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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 631-650 of 650 questions
631 Hard
631 of 650
Understanding +4 | -1
Let f:R→Rf:R \to R be a function defined by f(x)=−∣x∣3+∣x∣1+x2,f\left( x \right) = - \frac{{{{\left| x \right|}^3} + \left| x \right|}}{{1 + {x^2}}}, then the graph of f(x)f\left( x \right) lies in the
View answer and explanation
Correct option: D
III and IV quadrants
Option D (III and IV quadrants) is correct because the function f(x)f(x) is always negative or zero, causing its graph to lie in the third and fourth quadrants.
632 Medium
632 of 650
Application +4 | -1
Let f(x)=x−1x+1f(x)=\frac{x-1}{x+1}, then f{f(x)}f\{f(x)\} is
View answer and explanation
Correct option: B
−1x-\frac{1}{x}
Option B (−1x- \frac{{1}}{{x}}) is correct because when you compute f{f(x)}f\{f(x)\} with f(x)=x−1x+1f(x) = \frac{{x-1}}{{x+1}}, the result simplifies to −1x-\frac{{1}}{{x}}.
633 Easy
633 of 650
Application +4 | -1
A quadratic polynomial maps from [−2,3][-2,3] onto [0,3][0,3] and touches XX-axis at x=3x=3, then the polynomial is
View answer and explanation
Correct option: B
325(x2−6x+9)\frac{3}{25}\left(x^2-6 x+9\right)
Option B (325(x2−6x+9)\frac{3}{25}(x^2-6x+9)) is correct because the quadratic touches the x-axis at x=3 and maps [-2,3] onto [0,3].
634 Easy
634 of 650
Application +4 | -1
If f(ln⁡(1+∣x∣))=(1−ln⁡(1+∣x∣))17f(\ln (1+|x|))=(1-\ln (1+|x|))^{\frac{1}{7}}, then f(f(cos⁡x))f(f(\cos x)) is equal to:
View answer and explanation
Correct option: C
cos⁡x\cos x
Option C (cos⁡x\cos x) is correct because f(f(x))=xf(f(x)) = x, therefore f(f(cos⁡x))=cos⁡xf(f(\cos x)) = \cos x.
635 Hard
635 of 650
Analyzing +4 | -1
The domain of definition of
f(x)=log⁡0.5{−log⁡2(3x−13x+2)},  isf\left( x \right) = {\log _{0.5}}\left\{ { - {{\log }_2}\left( {\frac{{3x - 1}}{{3x + 2}}} \right)} \right\},\;\mathrm{is}
View answer and explanation
Correct option: C
(1/3,∞)\left( {1/3,\infty } \right)
Option C ( (1/3, ∞) ) is correct because it satisfies the conditions for the logarithm to be defined, ensuring the arguments are positive.
636 Easy
636 of 650
Application +4 | -1
For x∈Rx \in \mathbb{R}, two real valued functions f(x)f(x) and g(x)g(x) are such that, g(x)=x+1g(x)=\sqrt{x}+1 and fog(x)=x+3−xfog(x)=x+3-\sqrt{x}. Then f(0)f(0) is equal to
View answer and explanation
Correct option: B
5
Option B (5) is correct because by substituting t=x+1t = \sqrt{x} + 1 into the given functional equation, we can find f(t)f(t) and then evaluate f(0)f(0).
637 Medium
637 of 650
Understanding +4 | -1
Let f(x)=min⁡{x,  x2},  f\left( x \right) = \min \left\{ {x,\;{x^2}} \right\},\; for every x∈R.x \in R. . Then,
View answer and explanation
Correct option: A
f(x)={x,      x≥1x2,      0≤x<1x,      x<0f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\begin{array}{cccccccccccccccccccc} {x,\;\;\;x \ge 1}\\ {{x^2},\;\;\;0 \le x < 1} \end{array}}\\ {x,\;\;\;x < 0} \end{array}} \right.
Option A is correct because it accurately represents the minimum value between x and x² for different ranges of x, considering x >= 1, 0 <= x < 1, and x < 0.
638 Easy
638 of 650
Analyzing +4 | -1
If f(x)=4x4x+2f\left( x \right) = \frac{{{4^x}}}{{{4^x} + 2}}, then f(197)+f(297)+.…+f(9697)f\left( {\frac{1}{{97}}} \right) + f\left( {\frac{2}{{97}}} \right) + . \ldots + f\left( {\frac{{96}}{{97}}} \right) is equal to
View answer and explanation
Correct option: B
48
Option B (48) is correct because the sum can be paired such that each pair sums to 1.
639 Easy
639 of 650
Understanding +4 | -1
The function f:R→R,f(x)=x2,∀x∈Rf: R \rightarrow R, f(x)=x^{2}, \forall x \in R is
View answer and explanation
Correct option: D
Neither injection nor surjection
Option D (Neither injection nor surjection) is correct because the function f(x)=x2f(x) = x^2 is neither one-to-one (injective) nor onto (surjective) when mapping from R to R.
