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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Mixed difficulty

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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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361 Easy
361 of 650
Application +4 | -1
The domain of definition of

f(x)=log⁡10{log⁡10(1+x3)},  isf\left( x \right) = {\log _{10}}\left\{ {{{\log }_{10}}\left( {1 + {x^3}} \right)} \right\},\;\mathrm{is}
View answer and explanation
Correct option: B
(0,∞)\left( {0,\infty } \right)
Option B ((0,∞)(0,\infty)) is correct because the argument of the outer logarithm must be greater than 0, and the argument of the inner logarithm must also be greater than 0.
362 Easy
362 of 650
Application +4 | -1
Let f:R→R,g:R→Rf:R \to R,g:R \to R be two functions given by f(x)=2x−3,g(x)=x3+5.f\left( x \right) = 2x - 3,g\left( x \right) = {x^3} + 5. Then, (fog)−1x{\left( {fog} \right)^{ - 1}}x is equal to
View answer and explanation
Correct option: A
(x−72)1/3{\left( {\frac{{x - 7}}{2}} \right)^{1/3}}
Option A ((x−72)1/3{\left( {\frac{{x - 7}}{2}} \right)^{1/3}}) is correct because it represents the inverse of the composite function f(g(x)).
363 Easy
363 of 650
Application +4 | -1
The solution set of the inequality logsin(π3)(x2−3x+2)≥2\mathrm{lo}{\mathrm{g}_{\mathrm{sin}\left( {\frac{\pi }{3}} \right)}}\left( {{x^2} - 3x + 2} \right) \ge 2 is
View answer and explanation
Correct option: C
[12,1)∪(2,52]\left[ {\left. {\frac{1}{2},1} \right) \cup \left( {2,\frac{5}{2}} \right.} \right]
Option C [12,1)∪(2,52]\left[ {\left. {\frac{1}{2},1} \right) \cup \left( {2,\frac{5}{2}} \right.} \right] is correct because it represents the intersection of the solution set of the logarithmic inequality and the domain of the logarithm.
364 Easy
364 of 650
Application +4 | -1
Let f:A→Bf: \mathrm{A} \rightarrow \mathrm{B} be a function defined as f(x)=x−1x−2f(x)=\frac{x-1}{x-2}, where A=R−{2}A=R-\{2\} and B=R−{1}B=R-\{1\}. Then ff is
View answer and explanation
Correct option: D
invertible and f−1(y)=2y−1y−1f^{-1}(y)=\frac{2 y-1}{y-1}
The function is invertible, and its inverse is found by swapping x and y and solving for y, resulting in option D (f−1(y)=2y−1y−1f^{-1}(y)=\frac{2 y-1}{y-1}).
365 Medium
365 of 650
Understanding +4 | -1
Let RR be a relation in NN defined by R={(1+x,1+x2):x≤5,x∈N}R=\left\{\left(1+x, 1+x^2\right): x \leq 5, x \in N\right\}.
Which of the following is false?
View answer and explanation
Correct option: A
R={(2,2),(3,5),(4,10),(5,17),(6,25)}R=\{(2,2),(3,5),(4,10),(5,17),(6,25)\}
Option A (R={(2,2),(3,5),(4,10),(5,17),(6,25)}R=\{(2,2),(3,5),(4,10),(5,17),(6,25)\}) is false because the last element should be (6,26), not (6,25).
366 Easy
366 of 650
Understanding +4 | -1
The domain of the real function f(x)=14−x2f\left( x \right) = \frac{1}{{\sqrt {4 - {x^2}} }} is
View answer and explanation
Correct option: C
(-2, 2)
Option C ((-2, 2)) is correct because the expression inside the square root must be positive, leading to the inequality 4 - x² > 0.
367 Easy
367 of 650
Understanding +4 | -1
Let f:R→Rf: R \rightarrow R be a function defined by f(x)=x3+4f(x)=x^3+4, then ff is
View answer and explanation
Correct option: C
bijective
Option C (bijective) is correct because the function f(x) = x3 + 4 is both injective (one-to-one) and surjective (onto) over the real numbers.
