Question

( f(x)=frac{x(1+a cos x)-b sin x}{x^{3}} )
( f(0)=1 )
for ( (t+f(x)=f(0) )
( x rightarrow 0 )
then, 51 is continew
( lim _{x rightarrow 0} f(x)=lim _{x rightarrow 0} frac{x(1+operatorname{acos} x)-operatorname{bsin} x}{x^{3}} rightarrow frac{0}{0} f(0) )
( f(0)=lim _{x rightarrow 0} frac{(1+a cos x)+x(-a sin x)-b cos x}{3 x^{2}} )
( f(0)=lim _{x rightarrow 0} frac{1+a cos x-a x sin x-b cos x}{3 x^{2}} )
When we substitute ( x=0 ), denominato" is "O" but numerator is ( 1+a-b )
So, for it to be confinows it should be
( i n frac{0}{0} operatorname{fog} m cdot 30, quad 1+a-b=0 rightarrow 0 )

# Single Choice x(1-acosx) - bsin x X=0 Let f(0)=1. The value of a and b so that fis a continuous function are: and A 5/2, 3/2 B 5/2, -3/2 C -5/2, -3/2 D None of these

Solution