Limit Continuity and Differentiability 5 Question 2
2. If $f(x)=$
$$ \begin{array}{cc} q, & x=0 \\ \frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{3 / 2}}, & x>0 \end{array} $$
is continuous at $x=0$, then the ordered pair $(p, q)$ is equal to
(2019 Main, 10 April I)
(a) $-\frac{3}{2},-\frac{1}{2}$
(b) $-\frac{1}{2}, \frac{3}{2}$
(c) $\frac{5}{2}, \frac{1}{2}$
(d) $-\frac{3}{2}, \frac{1}{2}$
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Answer:
Correct Answer: 2. (c)
Solution:
- $g \circ f(x)=$
$$ f(x)+1, \quad \text { if } f(x)<0 $$
$$ \begin{array}{rr} f(x)-1}^{2}+b, & \text { if } f(x) \geq 0 \\ x+a+1, & \text { if } x<-a \end{array} $$
$$ \begin{aligned} & =(x+a-1)^{2}+b, \quad \text { if }-a \leq x<0 \\ & (|x-1|-1)^{2}+b, \quad \text { if } x \geq 0 \end{aligned} $$
As $g \circ f(x)$ is continuous at $x=-a$
$$ \begin{array}{r} \text { gof }(-a)=\operatorname{gof}\left(-a^{+}\right)=\operatorname{gof}\left(-a^{-}\right) \\ 1+b=1+b=1 \quad \Rightarrow \quad b=0 \end{array} $$
$\Rightarrow \quad 1+b=1+b=1 \quad \Rightarrow \quad b=0$
Also, $g \circ f(x)$ is continuous at $x=0$
$$ \begin{aligned} & \Rightarrow \quad \operatorname{gof}(0)=\operatorname{gof}\left(0^{+}\right)=\operatorname{gof}\left(0^{-}\right) \\ & \Rightarrow \quad b=b=(a-1)^{2}+b \Rightarrow a=1 \\ & \text { Hence, } \quad \text { gof }(x)=\begin{array}{cl} x+2, & \text { if } x<-1 \\ x^{2}, & \text { if }-1 \leq x<0 \\ (|x-1|-1)^{2}, & \text { if } x \geq 0 \end{array} \end{aligned} $$
In the neighbourhood of $x=0$, $g \circ f(x)=x^{2}$, which is differentiable at $x=0$.
