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Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity5 Marks
Question
The function f(x) is befined as follows: $\text{f(x)}=\begin{cases}\text{x}^2+\text{ax}+\text{b},&0\leq\text{x}<2\\3\text{x}+2,&2\leq\text{x}\leq4\\2\text{ax}+5\text{b},&4<\text{x}\leq8\end{cases}$ if is continuous on [0, 8] find the value of a and b.
✓

Answer

It is given that the f(x) is continuous on [0, 8]
f(x) is continuous at x = 2 and x = 4
Now, At x = 2
LHL = RHL = f(2) ....(A)
f(2) = 3 × 2 + 2 = 8 ....(i)
$\text{LHL}=\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(2-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(2-\text{h}^2)+\text{a}(2-\text{h})+\text{b}=4+2\text{a}+\text{b}$
From (A)
4 + 2a + b = 8
2a + b = 4 ....(B)
Now, At x = 4
LHL = RHL = f(4) ....(C)
f(4) = 3 × 4 + 2 = 14 ....(ii)
$\text{RHL}=\lim\limits_{\text{x}\rightarrow4^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(4+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}2\text{a}(4+\text{h})+5\text{b}=8\text{a}+5\text{b}$
From (c) we get
8a + 5b = 14 ....(D)
Solving (B) and (D) we get
a = 3 and b = -2

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