368 Easy
368 of 650
Understanding +4 | -1
Suppose that ff is continuous on [a,b][\mathrm{a}, \mathrm{b}] and that f(x)f(\mathrm{x}) is an integer for each x\mathrm{x} in [a,b][\mathrm{a}, \mathrm{b}]. Then in [a,b][\mathrm{a}, \mathrm{b}]
View answer and explanation
Correct option: D
f(x)f(x) is constant
Option C (0) is correct because the sum of the area vectors of the faces of a tetrahedron is always zero.
369 Easy
369 of 650
Understanding +4 | -1
Let A={1,2,3}  and  B={2,3,4}A = \left\{ {1,2,3} \right\}\;and\;B = \left\{ {2,3,4} \right\} then which of the following relations is a function from A  to  B?A\;to\;B?
View answer and explanation
Correct option: C
{(1,3),(2,3),(3,3)}\left\{ {\left( {1,\mathrm{ }3} \right),\mathrm{ }\left( {2,\mathrm{ }3} \right),\mathrm{ }\left( {3,\mathrm{ }3} \right)} \right\}
Option C ({(1,3),(2,3),(3,3)}\{(1, 3), (2, 3), (3, 3)\}) is correct because each element in A is mapped to a unique element in B, satisfying the definition of a function.
370 Medium
370 of 650
Understanding +4 | -1
f  :[−4,0]→Rf\;:\left[ { - 4,0} \right] \to R is given byf(x)=ex+sin⁡x,f\left( x \right) = {e^x} + \sin x, its even extension to [−4,4],\left[ { - 4,4} \right], is
View answer and explanation
Correct option: B
e−∣x∣−sin⁡∣x∣{e^{ - \left| x \right|}} - \sin \left| x \right|
Option B (e−∣x∣−sin⁡∣x∣{e^{ - \left| x \right|}} - \sin \left| x \right|) is correct because it defines the even extension of the given function over the specified interval.
371 Medium
371 of 650
Application +4 | -1
Let A≡{1,2,3,4},B≡{a,b,c}A \equiv\{1,2,3,4\}, B \equiv\{a, b, c\}, then number of functions from A→BA \rightarrow B, which are not onto i
View answer and explanation
Correct option: C
45
Option C (45) is correct because it calculates the total number of functions from A to B and subtracts the number of onto functions.
372 Easy
372 of 650
Understanding +4 | -1
If R denotes the set of all real numbers, then the function f:R→Rf:R \to R defined by f(x)=∣x∣f\left( x \right) = \left| x \right| is
View answer and explanation
Correct option: D
Neither one-one nor onto
Option D is correct because the absolute value function is neither one-to-one (since f(x) = f(-x)) nor onto (since the range is [0, ∞)).
373 Hard
373 of 650
Analyzing +4 | -1
If f(x)=cos⁡(sin⁡x)+sin⁡(cos⁡x),f\left( x \right) = \sqrt {\cos \left( {\sin x} \right)} + \sqrt {\sin \left( {\cos x} \right)} , then range of f(x)f\left( x \right) is
View answer and explanation
Correct option: B
[cos⁡1,1+sin⁡1]\left[ {\sqrt {\cos 1} ,1 + \sqrt {\sin 1} } \right]
Option D (None of these) is incorrect. The correct range is [cos⁡1,1+sin⁡1]\left[ {\sqrt {\cos 1} ,1 + \sqrt {\sin 1} } \right].
374 Easy
374 of 650
Understanding +4 | -1
Let f:(−1,1)→Bf:(-1,1) \rightarrow B, be a function defined by f(x)=tan⁡−12x1−x2f(x)=\tan ^{-1} \frac{2 x}{1-x^2}, then ff is both one- one and onto when BB is the interval
View answer and explanation
Correct option: B
(−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
Option B (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) is correct because the range of f(x)=tan⁡−12x1−x2f(x) = \tan^{-1} \frac{{2x}}{{1-x^2}} for x∈(−1,1)x \in (-1, 1) is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}).
