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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Medium

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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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Showing 331-353 of 353 questions
331 Medium
331 of 353
Analyzing +4 | -1
The range of the function

f(x)=x1+∣x∣,x∈Rf(x)=\frac{x}{1+|x|}, x \in R, is
View answer and explanation
Correct option: B
(−1,1)(-1,1)
Option B ((-1,1)) is correct because the function f(x) = x/(1+|x|) is bounded between -1 and 1, and it never reaches these values.
332 Medium
332 of 353
Application +4 | -1
The function f(x)=0f(x)=0 has eight distinct real solution and ff also satisfy f(4+x)=f(4−x)f(4+x)=f(4-x). The sum of all the eight solution of f(x)=0f(x)=0 is:
View answer and explanation
Correct option: B
32
Option B (32) is correct because the sum of the roots is calculated using the symmetry property of the function around x=4.
333 Medium
333 of 353
Application +4 | -1
f(x)=x(x−p)q−p+x(x−q)p−q,p≠q\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}(\mathrm{x}-\mathrm{p})}{\mathrm{q}-\mathrm{p}}+\frac{\mathrm{x}(\mathrm{x}-\mathrm{q})}{\mathrm{p}-\mathrm{q}}, \mathrm{p} \neq \mathrm{q}. What is the value of f(p)+f(q)f(p)+f(q) ?
View answer and explanation
Correct option: B
f(p+q)\mathrm{f}(\mathrm{p}+\mathrm{q})
Option B (f(p+q)f(p+q)) is correct because by substituting p and q into the function, we find that f(p) = p and f(q) = q, and then f(p+q) simplifies to p+q, which is equal to f(p) + f(q).
334 Medium
334 of 353
Application +4 | -1
The domain of the function f(x)=log⁡10(3x−x2)2f(x)=\sqrt{\log _{10} \frac{\left(3 x-x^2\right)}{2}} is
View answer and explanation
Correct option: D
[1,2][1,2]
Option D ([1,2]) is correct because it satisfies both conditions: the argument of the logarithm must be positive, and the logarithm itself must be non-negative.
335 Medium
335 of 353
Application +4 | -1
If log⁡616=k\log _6 16=k, then log⁡1824=\log _{18} 24=
View answer and explanation
Correct option: B
2(k+2)8−k\frac{2(k+2)}{8-k}
Option B (2(k+2)8−k\frac{{2(k+2)}}{{8-k}}) is correct because it follows from converting the logarithms to base 2 and then expressing log⁡1824\log_{{18}} 24 in terms of k.
336 Medium
336 of 353
Analyzing +4 | -1
If f(x)=1−x1+x,x≠0,−1  and  α=f(f(x))+f(f(1x)),f\left( x \right) = \frac{{1 - x}}{{1 + x}},x \ne 0, - 1\;and\;\alpha = f\left( {f\left( x \right)} \right) + f\left( {f\left( {\frac{1}{x}} \right)} \right), then
View answer and explanation
Correct option: C
∣α∣>2\left| \alpha \right| > 2
Option C is correct because after simplifying f(f(x))f(f(x)) and f(f(1/x))f(f(1/x)), we find that α=x+1/x\alpha = x + 1/x, and thus ∣α∣=∣x+1/x∣≥2|\alpha| = |x + 1/x| \ge 2.
337 Medium
337 of 353
Analyzing +4 | -1
If n{x}=x+[x]n\{x\}=x+[x], and n>1n>1 has exactly 5 solutions, then n\mathrm{n} is equal to ([⋅],{⋅}([\cdot],\{\cdot\} denotes the greatest integer function and fractional part of xx respectively)
View answer and explanation
Correct option: A
10,11
Option B ({x/1≤x<2}\{x / 1 \le x<2\}) is correct because it correctly identifies the interval that is excluded from the union of sets A and B to form the entire real line.
338 Medium
338 of 353
Application +4 | -1
Let A={1,2,3,4,5}A=\{1,2,3,4,5\}. The domain of the relation on AA defined by R={(x,y):y=2x−1}R=\{(x, y): y=2 x-1\} is
View answer and explanation
Correct option: A
{1,2,3}\{1,2,3\}
Option A ({1,2,3}\{1,2,3\}) is correct because the domain of the relation consists of the x-values for which y is also in set A.
339 Medium
339 of 353
Analyzing +4 | -1
If 3x+22x≥5x,{3^x} + {2^{2x}} \ge {5^x}, then the solution set for xx is
View answer and explanation
Correct option: A
(−∞,2]\left( { - \infty ,2} \right]
Option A ((−∞,2](- \infty, 2]) is correct because the inequality holds true for x less than or equal to 2.
340 Medium
340 of 353
Application +4 | -1
Let f:R→Rf: R \rightarrow R be continuous and periodic with period T>0T > 0. Then
View answer and explanation
Correct option: A
f(x0+T/2)=f(x0)f\left(x_0+T / 2\right)=f\left(x_0\right) for some x0∈[k,k+T/2],k∈Rx_0 \in[k, k+T / 2], k \in R
Option A (f(x0+T/2)=f(x0)f\left(x_0+T / 2\right)=f\left(x_0\right) for some x0∈[k,k+T/2],k∈Rx_0 \in[k, k+T / 2], k \in R) is correct because it applies the Intermediate Value Theorem to show the existence of a point where the function's value at x_0 + T/2 equals its value at x_0.
341 Medium
341 of 353
Understanding +4 | -1
Let f(x)f(x) and g(x)g(x) are two functions defined in (−a,a)(-a, a) such that f(x)f(x) and g(x)g(x) are both odd function. Then which one of the following is correct?
View answer and explanation
Correct option: D
fof (x)(x) is an odd function
Option D (fof (x)(x) is an odd function) is correct because the composition of two odd functions results in an odd function.
342 Medium
342 of 353
Application +4 | -1
The domain and range of the relation R\mathrm{R} given by R={(x,y):y=x+6xR=\left\{(x, y): y=x+\frac{6}{x}\right.; where x,y∈Nx, y \in N and x<6}x < 6\} is
View answer and explanation
Correct option: A
{1,2,3},{7,5}\{1,2,3\},\{7,5\}
Option A ({1,2,3},{7,5} ) is correct because it correctly identifies the domain and range of the given relation by substituting the values of x and finding the corresponding y values.
343 Medium
343 of 353
Application +4 | -1
Let f={(0,−1),(−1,3),(2,3),(3,5)}f=\{(0,-1),(-1,3),(2,3),(3,5)\} be a function from ZZ to ZZ defined by f(x)=ax+bf(x)=a x+b. Then,
View answer and explanation
Correct option: C
a=2,b=−1a=2, b=-1
Option C (a=2,b=−1a=2, b=-1) is correct because substituting these values into the function f(x)=ax+bf(x) = ax + b satisfies the given conditions.
344 Medium
344 of 353
Understanding +4 | -1
Which of the following functions is periodic?
View answer and explanation
Correct option: A
[x]−x[x]-x
Option A ([x]−x[x]-x) is correct because it is a periodic function with a period of 1.
345 Medium
345 of 353
Application +4 | -1
If a function FF is such that F(0)=2,F(1)=3F(0)=2, F(1)=3, F(x+2)=2F(x)−F(x+1)F(x+2)=2 F(x)-F(x+1) for x≥0x \geq 0, then F(5)F(5) is equal to
View answer and explanation
Correct option: D
13
The correct answer is D (13) because it's obtained by recursively applying the given function definition to find F(5).

