Answer

Verified

62.1k+ views

Hint:The given equation is a transcendental equation, to solve it we will make it an algebraic equation and then we will solve it using a basic algorithm approach and simplify it further.

Complete step-by-step answer:

Given equation is

$ \Rightarrow {e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$ ………………………………… (1)

Multiply the equation 1 by ${e^{\sin x}}$ both sides

$ \Rightarrow {e^{2\sin x}} - 4{e^{\sin x}} - 1 = 0$………………………………… (2)

The equation (2) is a quadratic equation in terms of ${e^{\sin x}}$

Replacing ${e^{\sin x}}{\text{ by x}}$ , we get

$ \Rightarrow {x^2} - 4x - 1 = 0$ …………………………………………… (3)

Using the formula of quadratic equation to find the value of $x$

$

a{x^2} + bx + c = 0 \\

x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\

$

On comparing equation (3) with the above formula, we get

\[a = 1,b = - 4,c = - 1\]

Substituting these values in the formula, we get

$

\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\

\Rightarrow x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 1 \times ( - 1)} }}{{2 \times 1}} \\

$

On simplifying the above equation further, we obtain

\[

\Rightarrow x = \dfrac{{4 \pm \sqrt {16 + 4} }}{{2 \times 1}} \\

\Rightarrow x = \dfrac{{4 \pm \sqrt {20} }}{2} \\

\Rightarrow x = 2 \pm \sqrt 5 \\

\]

Now, replacing \[x\] by ${e^{\sin x}}$ , we get

$ \Rightarrow {e^{\sin x}} = 2 \pm \sqrt 5 $………………………. (4)

Converting the equation (4) from exponential to logarithmic, we get

. As we know ,$\left[ {y = {{\log }_e}(x) \Rightarrow x = {e^y}} \right]$

Since $\log (2 - \sqrt 5 )$is not defined

\[ \Rightarrow \sin x = \log (2 + \sqrt 5 )\]

As we know ,$2 + \sqrt 5 > e \Rightarrow \log (2 + \sqrt 5 ) > 1$

$ \Rightarrow \sin x > 1$ , which is not possible

Hence, for the given equation no solution exists.

Option A is correct.

Note: To solve these types of questions basic arithmetic operation and logarithmic operations are required. As we know, $y = {\log _e}(x) \Rightarrow x = {e^y}$ . There are two methods to solve quadratic equations; one is by factorising and other is using square root property. In the question, we have used the square root property as the equation is not easily factored.

Complete step-by-step answer:

Given equation is

$ \Rightarrow {e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$ ………………………………… (1)

Multiply the equation 1 by ${e^{\sin x}}$ both sides

$ \Rightarrow {e^{2\sin x}} - 4{e^{\sin x}} - 1 = 0$………………………………… (2)

The equation (2) is a quadratic equation in terms of ${e^{\sin x}}$

Replacing ${e^{\sin x}}{\text{ by x}}$ , we get

$ \Rightarrow {x^2} - 4x - 1 = 0$ …………………………………………… (3)

Using the formula of quadratic equation to find the value of $x$

$

a{x^2} + bx + c = 0 \\

x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\

$

On comparing equation (3) with the above formula, we get

\[a = 1,b = - 4,c = - 1\]

Substituting these values in the formula, we get

$

\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\

\Rightarrow x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 1 \times ( - 1)} }}{{2 \times 1}} \\

$

On simplifying the above equation further, we obtain

\[

\Rightarrow x = \dfrac{{4 \pm \sqrt {16 + 4} }}{{2 \times 1}} \\

\Rightarrow x = \dfrac{{4 \pm \sqrt {20} }}{2} \\

\Rightarrow x = 2 \pm \sqrt 5 \\

\]

Now, replacing \[x\] by ${e^{\sin x}}$ , we get

$ \Rightarrow {e^{\sin x}} = 2 \pm \sqrt 5 $………………………. (4)

Converting the equation (4) from exponential to logarithmic, we get

. As we know ,$\left[ {y = {{\log }_e}(x) \Rightarrow x = {e^y}} \right]$

Since $\log (2 - \sqrt 5 )$is not defined

\[ \Rightarrow \sin x = \log (2 + \sqrt 5 )\]

As we know ,$2 + \sqrt 5 > e \Rightarrow \log (2 + \sqrt 5 ) > 1$

$ \Rightarrow \sin x > 1$ , which is not possible

Hence, for the given equation no solution exists.

Option A is correct.

Note: To solve these types of questions basic arithmetic operation and logarithmic operations are required. As we know, $y = {\log _e}(x) \Rightarrow x = {e^y}$ . There are two methods to solve quadratic equations; one is by factorising and other is using square root property. In the question, we have used the square root property as the equation is not easily factored.

Recently Updated Pages

Write a composition in approximately 450 500 words class 10 english JEE_Main

Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main

What is the common property of the oxides CONO and class 10 chemistry JEE_Main

What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main

If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main

The area of square inscribed in a circle of diameter class 10 maths JEE_Main

Other Pages

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main

Two blocks are in contact on a frictionless table One class 11 physics JEE_Main

Find the magnetic field at P due to arrangement as class 12 physics JEE_Main