If f(x)= … is continuous at x=0 , then a+2b is equal to:

$$f({0}^{-})=\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}(\frac{\mathrm{sin}ax\mathrm{cos}2x}{x}+\frac{\mathrm{sin}2x\mathrm{cos}ax}{x}+\frac{\mathrm{sin}x}{x})$$f(0−)=x→0−lim​(xsinaxcos2x​+xsin2xcosax​+xsinx​)

$$=\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}(a\mathrm{cos}2x+2\mathrm{cos}ax+1)$$=x→0−lim​(acos2x+2cosax+1)

$$=a+3$$=a+3

Also, $f(0)=b$f(0)=b

$f({0}^{+})=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{x}^{\frac{1}{3}}}{{x}^{\frac{1}{3}}}(\frac{{(1+3x)}^{\frac{1}{3}}-1}{x})$f(0+)=x→0lim​x31​x31​​(x(1+3x)31​−1​), this is $(\frac{0}{0})$(00​)form

Apply $\mathrm{LH}$LH Rule

$$\Rightarrow f({0}^{+})=1$$⇒f(0+)=1

$$\therefore a=-2$$∴a=−2

$$b=1$$b=1

$$\therefore a+2b=0$$∴a+2b=0