sin2x ന്ടെ Maclaurian Series വിപുലീകരണത്തിൽ x³ -ന്ടെ ഗുണാങ്കം ഏത് ? A-4/3B-2/3C4/3D8/3Answer: A. -4/3 Read Explanation: Maclaurian series expansionf(x)=f(0)+f′(0)x+f"(0)2!x2+f′′′(0)3!x3+.......f(x) = f(0) + f'(0)x + \frac{f"(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+.......f(x)=f(0)+f′(0)x+2!f"(0)x2+3!f′′′(0)x3+.......x³ -ന്ടെ ഗുണാങ്കം=f′′′(0)3!=\frac{f'''(0)}{3!}=3!f′′′(0)f(x)=sin2xf(x)=sin2xf(x)=sin2xf′(x)=2cos2xf'(x)=2cos2xf′(x)=2cos2xf′′(x)=−4sin2xf''(x)=-4sin2xf′′(x)=−4sin2xf′′′(x)=−8cos2xf'''(x)=-8cos2xf′′′(x)=−8cos2x=f′′′(0)3!=−8cos03!=−86=−43=\frac{f'''(0)}{3!}=\frac{-8cos0}{3!}=\frac{-8}{6}=\frac{-4}{3}=3!f′′′(0)=3!−8cos0=6−8=3−4 Read more in App
