Question

class 11 maths maths maths ​

Answer 1

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

Given equation is

$\rm :\longmapsto\: | {3}^{x} + 1 | + | {3}^{x} - 1| = 18$

We know,

$\rm :\longmapsto\: {3}^{x} > 0 \: \forall \: x \in \: real \: number$

$\bf\implies \: {3}^{x} + 1> 1$

$\bf\implies \: | {3}^{x} + 1| = {3}^{x} + 1$

So, given equation can be rewritten as

$\rm :\longmapsto\: {3}^{x} + 1 + | {3}^{x} - 1| = 18$

$\rm :\longmapsto\: {3}^{x} + | {3}^{x} - 1| = 18 - 1$

$\rm :\longmapsto\: {3}^{x} + | {3}^{x} - 1| = 17$

Now, we find the breaking point.

$\rm :\longmapsto\: {3}^{x} - 1 = 0$

$\rm :\longmapsto\: {3}^{x} = 1$

$\bf\implies \:x = 0$

Thus,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: | {3}^{x} - 1 | = \begin{cases} &\sf{ - ( {3}^{x} - 1) \: \: when \: x < 0} \\ &\sf{ \: \: \: \: {3}^{x} - 1 \: \: when \: x \: \geqslant 0 } \end{cases}\end{gathered}\end{gathered}$

Case :- 1

$\bf :\longmapsto\:When \: x < 0$

So, given equation

$\rm :\longmapsto\: {3}^{x} + | {3}^{x} - 1| = 17$

reduces to

$\rm :\longmapsto\: {3}^{x} - ( {3}^{x} - 1) = 17$

$\rm :\longmapsto\: {3}^{x} - {3}^{x} + 1 = 17$

$\rm :\longmapsto\:1 = 17$

$\bf\implies \:There \: is \: no \: solution \: in \: this \: case$

Case :- 2

$\bf :\longmapsto\:When \: x \geqslant 0$

The equation

$\rm :\longmapsto\: {3}^{x} + | {3}^{x} - 1| = 17$

reduces to

$\rm :\longmapsto\: {3}^{x} + {3}^{x} - 1 = 17$

$\rm :\longmapsto\: 2.{3}^{x} = 17 + 1$

$\rm :\longmapsto\: 2.{3}^{x} = 18$

$\rm :\longmapsto\: {3}^{x} = 9$

$\rm :\longmapsto\: {3}^{x} = {3}^{2}$

$\bf\implies \:x = 2$

So,

Number of roots of the equation

$\bf :\longmapsto\: | {3}^{x} + 1 | + | {3}^{x} - 1| = 18 \: is \: \: \: \red{1}$

Answer 2

Answer:

Number of roots of the equation is 18

Step-by-step explanation: