Question

Let $a,b \in R,(a \ne 0).$ If the function $f$ defined as
 $$
 f\left( x \right)=\left\{\begin{array}{ll}
   \dfrac{{2{x^2}}}{a},0 \leqslant x < 1 \\
  a,1 \leqslant x < \sqrt 2 \\
  \dfrac{{2{b^2} - 4b}}{{{x^3}}},\sqrt 2 \leqslant x < \infty \\
  \end{array}
\right.
$$
is continuous in the interval $[0,\infty )$, then the ordered pair $(a,b)$ is:
A. $( - \sqrt 2 ,1 - \sqrt 3 )$
B. $(\sqrt 2 , - 1 + \sqrt 3 )$
C. $\left( {\sqrt 2 ,1 \pm \sqrt 3 } \right)$
D. $( - \sqrt 2 ,1 + \sqrt 3 )$

Answer 1

Hint: In this question, we are given that $a,b \in R,(a \ne 0).$ and also that $f(x)$is continuous in the given interval, so we use the definition of the continuous function.A function f is said to be continuous at point $x = a$, if $f(a) = f(a - ) = f(a + )$ Where, $f(a - ) = \mathop {\lim }\limits_{h \to 0} f(a - h)$ and $f(a + ) = \mathop {\lim }\limits_{h \to 0} f(a + h)$ and from this we will get the required values of $a,b$.

Complete step-by-step answer:
Given $a,b \in R,(a \ne 0).$ $f(x)$ is continuous in the interval.
 $$
 f\left( x \right)=\left\{\begin{array}{ll}
   \dfrac{{2{x^2}}}{a},0 \leqslant x < 1 \\
  a,1 \leqslant x < \sqrt 2 -------(1) \\
  \dfrac{{2{b^2} - 4b}}{{{x^3}}},\sqrt 2 \leqslant x < \infty \\
  \end{array}
\right.
$$
Here we are given that function is continuous in the interval $[0,\infty )$, but clearly looking at the definition of $f(x)$ function, we can see that $1,\sqrt 2 $ are the two doubtful points.
We know that the function is said to be continuous at $x = a$ if,
$f(a + ) = f(a) = f(a - )$
Now at $x = 1$, we know that it is continuous so we can write that
$f(1 + ) = f(1) = f(1 - )$
Firstly $f(1) = a$by using (1)
Now, we will consider $f(1 - ) = \mathop {\lim }\limits_{h \to 0} f(1 - h)$.
And for this limit value we will take $f(x)$ will be $x < 1$
Now from (1) $f(x) = \dfrac{{2{x^2}}}{a}$,
So,
$f(1 - ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{2{{(1 - h)}^2}}}{a}$
            $ = \dfrac{2}{a}$
So, we get $f(1) = a = \dfrac{2}{a} = f(1 - )$
$a = \pm \sqrt 2 $$ - - - - - (2)$
Now we also know that function is continuous at $x = \sqrt 2 $, so
$f(\sqrt 2 + ) = f(\sqrt 2 ) = f(\sqrt 2 - )$
Now
$f(\sqrt 2 ) = \dfrac{{2{b^2} - 4b}}{{{{(\sqrt 2 )}^3}}}$
Now consider $f(\sqrt 2 + ) = \mathop {\lim }\limits_{h \to 0} f(\sqrt 2 + h)$
And now for this limit value we will take $f(x)$ for $x > \sqrt 2 $
So, from (1) $f(x) = \dfrac{{2{b^2} - 4b}}{{{x^3}}}$.
Now, $f(\sqrt 2 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{2{b^2} - 4b}}{{{{(\sqrt 2 + h)}^3}}}$
                $ = \dfrac{{2{b^2} - 4b}}{{{{(\sqrt 2 )}^3}}}$
$f(\sqrt 2 + ) = \dfrac{{b(b - 2)}}{{\sqrt 2 }}$
Now consider $f(\sqrt 2 - ) = \mathop {\lim }\limits_{h \to 0} f(\sqrt 2 - h)$
So, for this limit value we will use the function $f(x)$ for $x < \sqrt 2 $.
Now from (1) \[f(x) = a\].
Now $f(\sqrt 2 - ) = a$
Therefore $f(\sqrt 2 + ) = \dfrac{{b(b - 2)}}{{\sqrt 2 }}$$ = a = f(\sqrt 2 - )$
$a\sqrt 2 = b(b - 2)$
Now we also have $a = \pm \sqrt 2 $
Using it we get that
$b(b - 2) = \pm 2$
Now we get
$b(b - 2) = 2$$ - - - - - (3)$
$b(b - 2) = - 2$$ - - - - - - - (4)$
Taking (3), we get
${b^2} - 2b - 2 = 0$
By using the quadratic equation formula, we get that
$b = \dfrac{{2 \pm \sqrt {4 + 4.2} }}{2}$
$b = 1 \pm \sqrt 3 $$ - - - - - - (5)$
Taking (4), we get that
${b^2} - 2b + 2 = 0$
$b = \dfrac{{2 \pm \sqrt {4 - 4.2} }}{2}$
$b = \dfrac{{2 \pm \sqrt { - 4} }}{2}$
Hence it gets to us the non-real values.
Hence ordered pair is $\left( {\sqrt 2 ,1 \pm \sqrt 3 } \right)$

So, the correct answer is “Option C”.

Note:In the above question, we have taken two terms $f(1),f(1 - )$ from the equation such that $f(1 + ) = f(1) = f(1 - )$.This is not necessary to take $f(1),f(1 - )$ from the three, we can take any two and solve to get the value of $a$ or $b$. Similarly, we have taken $f(\sqrt 2 - ),f(\sqrt 2 + )$ from the equation such that $f(\sqrt 2 - ) = f(\sqrt 2 ) = f(\sqrt 2 + )$.From the above equation, we can take any two values to make an equation and get the values of $a,b$.