Quadratic EquationHard
Question
The equation esinx - e-sinx - 4 = 0 has :
Options
A.exactly four real roots.
B.infinite number of real roots.
C.no real roots.
D.exactly one real root.
Solution
Given esinx - e-sinx = 4
let esinx = y
⇒ y2 - 4y - 1 = 0
but we know that
so esinx ≠ 2 + √ 5 and 2 - √5
so No real solution of given equation.
let esinx = y
⇒ y2 - 4y - 1 = 0 but we know that
so esinx ≠ 2 + √ 5 and 2 - √5
so No real solution of given equation.
Create a free account to view solution
View Solution FreeMore Quadratic Equation Questions
If the equation ax2 + 2bx − 3c = 0 has no real roots and < a + b, then-...If roots of x2 − (a − 3)x + a = 0 are such that both of them is greater than 2, then-...If one of the roots of x(x + 2) = 4 − (1− ax2) tends ∞, then a will tend to -...If α, β are roots of the equation (3x + 2)2 + p ( 3x + 2) + q = 0, then roots of x2 + px + q = 0 are -...If the roots of both the equations px2 + 2qx + r = 0 and qx2 − 2x + q = 0 are real, then -...