Let `f(x)={{:(,(1)/(|x|),"if "|x| gt 2,"then "f(x)is),(,a+bx^(2), , "if"|x| le2):}` is differentiable at x=-2 forA. `a=(3)/(4), b=(1)/(6)`B. `a=(3)/(4), b=(1)/(16)`C. `a=-(1)/(4), b=(1)/(16)`D. `a=(1)/(4), b=-(1)/(16)`

Let `f(x)={{:(,(1)/(|x|),"if "|x| gt 2,"then "f(x)is),(,a+bx^(2), , "i

1.	Let `f(x)={{:(,(1)/(\|x\|),"if "\|x\| gt 2,"then "f(x)is),(,a+bx^(2), , "if"\|x\| le2):}` is differentiable at x=-2 forA. `a=(3)/(4), b=(1)/(6)`B. `a=(3)/(4), b=(1)/(16)`C. `a=-(1)/(4), b=(1)/(16)`D. `a=(1)/(4), b=-(1)/(16)`
Answer» Correct Answer - B We have `f(x)={{:(,-(1)/(x),"if "x lt -2),(,a+bx^(2),"if"-2 le x lte 2),(,(1)/(x),"if"x gt 2):}` Since, f(x) is differentiable at x=-2. So, it is continuous there at `therefore underset(x to -2^(-))lim f(x)=underset(x to -2^(+))lim f(x)=f(-2)` `Rightarrow underset(x to -2^(-))lim f(x)=underset(x to -2^(+))lima+bx^(2)=a+b(-2)^(2)` `Rightarrow (1)/(2)=a+4b` As f(x) is differentiable at x=-2. Therefore `underset(x to -2^(-))lim (f(x)-f(-2))/(x-(-2))=underset(x to -2^(+))lim (f(x)-(f-2))/(x-(-2))` `underset(x to -2^(-))lim (-(1)/(x)-(a+4b))/(x+2) =underset(x to -2^(+))lim ((a+bx^(2))-(a+4b))/(x+2)` `underset(x to -2^(-))lim (-(1)/(x)-(1)/(2))/(x+2)=b underset(x to -2^(+))lim (x^(2)-4)/(x+2)` `Rightarrow underset(x to -2^(-))lim -(1)/(2x)=b underset(x to =2^(+))lim (x-2)` `Rightarrow (1)/(4)=b(-4) Rightarrow b=-(1)/(16)...(ii)` Solving (i) and (ii), we get `a=(3)/(4) and b=-(1)/(16)`