The function $f(x)=\left\{\begin{array}{c}\frac{x^{2}}{a}, \text { if } 0 \leq x<1 \\ a, \text { if } 1 \leq x<\sqrt{2} \\ \frac{2 b^{2}-4 b}{x^{2}}, \text { if } \sqrt{2} \leq x<\infty\end{array}\right.$ Is continuous on $[0, \infty)$. Find the most suitable values of $a$ and $b$.
The function $f(x)=\left\{\begin{array}{c}\frac{x^{2}}{a}, \text { if } 0 \leq x<1 \\ a, \text { if } 1 \leq x<\sqrt{2} \\ \frac{2 b^{2}-4 b}{x^{2}}, \text { if } \sqrt{2} \leq x<\infty\end{array}\right.$ Is continuous on $[0, \infty)$. Find the most suitable values of $a$ and $b$.

A real function f is said to be continuous at x = c, where c is any point in the domain of f if

https://gradeup-question-images.grdp.co/liveData/PROJ23776/1543472595240417.png

h is a very small positive number. i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.

A function is continuous at x = c if

RD Sharma Solutions for Class 12 Maths Chapter 9 Continuity Image 216
RD Sharma Solutions for Class 12 Maths Chapter 9 Continuity Image 217

From equation 1, f(x) is changing its expression at x = 1

Given, f (x) is continuous everywhere

RD Sharma Solutions for Class 12 Maths Chapter 9 Continuity Image 218

Also from equation 1, f(x) is also changing its expression at x = √2

Given, f (x) is continuous everywhere

RD Sharma Solutions for Class 12 Maths Chapter 9 Continuity Image 219

$\therefore \mathrm{b}^{2}-2 \mathrm{~b}=\mathrm{a}, \ldots \ldots \ldots \ldots \ldots$ Equation 3
From equation $2, a=-1$
$b^{2}-2 b=-1$
$\begin{array}{l}
\Rightarrow b^{2}-2 b+1=0 \\
\Rightarrow(b-1)^{2}=0
\end{array}$
$\therefore b=1$ when $a=-1$
Putting $a=1$ in equation 3 :
$b^{2}-2 b=1$
$\Rightarrow b^{2}-2 b-1=0$
$\Rightarrow b=\frac{-(-2) \pm \sqrt{(-2)^{2}-4(-1)}}{2}=\frac{2 \pm \sqrt{8}}{2}=1 \pm \sqrt{2}$
Thus,
For $a=-1 ; b=1$
For $a=1 ; b=1 \pm \sqrt{2}$