Consider two function `y=f(x) and y=g(x)` defined as
`f(x)={{:(ax^(2)+b,,0lexle1),(bx+2b,,1ltxle3),((a-1)x+2c-3,,3ltxle4):}`
`and" "g(x)={{:(cx+d,,0lexle2),(ax+3-c,,2ltxlt3),(x^(2)+b+1,,3gexle4):}`
`lim_(xrarr2) (f(x))/(|g(x)|+1)` exists and f is differentiable at x = 1. The value of limit will be

Options

A. `-2`
B. `-1`
C. 0
D. 2

Answer

C

Explanation

Given two functions : `f(X)={{:(ax^(2)+b,,0lexle1),(bx+2b,, 1ltxle3),((a-1)x+2x-3,,3ltxle4):}`
and `g(x)={{:(cx+d,, 0le xle2),(ax+3-c,,2ltxlt3),(x^(2)+b+1,,xlexle4):}`
Condition for continuity of `f(x):f(1^(-))=f(1)=f(1^(+))and f(3^(-))=f(3)=f(3^(+))`
`rArr" "a+b=3b and 5b=3a+2c-6`
`rArr" "a=2b and c=3-(b)/(2)`
condition for continuity of g(x),
`g(2)=g(2^(-))=g(2^(+))and g(3^(-))=g(3)=g(3^(+))`
`rArr" "2c+d=2a+3-c and 3a+3 -c =10+b`
`rArr" "3c+d-2a=3 and b+c -3a =-7`
Also `f'(x)={{:(2ax,,0ltxlt1),(b,,1ltxlt3),(a-1,,3ltxlt4):}andg'(x)={{:(c,,0ltxlt2),(a,,2ltxlt3),(2x,,3ltxlt4):}`
`LHL=underset(hrarr0)(lim)(f(2-h))/(|g(2-h)|+1)underset(hrarr0)(lim)(b(4-h))/(|(c(2-h)+d)|+1)`
`=(4b)/(|2c+d|+1)`
`RHL=underset(hrarr0)(lim)(f(2+h))/(|g(2+h)|+1)underset(hrarr0)(lim)(b(4+h))/(|(a(2+h)+3-c)|+1)`
`=(4b)/(|2a+3-c|+1)`
`because" f is differentiable at x = 1 i.e., a = b= 0"`
So, LHL = RHL = 0 `because` (it is given that limit exists)