The equation x^3-3x+a=0 will have exactly one root if
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If the equation x3−3x+a=0 has 3 distinct roots between 0 and 1 then the integer value of a is
Can we solve it without Calculus.
MyTry:: Let x=2cosϕ, where −1≤cosϕ≤1⇒0≤ϕ≤π(Principle value, Inverse exists)
8cos3ϕ−6cosϕ+a=0⇒4cos3ϕ−3cosϕ=−a2⇒cos3ϕ=−a2
Now −1≤cos3ϕ≤1, So −1≤−a2≤1⇒−2≤a≤2
So we get a={−2,−1,0,1,2}
But answer is no integer value of a
Thanks
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