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The equation esin x - e-sin x - 4 = 0 has
- a)no solution
- b)two solution
- c)three solution
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The equation esin x- e-sin x- 4 = 0 hasa)no solutionb)two solutionc)th...
E^sinx - e^-sinx -4=0
=> e^(sinx +sinx) -1 -4sinx = 0
=> e^2sinx -4sinx -1=0
let e^sinx = y
=> y² -4y -1 = 0
=> y² -4y +4 -4-1=0
=> (y-2)²=5
=> y=(±√5) +2
=> e^sinx = (±√5)+2
First of all e^sinx can't be negative.
Secondly, minimum value of sinx is -1. So minimum value of e^sinx will be 1/e = 0.367(approx.)
Similarly maximum value of sin x is + 1. So the maximum value of e^sin x will be e i.e. 2.718(approx.).
As both the values that we have obtained, are out of the range of values defined for e^sinx , the equation has no solution.
=> e^(sinx +sinx) -1 -4sinx = 0
=> e^2sinx -4sinx -1=0
let e^sinx = y
=> y² -4y -1 = 0
=> y² -4y +4 -4-1=0
=> (y-2)²=5
=> y=(±√5) +2
=> e^sinx = (±√5)+2
First of all e^sinx can't be negative.
Secondly, minimum value of sinx is -1. So minimum value of e^sinx will be 1/e = 0.367(approx.)
Similarly maximum value of sin x is + 1. So the maximum value of e^sin x will be e i.e. 2.718(approx.).
As both the values that we have obtained, are out of the range of values defined for e^sinx , the equation has no solution.
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Community Answer
The equation esin x- e-sin x- 4 = 0 hasa)no solutionb)two solutionc)th...
The given equation can be written as
e2sin x - 4e sinx -1 = 0
⇒
⇒ sin x = In (2 +√5)
[In (2 - √5) not defined as 2-√5 is - ve]
Now, 2 + √5 > e ⇒ = In (2 + √5) > 1 = sin x > 1
which is not possible ∴ No real solution.
e2sin x - 4e sinx -1 = 0
⇒
⇒ sin x = In (2 +√5)
[In (2 - √5) not defined as 2-√5 is - ve]
Now, 2 + √5 > e ⇒ = In (2 + √5) > 1 = sin x > 1
which is not possible ∴ No real solution.
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