375 Easy
375 of 650
Analyzing +4 | -1
If sides of △ABC\triangle A B C are a,ba, b and cc such that 2b=a+c2 b=a+c, then exhaustive range of bc\frac{b}{c} is
View answer and explanation
Correct option: C
(23,2)\left(\frac{2}{3}, 2\right)
Option C ((23,2)\left(\frac{{2}}{{3}}, 2\right)) is correct because the given condition 2b=a+c2b = a+c and triangle inequalities lead to the range of bc\frac{{b}}{{c}} being between 23\frac{{2}}{{3}} and 2.
376 Easy
376 of 650
Application +4 | -1
Find the number of integers not in the domain of f(x)=cos⁡−1(2−x2x)f(x)=\cos ^{-1}\left(\frac{2-x}{2 x}\right).
View answer and explanation
Correct option: A
2
Option A (2) is correct because there are two integers that make the argument of the inverse cosine function outside the range of [-1, 1].
377 Easy
377 of 650
Analyzing +4 | -1
Solution set of inequality log⁡ex−2x−3{\log _e}\frac{{x - 2}}{{x - 3}} is
View answer and explanation
Correct option: B
(−∞,  2)\left( { - \infty ,\;2} \right)
Option B ((−∞,  2)\left( { - \infty ,\;2} \right)) is correct because the inequality is satisfied when the argument of the logarithm is between 0 and 1.
378 Medium
378 of 650
Application +4 | -1
The domain of the function f(x  )=log⁡2x−1(x−1)f\left( {x\;} \right) = {\log _{2x - 1}}\left( {x - 1} \right) is
View answer and explanation
Correct option: A
(1,  ∞)\left( {1,\;\infty } \right)
Option D (None of these) is incorrect. Option A (1,  ∞)\left( {1,\;\infty } \right) is correct because the argument of the logarithm and the base must be positive and the base cannot be 1.
379 Medium
379 of 650
Understanding +4 | -1
If f:R→R&g:R→Rf: \mathbb{R} \rightarrow \mathbb{R} \& \quad g: \mathbb{R} \rightarrow \mathbb{R} be two given functions, then 2min⁡{f(x)−g(x),0}2 \min \{f(x)-g(x), 0\} equals
View answer and explanation
Correct option: D
f(x)−g(x)−∣g(x)−f(x)∣f(x)-g(x)-|g(x)-f(x)|
Option D is correct because it uses the identity min(a, b) = (a+b-|a-b|)/2 and simplifies the expression.
380 Medium
380 of 650
Understanding +4 | -1
Which of the following statement is true (T)(\mathrm{T}) or false (F)(\mathrm{F}) for the function f:R→Rf: \mathrm{R} \rightarrow \mathrm{R} ?
(i) If ∣f(x)−f(y)∣≤30∣x−y∣,∀x,y∈R|f(x)-f(y)| \leq 30|x-y|, \forall x, y \in \mathrm{R}, then ff is a continuous function on R\mathrm{R}.
(ii) If ∣f(x)−f(y)∣≤30∣x−y∣,∀x,y∈R|f(x)-f(y)| \leq 30|x-y|, \forall x, y \in \mathrm{R}, then ff is differentiable on R\mathrm{R}.
(iii) If ∣f(x)−f(y)∣≤21∣x−y∣2,∀x,y∈R|f(x)-f(y)| \leq 21|x-y|^2, \forall x, y \in \mathbb{R}, then ff is differentiable on R\mathrm{R}.
(iv) If ∣f(x)−f(y)∣≤21∣x−y∣2,∀x,y∈R|f(x)-f(y)| \leq 21|x-y|^2, \forall x, y \in \mathrm{R}, then ff is a constant function.
View answer and explanation
Correct option: B
TFTT
Option B (TFTT) is correct because statement (i) is true, statement (ii) is false, and statements (iii) and (iv) are true based on the given conditions.
381 Easy
381 of 650
Analyzing +4 | -1
The set of all real xx satisfying the inequality 3−∣x∣4−∣x∣>0\frac{{3 - \left| x \right|}}{{4 - \left| x \right|}} > 0
View answer and explanation
Correct option: A
[−3,  3]∪(−∞,  −4)∪(4,  ∞)\left[ { - 3,\;3} \right] \cup \left( { - \infty ,\; - 4} \right) \cup \left( {4,\;\infty } \right)
The correct answer is A. By analyzing the inequality and considering the cases where the numerator and denominator have the same sign, we find the solution set to be (-∞, -4) U [-3, 3] U (4, ∞).
