If `f(x)=(3x^2+a x+a+1)/(x^2+x-2),` then which of the following can be correct `("lim")_(xvec1)f(x)e xi s t s a=-2` `("lim")_(xvec-2)f(x)e xi s t s a=13` `("lim")_(xvec1)f(x)=4/3` `("lim")_(xvec-2)f(x)=-1/3`

Options

A. `"ln "a_(1)`
B. `e^(a_(n))`
C. a_(1)`
D. `a_(n)`

Answer

A, B, C, D

Explanation

`f(x)=(3x^(2)+ax+a+1)/((x+2)(x-1))`
As `xto1,D^(r)to0," Hence as "xto 1,N^(r)to0.` Therefore,
`3+2a+1=0" or "a=-2`
As `xto-2,D^(r)to0." Hence as "xto-2,N^(r)to0.` Therefore,
`12-2a+a+1=0" or "a=13`
Now, `underset(xto1)limf(x)=underset(xto1)lim(3x^(2)-2x-1)/((x+2)(x-1))=underset(xto1)lim((3x+1
)(x-1))/((x+2)(x-1))=(4)/(3)`
Now, `underset(xto-2)lim(3x^(2)+13x+14)/((x+2)(x-1))=underset(xto-2)lim((3x+7)(x+2))/((x+2)(x-1))=-(1)/(3)`