Let $f:N \to Y$ be a function defined as $f\left( x \right) = 4x + 3$ where $Y = \left\{ {y \in N:y = 4x + 3\;{\rm{for\;some}}\;x \in N} \right\}$ Show that f is invertible and its inverse is

$g\left( y \right) = \frac{{3y + 4}}{3}$

$g\left( y \right) = 4 + \frac{{y + 3}}{4}$

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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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481 Easy

481 of 650

Understanding +4 | -1

The range of the function

f(x)=\frac{x-2}{2-x}

when

x \neq 2

View answer and explanation

Correct option: C

\{-1\}

Option C (

\{-1\}

) is correct because the function simplifies to -1 for all x not equal to 2.

482 Easy

482 of 650

Application +4 | -1

If the function

f:N \to N

is defined by

f\left( x \right) = \sqrt x

, then

\frac{{f\left( {25} \right)}}{{f\left( {16} \right) + f\left( 1 \right)}}

is equal to

View answer and explanation

Correct option: D

1

The correct answer is D (1) because substituting the values into the function and simplifying results in 1.

483 Medium

483 of 650

Analyzing +4 | -1

[x]

denotes the greatest integer

\le x

, then

\left[ {\frac{2}{3}} \right] + \left[ {\frac{2}{3} + \frac{1}{{99}}} \right] + \left[ {\frac{2}{3} + \frac{2}{{99}}} \right] + ... + \left[ {\frac{2}{3} + \frac{{98}}{{99}}} \right]

is equal to

View answer and explanation

Correct option: C

Option C is correct. The value inside the greatest integer function becomes greater than or equal to 1 when r >= 33. Therefore, the sum is equal to the number of terms where r >= 33, which is 98 - 33 + 1 = 66.

484 Easy

484 of 650

Application +4 | -1

f:\left[ {2,\;3} \right] \to R

is defined by

f\left( x \right) = {x^3} + 3x - 2

, then the range

f\left( x \right)

is contained in the interval

View answer and explanation

Correct option: B

\left[ {12,\mathrm{ }34} \right]

Option B ([12, 34]) is correct because the range is determined by evaluating the function at the endpoints of the interval [2, 3] since the function is monotonically increasing.

485 Easy

485 of 650

Application +4 | -1

Let

f(x)=\frac{1-x}{1+x}

. Then

f o f(\cos x)

is equal to

View answer and explanation

Correct option: B

\cos x

Option B (cos x) is correct because applying the function f twice results in the original input, cos x.

486 Easy

486 of 650

Application +4 | -1

f\left( x \right) = 2{x^6} + 3{x^4} + 4{x^2}

, then

f'\left( x \right)

View answer and explanation

Correct option: B

An odd function

Option B (An odd function) is correct because the derivative of an even function is always an odd function.

487 Easy

487 of 650

Application +4 | -1

If the function

f(x)

on the domain

\left[\frac{1}{2}, \infty\right)

is defined by

f(x)=2^{x(x-1)}

then

f^{-1}(x)

equals:

View answer and explanation

Correct option: A

\frac{1}{2}\left(1+\sqrt{1+4 \log _2 x}\right)

Option A (

\frac{1}{2}\left(1+\sqrt{1+4 \log _2 x}\right)

) is correct because it is the correct inverse function obtained by solving for x in terms of y and considering the domain.

488 Easy

488 of 650

Analyzing +4 | -1

The domain of the function

f\left( x \right) = \frac{{{{\sin }^{ - 1}}\left( {x - 3} \right)}}{{\sqrt {9 - {x^2}} }},

View answer and explanation

Correct option: B

\left[ {2,\;3} \right)

Option B is correct because it satisfies both conditions: -1 <= x-3 <= 1 and 9 - x² > 0, considering the denominator cannot be zero.

489 Easy

489 of 650

Application +4 | -1

f\left( x \right)

is an even function and

f'\left( x \right)

exists, then

f'\left( e \right) + f'\left( { - e} \right)

View answer and explanation

Correct option: B

= \mathrm{ }0

Option B (= 0) is correct because if f(x) is an even function, its derivative f'(x) is an odd function. Therefore, f'(-e) = -f'(e), and their sum is zero.

490 Easy

490 of 650

Analyzing +4 | -1

The function

f:R \to R

given by

f\left( x \right) = {x^2} + x,

, is

View answer and explanation

Correct option: D

Many one and into

Option D (Many one and into) is correct because the function is a parabola and doesn't cover the entire range of real numbers, and multiple x values can map to the same y value.

