Question

The equation ${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$ has:
$\left( a \right)$ Infinite number of real roots.
$\left( b \right)$ No real roots
$\left( c \right)$ Exactly one real roots
$\left( d \right)$ Exactly four real roots.

Answer 1

Hint: In this particular question use the concept that assume ${e^{\sin x}}$ to any other variable so that the equation converts into a quadratic equation then simplify this equation using quadratic formula so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given equation
${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$
Let, ${e^{\sin x}} = t$................ (1)
Now substitute this value in the above equation we have,
$ \Rightarrow t - {t^{ - 1}} - 4 = 0$
Now simplify this equation we have,
$ \Rightarrow t - \dfrac{1}{t} - 4 = 0$
$ \Rightarrow {t^2} - 1 - 4t = 0$
$ \Rightarrow {t^2} - 4t - 1 = 0$
So the above equation is a quadratic equation which cannot be factorize as the discriminant D = $\sqrt {{b^2} - 4ac} = \sqrt {16 - 4\left( { - 1} \right)} = \sqrt {20} $ so we use quadratic formula so we have,
$ \Rightarrow t = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where a = 1, b = -4, c = -1 so we have,
$ \Rightarrow t = \dfrac{{4 \pm \sqrt {{4^2} - 4\left( { - 1} \right)} }}{{2\left( 1 \right)}} = \dfrac{{4 \pm \sqrt {20} }}{2} = \dfrac{{4 \pm 2\sqrt 5 }}{2} = 2 \pm \sqrt 5 $
$ \Rightarrow t = 2 + \sqrt 5 = 4.236$, $t = 2 - \sqrt 5 = - 0.236$
Now from equation (1) we have
$ \Rightarrow {e^{\sin x}} = 4.236,{e^{\sin x}} = - 0.236$................. (2)
Now as we know that $ - 1 \leqslant \sin x \leqslant 1$, so ${e^{\sin x}}$ can be vary in the range of $\left[ {{e^{ - 1}},e} \right] = \left[ {0.3678,2.718} \right]$
$ \Rightarrow {e^{\sin x}} = \left[ {0.3678,2.718} \right]$ (Maximum possible range)
But from equation (2) both the values are outside the above range.
So no solution is possible.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the range of sin x which is stated above and always recall the quadratic formula to solve the complex quadratic equation which is given as, $t = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, so first convert the equation in to quadratic equation as above and then apply quadratic formula as above we will get the required answer.