Question

A function $f\left(x\right)$ is defined as $f\left(x\right)=\frac{Asinx+sin2x}{x^3},(x\ne 0).$ If the function is continuous at $x=0$, then-

• A. $A=-2$
• B. $F\left(0\right)=-1$
• C. $A=1$
• D. $F\left(0\right)=1$

Answer 1

Answer: (A)

$A=-2$

$f\left(x\right)=\frac{Asinx+sin2x}{x^3}$

$sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$

$sin2x=2x-\frac{(2x{)}^3}{3!}+\frac{(2x{)}^5}{5!}-\frac{(2x{)}^7}{7!}+...$

$\Rightarrow \frac{Asinx+sin2x}{x^3}=\frac{(Ax+2x)}{x^3}-\frac{\left(A{x}^3\right)+(2x{)}^3}{x^3.3!}+\frac{(A{x}^5+(2x{)}^5)}{x^{5}5!}-....$

$=\frac{(A+2)}{x^2}-\frac{(A+8)}{8!}+\frac{(A+2^5)}{5!}{x}^2-...$

$\Rightarrow$ For f(x) to be continuous at $x=0$

we should have

$A+2=0$

$\Rightarrow A=-2$