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346 Medium
346 of 353
Understanding +4 | -1
If R={(x,y):x,y∈Z,x2+y2≤4}R=\left\{(x, y): x, y \in Z, x^2+y^2 \leq 4\right\} is a relation in ZZ, then domain of RR is
View answer and explanation
Correct option: C
{−2,−1,0,1,2}\{-2,-1,0,1,2\}
Option C ({-2, -1, 0, 1, 2}) is correct because the domain of R consists of all integer values of x for which there exists an integer y such that x2+y2≤4x^2 + y^2 \leq 4.
347 Medium
347 of 353
Application +4 | -1
Given the function f(x)=ax+a−x2,(a>2)f(x)=\frac{a^{x}+a^{-x}}{2},(a>2). Then f(x+y)+f(x−y)=f(x+y)+f(x-y)=
View answer and explanation
Correct option: A
2f(x)⋅f(y)2 f(x) \cdot f(y)
Option A (2f(x)⋅f(y)2 f(x) \cdot f(y)) is correct because the expression f(x+y)+f(x−y)f(x+y) + f(x-y) simplifies to 2f(x)f(y)2f(x)f(y) using the given function definition.
348 Medium
348 of 353
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}}.
349 Medium
349 of 353
Application +4 | -1
If g:(−∞,∞)→(−π2,π2),g(x)=2tan⁡−1(eπ)−π2g:(-\infty, \infty) \rightarrow\left(\frac{-\pi}{2}, \frac{\pi}{2}\right), g(x)=2 \tan ^{-1}\left(e^\pi\right)-\frac{\pi}{2} and ff is inverse of gg, then f(0)=……f(0)=\ldots \ldots
View answer and explanation
Correct option: A
1
Option A (1) is correct because after finding the inverse function and differentiating it, substituting x=0 gives the value 1.
350 Medium
350 of 353
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.
351 Medium
351 of 353
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.
352 Medium
352 of 353
Analyzing +4 | -1
Let f(x)={x∣x∣,x⩽−1[x+1]+[1−x],−1<x<1x∣x∣,x⩾1f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {x\left| x \right|,}&{x \leqslant - 1} \\ {\left[ {x + 1} \right] + \left[ {1 - x} \right],}&{ - 1 < x < 1} \\ {x\left| x \right|,}&{x \geqslant 1} \end{array}} \right. Where [ ] represents the greatest integer function. Then f(x)f(x) is
View answer and explanation
Correct option: A
Even
Option A (Even) is correct because f(-x) = f(x) for all x in the domain, satisfying the condition for an even function.
353 Medium
353 of 353
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.
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