Question

please solve this question ..........please ​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

DOMAIN

$f(x) = \sqrt{ {x}^{2} - 16 }$

Here f (x) is well defined when

${x}^{2} - 16 \geqslant 0$

$\implies \: {x}^{2} \geqslant 16$

$\implies \: x \geqslant 4 \: \: \: or \: \: \: x \leqslant - 4$

So the required Domain of the function is

$= \{ \: x \in \mathbb{R \:}: x \geqslant 4 \: \: or \: \: x \leqslant - 4 \: \: \}$

$= ( - \infin , - 4 \: ] \: \cup \: \: [ \: 4 \: , \infin \: )$

RANGE

Let

$y = \sqrt{ {x}^{2} - 16}$

$\implies \: {y}^{2} = {x}^{2} - 16$

$\implies \: {x}^{2} = {y}^{2} + 16$

We know that

$for \: any \: \: x \in \mathbb{R}$

${x}^{2} \geqslant 0$

$\implies \: { {y}}^{2} + 16 \geqslant 0$

$\implies \: { {y}}^{2} \geqslant - 16$

Hence the required Range is

$= \: [ \: 0 \: , \infin \: ]$

Answer 2

Answer:

To Prove ∑nr=03r×Crn=4n by binomial expansion

(a+b)n=C0nanb0+C1nan−1b1....Cnna0bn=

∑nr=0Crn×an−r×br

if

we compare the two terms

∑nr=03r×Crn=∑nr=0Crn×an−r×br

hence if we make a=1 and b=3

we get ∑nr=0Crn×1n−r×3r

which is same as given term,

∑nr=03r×Crn

putting the same values in (a+b)n=(1+3)n=4n