If the function f (x) is continuous in [0,8], where f(x) = ll x^2+a x+6, & 0 x< 2 3 x+2, & 2 x 4 2 a x+5 b, & 4< x 8 . , then

If the function f (x) is continuous in [0,8], where f(x) = ll x^2…

MCQ

If the function $f (x)$ is continuous in $[0,8]$, where
$f(x)$ $=\left\{\begin{array}{ll}x^2+a x+6, & 0 \leq x< 2 \\3 x+2, & 2 \leq x \leq 4 \\2 a x+5 b, & 4< x \leq 8\end{array}\right. \text {, then }$

A
$a=1, b=\frac{22}{5}$
✓
$a=-1, b=\frac{22}{5}$
C
$a=1, b=\frac{-22}{5}$
D
$a=-1, b=\frac{-22}{5}$

✓

Answer

Correct option: B.

$a=-1, b=\frac{22}{5}$

(B)
Since $f(x)$ is continuous in $[0,8]$.
$\therefore \quad$ it is continuous at $x=2$ and $x=4$.
$\therefore \quad \lim _{x \rightarrow 2^{-}} f (x)=\lim _{x \rightarrow 2^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 2^{-}}\left(x^2+a x+6\right)=\lim _{x \rightarrow 2^{+}}(3 x+2)$
$\Rightarrow(2)^2+2 a+6=3(2)+2$
$\Rightarrow 10+2 a =8$
$\Rightarrow a=-1$ ...(i)
Also, $\lim _{x \rightarrow 4^{-}} f (x)=\lim _{x \rightarrow 4^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 4^{-}}(3 x+2)=\lim _{x \rightarrow 4^{+}}(2 a x+5 b)$
$\Rightarrow 3(4)+2=2 a (4)+5 b$
$\Rightarrow 14=8 a+5 b$
$\Rightarrow b=\frac{22}{5} \quad\ldots[From (i)]$

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