Question

Let `alpha, beta in R` such that `lim_(x ->0) (x^2sin(betax))/(alphax-sinx)=1` . Then `6(alpha + beta)` equals
A. 6
B. 7
C. 2
D. 12

Answer 1

Correct Answer - B
We find that
`lim_(xto0)(x^2sinbetax)/(alphax-sinx)`
` =lim_(xto0) (x(sin betax))/(a-(sinx)/(x))=(0)/(alpha-1)=0, if alpha ne 1`.
But, it is given that `lim_(xto 0) (x^2sinbetax)/(alphax-sin)=1.So, alpha=1`
Now,
`lim_(xto0) (x^2sinbetax)/(alphax-sinx)=1`
`rArr lim_(xto0) (beta^3((sinbetax)/(betax))=1`
`rArr lim_(xto0) (((sinbeta x)/(beta x)))/(((x-sinx)/(x^3)))=(1)/(beta)`
`rArr 6=(1)/(beta) rArr beta =(1)/(beta)" " [because lim_(xto0) (x-sinx)/(x^3)=(1)/(6)]`
Hence, `6(alpha+beta) =6 (1+(1)/(6))=7`.