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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability3 Marks
Question
If function $f(x)=\left\{\begin{array}{cc}\frac{x^2-2 x-3}{x+1}, & : x \neq 1 \\ \lambda & : x=-1\end{array}\right.$ is continuous at $x = - 1$ then find value of $\lambda$.
✓

Answer

at $x=-1$
$
f(x)=\lambda \quad \therefore \quad f(-1)=\lambda
$
value of R.H.L. at $x=-1$$
\begin{aligned}
\lim _{h \rightarrow 0} f(-1+h) & =\lim _{h \rightarrow 0}\left[\frac{(-1+h)^2-2(-1+h)-3}{-1+h+1}\right] \\
& =\lim _{h \rightarrow 0}\left[\frac{1-2 h+h^2+2-2 h-3}{h}\right] \\
& =\lim _{h \rightarrow 0} \frac{h^2-4 h}{h} \\
& =\lim _{h \rightarrow 0}(h-4)=-4
\end{aligned}
$
$\therefore$ function is continuous at $x=-1$
$\therefore \quad f(-1)=\lim _{h \rightarrow 0} f(-1+h)$
$\Rightarrow \quad \lambda=-4 \quad \therefore \lambda=-4$

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