Question

The number of real solutions of the equation sin(e) = 2018^x+ 2018^-x
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Answer 1

Step-by-step explanation:

Given The number of real solutions of the equation                             sin(e)^x = 2018^x+ 2018^-x

• Given sin (e^x) = 2018^x + 2018^-x
• Now consider the right hand side we get
• 2018^x + 2018^-x is a point a quantity for all x belongs to R.
• So AM ≥ GM
• So 2018^x + 2018^-x / 2 ≥ (2018^x . 2018^-x)^1/2  
• so 2018^x + 2018^-x ≥ 2
• Therefore minimum value of 2018^x + 2018^-x = 2
• Now we have sin (e^x)
• Range of sin will be (-1,1)
• But maximum value of sin e^x is 1
• Now the maximum value of the function sin e^x is 1 and minimum value is 2, so it will not be equal to the value of x, since minimum value is 2 and maximum value is 1.
• So the number of real solutions for the given equation is 0.

Reference link will be

https://brainly.in/question/8513664