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JEE • MATHEMATICS • FUNCTIONS

Functions practice · Medium

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Showing 181-210 of 353 questions

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181 Medium
181 of 353
Analyzing +4 | -1
Let f(x)=∣a−10axa−1ax2axa∣, a∈R.f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} a&{ - 1}&0 \\ {ax}&a&{ - 1} \\ {a{x^2}}&{ax}&a \end{array}} \right|,\,a \in R. Then the sum of the square of all the values of a,a, for which 2f′(10)−f′(5)+100=0,2f'\left( {10} \right) - f'\left( 5 \right) + 100 = 0, is
View answer and explanation
Correct option: C
125
The correct answer is C (125) because it is the sum of the squares of all values of a that satisfy the given condition involving the derivative of the determinant.
182 Medium
182 of 353
Application +4 | -1
If a,b,c,ka, b, c, k are real constants and α,β,γ\alpha, \beta, \gamma are real variables and atan⁡α+btan⁡β+ctan⁡γ=ka \tan \alpha+b \tan \beta+c \tan \gamma=k then the minimum value of tan⁡2α+tan⁡2β+tan⁡2γ\tan ^2 \alpha+\tan ^2 \beta+\tan ^2 \gamma is
View answer and explanation
Correct option: D
k2a2+b2+c2\frac{k^2}{a^2+b^2+c^2}
Option D (k2a2+b2+c2\frac{{k^2}}{{a^2+b^2+c^2}}) is correct because it is derived from Cauchy-Schwarz inequality applied to the given equation.
183 Medium
183 of 353
Application +4 | -1
If f:R→Sf: R \rightarrow S defined by f(x)=sin⁡x−3cos⁡x+1f(x)=\sin x-\sqrt{3} \cos x+1 is onto, then the interval of SS is
View answer and explanation
Correct option: A
[−1,3][-1,3]
Option A ([-1,3]) is correct because the range of the function f(x) = sin(x) - √3cos(x) + 1 is [-1, 3] when it's onto.
184 Medium
184 of 353
Application +4 | -1
The range of the function f(x)=x2+2x+2f(x)=x^{2}+2 x+2 is
View answer and explanation
Correct option: D
[1,∞)[1, \infty)
Option D ([1, ∞)) is correct because the range of the quadratic function is determined by finding the vertex of the parabola and considering the direction it opens.
185 Medium
185 of 353
Application +4 | -1
If f:R→Rf: R \rightarrow R and g:R→Rg: R \rightarrow R defined by f(x)=2x+3f(x)=2 x+3 and g(x)=x2+7g(x)=x^2+7, then the value of xx for which f(g(x))=25f(g(x))=25 is
View answer and explanation
Correct option: B
±2\pm 2
Option B (±2\pm 2) is correct because it is the solution to the equation f(g(x)) = 25.
186 Medium
186 of 353
Application +4 | -1
If f(x+y,  x−y)=xyf\left( {x + y,\;x - y} \right) = xy, then the arithemetic mean of f(x,  y)  and  f(y,  x)f\left( {x,\;y} \right)\;and\;f\left( {y,\;x} \right) is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the arithmetic mean simplifies to zero after substituting and simplifying.
187 Medium
187 of 353
Remembering +4 | -1
A point c\mathrm{c} in the domain of a function f\mathrm{f} is called a critical point of f\mathrm{f} if