382 Easy
382 of 650
Application +4 | -1
The domain of sin⁡−1[log⁡3(x3)]\sin ^{-1}\left[\log _{3}\left(\frac{x}{3}\right)\right] is
View answer and explanation
Correct option: A
[1,9][1,9]
Option A ([1,9]) is correct because the domain is found by solving the inequality -1 <= log3(x/3) <= 1.
383 Easy
383 of 650
Understanding +4 | -1
If f(x)=acos⁡(bx+c)+df(x)=a \cos (b x+c)+d, then range of f(x)f(x) is
View answer and explanation
Correct option: D
[d−a,d+a][d-a, d+a]
Option D ([d−a,d+a][d-a, d+a]) is correct because the range of cos⁡(bx+c)\cos(bx+c) is [-1, 1], so the range of acos⁡(bx+c)a\cos(bx+c) is [-a, a], and adding d shifts the range to [d-a, d+a].
384 Easy
384 of 650
Application +4 | -1
If f(x)f(x) is an odd function, f(1)=3f(1)=3, and f(x+2)f(x+2) =f(x)+f(2)=f(x)+f(2), then the value of f(3)f(3) is
View answer and explanation
Correct option: A
9
Option A (9) is correct because by using the properties of odd functions and the given functional equation, we can deduce that f(3) = 9.
385 Easy
385 of 650
Application +4 | -1
If from R→R,f(x)=(x+1)2,g(x)=x2+1R \rightarrow R, f(x)=(x+1)^2, g(x)=x^2+1, then (( fog )(−3))(-3) equals
View answer and explanation
Correct option: A
121
Option A (121) is correct because when you evaluate the composite function f(g(x))f(g(x)) at x=−3x = -3, the result is 121.
386 Medium
386 of 650
Application +4 | -1
If f(x)={x}+{x+1}+{x+2}……..x+99}f(x)=\{x\}+\{x+1\}+\{x+2\} \ldots \ldots . . x+99\}, then the value of [f(2)][f(\sqrt{2})] is, where {\{.}\} denotes fractional part function & [.][.] denotes the greatest integer function

View answer and explanation
Correct option: C
41
Option B (41) is correct because the fractional part of 2\sqrt{2} is used to evaluate the function, and then the greatest integer function is applied.
387 Easy
387 of 650
Understanding +4 | -1
If f:R→R  and g:R→Rf:R \to R\;and\,g:R \to R are defined by f(x)=x−[x]  and  g(x)=[x]f\left( x \right) = x - \left[ x \right]\;and\;g\left( x \right) = \left[ x \right] for x∈Rx \in R, where [x]\left[ x \right] is the greatest integer not exceeding x, then for every x∈R,  f(g(x))x \in R,\;f\left( {g\left( x \right)} \right) is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because f(g(x))=f([x])=[x]−[[x]]=[x]−[x]=0f(g(x)) = f([x]) = [x] - [[x]] = [x] - [x] = 0.
388 Medium
388 of 650
Application +4 | -1
If f:R→Rf: R \rightarrow R is given by f(x)=3x−5f(x)=3 x-5, then f−1(x)f^{-1}(x) ?
View answer and explanation
Correct option: B
Is given by x+53\frac{x+5}{3}
Option B (x+53\frac{{x+5}}{{3}}) is correct because it represents the inverse function obtained by swapping xx and yy and solving for yy.
389 Medium
389 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
The correct answer is B (2+x22+x^2). By substituting y=1+xy = 1 + \sqrt{x} into the given equation, we can find the expression for f(x)f(x).
390 Easy
390 of 650
Application +4 | -1
Let ff be a function satisfying 2f(x)−3f(1/x)=x22 f(x)-3 f(1 / x)=x^2 for any x≠0x \neq 0. Then the value of f(2)f(2) is
View answer and explanation
Correct option: B
−7/4-7 / 4
Option B (-7/4) is correct because it is the solution obtained by solving the system of equations derived from the given functional equation.
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