Neither (S1) nor (S2) is true) because the function is not defined for any real number x, making both the domain and range empty sets.
17 Easy
17 of 478
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.
18 Easy
18 of 478
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.
19 Easy
19 of 478
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.
20 Easy
20 of 478
Application +4 | -1
If f(x)=(2x+5)(x−3)f(x)=\frac{(2 x+5)}{(x-3)} and f(g(x))=1(x+1)f(g(x))=\frac{1}{(x+1)}, then ∣9 g(4)14∣\left|\frac{9 \mathrm{~g}(4)}{14}\right| is
View answer and explanation
Correct option: D
2
Option D (2) is correct because solving the functional equation f(g(x))f(g(x)) allows us to find g(4)g(4) and then calculate the absolute value.
21 Easy
21 of 478
Application +4 | -1
If log⁡1/2([x]2−4[x]+5)≥0\log _{1 / 2}\left([x]^2-4[x]+5\right) \geq 0, then xx belongs to
View answer and explanation
Correct option: C
[2,3)[2,3)
The correct answer is C ([2,3)) because the inequality simplifies to ([x]-2)2 <= 0, implying [x] = 2, which means 2 <= x < 3.
22 Easy
22 of 478
Application +4 | -1
If α≤log⁡5[2(sin⁡x−cos⁡x)+3]≤β\alpha \leq \log _{\sqrt{5}}[\sqrt{2}(\sin x-\cos x)+3] \leq \beta then
View answer and explanation
Correct option: C
α=0,β=2\alpha=0, \beta=2
The correct answer is C (α=0,β=2\alpha=0, \beta=2) because by finding the range of sin⁡x−cos⁡x\sin x - \cos x and then the range of the entire expression, we can determine the bounds of the logarithm.
23 Easy
23 of 478
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.
24 Easy
24 of 478
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.
25 Easy
25 of 478
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 Easy
26 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=24^4P_3 = 24.
27 Easy
27 of 478
Application +4 | -1
Let f(x)=x5+2ex/4f(x)=x^5+2 e^{x/4} for all x∈Rx \in R. Consider a function g(x)g(x) such that (gof)(x)=x(gof)(x)=x for all x∈Rx \in R. Then the value of 8g′(2)8 {g}^{\prime}(2) is:
View answer and explanation
Correct option: A
16
Option A (16) is correct because it is derived from finding the derivative of the inverse function and evaluating at the given point.
28 Easy
28 of 478
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 Easy
29 of 478
Understanding +4 | -1
E,I,R,OE, I, R, O denote respectively the sets of all equilateral, isosceles, right angled and obtuse angled triangles in a plane, then which of the following is not true.
View answer and explanation
Correct option: C
E∩I=ϕ,E∩O≠ϕE \cap I=\phi, E \cap O \neq \phi
Option A (α=β\alpha=\beta) is correct because the coefficients of xrx^r and xn−rx^{n-r} in the expansion of (1+x)n(1+x)^n are equal due to the symmetry of binomial coefficients.
30 Easy
30 of 478
Understanding +4 | -1
If f(x)f(x) satisfies the relation f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y) for all x,y∈Rx, y \in R and f(1)=5f(1)=5 then f(x)f(x) is
View answer and explanation
Correct option: B
An odd function
Option B (An odd function) is correct because the given relation f(x+y)=f(x)+f(y)f(x+y) = f(x) + f(y) with f(1)=5f(1) = 5 implies that f(x)=5xf(x) = 5x, which is a linear function passing through the origin, and thus an odd function.
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