For what values of a and b is the function f(x)=x^2-4 x-2 , for x<2 =a x^2-b x+3, for 2 x<3 =2 x-a+b, for x 3 continuous for every x on R ?

f(x) is continuous for every x on R.....(given) f(x) is continuous at x=2 and x=3. As f(x) is continuous at x=2, _ x 2^ - f(x)= _ x 2^ + f(x) _ x 2^ - x^2-4 x-2 = _ x 2^ + (a x^2-b x+3 ) _ x 2^ - (x-2)(x+2) x-2 = _ x 2^ + (a x^2-b x+3 ) _ x 2^ - (x+2)= _ x 2^ + (a x^2-b x+3 ) [ l x 2, x 2 x-2 0 ] 2+2=a(2)^2-b(2)+3 (i) 4 a-2 ~b+3=4 4 a-2 ~b=1 As f(x) is continuous at x=3, _ x 3^ - f(x)= _ x 3^ + f(x) _ x 3^ - (a x^2-b x+3 )= _ x 3^ + (2 x-a+b) a(3)^2-b(3)+3=2(3)-a+b 9 a-3 b+3=6-a+b 10 a-4 b=3 .( (ii) Multiplying (i) by 2, we get 8 a-4 b=2 ...(iii) Subtracting (ii) from (iii), we get -2 a=-1 a=1 2 Substituting a=1 2 in (i), we get 4 (1 2 )-2 b=1 2-2 b=1 1=2 b b=1 2 a=1 2 and b=1 2

For what values of a and b is the function f(x)=x^2-4 x-2 , for x…

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Question

For what values of $a$ and $b$ is the function
$ f(x)=\frac{x^2-4}{x-2} \text {, for } x<2 \\
=a x^2-b x+3, \text { for } 2 \leq x<3 \\
=2 x-a+b, \text { for } x \geq 3$
continuous for every $x$ on $R$ ?

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Answer

$f(x)$ is continuous for every $x$ on $R.....($given$)$
$\therefore \mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=2$ and $\mathrm{x}=3$.
As $f(x)$ is continuous at $x=2$,
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)$
$\therefore \lim _{x \rightarrow 2^{-}} \frac{x^2-4}{x-2}=\lim _{x \rightarrow 2^{+}}\left(a x^2-b x+3\right)$
$\therefore \lim _{x \rightarrow 2^{-}} \frac{(x-2)(x+2)}{x-2}=\lim _{x \rightarrow 2^{+}}\left(a x^2-\mathrm{b} x+3\right)$
$\therefore \lim _{x \rightarrow 2^{-}}(x+2)=\lim _{x \rightarrow 2^{+}}\left(a x^2-\mathrm{b} x+3\right) \ldots\left[\begin{array}{l}
\because x \rightarrow 2, x \neq 2 \\
\therefore x-2 \neq 0
\end{array}\right] \\
\therefore 2+2=\mathrm{a}(2)^2-\mathrm{b}(2)+3 \ldots \text { (i) }$
$\therefore 4 \mathrm{a}-2 \mathrm{~b}+3=4$
$\therefore 4 \mathrm{a}-2 \mathrm{~b}=1 \ldots$
As $\mathrm{f}(x)$ is continuous at $x=3$, $\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)$
$\therefore \lim _{x \rightarrow 3^{-}}\left(a x^2-b x+3\right)=\lim _{x \rightarrow 3^{+}}(2 x-a+b)$
$\therefore a(3)^2-b(3)+3=2(3)-a+b$
$ \therefore 9 a-3 b+3=6-a+b$
$ \therefore 10 a-4 b=3 \ldots .(\text { (ii) }$
Multiplying $(i)$ by $2$, we get
$8 a-4 b=2 \text {...(iii) }$
Subtracting $(ii)$ from $(iii),$ we get
$-2 \mathrm{a}=-1$
$ \therefore \mathrm{a}=\frac{1}{2}$
Substituting $a=\frac{1}{2}$ in $(i),$ we get
$4\left(\frac{1}{2}\right)-2 b=1$
$ \therefore 2-2 b=1$
$ \therefore 1=2 b$
$ \therefore b=\frac{1}{2}$
$ \therefore a=\frac{1}{2}$ and $b=\frac{1}{2}$