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Question: The equation \({e^{\sin x}} - {e^{ - \sin x}} - 4 = 0\) has: \(\left( a \right)\) Infinite number ...

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

Explanation

Solution

In this particular question use the concept that assume esin⁡x{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
esin⁡x−e−sin⁡x−4=0{e^{\sin x}} - {e^{ - \sin x}} - 4 = 0
Let, esin⁡x=t{e^{\sin x}} = t................ (1)
Now substitute this value in the above equation we have,
⇒t−t−1−4=0\Rightarrow t - {t^{ - 1}} - 4 = 0
Now simplify this equation we have,
⇒t−1t−4=0\Rightarrow t - \dfrac{1}{t} - 4 = 0
⇒t2−1−4t=0\Rightarrow {t^2} - 1 - 4t = 0
⇒t2−4t−1=0\Rightarrow {t^2} - 4t - 1 = 0
So the above equation is a quadratic equation which cannot be factorize as the discriminant D = b2−4ac=16−4(−1)=20\sqrt {{b^2} - 4ac} = \sqrt {16 - 4\left( { - 1} \right)} = \sqrt {20} so we use quadratic formula so we have,
⇒t=−b±b2−4ac2a\Rightarrow t = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}, where a = 1, b = -4, c = -1 so we have,
⇒t=4±42−4(−1)2(1)=4±202=4±252=2±5\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
⇒t=2+5=4.236\Rightarrow t = 2 + \sqrt 5 = 4.236, t=2−5=−0.236t = 2 - \sqrt 5 = - 0.236
Now from equation (1) we have
⇒esin⁡x=4.236,esin⁡x=−0.236\Rightarrow {e^{\sin x}} = 4.236,{e^{\sin x}} = - 0.236................. (2)
Now as we know that −1⩽sin⁡x⩽1 - 1 \leqslant \sin x \leqslant 1, so esin⁡x{e^{\sin x}} can be vary in the range of [e−1,e]=[0.3678,2.718]\left[ {{e^{ - 1}},e} \right] = \left[ {0.3678,2.718} \right]
⇒esin⁡x=[0.3678,2.718]\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=−b±b2−4ac2at = \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.