\(0\)
\(6\)
\(9\)
\(∞\)
\(3\)
\(\dfrac{1}{2}(x^3+1)=(2x-1)^{⅓}\)
\(⇒\)\([\dfrac{1}{2}(x^3+1)]^{3}=2x-1\)
\(⇒[\dfrac{1}{8}(x^9+3x^6+3x^3+1)]=2x-1\)
\(⇒[\dfrac{1}{8}(x^9+3x^6+3x^3+1)]=2x-1\)
\(⇒[\dfrac{1}{8}(x^9+3x^6+3x^3-2x+9)=0\)
Hence this polynomial of degree \(9\) has \(9\) solutions.(_Ans.)
If sin y = sin 3t and x = sin t, then \(\frac{dy}{dx}\) =