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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability3 Marks
Question
$f(x)=\left\{\begin{array}{c}5 x^2-4, \text { if } x \leq 1 \\ 4 x^2-3 x \text {, if } x>2\end{array}\right.$ Examine the continuity.
✓

Answer

value of L.H.L. at $x=1$.$
\begin{aligned}
\lim _{x \rightarrow 1^{-}} f(x) & =\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}[5(1-h)-4] \\
& =\lim _{h \rightarrow 0}(5-5 h-4)=\lim _{h \rightarrow 0}(1-5 h)=1
\end{aligned}
$
value of R.H.L. at $x=1$$
\begin{aligned}
\lim _{x \rightarrow 1^{+}} f(x) & =\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}\left[4(1+h)^2-3(1+h)\right] \\
& =\lim _{h \rightarrow 0}\left(4+4 h^2+8 h-3-3 h\right) \\
& =\lim _{h \rightarrow 0}\left(4 h^2+5 h+1\right)=1
\end{aligned}
$value of function at $x=1$$
\begin{aligned}
& & f(x)=5 x-4 \\
\therefore & & f(1)=5 \times 1-4=5-4=1 \\
\because & & f(1)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} f(1+h)
\end{aligned}
$
Hence function is continuous at $x=1$ and except $x=$ 1 this function is also continuous due to universal continuous.

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