I. f′(c)=0\mathrm{f}^{\prime}(\mathrm{c})=0

II. fis not differentiable at c\mathrm{c}.

Choose the correct option
View answer and explanation
Correct option: A
Either I or II are true
Option A (Either I or II are true) is correct because a critical point occurs where the derivative is zero or the function is not differentiable.
188 Medium
188 of 353
Application +4 | -1
If f(x)=1−1xf(x)=1-\frac{1}{x}, then f{f(1x)}f\left\{f\left(\frac{1}{x}\right)\right\} is
View answer and explanation
Correct option: C
xx−1\frac{x}{x-1}
Option C (xx−1\frac{{x}}{{x-1}}) is correct because after evaluating the composite function f{f(1x)}f\{f(\frac{{1}}{{x}})\}, the expression simplifies to xx−1\frac{{x}}{{x-1}}.
189 Medium
189 of 353
Application +4 | -1
If 3f(x)−f(1x)=log⁡x43 f(x)-f\left(\frac{1}{x}\right)=\log x^{4}, then f(e−x)f\left(e^{-x}\right) is
View answer and explanation
Correct option: D
−x-x
Option D (-x) is correct because it is the result of substituting e(-x) into the derived function f(x) after solving the given functional equation.
190 Medium
190 of 353
Understanding +4 | -1
The function f(x)={1,      x∈Q0,      x∉Qf\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1,\;\;\;x \in Q}\\ {0,\;\;\;x \notin Q} \end{array}} \right. is
View answer and explanation
Correct option: D
Periodic with indeterminate period
Option D (Periodic with indeterminate period) is correct because the function repeats for every rational number, but there's no smallest positive rational number.
191 Medium
191 of 353
Analyzing +4 | -1
Let F:R→RF:R \to R and g:R→Rg:R \to R be two functions defined by f(x)=log⁡e(x2+1)−e−x+1f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1 and g(x)=1−2e2xex.g(x) = \frac{{1 - 2{e^{2x}}}}{{{e^x}}}.
Then for which of the following range of α,\alpha , the inequality f(g((α−1)23))>f(g(α−53))f\left( {g\left( {\frac{{{{(\alpha - 1)}^2}}}{3}} \right)} \right) > f\left( {g\left( {\alpha - \frac{5}{3}} \right)} \right) holds?
View answer and explanation
Correct option: A
(2, 3)
Option A ((2, 3)) is correct because by determining that f(x) is increasing and g(x) is decreasing, the inequality can be simplified to a quadratic inequality, which gives the solution (2, 3).
192 Medium
192 of 353
Application +4 | -1
If f(x)=cos⁡(log⁡x)f(x)=\cos (\log x), then f(x2)f(y2)−12[f(x22)+f(x2y2)]f\left(x^{2}\right) f\left(y^{2}\right)-\frac{1}{2}\left[f\left(\frac{x^{2}}{2}\right)+f\left(\frac{x^{2}}{y^{2}}\right)\right] has the value
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the given expression simplifies to 12[cos⁡log⁡x2y2−cos⁡log⁡x22]\frac{{1}}{{2}}\left[\cos \log x^2 y^2-\cos \log \frac{{x^2}}{{2}}\right], which doesn't match any of the provided options.
193 Medium
193 of 353
Application +4 | -1
If f(x)=cos⁡(log⁡x)f(x)=\cos (\log x), then the value of f(x)⋅f(4)−12[f(x4)+f(4x)]f(x) \cdot f(4)-\frac{1}{2}\left[f\left(\frac{x}{4}\right)+f(4 x)\right]
View answer and explanation
Correct option: C
0
The correct answer is 0 because when substituting f(x)=cos⁡(log⁡x)f(x) = \cos(\log x) into the expression and simplifying using trigonometric identities, the result is 0.
194 Medium
194 of 353
Analyzing +4 | -1
The maximum value of 4  sin⁡2x−12  sin⁡  x+74\,\,{\sin ^2}x - 12\,\,\sin \,\,x + 7 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the maximum value of the given expression is 23, which is not among the provided options.
195 Medium
195 of 353
Application +4 | -1
If f(x)=log⁡(1+x1−x)f(x)=\log \left(\frac{1+x}{1-x}\right) and g(x)=3x+x31+3x2g(x)=\frac{3 x+x^3}{1+3 x^2}, then f(g(x))f(g(x)) is equal to
View answer and explanation
Correct option: C
3f(x)3 f(x)
Option C (3f(x)3 f(x)) is correct because substituting g(x) into f(x) and simplifying results in 3 times f(x).

