Relations and Functions - Class 11 Mathematics - Chapter 2 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Relations and Functions | NCERT | Mathematics | Class 11
If $A = \{2, 4, 6, 8\}$ and $B = \{1, 2, 3, 4, 5\}$, then $A - B =$
A $\{1,3,5\}$
B $\{\}$
C $\{6, 8\}$
D $\{2,4\}$
The correct answer is Option C: $\{6, 8\}$.
To find the set difference $A - B$, we remove all elements of set $B$ from set $A$:
Given: $$ A = {2, 4, 6, 8} $$ $$ B = {1, 2, 3, 4, 5} $$
The elements of $A$ that are not in $B$ are $6$ and $8$. Thus: $$ A - B = {6, 8} $$
If $f(x) + f(x+a) + f(x+2a) + \ldots + f(x+na) = \text{constant}$; $\forall x \in$ R and a $> 0$ and $f(x)$ is periodic, then the period of $f(x)$ is
A $(n+1)a$
B $e^{(n+1)a}$
C $na$
D $e^{n a}$
Solution The correct answer is A: $(n+1)a$.
Let's consider the function sum: $$ f(x) + f(x+a) + f(x+2a) + \ldots + f(x+na) = k $$ Here, $k$ is a constant for all values of $x$ and $a > 0$.
To find the period of the function, substitute $x$ with $x+a$ in the sum: $$ f(x+a) + f(x+2a) + \ldots + f(x+(n+1)a) = k $$
Comparing both expressions, we notice that each term in the second sum shifts by $a$. The last term in the first sum, $f(x+na)$, correspondingly, in the second sum becomes $f(x+(n+1)a)$, and $f(x)$ is not present. Therefore, we equate: $$ f(x) = f(x+(n+1)a) $$
This tells us that the function repeats itself every $(n+1)a$, indicating that the period $T$ of $f(x)$ is: $$ T = (n+1)a $$
Thus, the period of $f(x)$ is $(n+1)a$.
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