640 Medium
640 of 650
Application +4 | -1
Let [x]\left[ x \right] denote the greatest integer ≤x\le x. If f(x)=[x]  and  g(x)=∣x∣f\left( x \right) = \left[ x \right]\;and\;g\left( x \right) = \left| x \right|, then the value of f(g(85))−g(f(−85))f\left( {g\left( {\frac{8}{5}} \right)} \right) - g\left( {f\left( { - \frac{8}{5}} \right)} \right) is
View answer and explanation
Correct option: D
-1
Option D (-1) is correct because it is the result of evaluating the functions f(x) and g(x) at the specified points and then subtracting the results.
641 Easy
641 of 650
Analyzing +4 | -1
Which of the following functions has period π\pi ?
View answer and explanation
Correct option: A
∣−tan⁡x∣+cos⁡2  x\left| { - \tan x} \right| + \cos 2\;x
Option A (∣−tan⁡x∣+cos⁡2  x\left| { - \tan x} \right| + \cos 2\;x) is correct because both |tan(x)| and cos(2x) have periods related to or equal to π\pi, and their combination results in a period of π\pi.
642 Easy
642 of 650
Application +4 | -1
If f(y)=log⁡yf(y)=\log y, then f(y)+f(1y)f(y)+f\left(\frac{1}{y}\right) is equal to:
View answer and explanation
Correct option: C
0
The correct answer is C (0) because the sum of log y and log(1/y) simplifies to log 1, which is 0.
643 Easy
643 of 650
Understanding +4 | -1
If xx is real number and ∣x∣<3|x| < 3, then
View answer and explanation
Correct option: B
−3<x<3-3 < x < 3
Option B (-3 < x < 3) is correct because it represents the range of values for x that satisfy the absolute value inequality |x| < 3.
644 Easy
644 of 650
Application +4 | -1
Let f:(4,6)→(6,8)f:(4,6) \rightarrow(6,8) be a function defined by f(x)=x+[x2]f(x)=x+\left[\frac{x}{2}\right], where [.] denotes the greatest integer function, then f−1(x)f^{-1}(x) is equal to
View answer and explanation
Correct option: C
x−2x-2
Option C (x−2x-2) is correct because it is the inverse function of f(x) = x + 2 on the given domain.
645 Easy
645 of 650
Remembering +4 | -1
If sin⁡−1x=y\sin ^{-1} x=y, then
View answer and explanation
Correct option: B
−π2≤y≤π2-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}
Option B (−π2≤y≤π2-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}) is correct because the range of the principal value of sin⁡−1x\sin^{-1} x is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

Explore Other Chapters in MATHEMATICS

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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
646 Easy
646 of 650
Analyzing +4 | -1
Let f(x)={0,      x=0x2sin⁡π/2x,      ∣x∣<1x∣x∣,      ∣x∣≥1.f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\begin{array}{cccccccccccccccccccc} {0,\;\;\;x = 0}\\ {{x^2}\sin \pi /2x,\;\;\;\left| x \right| < 1} \end{array}}\\ {x\left| x \right|,\;\;\;\left| x \right| \ge 1} \end{array}} \right.. Then, f(x)f\left( x \right) is
View answer and explanation
Correct option: B
An odd function
The correct answer is B (An odd function) because f(-x) = -f(x) for all x, satisfying the definition of an odd function.
647 Easy
647 of 650
Application +4 | -1
The domain of definition of the function f(x)=7−xPx−3,f\left( x \right){ = ^{7 - x}}{P_{x - 3}}, is
View answer and explanation
Correct option: C
{3,  4,  5}\left\{ {3,\;4,\;5} \right\}
Option C ({3,  4,  5}\left\{ {3,\;4,\;5} \right\}) is correct because the domain is determined by the conditions 7-x >= 0, x-3 >= 0, and 7-x >= x-3, which lead to x being 3, 4, or 5.
648 Easy
648 of 650
Understanding +4 | -1
Let A={1,2,3,4}A=\{1,2,3,4\} and R:A→AR: A \rightarrow A be the relation defined by R={(1,1),(2,3),(3,4),(4,2)}R=\{(1,1),(2,3),(3,4),(4,2)\}. The correct statement is :
View answer and explanation
Correct option: C
RR is an onto function.
Option C is correct because the range of R is equal to its codomain A, making it an onto function.
649 Medium
649 of 650
Application +4 | -1
Let f(x+1x)=x2+1x2,x∈R−{0}f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}, x \in R-\{0\}, then f(x)f(x) is equal to
View answer and explanation
Correct option: C
x2−2x^2-2, when ∣x∣≥2|x| \geq 2
Option C (x2−2x^2-2, when ∣x∣≥2|x| \geq 2) is correct because by substituting z=x+1xz = x + \frac{{1}}{{x}} into the equation f(x+1x)=x2+1x2f(x + \frac{{1}}{{x}}) = x^2 + \frac{{1}}{{x^2}}, we find that f(z)=z2−2f(z) = z^2 - 2, which translates to f(x)=x2−2f(x) = x^2 - 2 when ∣x∣≥2|x| \geq 2.
650 Easy
650 of 650
Application +4 | -1
A quadratic function f(x)f(x) satisfies f(x)≥0f(x) \geq 0 for all real xx. If f(2)=0f(2)=0 and f(4)=12f(4)=12, then the value of f(6)f(6) is:
View answer and explanation
Correct option: C
48
Option C (48) is correct because by using the given conditions, we can find the quadratic function and then evaluate it at x=6.
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