Question

(i) 5 – 4 sin 3x find domain and range​

Answer 1

-1< sin 3 x<1

-4<4 sin 3x<4

4>-4 sin 3x>-4

5+4>5-4 sin 3x>5-4

9>5-4sin 3x>1

1,9

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = 5 - 4 \: sin3x$

Domain :-

Domain of a function is defined as set of all those real numbers for which f(x) is defined.

So,

$\rm :\longmapsto\:Domain \: of \: f(x) : \: x \: \in \: R$

Range :-

The range of a real function is the set of all those real number values taken f(x) at points in its domain.

We know,

$\rm :\longmapsto\: - 1 \leqslant sin3x \leqslant 1$

On multiply by - 4, we get

$\rm :\longmapsto\: 4 \geqslant - 4sin3x \geqslant - 4$

$\rm :\longmapsto\: - 4 \leqslant - 4sin3x \leqslant 4$

On adding 5 in each term, we get

$\rm :\longmapsto\: 5- 4 \leqslant 5 - 4sin3x \leqslant 5 + 4$

$\rm :\longmapsto\: 1 \leqslant 5 - 4sin3x \leqslant 9$

$\rm :\longmapsto\: 1 \leqslant f(x) \leqslant 9$

$\bf\implies \:f(x) \: \in \: [1, \: 9 ]$

Additional Information :-

Question :-

Find the range of the function

$\rm :\longmapsto\:f(x) = \dfrac{1}{2 + 3sin2x}$

Solution :-

We know that,

$\rm :\longmapsto\: - 1 \leqslant sin2x \leqslant 1$

On multiply by 2, on both sides, we get

$\rm :\longmapsto\: - 2 \leqslant 2sin2x \leqslant 2$

On Adding 3 in each term, we get

$\rm :\longmapsto\: 3- 2 \leqslant 2sin2x + 3\leqslant 2 + 3$

$\rm :\longmapsto\: 1 \leqslant 2sin2x + 3\leqslant 5$

$\rm :\longmapsto\:\dfrac{1}{5} \leqslant \dfrac{1}{3 + 2sin2x} \leqslant 1$

$\rm :\longmapsto\:\dfrac{1}{5} \leqslant f(x) \leqslant 1$

$\bf\implies \:f(x) \: \in \: \bigg[\dfrac{1}{5} , \: 1]$

Hence,

Range of

$\bf\implies \:\dfrac{1}{3 + 2sin2x} \: \in \: \bigg[\dfrac{1}{5} , \: 1]$