491 Medium

491 of 650

Application +4 | -1

f\left( x \right) = \sqrt {{{\sin }^{ - 1}}\left( {{{\log }_2}x} \right)}

exists for

View answer and explanation

Correct option: B

x \in \left[ {1,\;2} \right]

Option B (

x \in \left[ {1,\;2} \right]

) is correct because the function exists when the argument of the square root is non-negative and the argument of the inverse sine lies between -1 and 1.

492 Easy

492 of 650

Application +4 | -1

The set of all values of

x

satisfying the inequality

\log _2 \sqrt{x}-2\left(\log _{1 / 4} x\right)^2+1>0

is the interval

View answer and explanation

Correct option: C

\left(\frac{1}{2}, 4\right)

Option C (

\frac{1}{2}, 4

) is correct because it represents the interval of x values that satisfy the given logarithmic inequality after simplification and solving.

493 Easy

493 of 650

Analyzing +4 | -1

Foe real

x

, let

f\left( x \right) = {x^3} + 5x + 1

, then

View answer and explanation

Correct option: C

f

is one-one and onto

R

Option C (

f

is one-one and onto

R

) is correct because the function is strictly increasing and continuous over the real numbers.

494 Medium

494 of 650

Analyzing +4 | -1

If the function

f\;:R \to A

given by

f\left( x \right) = \frac{{{x^2}}}{{{x^2} + 1}}

is a surjection, then

A =

View answer and explanation

Correct option: D

\left[ {0,\;1} \right)

Option D

\left[ {0,\;1} \right)

is correct because the range of the function, excluding 1, makes it a surjection.

495 Easy

495 of 650

Application +4 | -1

f\left( x \right) = \log \left( {\frac{{1 + x}}{{1 - x}}} \right),

then

f\left( {\frac{{2x}}{{1 + {x^2}}}} \right)

is equal to

View answer and explanation

Correct option: C

2f\left( x \right)

Option C (2f(x)) is correct because by substituting (2x)/(1+x²) into f(x) and using logarithm properties, we can simplify the expression to 2f(x).

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496 Medium

496 of 650

Application +4 | -1

Let

f: R \rightarrow R

be defined by

f(x)=2 x+|x|

, then

f(2 x)+f(-x)-f(x)=

View answer and explanation

Correct option: B

2|x|

Option B (

2|x|

) is correct because substituting the given function

f(x)=2 x+|x|

into the expression

f(2 x)+f(-x)-f(x)

and simplifying results in

2|x|

497 Medium

497 of 650

Understanding +4 | -1

The function

f(x)=\log \left(x+\sqrt{x^{2}+1}\right)

, is

View answer and explanation

Correct option: C

an odd function

Option C is correct because f(-x) = -f(x), which satisfies the condition for an odd function.

498 Hard

498 of 650

Understanding +4 | -1

f\left( { - x} \right) = - f\left( x \right),

then

f\left( x \right)

View answer and explanation

Correct option: B

An odd function

Option B (An odd function) is correct because the given condition

f(-x) = -f(x)

is the definition of an odd function.

499 Easy

499 of 650

Understanding +4 | -1

Let

A

and

B

be non empty sets in

R

and

f: A \rightarrow B

is a bijective function.

Statement 1 :

\mathrm{f}

is an onto function.

Statement 2: There exists a function

g: B \rightarrow A

such that fog

=I_{B}

View answer and explanation

Correct option: D

Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1 .

Option D is correct. Statement 1 is true because a bijective function is onto. Statement 2 is also true because a bijective function has an inverse, but Statement 2 doesn't explain Statement 1.

500 Medium

500 of 650

Application +4 | -1

f(x)

is defined on

\left[ {0,\;1} \right],

then the domain of

f\left( {3{x^2}} \right),

View answer and explanation

Correct option: B

\left[ { - 1/\sqrt 3 ,1/\sqrt 3 } \right]

Option B is correct because the domain of f(3x²) is found by ensuring 0 <= 3x² <= 1, which leads to the interval [-1/sqrt(3), 1/sqrt(3)].

