Question

Important Question for jee mains.
mathamatics :-Practice set 1 ​

Answer 1

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

Given function is

$\rm :\longmapsto\:f(x) = \sqrt[3]{1 - 3x} + 3 {cos}^{ - 1}\bigg(\dfrac{2x - 1}{3} \bigg) + {e}^{3tanx}$

Consider,

$\rm :\longmapsto\: \sqrt[3]{1 - 3x}$

Let assume that

$\rm :\longmapsto\:g(x) = \sqrt[3]{1 - 3x}$

We know,

Domain is the set of those values where function is well defined.

Since, cube root is defined for all real numbers.

So,

$\bf\implies \:Domain \: of \: g(x), \: D_g \: = \: R$

Now,

Consider,

$\rm :\longmapsto\:h(x) = 3 {cos}^{ - 1}\bigg(\dfrac{2x - 1}{3} \bigg)$

We know,

$\rm :\longmapsto\: {cos}^{ - 1}x \: is \: defined \: when \: - 1 \leqslant x \leqslant 1$

So,

In order that, h(x) is defined

$\rm :\longmapsto\: - 1 \leqslant \dfrac{2x - 1}{3} \leqslant 1$

$\rm :\longmapsto\: - 3 \leqslant 2x - 1 \leqslant 3$

$\rm :\longmapsto\: - 3 + 1 \leqslant 2x - 1 + 1 \leqslant 3 + 1$

$\rm :\longmapsto\: - 2 \leqslant 2x \leqslant 4$

$\rm :\longmapsto\: - 1\leqslant x \leqslant 2$

Hence,

$\bf\implies \:Domain \: of \: h(x), \: D_h \: = \: - 1 \leqslant x \leqslant 2$

Consider,

$\rm :\longmapsto\:p(x) = {e}^{3tanx}$

We know that,

$\rm :\longmapsto\:tanx \: is \: defined \: when \: x \: \ne \: odd \: multiples \: of \: \dfrac{\pi}{2}$

So,

Domain of p(x) is defined as

$\rm :\longmapsto\:Domain = R - \{(2n - 1)\dfrac{\pi}{2} \} \: where \: n \: is \: natural \: number$

So,

Domain of f(x) =

$\rm \:  =  \:  \:D_g \: \cup \: D_h \: \cap \: D_p$

$\rm \:  = x \:  \in \: \: - 1 \leqslant x \leqslant 2 - \bigg \{\dfrac{\pi}{2}\bigg \}$

$\rm \:  =  \:  \:x \: \in \: [ - 1, \: 2] - \bigg \{\dfrac{\pi}{2} \bigg\}$

So,

• Option (d) is correct.