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Continuity and Differentiability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSContinuity and Differentiability4 Marks
Question
If given function is continuous at $x=2$. Find the value of $k$ :$
f(x)=\left\{\begin{array}{cc}
\frac{x^3+x^2-16 x+20}{(x-2)^2}, & \text { when } x \neq 2 \\
k, & \text { when } x=2
\end{array}\right.
$
✓

Answer

Given function
$f(x)=\left\{\begin{array}{cc}
\frac{x^3+x^2-16 x+20}{(x-2)^2}, & \text { when } x \neq 2 \\
k, & \text { when } x=2
\end{array}\right.$
value of function at $x=2$
$f(2)=k$
value of R.H.L.
$\begin{aligned} \lim _{x \rightarrow 2^{+}} f(x) & =\lim _{h \rightarrow 0} f(2+h) \\ & =\lim _{h \rightarrow 0} \frac{(2+h)^3+(2+h)^2-16(2+h)+20}{(2+h-2)^2}\end{aligned}$
$\begin{array}{c}=\lim _{h \rightarrow 0} \frac{8+h^3+12 h+6 h^2+4+h^2+4 h-32-16 h+20}{h^2} \\ =\lim _{h \rightarrow 0} \frac{h^3+7 h^2}{h^2}=\lim _{h \rightarrow 0} \frac{h^2(h+7)}{h^2}=7\end{array}$
because function is continuous at $x=2$.
$\begin{aligned}
& \therefore & f(2) & =\lim _{h \rightarrow 0} f(2+h) \\
\Rightarrow & & k & =7
\end{aligned}$

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