If a function $\mathrm{f}(\mathrm{x})$ defined by $f(x)=\left\{\begin{array}{c}a x^2+b x+c, x \leq-1 \\ 2…
If a function $\mathrm{f}(\mathrm{x})$ defined by
$f(x)=\left\{\begin{array}{c}a x^2+b x+c, x \leq-1 \\ 2 x^2+4 x+1,-1 < x < 1 \\ c x^2+b x+a, x \geq 1\end{array}\right.$
is continuous on $\mathbb{R}$, and $\lim _{x \rightarrow \frac{3}{2}} f(x)=14$, then
$\lim _{x \rightarrow-2} f(x)=\quad x \rightarrow \frac{3}{2}$
6
-8
5
1
Solution
$\lim _{x \rightarrow \frac{3}{2}} f(x)=14 \Rightarrow$ C. $\frac{9}{4}+\frac{3 b}{2}+a=14$
$\Rightarrow 9 c+6 b+4 a=56$ ...(i)
Right hand limit at $x=-1=f(-1)$
$\begin{aligned} & \Rightarrow \lim _{x \rightarrow-1^{+}} f(x)=a(-1)^2+b(-1)+c \\ & \Rightarrow 2(-1)^2+4(-1)+1=\mathrm{a}-\mathrm{b}+\mathrm{c}\end{aligned}$
$\Rightarrow a-b+c=-1$ ...(ii)
Left hand limit at $x=1=f(1)$
$\begin{aligned} & \Rightarrow \lim _{x \rightarrow 1^{-}} f(x)=c+b+a \\ & \Rightarrow 2(1)^2+4(1)+1=a+b+c\end{aligned}$
$\Rightarrow a+b+c=7$ ...(iii)
[OBJEFrom equation (ii) \& (iii):
$a+c=3 \Rightarrow c=3-a$
From equation (iii): $b+3=7 \Rightarrow b=4$
Now from (i): $9(3-a)+6 \times 4+4 a=56$
$\Rightarrow a=-1$ and $\mathrm{c}=3-(1)=4$
Now, $\lim _{x \rightarrow-2} f(x)=4 a-2 b+c$
$=4 \times(-1)-2 \times 4+4=-8$