If f: R R is defined by f(x) = x + x, then f(f(x)) is: Option 1: 2 x + 2x Option 2: ^2 x + x^2 Option 3: ( x + x) + x + x Option 4: ^2 x + 2 x + x

Answer: ( x + x) + x + x. Solution: Let u = f(x) = x + x. Then f(f(x)) = f(u) = u + u = ( x + x) + x + x.

Algebra

Relations & Functions

Easy

core

If $f: R \to R$ is defined by $f(x) = \sin x + x$ , then $f(f(x))$ is:

$2\sin x + 2x$

$\sin^2 x + x^2$

$\sin(\sin x + x) + \sin x + x$

$\sin^2 x + 2\sin x + x$

✅ Correct Option: 3

Let $u = f(x) = \sin x + x$ . Then $f(f(x)) = f(u) = \sin u + u = \sin(\sin x + x) + \sin x + x$ .

2022: 23 Aug Shift 1

If $n(A) = 3$ , $n(B) = 2$ , then number of all possible surjective function from set A to set B are :

2025: 15 May Shift 2

A relation R in the set A = {1, 2,3, 4} is given by R = {(1,1), (2,2), (1,2), (2,3), (3,4), (4,4), (1,3), (2,4), (1,4)} is

2026: 21 May Shift 1

If $A = R - \{2\}$ and $B = R - \{1\}$ and function $f: A \rightarrow B$ defined by $f(x) = \frac{x-1}{x-2}$ , then $f$ is

2026: 30 May Shift 1

The signum function $f:R\to R$ , defined by

$f(x)=\begin{cases}-1, & x<0\\0, & x=0\\1, & x>0\end{cases}$ is: (Where R is set of real numbers)

2025: 13 May Shift 2

Match List-I with List-II

Let $f: A \rightarrow B$ be a function given by $f(x) = x^2$

List-I	List-II
Domain and Co-domain	Kind
(A) $A = \mathbb{R}$ and $B = \mathbb{R}$	(I) $f$ is both one-one and onto
(B) $A = \mathbb{R}$ and $B = [0, \infty]$	(II) $f$ is one-one but not onto
(C) $A = B = [0, \infty]$	(III) $f$ is not one-one but onto
(D) $A = [0, \infty]$ and $B = \mathbb{R}$	(IV) $f$ is neither one-one nor onto

Choose the correct answer from the options given below:

2025: 19 May Shift 1

The relation R in $\mathbb{R}$ (set of real numbers) is defined by $R = \{(a,b): a \leq b^3\}$ , then R is