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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
196 Medium
196 of 353
Analyzing +4 | -1
If f(x)=[x]f(x)=[x] and g(x)=x−[x]g(x)=x-[x], then which of the following is the zero function?
View answer and explanation
Correct option: D
(f∘g)(x)(f \circ g)(x)
Option D ((f∘g)(x)(f \circ g)(x)) is the zero function because g(x)g(x) always returns a value between 0 and 1, and f(x)f(x) (the floor function) of that value is always 0.
197 Medium
197 of 353
Understanding +4 | -1
The period of f(x)=x−[x]f(x)=x-[x], if it is periodic, is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the function repeats its values after an interval of 1, making 1 the period.
198 Medium
198 of 353
Understanding +4 | -1
Which pairs of function is identical ?
View answer and explanation
Correct option: B
f(x)=sin⁡2x+cos⁡2x;g(x)=1f(x)=\sin ^{2} x+\cos ^{2} x ; g(x)=1
Option B is correct because both functions have the same domain (all real numbers) and the same range (always equal to 1).
199 Medium
199 of 353
Application +4 | -1
Let E={1,2,3,4}\mathrm{E}=\{1,2,3,4\} and F={1,2}\mathrm{F}=\{1,2\}. Then the number of onto functions from E\mathrm{E} to Fis\mathrm{Fis}
View answer and explanation
Correct option: A
14
Option A (14) is correct because the number of onto functions from E to F is calculated using the formula for surjections, resulting in 14.
200 Medium
200 of 353
Application +4 | -1
let (5+26)=p+f(5+2 \sqrt{6})=p+f where nεNn \varepsilon N and 0<f<10 < f < 1 then the values of (f2−\left(f^2-\right. f+pf−p)f+p f-p)
View answer and explanation
Correct option: B
a negative integer
Option B (a negative integer) is correct because the expression simplifies to -1, which is a negative integer, by using the conjugate and properties of binomial expansion.
201 Medium
201 of 353
Understanding +4 | -1
If ff and gg are two functions defined as f(x)=x+2f(x)=x+2, x≤0;g(x)=3,x≥0x \leq 0 ; g(x)=3, x \geq 0, then the domain of f+gf+g is
View answer and explanation
Correct option: A
{0}\{0\}
Option A ({0}\{0\}) is correct because the domain of the sum of two functions is the intersection of their individual domains, and in this case, it's only {0}.
202 Medium
202 of 353
Application +4 | -1
In a function f(x)f(x) is defined for x∈[0,  1],x \in \left[ {0,\;1} \right], then the function f(2  x+3)f\left( {2\;x + 3} \right) is defined for
View answer and explanation
Correct option: B
x∈[−3/2,  −1]x \in \left[ { - 3/2,\; - 1} \right]
The correct answer is B (x ∈ [-3/2, -1]) because it is the interval for x that makes 2x + 3 fall within the defined domain [0, 1] of f(x).
203 Medium
203 of 353
Application +4 | -1
The domain of definition of the function

f(x)=log⁡10(5x−x24),isf\left( x \right) = \sqrt {{{\log }_{10}}\left( {\frac{{5x - {x^2}}}{4}} \right)} ,\mathrm{is}
View answer and explanation
Correct option: A
[1,4]\left[ {1,4} \right]
Option A [1,4]\left[ {1,4} \right] is correct because the domain is found by ensuring the argument of the logarithm is positive and greater than or equal to 1.
204 Medium
204 of 353
Application +4 | -1
If ff is an exponential function and gg is a logarithmic function, then f∘g(1)f \circ g(1) will be
View answer and explanation
Correct option: B
log⁡ee\log _e e
Option B (log⁡ee\log_e e) is correct because the exponential and logarithmic functions are inverses, and evaluating at 1 simplifies to log⁡ee=1\log_e e = 1.
205 Medium
205 of 353
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: C
13(x+4)\frac{1}{3}(x+4)
Option C (13(x+4)\frac{{1}}{{3}}(x+4)) is correct because it represents the inverse function obtained by swapping xx and yy and solving for yy.
206 Medium
206 of 353
Understanding +4 | -1
The range of the function f(x)=∣x∣f(x)=|x| is
View answer and explanation
Correct option: C
[0,∞)[0, \infty)
Option C ([0,∞)[0, \infty)) is correct because the absolute value of any real number is always non-negative.
207 Medium
207 of 353
Application +4 | -1
If the domain of the function f(x)=[x]1+x2f(x)=\frac{[x]}{1+x^{2}}, where [x][x] is greatest integer ≤x\leq x, is [2,6)[2,6), then its range is
View answer and explanation
Correct option: A
(537,25]\left(\frac{5}{37}, \frac{2}{5}\right]
The correct answer is C ((526,25]\left(\frac{{5}}{{26}}, \frac{{2}}{{5}}\right]) because it represents the range of the function over the given domain, considering the greatest integer function.
208 Medium
208 of 353
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).
209 Medium
209 of 353
Application +4 | -1
If f(x)=cos⁡(log⁡x)f(x)=\cos (\log x), then f(x)f(y)−{(1/2)[f(x/y)+f(xy)]}f(x) f(y)-\{(1 / 2)[f(x / y)+f(x y)]\} is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the expression simplifies to 0, which is not among the given options.
210 Medium
210 of 353
Analyzing +4 | -1
If f(x+y)=f(x)f(y)∀x,y∈Rf(x+y)=f(x) f(y) \forall x, y \in R and f(0)≠0f(0) \neq 0, then the function ϕ(x)=f(x)1+(f(x))2+sin⁡(π[x2+x]1+sin⁡(cos⁡x)\phi(x)=\frac{f(x)}{1+(f(x))^2}+\frac{\sin \left(\pi\left[x^2+x\right]\right.}{1+\sin (\cos x)} (where [] denote the greatest integer function) is
View answer and explanation
Correct option: A
An even function
Option A (An even function) is correct because the function φ(x) is shown to satisfy φ(-x) = φ(x), indicating it is an even function.
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