501 Hard

501 of 650

Application +4 | -1

Suppose

f:\left[ { - 2,2} \right] \to R

is defined by

f\left( x \right) = \left\{ {\;\begin{array}{cccccccccccccccccccc} { - 1\;\mathrm{for} - 2 \le x \le 0}\\ {x - 1\;\mathrm{for}\;0 \le x \le 2} \end{array}} \right.

then

\left\{ {x \in \left[ { - 2,2} \right]:x \le 0\;\mathrm{and}\;f\left( {\left| x \right|} \right) = x} \right\} =

View answer and explanation

Correct option: C

\left\{ { - 1/2} \right\}

Option C ({-1/2}) is correct because it is the only value that satisfies the given conditions and falls within the specified interval.

502 Medium

502 of 650

Application +4 | -1

f\left( {2x + 3} \right) = \sin x + {2^x},

then

f\left( {4m - 2n + 3} \right)

is equal to

View answer and explanation

Correct option: D

\sin \left( {2m - n} \right) + {2^{2m - n}}

Option D (

\sin \left( {2m - n} \right) + {2^{2m - n}}

) is correct because it is obtained by substituting x = 2m - n into the given function.

503 Easy

503 of 650

Analyzing +4 | -1

A=\left\{3^{\mathrm{n}}: n \in \mathbf{N}, n \leq 6\right\}, B=\left\{9^{\mathrm{n}}: n \in \mathbf{N}, n \leq 4\right\}

then which of the following is false

View answer and explanation

Correct option: A

A \Delta B=\{6561\}

Option A (

A \Delta B=\{6561\}

) is false because the symmetric difference

A \Delta B

should include all elements in A or B but not in both, which is {3, 27, 243, 6561}, not just {6561}.

504 Easy

504 of 650

Application +4 | -1

The domain of the function

f(x)=\sin ^{-1}\left[\log _{2}(x / 2)\right]

View answer and explanation

Correct option: A

[1,4]

Option A ([1,4]) is correct because the domain is found by solving the inequality -1 <= log2(x/2) <= 1.

505 Easy

505 of 650

Application +4 | -1

Range of the function

f(x)=9-7 \sin x

View answer and explanation

Correct option: B

[2,16]

Option B (

[2,16]

) is correct because the minimum value of the function is 9 - 7(1) = 2 and the maximum value is 9 - 7(-1) = 16.

506 Medium

506 of 650

Understanding +4 | -1

f\left( x \right) = \frac{1}{{\sqrt {\left| x \right| - x} }}

then, domain of

f\left( x \right)

View answer and explanation

Correct option: A

\left( { - \infty ,\;0} \right)

Option A (

\left( { - \infty ,\;0} \right)

) is correct because the function is defined only when

|x| > x

, which occurs when x is negative.

507 Easy

507 of 650

Understanding +4 | -1

Number of solution for

\left|3 x^2-2\right|=[-2 \pi]

is ([-] denotes greatest integer)

View answer and explanation

Correct option: A

Option A (0) is correct because the left-hand side is always non-negative, while the right-hand side is negative, resulting in no solutions.

508 Easy

508 of 650

Analyzing +4 | -1

Let

f(x)=\frac{x-3}{x+1}, x \neq-1

. Then

f^{2010}

(2014) (where

f^n(x)=f o f \ldots

(x)(n

times)) is

View answer and explanation

Correct option: D

2014

Option D (2014) is correct because the function repeats every 3 iterations, and 2010 is a multiple of 3, so f²⁰¹⁰(2014) = 2014.

509 Medium

509 of 650

Understanding +4 | -1

x,y \in R,

then

\frac{1}{2}\left( {x + y + \left| {x - y} \right|} \right) = x

holds iff

View answer and explanation

Correct option: D

x \ge y

Option D (

x \ge y

) is correct because the equation holds true only when x is greater than or equal to y, ensuring the absolute value simplifies correctly.

510 Medium

510 of 650

Application +4 | -1

Let

f:N \to Y

be a function defined as

f\left( x \right) = 4x + 3

where

Y = \left\{ {y \in N:y = 4x + 3\;\mathrm{for\;some}\;x \in N} \right\}

Show that f is invertible and its inverse is

View answer and explanation

Correct option: A

g\left( y \right) = \frac{{y - 3}}{4}

Option A (

g(y) = \frac{{y - 3}}{4}

) is correct because it is the inverse function obtained by solving for x in the equation y = 4x + 3.

Revision notes and key takeaways

Review FUNCTIONS in your textbook or class notes, then use this set to check your understanding.

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